Mensuration (lb + bh + lh)

(vi) Diagonal = √

SOLIDS CUBE
A solid has three dimensions, namely length, breadth or A cube is a solid figure having six faces. All the faces of a
width, and height or thickness. The plane surfaces that cube are equal squares (let us say of the side ‘a’).
bind it are called its faces and the solid so generated is Therefore, the length, breadth, and height of a cube are
known as polyhedron. equal.

The volume of any solid figure is the amount of space (i) Volume = a3
enclosed within its bounding faces. A solid has edges, (ii) Lateral surface area (LSA) or area of the four
vertices, and faces, which are shown in the figure. walls = 4a2
A solid has the following two types of surface areas: (iii) Total surface area (TSA) = 6a2
Lateral Surface Area Lateral surface area (LSA) of a (iv) Diagonal = a √
solid is the sum of the areas of all the surfaces it has
except the top and the base. RIGHT CIRCULAR CYLINDER
Total Surface Area Total surface area (TSA) of a solid
is the sum of the LSA and the areas of the base and the
top.

Note: In case of solids, like the cube and cuboid, the LSA
consists of plane surface areas (i.e., area of all surfaces
except the top and base), whereas in case of solids, like
cone and cylinder, it consists of curved surface areas
In the above figure, r is the radius of the base and h is the
(CSA). Therefore, for such solids, the LSA is also called
height of a right circular cylinder. A cylinder is generated
CSA.
by rotating a rectangle or a square by fixing one of its
sides.
Euler’s Rule
(i) Volume = area of base × height
(ii) Volume = r2 h
Euler’s rule states that for any regular solid:
(iii) Curved surface area (CSA) = Perimeter of base ×
Number of faces (F) + Number of vertices (V) = Number
height
of edges (E) + 2
(iv) LSA = 2 rh
(v) Total surface area (TSA) = LSA + area of the top +
CUBOID area of the base
A cuboid is a rectangular solid having six rectangular (vi) TSA = 2 rh + r2 + r2
faces. The opposite faces of a cuboid are equal (vii) TSA = 2 r(r + h)
rectangles. A cuboid has a length (l), breadth (b), and
height (h). Some Important Deductions

Figure 1 Figure 2
If the above rectangular sheet of paper (ABCD) is rolled
In Figure 2, ED is the diagonal of the cuboid. Moreover, along its length to form a cylinder, then the radius (r) of
the area of the surface GDCH is x, the area of the surface the cylinder will be (L/2n) and its height will be b and
HEBC is y, and the area of the surface GFEH is z.
volume of this cylinder = , where l is the length of the
(i) Volume = Area of base × height = lbh rectangle.
(ii) Volume = √
(iii) Volume = xh = yl = zb
(iv) Lateral surface area (LSA) or area of the four walls
= 2 (l + b) h
(v) Total surface area (TSA) = 2(x + y + z) = 2

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If the above rectangular sheet of paper (ABCD) is rolled
along its breadth to form a cylinder, then the radius (r) of
the cylinder will be and its height will be L. Volume of
this cylinder = .
However, for the sake of representing the formula, we
will use another form of frustum right now as given
PRISM below:
A prism is a solid having identical and parallel top and
bottom faces, that is, they will be identical polygons of
any number of sides. The side faces of a prism are
rectangular and are known as lateral faces. The distance
between two bases is known as the height or the length
of the prism.
In the above figure, r is the radius of the base, h is the
vertical height of the frustum, and l is the slant height of
the frustum.
(i) Volume =
(ii) Slant height = = √
(iii) Curved surface area (CSA) = (R + r)
(iv) Total surface area (TSA) = CSA + Area of the top +
area of the base
(i) Volume = Area of base × Height TSA = (R + r) + r2 + R2
(ii) Lateral surface area (LSA) = Perimeter of the base × TSA = (R + r + r2 + R2)
Height (v) To find the height (H) of original cone.
(iii) Total surface area (TSA) = LSA + (2 × Area of the
H=
base)

RIGHT CIRCULAR CONE PYRAMID

In the above figure, ‘r’ is the radius of the base, h is the

height, and l is the slant height of the right circular cone.
(i) Volume = × Area of the base height × height
Volume = h
A pyramid is a solid having an n-sided polygon at its
(ii) Slant height = l = √ base. The side faces of a pyramid are triangular with the
(iii) Curved surface area (CSA) = rl top as a point.
(iv) Total surface area (TSA) = (CSA + Area of the base)
TSA = rl + r2

Frustum of Cone
A cone whose top portion is sliced o by a plane which is
parallel to the base is called frustum of cone.
Formation of frustum:

In the above figures, OM is the height of the pyramid.

(i) Volume = × Area of the base × Height

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(ii) Lateral surface area (LSA) = × (Perimeter of the
Tetrahedron
base) × Slant Height
A tetrahedron is a solid with four faces. All the faces of a
(iii) Total surface area (TSA) = LSA + Area of the
tetrahedron are equilateral triangles. A tetrahedron has
base
four vertices and six edges.
Frustum of Pyramid
A pyramid whose top portion is sliced o by a plane that is
parallel to the base is called the frustum of a pyramid.

Octahedron
An octahedron is a solid that has eight faces. All the faces
of an octahedron are equilateral triangles. An octahedron
has six vertices and 12 edges.
In the above figure, a1 is the area of the top face of the
frustum, a2 is the area of the bottom face of the frustum,
h is the height of the frustum, and l is the slant height of
the frustum.
(i) Volume = h √
(ii) Lateral surface area (LSA) = l
where P1 and P2 are perimeters of the top and
the bottom faces.
(iii) Total surface area (TSA) = LSA + a1 + a2

SPHERE Inscribed and Circumscribed Solids

If a sphere of the maximum volume is inscribed in a cube
of edge ‘a’, then the radius of the sphere = .
If a cube of the maximum volume is inscribed in a sphere
of radius ‘r’, then the edge of the cube = × r .
√
If a cube of the maximum volume is inscribed in a
hemisphere of radius ‘r’, then the edge of the cube =
√ .
In the above figure, r is the radius of the sphere.
Some Important Deductions
(i) Volume =
(ii) Surface area = 4

HEMISPHERE

If a cone is made by a sector of a circle (AOBD), then the

following two things must be remembered:
The area of the sector of a circle (AOBD) = The CSA of the
(i) Volume = cone
Radius of the circle (r) = Slant height (l) of the cone
(ii) Curved surface area (CSA) = 2
(iii) Total surface area (TSA) = LSA + Area of the top face
(read circle)
FORMULAE
TSA = 2 + 1. Cuboid
TSA = 3

SOME MORE SOLIDS

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 Figure
Volume  Area of the base × Height
Curved/Lateral surface area  Perimeter of
the base ×
Height
Figure Total Surface area  Lateral surface area +
Nomenclature  l = length, b = breadth, 2(Area of
h = height base)
Volume  lbh
Curved/Lateral surface area  2(l + b) h 6. Right pyramid
Total Surface area  2 (lb + bh + hl)

3.Cube

Figure
Figure Nomenclature  height
Nomenclature  a = edge/side Volume  1/3 area of the base × Height
Volume  Curved/Lateral surface area  1/2
Curved/Lateral surface area  ×Perimeter of the
Total Surface area  base × Slant height
Total Surface area  Lateral surface area +
Area of
3. Right circular cylinder base

7. Sphere

Figure
Nomenclature  R = radius of base, h = height
of the cylinder
Volume  h
Curved/Lateral surface area  Figure
Total Surface area  2 Nomenclature  r = radius
Volume 
4. Right circular cone Total Surface area 

8. Hemisphere

Figure
Nomenclature  r = radius, h = height,
l = slant height, Figure
l=√ Nomenclature  r = radius
Volume 
Volume 
Curved/Lateral surface area 
Curved/Lateral surface area 
Total Surface area 
Total Surface area 
5. Right triangular prism 9. Spherical Shell

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 Figure Q6.
Nomenclature  r = inner radius, R = outer The perimeter of a square and a circular field are the
radius same. If the area of the circular field is 3850 sq meter.
Volume  What is the area (in m2) of the square?
Total Surface area  4 (a) 4225
(b) 3025
10. Frustum of a Cone (c) 2500
(d) 2025.
Q7.
The perimeter of the top of a rectangular table is 28m.,
whereas its area is 48 m2. What is the length of its
diagonal?
(a) 5 m
Figure
(b) 10 m
Total Surface area  Lateral surface area +
Area of top + Area of base (c) 12 m
(d) 12.5 m
Q8.
Q1. The breadth of a rectangular hall is three-fourth of its
If the length of the diagonal AC of a square ABCD is 5.2 length. If the area of the floor is 192 sq. m., them the
cm, then the area of the square is difference between the length and breadth of the hall is:
(a) 15. 12 sq. cm (a) 8 m
(b) 13. 52 sq. cm (b) 12 m
(c) 12.62 sq. cm (c) 4 m
(d) 10.00 sq. cm (d) 22 m
Q2. Q9.
The length of the diagonal of a square is 'a' cm. Which of The diagonal of a square is 4√2 cm. The diagonal of
the following represents the area of the square (in sq. another square where area of double that of the first
cm) ? square is
(a) 2a (a) 8√2 cm
(b) a/√2 (b) 16 cm
(c) a2/2 (c) √32 cm
(d) a2/4 (d) 8 cm
Q3. Q10.
The breadth of a rectangular hall is three-fourth of its The diagonal of a square A is (a +b). The diagoanal of a
length. If the area of the floor is 768 sq. m. , then the square whose area is twice the area of square A is
difference between the length and breadth of the hall is (a) 2(a +b)
(a) 8 meter (b) 2(a+b)2
(b) 12 meters (c) √2(a-b)
(c) 24 meters (d) √2 (a+b)
(d) 32 meters Q11.
Q4. The length of a rectangular garden is 12 metres and its
Find the length of the largest rod that can be placed in a breadth is 5 metres. Find the length of the diagonal of a
room 16m long, 12m broad and 32/3 high square garden having the same area as that of the
(a) 123 meters rectangular garden
(b) 68 meter (a) 2√30 m
(c) 68/3 meter (b) √13 m
(d) 45/2 meter (c) 13 m
Q5. (d) 8√15 m
Between a square parameter 44cm and a circle of Q12.
circumference 44 cm, which figure has larger area and The areas of a square and a rectangle are equal. The
by how much length of the rectangle is greater than the length of any of
(a) Square 33cm2 the square by 5 cm and the breadth is less by 3 cm. Find
(b) Circle 33 cm2 the perimeter of the rectangle.
(c) Both have equal area. (a) 17 cm
(d) Dwustr 495 cm2 (b) 26 cm

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(c) 30 cm The difference between the length and breadth of a
(d) 34 cm rectangle is 23 m. If it’s perimeter is 206 m , then its area
Q13. is
The perimeter of a rectangle is 160 meter and the (a) 1520m2
difference of two sides is 48 metre. Find the side of a (b)2420m2
square whose area is equal to the area of this rectangle, (c) 2480m2
(a) 32 m (d) 2520m2
(b) 8 m Q20.
(c) 4 m The area (in m2) of the square which has the same
(d) 16 m perimeter as a rectangle whose length is 48 m and is 3
Q14. times its breadth is
The perimeter of two squares is 24 cm and 32 cm. The (A) 1000
perimeter (in cm) of a third square equal in area to the (b) 1024
sum of the areas of these squares is (c) 1600
(a) 45 (d) 1042
(b) 40 Q21.
(c) 32 The perimeter of two squares is 40 cm and 32 cm. The
(d) 48 perimeter of a third square whose area is the differences
Q15. of the two squares is
A wire when bent in the form of a square encloses an (a) 24 cm
area of 484 sq. cm. What will be the enclosed area? When (b) 42 cm
the wire is bent into the form of a circle? (c) 40 cm
(a) 125 cm2 (d) 20 cm
Q22.
(b) 230 cm2
The perimeter of five squares are 24 cm 32 cm ,40 cm ,
(c) 550 cm2
76 cm and 80 cm respectively the perimeter of another
(d) 616 cm2
square equal in area to sum of the areas of these squares
Q16.
is
Find the length of the longest rod that can be placed in a
(a) 31 cm
hall of 10m length , 6 m breadth and 4 m height
(b) 62 cm
(a) 2√38 m
(c) 124 cm
(b) 4√38 m
(d) 961 cm
(c) 2√19 m
Q23.
(d) √152 m
There is a rectangular tank of length 180 m and breadth
Q17.
120 m in a circular field of the area of the land portion of
The differences of the area of two squares drawn on two
the field is 40000 m2, what is the radius of the field ?
line segments of different lengths is 32sq. cm. Find the
(a) 130 m
length of the greater line segment if one is longer than
(b) 135 m
the other by 2 cm .
(c) 140 m
(a) 7 cm
(d) 145 m
(b) 9 cm
Q24.
(c) 11 cm
The length of a rectangular hall is 5m more than its
(d) 16 cm
breadth. The area of the hall is 750m2. The length of the
Q18.
hall is
A took 15 sec to cross a rectangular field diagonally
(a) 15 m
walking at the ratio diagonally walking at the ratio of
(b) 22.5 m
52m/min and B took the same time to cross the same
(c) 25 m
field along its sides walking at the rate of 68m/min. The
(d) 30 m
area of the field is
Q25.
(a) 30m2
A cistern 6 m long and 4 m wide contains water up to
(b) 40 m2
depth of 1m 25 cm. The total area of the wet surface is
(c) 50 m2
(a) 55m2
(d) 60 m2
(b) 53.52
Q19.
(c) 50m2
(d) 49m2

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Q26. (b)120
If the length and breadth of a rectangle are in the same (c)150
ratio 3:2 and its perimeter is 20 cm, then the area of the (d)200
rectangle (in cm3 ) is Q33.
(a) 24 cm2 A street of width 10 meters surrounds from outside a
(b) 36cm2 rectangular garden whose measurement is 200 m x 180
(c) 48cm2 m. the area of the path (in square meters) is
(d) 12 cm2 (a) 8000
Q27. (b) 7000
The perimeter of a rectangle and a square are 160 m (c) 7500
each. The area of the rectangle is less than that of the (d) 8200
square by 100 sq. m. The length of the rectangle is Q34.
(a) 30 m The area of the square inscribed in a circle of radius 8 cm
(b) 60 m is
(c) 40 m (a) 256 sq. cm
(d) 50 m (b) 250 sq. cm
Q28. (c) 128 sq. cm
A path of uniform width runs round the inside of a (d) 125 sq. cm
rectangular field 38 m long and 32 m wide , if the path Q35.
occupies 600m2, then the width of the path is Area of square with diagonal 8√2 cm is
(a) 30 m (a) 64 cm2
(b) 5 m
(b) 29 cm2
(c) 18.75 m
(c) 56 cm2
(d) 10 m
(d) 128 cm2
Q29.
Q36.
The perimeter of the floor of a room is 18 m. What is the
If the area of a rectangle be (x2 +7x +10) sq. cm, then one
area of the walls of the room if the height of the room is 3
of the possible perimeter of it is
m?
(a) (4x + 14) cm
(a) 21m2 (b) (2x+14) cm
(b) 42m2 (c) (x+ 14 cm)
(c) 54 m2 (d) (2x+7) cm
(d) 108 m2 Q37.
Q30. If the perimeter of a square and a rectangle are the same.
A copper wire is bent in the shape of a square of area 81 then the area P and Q enclosed by them would satisfy
cm2. If the same wire is bent in the form of a semicircle , the condition
the radius ( in cm) of the semicircle is (take pie = 22/7) (a) P < Q
(a)126 (b) P ≤ Q
(b)14 (c) P > Q
(c)10 (d) P = Q
(d)7 Q38.
Q31. A cube of edge 6 cm is painted on all sides and then cut
A copper wire is bent in the form of a square with an into unit cubes. The number of unit cubes with no sides
area of 121 cm2.if the same wire is bent in the form of a painted is
circle , the radius ( in cm) of the circle is (take pie = (a)0
22/7) (b) 64
(a)7 (c) 186
(b)14 (d) 108
(c)8 Q39.
(d)12 The length of diagonal of a square is 15√2 cm. Its area is
Q32. (a)112.5 cm2
Water flows into a tank which is 200m long and 150 m (b)150 cm2
wide through a pipe of cross-section 0.3m x 0.2m at (c)255√2/2 cm2
20km/h. then the time (in hours) for the water level in (d)225 cm2
the tank to reach 8 m is Q40.
(a)50

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A kite in the Shape of a square with a diagonal 32 cm (c) 4 πa units
attached to an equilateral triangle of the base 8 cm. (d) 2a/ π units
Approximately how much paper has been used to make Q47.
it ?n (Use √3 =1.732) The perimeter and length of a rectangle are 40 m and 12
(a) 539.712 cm2 m respectively Its breadth will be
(b) 538.721 cm2 (a) 10m
(c) 540.712 cm2 (b) 8 m
(d) 539.217 cm2 (c) 6 m
Q41. (d)3m
A lawn is in the form of a rectangle having its breadth Q48.
and length in the ratio 3 : 4. The area of the lawn is 1/12 If each edge of square are be doubled then the increase
hectare. The breadth of the lawn is percentage in its area is
(a) 25 meters (a)200%
(b) 50 meters (b)250%
(c) 75 meters s (c)280%
(d) 100 meters (d)300%
Q42. Q49.
The area of a rectangle is thrice that of a square. The An elephant of length 4 m is at one corner of rectangular
length of the rectangle is 20 cm and the breadth of the cage of size (16 m x 30m) and faces towards the
rectangle is 3/2 times that of the side of the square. The diagonally opposite corner. If the elephant starts moving
side the square (in cm) is towards the diagonally opposite corner it takes 15
(a) 10 seconds to reach this corner. Find the speed of the
elephant
(b)20
(a) 1 m/sec
(c) 30
(b) 2m/sec
(d) 60
(c) 1.87m/sec
Q43.
(d) 1.5 m/sec
The length and breadth of a rectangular field are in the
Q50.
ratio 7 : 4. A part n4 m wide running all around outside
A circle is inscribed in a square of side35cm. The area of
has an area of 416 m2 .the breadth (in m) of the field is
the remaining portion of the square which is not
(a) 28
enclosed by the circle is
(b) 14
(a) 962.5 cm2
(c) 15
(b)262.5cm2
(d) 16
(c) 762.5cm2
Q44.
(d)562.4 cm2
How many tiles, each 4 decimeter square will be
Q51.
required to cover the floor of a room 8 m long and 6 m
If the side of a square is 1/2 (x+1) units and its diagonal
broad?
is (3-X )/√2 units, then the length of the side of the
(a) 200
square would be
(b) 260
(a) 4/3 units
(c) 280
(b) 1umit
(d) 300
(c)1/2 units
Q45.
(d) 2 units
A godown is 15 m long and 12 m broad, The sum of the
Q52.
area of the floor and the celling is equal to the sum of
A rectangular carpet has an area of 120 m 2 and a
areas of the four walls. The volume (in m 3) of the
perimeter of 46 meter. The length of its diagonal is:
godown is :
(a) 17 meter
(a) 900
(b) 21 meter
(b) 1200
(c) 13 meter
(c)1800
(d) 23 meter
(d) 720
Q53.
Q46.
If the length of a diagonal of a square is 6√2 cm, then its
Length of a side of a square inscribed in circles is a√2
area will be
units. The circumference of the circle is
(a) 24√2 cm2
(a) 2πa units
(b) 24 cm2
(b) πa units

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(c) 36 cm2 The diameter of a toy wheel is 14cm, What is the
(d) 72 cm2 distance travelled by it in 15 revolutions?
Q54. (a) 880 cm
The length of a room is 3 meter more than its breadth. If (b) 660 cm
the area of a floor of the room is 70 meter2, then the (c) 600 cm
perimeter of the floor will be (d) 560 cm
(a) 14 miter Q62.
(b) 28 miter A can go round a circular path 8 times in 40 minutes. If
(c) 34 miter the diameter of the circle is increased to 10 times the
(d) 17 miter original diameter, the time required by A to go round the
Q55. new path once travelling at the same speed as before is :
The length of a rectangle is twice the breadth. If area of (a)25min
the rectangle be 417.605 sq. m., then length is (b)20min
(a) 29.08 miter (c)50min
(b) 29.80 miter (d)100 min
(c) 29.09 miter Q63.
(d) 28.90 miter The base of a triangle is 15 cm and height is 12 cm. the
Q56. height of another triangle of double the area having the
The area of a sector of a circle of radius 5 cm, formed by base 20 cm is
an arc of length 3.5 cm is : (a) 9cm
(a) 8.5 cm2 (b) 18 cm
(b) 8.75 cm2 (c) 8 cm
(c) 7.75 cm2 (d) 12.5 cm
(d) 7.50 cm2 Q64.
Q57. If a wire is bent into the shape of a square, the area of the
The radius of a circular wheel is 1.75 m. The number of square is 81sq. cm, When the wire is bent into a
revolutions it will make in travelling 11 km is (use pie= semicircular shape, the area of the semicircle is : (use
22/7) pie= 22/7)
(a)800 (a) 154 cm2
(b)900 (b) 77 cm2
(c)1000 (c) 44 cm2
(d)1200 (d) 22 cm2
Q58. Q65.
The radius of a wheel is 21cm, how many revolutions If the area of a triangle with base 12 cm is equal to the
will it make in travelling 924 meters? (use pie= 22/7) area of square with side 12 cm, the altitude of the
(a) 7 triangle will be
(b)11 (a) 12 cm
(c) 200 (b) 24 cm
(d) 700 (c) 18 cm
Q59. (d) 36 cm
The area (in sq., cm) of the largest circle that can be Q66.
drawn inside a square of side 28 cm is : The sides of a triangle are 3cm, 4 cm and 5 cm. The area
(a) 17248 (in cm2) of the triangle formed by joining the mid points
(b) 784 of this triangle is :
(c) 8624 (a) 6
(d) 616 (b)3
Q60. (c) 3/2
The area of the ring between two concentric circles, (d)3/4
whose circumference are 88 cm and 132 cm, is
Q67.
(a) 78 cm2
Three circle of radius 3.5 cm each are placed in such a
(b) 770 cm2
way that each touches the other two. The area of portion
(c) 715 cm2
enclosed by the circles is ,
(d) 660 cm2
(a) 1.975 cm2
Q61.
(b) 1.967 cm2

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(c) 19.68 cm2 (c) 30 minutes
(d) 21.22 cm2 (d) 40 minutes
Q68. Q75.
The area of a circular garden is 2464 sq. m. how much Find the diameter of a wheel that makes 113 revolutions
distance will have to be covered if you like to cross the to go 2 km 26 decameters.(Take pie = 22/7 )
garden along its diameter?(use pie = 22) (a)56/13 m
(a) 56 m (b)70/11m
(b) 48 m (c)136/11 m
(c) 28 m (d) 140/11 m
(d) 24 m Q76.
Q69. The radius of a circular wheel is 1.75 m. the number of
Four equal circles each of radius 'a' units touch one revolutions that it will make in travelling 11 km. is
another. The area enclosed between them (pie= 22/7).In (a) 1000
square units, is (b) 10,000
(a)3a2 (c) 100
(b)6 a2 /7 (d) 10
(c) 41 a2 Q77.
(d) a2/7 The circumference of a circle is 100 cm. The side of a
Q70. square inscribed in the circle is
The area of the greatest circle inscribed inside a square 100√2/π cm
of side 21 cm is (take pie =22/7) (b)50√2/ π cm
(a) 703/2 cm2 (c)100/ π cm
(b) 701/2 cm2 (d) 50√2 cm
(c)693/2 cm2 Q78.
(d) 695/2 cm2 A path of uniform width surrounds a circular park, The
Q71. difference of internal and external circumference of this
The area of an equilateral triangle is 400 √3 sq. m. Its circular path is 132 meters Its width is :
perimeter is : (a) 22 m
(a) 120 m (b) 20 m
(c) 21 m
(b) 150 m
(d) 24 m
(c) 90 m
Q79.
(d) 135 m
Four equal sized maximum circular plates are cut off
Q72.
from a square paper sheet of area 784 sq. cm. The
From a point in the interior of an equilateral triangle, the
circumference of each plate is
perpendicular distance of the sides are √3 cm, 2√3 cm
(a)22cm
and 5√3 cm, The perimeter ( in cm) of the triangle is
(b)44cm
(a) 64 (c)66cm
(b) 32 (d)88cm
(c) 48 Q80.
(d) 24 The circum radius of an equilateral triangle is 8 cm The
Q73. in- radius of the triangle is
The perimeter of a triangle is 30 cm and its area is 30 (a)3.25 cm
cm2. If the largest side measures 13 cm, What is the (b) 3.50 cm
length of the smallest side of the triangle? (c) 4cm
(a) 3 cm (d) 4.25 cm
(b) 4 cm Q81.
(c) 5 cm Three coins of the same size (radius 1 cm) are placed on
(d) 6 cm a table such that each of them touches the other two. The
Q74. area enclosed by the coins is
Diameter of a wheel is 3 meter, The wheel revolves 28 (a) (π /2 -√3) cm2
times in a minute. To cover 5. 280 km distance, the wheel (b) (√3 - π /2) cm2
will take ( Take pie = 22/7 ): (c) (2√3- π /2) cm2
(a) 10 minutes (d) ( 3√3 - π /2) cm2
(b) 20 minutes Q82.

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The area of the largest triangle that can be inscribed in a and 8 cm respectively, the length of the side of the
semicircle of radius r cm, is triangle is :
(a) 2r cm2 (a) 7 cm
(b) r2 cm2 (b) 10. 5 cm
(c) 2cm2 (c) 14√3 cm
(d)1 /2 r2cm2. (d) 14√3 /3cm
Q83. Q90.
The area of the greatest circle, which can be inscribed in In an isosceles triangle, the measure of each of equal
a square whose perimeter is 120 cm, is sides is 10 cm and the angle between them is 45°, then
(a)22/7 x(15)2 cm2 area of the triangle is
(b) 22/7 x(7/2)2 cm2 (a) 25 cm2
(c) 22/7 x(15/2)2 cm2 (b) 25/2(√2) cm2
(d) 22/7 x(9/2)2 cm2 (c) 25√2 cm2
Q84. (d) 2√3 cm2
The area of the incircle of an, equilateral triangle of side Q91.
42 cm is (take π=22/7): The area of circle whose radius is 6 cm is trisected by
(a)231 cm2 two concentric circles. The radius of the smallest circle is
(b) 462 cm2 (a) 2√3 cm
(c)22√3cm2 (b) 2√6 cm
(d) 924 cm2 (c) 2 cm
Q85. (d) 3 cm
The number of revolution a wheel of diameter 40 cm Q92.
makes in traveling a distance of 176 m, is (take π=22/7): The area of an equilateral triangle inscribed in a circle is
(a) 140 4√3 cm2. The area of the circle is
(b) 150 (a)16/3 π cm2
(c) 160 (b)22/3 π cm2
(d) 166 (c)28/3 π cm2
Q86. (d)42/3 π cm2
The length of the perpendiculars drawn from any point Q93.
in the interior of an equilateral triangle to the respective If the difference between the circumference and
sides are p1, p2, and p3.The length of each side of the diameter of a circle is 30 cm, then the radius of the
triangle is circle must be :
(a)2/√3(p1+p2+ p3 ) (a) 6 cm
(b)1/3(p1+p2+ p3 )
(b)7 cm
(c)1/√3(p1+p2+ p3 )
(c) 5 cm
(d)4/√3(p1+p2+ p3 )
(d) 8 cm
Q87.
Q94.
A circle is inscribed in a square, An equilateral triangle
The base and altitude of a right angled triangle are 12 cm
of side 4√3cm is inscribed in that circle. The length of the
and 5 cm respectively. The perpendicular distance of its
diagonal of the square (in cm) is
hypotenuse from the opposite vertex is
(a) 4√2
(b) 8 (a)56/13 cm
(c) 8√2 (b) 60/13cm
(d) 16 (c)5cm
Q88. (d) 7cm
The hypotenuse of a right angle isosceles triangle is 5 cm. Q95.
its area will be
(a) 5 sq. cm
(b) 6.25 sq. cm
(c) 6. 50 sq. cm
(d) 12.5 sq. cm
Q89.
From a point within an equilateral triangle,
perpendiculars drawn to the three sides are 6 cm, 7 cm

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The area of the shaded region in the figure given below is Q101.
A right triangle with sides 3 cm,4cm and 5 m is rotated
about the side 3 cm to from a cone. The volume of the
cone so formed is
(a) 16 π cm3
(b) 12 π cm3
(c)15 π cm3
(d)20 π cm3
Q102.
ABC is an equilateral triangle of side 2 cm with A, B, C as
center and radius 1 cm, three arcs are drawn. The area of
the region with in the triangle bounded by the three
arcs is
(a)( 3√3- π/2)
(b) (√3 - 3π/2)
(c) √3 - π/2 )
(a)a2/2(π/2 -1) sq. units (d) (π/2- √3)
(b) a2 (π -1) sq. units Q103.
(c) a2 (π/2 -1) sq. units The circumference of a circle is 11 cm and the angle of a
(d) a2/b2(π -1) sq. units sector of the circle is 60°. The area of the sector is (use π
Q96. =22/7)
The area of a circle is increased by 22 cm , if its radius (a)77/48cm2
is increased by 1 cm . the original radius of circle is (b)125/48 cm2
(a)6cm (c)75/48 cm2
(b)3.2cm (d)123/48 cm2
(c)3cm Q104.
(d)3.5cm If the difference between areas of the circum circle and
Q97. the incircle of an equilateral triangle is 44 cm2, then the
The area of the largest circle, that can be drawn inside a area of the triangle is ( Take π =22/7)
rectangle with sides 148 cm . by 14 cm is (a) 28 cm2
(a) 49 cm2 (b) 7√3 cm2
(b) 154 cm2 (c) 14√3 cm2
(c)378 cm2 (d) 21 cm2
(d) 1078 cm2 Q105.
Q98. If the area of a circle inscribed in a square is 9 π cm2,
A circle is inscribed in an equilateral triangle of side 8 then the area of the square is
cm. The area of the portion between the triangle and the (a) 24 cm2
circle is (b) 30 cm2
(a) 11 cm2 (c) 36 cm2
(b) 10.95 cm2 (d) 81 cm2
(c)10 cm2 Q106.
(d)10.50 cm2 The sides of a triangle are 6 cm, 8 cm and 10 cm. The
Q99. area of the greatest square that can be inscribed in it, is
In a triangular field having sides 30m, 72m and 78m, the (a)18 cm2
length of the altitude to the side measuring 72m is (b)15 cm2
(a) 25 m (c) 2304/49 cm2
(b) 28 m (d)576/49cm2
(c)30 m Q107.
(d) 35 m The length of a side of an equilateral triangle is 8 cm. the
Q100. area of region lying between the cumcircle and the
If the perimeter of a right-angledisosceles triangle is incircle of the triangle is ( use π =22/7)
(4√2 +4) cm,the length of the hypotenuse is (a)351/7 cm2
(a)4cm (b)352/7 cm2
(b)6cm (c)526/7 cm2
(c)8cm (d)527/7 cm2
(d)10 cm

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Q108. The area of an equilateral triangle is 4√3 cm2. The length
A wire, when bent in the form of a square, encloses a of each side of the triangle is
region having area 121 cm2. If the same wire is bent into (a)3cm cm
the form of a circle , then the area of the circle is (use π (b)2√2 cm
=22/7) (c)2√3 cm
(a)144 cm2 (d)4cm
(b)180cm2 Q116.
(c)154cm2 An equilateral triangle of side 6 cm has its corners cut
(d) 176cm2 off to form a regular hexagon In. Area (in cm2) of this
Q109. regular hexagon will be
If the perimeter of a semicircular field is 36 m. Find its (a)3√3
radius (use π =22/7) (b)3√6
(a) 7 m (c)6√3
(b) 8 m (d)5√3/2
(c) 14 m Q117.
(d) 16 m A 7 m wide road runs outside around a circular park,
Q110. whose circumference is 176 m, the area of the road is
The perimeter (in meters) of a semicircle is numerically (use π =22/7)
equal to its area ( in square meters). The length of its (a)1368 cm2
diameter is (use π =22/7) (b)1472 cm2
(a) 36/11m (c)1512 cm2
(b)61/11m (d)1760 cm2
(c)72/11m Q118.
(d)68/11m The length (in cm) of a chord of a circle of radius 13 cm
Q111. at a distance of 12 cm from its center is
One acute angle of a right angled triangle is double the (a) 5
other. If the length of its hypotenuse is 10 cm, then its (b) 8
area is (c) 10
(a)25/2 (√3)cm2 (d) 12
(b)25 cm2 Q119.
(c) 25√ 3cm2 The four equal circles of radius 4 cm drawn on the four
(d)75/2 cm2 corners of a square touch each other externally. Then the
Q112. area of the portion between the square and the four
If a triangle with base 8 cm has the same area as a circle sectors is
with radius 8 cm, then the corresponding altitude (in (a) 9(π - 4) sq. cm
cm) of the triangle is (b) 16 (4 - π) sq.cm
(a)12 π (c) 99(π - 4) sq. cm
(b)20 π (d) 169 (π - 4) sq.cm
(c)16 π Q120.
(d)32 π If the four equal circles of radius 3 cm touch each other
Q113. externally, then the area of the region bounded by the
The measure (in cm) of sides of a right angled triangle four circles is
are given by consecutive integers its area (in cm2)is (a) 4 ( 9- π ) sq. cm
(a)9 (b) 9 (4 - π) sq. cm
(b)8 (c) 5(6 - π) sq. cm
(c)5 (d) 6(5 — π) sq.cm
(d)6 Q121.
Q114. The length of each side of an equilateral triangle is 14√3
The area of a right-angled isosceles triangle having cm. The area of the incircle (in cm3) is
hypotenuse 16√2cm is (a) 450
(a) 144 cm2 (b) 308
(b) 128 cm2 (c) 154
(c) 112 cm2 (d) 77
(d) 110 cm2 Q122.
Q115.

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 Area of the incircle of an equilateral triangle with side 6 A straight line parallel to the base BC of the triangle ABC
cm is intersects AB and AC at the points D and E respectively.
(a) π /2 sq. cm If the area of the AABE be 36 sq., cm, then the area of the
(b) √3 π sq. cm ∆ABC is
(c) 6 π sq. cm (a) 18 sq.cm
(d) 3 π sq. cm (b) 36 sq.cm
Q123. (c) 18 cm
A copper wire is bent in the form of an equilateral (d) 36 cm
triangle and has area 121√3 cm2, If the same wire is bent Q130.
into the form of a circle, the area (in cm2) enclosed by the The length of two sides of an isosceles triangle is 15 and
wire is ( take pie = 22/7) 22 respectively. What are the possible values of
(a) 364.5 perimeter
(b) 693.5 (a)52 or 59
(c) 346.5 (b)52 or 60
(d) 639.5 (c)15 or 37
Q124. (d)37 or 29
At each corner of a triangular field of sides 26 m 28 m Q131.
and 30 m, a cow is tethered by a rope of length 7m, the The diameter of a wheel is 98 cm. the number of
area (in m) ungrazed by the cows is revolutions in which it will have to cover a distance of
(a) 336 1540 m is
(b) 259 (a) 500
(c) 154 (b) 600
(d) 77 (c) 700
Q125. (d) 800
In an equilateral triangle ABC, P&Q are midpoint of sides Q132.
AB & AC respectively such that PQ || BC. If PQ = 5 cm The wheel of a motor car makes 1000 revolutions in
then find the length of BC. moving 440 m . The diameter (in meter) of the wheel is
(a) 5 cm (a) 0.44
(b) 0.14
(b) 10 cm
(c) 0.24
(c) 15 cm
(d) 0.34
(d) 12 cm
Q133.
Q126.
A bicycle wheel makes 5000 revolutions in moving 11
ABC is an equilateral triangle P and Q are two points on
km . Then the radius of the wheel (in cm ) is (take
AB and AC respectively such that PQ||BC. If PQ = 5 cm
pie=22/7 )
,then find area of ∆APQ
(a) 70
(a)25/4 sq.cm
(b) 35
(b)25√3 sq.cm
(c) 17.5
(c) 25√3 /4 sq.cm
(d) 140
(d) 25√3 sq.cm
Q134.
Q127.
Three circles of diameter 10 cm each are bound together
The area of a circle with circumference 22cm is
by a rubber band as shown in the figure. the length of the
(a) 38.5 cm2 rubber band (in cm) if it is stretched is
(b) 39 cm2
(c) 36.5 cm2
(d) 40 cm2
Q128.
In ∆ ABC, O is the centroid and AD, BE, CF are three
medians and the area of ∆ ACO = 15 cm2 then area of
quadrilateral BDOF is
(a) 20 cm2
(b) 30 cm2
(c) 40 cm2
(d) 25 cm2
Q129.

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 (b) (16 - 4 π)sq. cm
(c) (16 - 8 π) sq. cm
(d) (4 - 2 π ) sq. cm
Q140.
If the numerical value of the perimeter of an equilateral
triangle is √3 times the area of it, then the length of each
side of the triangle is
(a) 2 units
(b) 3 units
(c) 4 units
(d) 6 units
Q141.
Each side of an equilateral triangle is 6 cm. Find its area
(a) 9√3 sq. cm
(b) 6√3 Sq. cm
(c)4√3 sq. cm
(a) 30
(d) 8√3Sq. cm
(b) 30+ 10 π
Q142.
(c) 10
The length of three medians of a triangle are 9 cm, 12
(d) 60+20 π
cm and 15 cm, The area (in sq. cm) of the triangle is
Q135.
(a) 24
If chord of length 16cm on is at a distance of 15 cm from
(b) 72
the center of circle then the length of the chord of the
(c) 48
same circle which is at a distance of 8 cm from the
(d) 144
center is equal to
Q143.
(a) 10 cm
The area of the triangle formed by the straight line 3x
(b) 20 cm
+2y = 6 and the co-ordinate axes is
(c) 30 cm
(a) 3 square units
(d) 40cm
(b) 6 square units
Q136.
(c) 4 square units
A semicircular shaped window has diameter of 63 cm, its
(d) 8 square units
perimeter equals (take pie =22/7 )
Q144.
(a) 126 cm
If the length of each side of an equilateral triangle is
(b) 162 cm
increased by 2 units, the area is found to be increased by
(c) 198 cm
3 + √3 square unit. The length of each side of the triangle
(d) 251 cm
is
Q137.
(a) √3 unit
In an equilateral triangle ABC of side 10 cm, the side BC
(b)3 units
is trisected at D & E. Then the length (in cm) of AD is
(c) 3√3 units
(a)3√7
(d) 3√2 units
(b)7√3
Q145.
(c)10√7/3
What is the area of the triangle whose sides are 9cm,
(d)7√10/3
10cm and 11cm?
Q138.
(a) 30 cm2
The perimeter of a triangle is 40cm and its area is 60
(b)60 cm2
cm2. If the largest side measures 17 cm, then the length (
(c) 30√2 cm2
in cm ) of the smallest side of the triangle is
(d)62√2 cm2
(a) 4
Q146.
(b) 6
The area of an isosceles triangle is 4 square units if the
(c) 8
length of the unequal side is 2 unit, the length of each
(d) 15
equal side is
Q139.
(a) 4 units
From four corners of a square sheet of side 4 cm four
(b) 2√3 units
pieces each in the shape of arc of a circle with radius 2
(c) √17 units
cm are cut out. The area of the remaining portion is :
(d) 3√2 units
(a) (8 - π) sq. cm

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Q147. and radius '2' units meet other two at D and E. Then the
What is the area of a triangle having perimeter 32 cm, area of the quadrilateral ABDE is
one side 11 cm and difference of other two sides 5 cm? (a)2√2 sq. units
(a) 8√30 cm2 (b)3√3 sq. units
(b) 5√35 cm2 (c)3√2 sq. units
(c) 6√30 cm2 (d)2√3 sq. units
(d) 8√2 cm2 Q155.
Q148. If the perimeter of a right angled triangle is 56 cm and
Area of equilateral triangle having side 2cm is area of the triangle is 84 sq. cm, then the length of the
(a)4cm2 hypotenuse is (in cm)
(b)√3cm2 (a) 25
(c)3cm2 (b) 50
(d)√6cm2 (c) 7
Q149. (d) 24
The area of a circle is increased by 22 cm2. when its Q156.
radius is increased by 1 cm. The original radius of the If the length of each median of an equilateral triangle is
circle is 6√3 cm, the perimeter of the triangle is
(a) 3 cm (a) 24 cm
(b) 5 cm (b) 32 cm
(c) 7cm (c) 36 cm
(d) 9 cm (d) 42 cm
Q150. Q157.
The radii of two circles are and 5cm and 12 cm. The The area of an equilateral triangle is 4√3 sq. cm. Its
area of a third circle is equal to the sum of the area of the perimeter is
two circles. The radius of the third circle is (a) 12 cm
(a) 13 cm (b) 6 cm
(b)21cm (c) 8 cm
(c) 30 cm (d)3√3
(d)17cm Q158.
Q151. A gear 12 cm in diameter is turning a gear 18 cm in
The perimeter of a semicircle path is 36 cm. find the diameter. When the smaller gear has 42 revolutions. how
area of this semicircle path many has the larger one made?
(a)40 sq. m (a) 28
(b)54 sq. m (b) 20
(c)53 sq. m (c) 15
(d) 77 sq. m (d) 24
Q152. Q159.
The area of a circle inscribed in a square of area 2m 2 is The perimeter of a semicircular area is 18 cm , then the
(a) π/4 m2 radius is :( using pie =22/7)
(b) π /2 m2 (a) 16/3 cm
(c) π m2 (b) 7/2 cm
(d) 2π m2 (c) 6 cm
Q153. (d) 4 cm
Three circles of radii 4 cm, 6 cm and 8 cm touch each Q160.
other pair wise externally, The area of the triangle A circle and a rectangle have the same perimeter. The
formed. by the line-segments joining the centres of the sides of the rectangle are 18 cm and 26 cm. The area of
three circles is the circle is (Take pie =22/7 )
(a) 144√13 sq. cm (a) 125 cm2
(b) 12√105 Sq. cm (b) 230 cm2
(c) 6√6 sq. cm (c) 550 cm2
(d) 24√16 sq. cm (d) 616 cm2
Q154. Q161.
Two circles with center A and B and radius 2 units touch The area of a circle is 38.5 sq. cm. Its circumference (in
each other externally at 'C', A third circle with center 'C' cm) is (use pie =22/7)
(a) 22

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(b) 24 (d) 13872
(c) 26 Q169.
(d) 32 The altitude draw In o the base of an isosceles triangle is
Q162. 8 cm and its perimeter is 64 cm. the area (in cm2)of the
A circle is inscribed in a square whose length of the triangle is
diagonal is 12√2 cm. An equilateral triangle is inscribed (a)240
in that circle. The length of the side of the triangle is (b)180
(a) 4 √3 cm (c)260
(b) 8√3 cm (d)120
(c) 6√3 cm Q170.
(d) 11√3 cm Three circle of radius a, b, c touch each other externally .
Q163. the area of the triangle formed by joining their center
The area (in sq. unit) of the triangle formed in the first (a) √(a+b+c)abc
quadrant by the line 3x +4y = 12 is (b) (a + b + c)√(a+b+c)
(a) 8 (c) ab + bc + ca
(b) 12 (d) None of the above
(c) 6 Q171.
(d) 4 The radii of two circles are 10 cm and 24 cm. The radius
Q164. of a circle whose area is the sum of the area of these two
The height of an equilateral triangle is 15 cm. the area of circles is
the triangle is (a) 36 cm
(a) 50√3 sq. cm
(b) 17 cm
(b) 70√3 sq. cm
(c) 34 cm
(c) 75√3 sq. cm
(d) 26 cm
(d) 150√3 sq. cm
Q172.
Q165.
A circle is inscribed in an equilateral triangle and a
The area of an equilateral triangle is 9√3 m, The length
square is inscribed in that circle. The ratio of the areas of
(in cm) of median is
the triangle and the Square is
(a)2√3
(a)√3:4
(b)3√3
(b) √3:8
(c)3√2
(c) 3√3:2
(d)2√2
(d) 3√3:1
Q166.
Q173.
The side of a triangle are 16 cm, 12 cm and 20 cm. Find
If area of an equilateral triangle is a and height b, then
the area,
the value of b2/a is
(a) 64 cm2
(a)3
(b) 112 cm2
(b)1/3
(c) 96 cm2
(c) √3
(d) 81 cm2
(d)1/√3
Q167.
Q174.
360 sq. cm and 250 sq. cm are the area of two similar
ΔABC is similar to ΔDEF such that BC = 3 cm, EF = 4 cm
triangles. If the length of one of the sides of the first
and area of ΔABC = 54 cm2, then the area of ΔDEF is : =
triangle be 8 cm , then the length of the corresponding
(а) 66 сm2
side of the second triangle is
(b) 78 cm2
(a)31/5cm
(c) 96 cm2
(b) 19/3cm
(d) 54 cZm2
(c) 20/3 cm
Q175.
(d) 6 cm
The area of two similar triangles ABC and DEF are 20cm2
Q168.
and 45cm2 respectively. If AB =5 cm, then DE is equal to
The perimeter of an isosceles triangle is 544 cm and each
(a) 6.5 cm
of the equal sides is 5/6 times the base .what is the area
(b) 7.5 cm
(in cm2) of the triangle?
(c) 8.5 cm
(a) 38172
(d) 5.5 cm
(b) 18372
Q176.
(c) 31872

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 C1 and C2 are two concentric circles with center at O, (b) π [(3.5)2 +(4.5) 2+ (5.5) 2 ]
Their radii are 12 cm, and 3 cm, respectively B and C are (c) 27 π
the point of contact of two tangents drawn to C2 from a (d) 13.5
point A lying on the circle C1 Then, the area of the Q183.
quadrilateral ABOC is Three sides of a triangle field are of length 15 m, 20m
(a)9√15/2 sq.cm and 25m long respectively. Find the cost of sowing seeds
(b) 12√15 sq. cm in the field at the rate of 5 rupees per m
(c) 9√15 sq. cm (a) Rs. 800
(d) 6√15 sq. cm (b) Rs. 600
Q177. (c)Rs.750
From a point P which is at a distance of 13 cm from (d) RS.150
center O of a circle of radius 5 cm in the same plane, a Q184.
pair of tangents PQ and PR are drawn to the circle Area A chord of length 30 cm is at a distance of 8 cm from the
of quadrilateral PQOR is center of a circle. The radius of the circle is:
(a) 65 cm2 (a) 17 cm
(b) 60 cm2 (b) 23 сm
(c) 30 cm2 (c) 21 cm
(d) 90 cm2 (d) 19 cm
Q178. Q185.
A circular road runs around a circular ground. If the The radius of the incircle of a triangle whose sides are
difference between the circumference of the outer circle 9cm ,12 cm and 15 cm is
and the inner circle is 66 meters, the width of the road is (a) 9 cm
: (b) 13 cm
(a) 10.5 meters (c)3 cm
(b) 7 meters (d) 6 cm
(c) 5.25 meters Q186.
(d) 21 meters The ratio of in radius' circumradius of a square is :
Q179. (a) 1:√2
The difference of perimeter and diameter of a circle is X (b) √2:√3
unit. The diameter of the circle is (c) 1 : 3
(a)X/(π -1) unit (d) 1 : 2
(b) X/( π+1) unit Q187.
(c) X/ π unit Three circles of equal radius 'a' cm touch each other. The
(d) (X/ π) -1 unit area of the shaded region is :
Q180.
The area of the circum circle of an equilateral triangle is
3π sq.cm2. The perimeter of the triangle is
(a) 3√3cm
(b) 9 cm
(с) 18 cm
(d) 3 cm
Q181.
A horse is tied to a post by a rope. If the horse moves
along a circular path always keeping the rope stretched
and describes 88 meters when it has traced out 72° at
the center, the length of the rope is (Take pie=22/7 )
(a) 70 m
(b) 75 m
(c) 80 m
(d) 65 m
Q182.
Three circles of radii 3.5 cm, 4.5 cm and 5.5 cm touch
each other externally. Then the perimeter of the triangle
formed by joining the centres of the circles, in cm is
(a) 27
(a)( √3+π /2)a2sq. cm

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(b) ( 6√3- π /2)a2sq. cm hypotenuse use from the opposite vertex. then p2 is
(c) ( √3- π )a2sq. cm equal to
(d) ( 2√3- π /2)a2sq. cm (a)a2+b2
Q188. (b) (1/a2)+(1/b2)
ABC is a right angled triangle. B being the right angle (c)(a2b2)/(a2+b2)
Mid- points of BC and AC are respectively B and A' Area (d) a2 -b2
of ∆A'B'C is Q195.
(a) 1/2 (area of ∆ABC) A is the center of circle whose radius is 8 and B is the
(b) 2/3( агеa of ∆ABC) center of a circle whose diameter is 8. If these two circles
(c) 1/4(area of ∆ABC touch externally, then the area of the circle with
(d) 1/8( area of ∆ABC) diameter AB is
Q189. (a) 36π
A wire of length 44 cm is first bent to form a circle and (b) 64π
then rebent to form a square. The difference of the two (c)144π
enclosed areas is (d) 256π
(a) 44 cm2 Q196.
(b) 33 cm2 If the numerical value of the height and the area of an
(c) 55 cm2 equilateral triangle be same, then the length of each side
(d) 66 cm2 of the triangle is
Q190. (a) 2 units
∠ACB is an angle in the semicircle of diameter AB = 5 cm (b) 4 units
and AC : BC = 3: 4. The area of the triangle ABC is (c) 5 units
(a) 6√2 sq. cm (d) 8 units
(b) 4 sq. cm Q197.
(c) 12 sq.cm If the length of a side of the square is equal to that of the
(d) 6 sq. cm diameter of a circle, then the ratio of the area of the
Q191. square and that of the circle (π=22/7)
If the lengths of the sides AB, BC and CA of a triangle ABC (a)14:11
are 10 cm, 8 cm and 6 cm respectively and If M is the (b)7:11
mid-point of BC and MN || AB to cut AC at N. then area of (c)11:14
the trapezium ABMN is equal to (d)11:7
(a) 18 sq. cm Q198.
(b) 20 sq. cm The median of an equilateral triangle is 6√3 cm. the
(c) 12 sq. cm area (in cm2)of the triangle is
(d) 16 sq. cm (a)72
Q192. (b)108
In an equilateral triangle of side 24 cm, a circle is (c)72√3
inscribed touching its sides, The area of the remaining (d)36√3
portion of the triangle is(√3 =1.732) Q199.
(a) 98.55 sq. cm In the numerical value of the circumference and area of a
(b) 100 sq. cm circle is same then the area is
(c) 101 Sq. cm (a) 6 π sq. unit
(d) 95 sq. cm (b) 4 π sq. unit
Q193. (c) 8 π sq. unit
Two sides of a plot measuring 32 m and 24 m and the (d) 12 π sq. unit
angle between them is a perfect right angle. The other Q200.
two sides measure 25 m each and the other three angles The area of an equilateral triangle is 48 sq. cm. The
are not right angles. The area of the plot in m 2 is length of the side is
(a) 768 (a) √8 x 4 cm
(b)534 (b) 4√3 cm
(c) 696.5 (c) 8 cm
(d)684 (d) 8 ∜3 cm
Q194. Q201.
a and b are two sides adjacent to the right angled
triangle and p is the perpendicular drawn to the

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The external fencing of a circular path around a circular The perimeters of a circle, a square and an equilateral
plot of land is 33m more than its interior fencing. The triangle are same and their areas are C. S and T
width of the path around the plot is respectively. Which of the following statement is true ?
(a) 5.52 m (a) C = S = T
(b) 5.25m (b) c > S > Т
(c) 2.55 m (c) C < S < T
(d) 2.25 m (d)S < C < T
Q202. Q209.
The perimeter of a triangle is 54 m and its sides are in A horse takes 5/2 seconds to complete a round around a
the ratio 5 : 6 : 7. The area of the triangle is circular field. If the speed of the horse was 66 m/sec,
(a)18 m2 then the radius of the field is, (Given it pie 22/ 7)
(b) 54√6 m2 (a) 25.62m
(c)27√2 m2 (b) 26.52 m
(d) 25 m2 (c) 25.26m
Q203. (d)26.25m
A circular wire of diameter 112 cm is cut and bent in the Q210.
form of a rectangle whose sides are in the ratio of 9 : 7. The diameter of the front wheel of an engine is 2x cm
The Smaller Side of the rectangle is and that of rear wheel is 2y cm to cover the same
(a) 77cm distance, find the number of times the rear wheel Will
(b)97 cm revolve when the front wheel revolves 'n' times ,
(c) 67 cm (a)n/xy times
(d) 84 cm (b)yn/x times
Q204. (c)nx/y times
If the perimeter of an equilateral triangle be 18 cm, then (d) xy/n times
the length of each median is Q211.
(a)3√2cm A bicycle wheel has a diameter (including the tyre) of 56
(b)2√3cm cm. The number of times the wheel will rotate to cover a
(c)3√3 distance of 2.2 km is (assume pie 22 /7)
(d) 2√2cm (a) 625
Q205. (b) 1250
Two equal maximum sized circular plates are cut off (c) 1875
from a circular paper sheet of circumference 352 cm. (d) 2500
Then the circumference of each circular plate is Q212.
(a) 176 cm If the altitude of an equilateral triangle is 12√3 cm, then
(b) 150 cm its area would be ;
(c) 165 cm (a)36√2 cm2
(d) 180 cm (b)144√3 cm2
Q206. (c)72 cm2
The in radius of an equilateral triangle is √3 cm, then the (d)12 cm2
perimeter of that triangle is Q213.
(a) 18 cm Let C1 and C2 be the inscribed and circumscribed circles
(b) 15 cm of a triangle with sides 3 cm, 4 cm and area of 5 cm then
(c) 2 cm area of C1 / area of C2 is
(d) 6 cm (a) 9/25
Q207. (b) 16/25
The difference between the circumference and diameter (c) 9/16
of a circle is 150 m. The radius of that circle is( take pie = (d) 4/25
22/7) Q214.
(a) 25 meter A circular swimming pool is surrounded by a concrete
(b) 35 meter wall 4m.wide, If the area of the concrete wall
(c) 30 meter surrounding the pool 11/25radius(in m) of the pool :
(d) 40 meter (a) 8
Q208.
(b) 16
(c) 30
(d) 20

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Q215. A piece of wire when bent to form a circle will have a
The sides of a triangle having area 7776 sq., cm are in the radius of 84 cm. If the wire is bent to form a square, the
ratio 3: 4: 5. The perimeter of the triangle is: length of a side of the square is -
(a) 400 cm (a) 152 cm
(b) 412 cm (b) 168 cm
(c) 424 cm (c) 132 cm
(d) 432 cm (d) 225 cm
Q216. Q222.
The perimeter of a sheet of paper in the shape of a The area of a circle is 324, sq. cm, The length of its
quadrant of a circle is 75 cm. Its area would be ( π = longest chord (in cm.) is
22/7) (a) 36
(a) 512.25 cm2 (b) 38
(c) 28
(b) 345.5 cm2
(d) 32
(c) 100 cm2
Q223.
(d) 693 cm=2
The circumference of a triangle is 24 cm and the
Q217.
circumference of its in-circle is 44 cm. Then the area of
A circle is inscribed in an equilateral triangle of side 8m.
the triangle is (taking π = 22/ 7)
The approximate area of the unoccupied space inside the
(a) 56 square cm
triangle is :
(b) 48 square cm
(a) 21 m2
(c) 84 square cm
(b) 11m2
(d) 68 square
(c) 20 m2
Q224.
(d) 22 m2
If the length of each of two equal sides of an isosceles
Q218.
triangle is 10 cm. and the adjacent angle is 45°, then the
In the figure, OED and OBA are sectors of a circle with
area of the triangle is
centre O. The area of the shaded portion.
(a) 20√2 square cm
(b) 25√2 square cm
(c) 12√2 Square cm
(d) 15√2 square cm
Q225.
The inner-radius of a triangle is 6 cm, and the sum of the
lengths of its sides is 50 cm. The area of the triangle (in
sq. cm ) is
(a) 11/16m2 (a) 150
(b) 11/8 m2 (b) 300
(c) 11/2 m2 (c) 50
(d) 11/4 m2 (d)56
Q219. Q226.
If the circumference of a circle is then the diameter of the One of the angles of a right angled triangle is 15°, and the
circle is 30/ π, then the diameter of the circle is: hypotenuse is 1 m. the area of the triangle ( in sq. cm) is
(a) 30 (a) 1220
(b) 15/ π (b) 1250
(c) 60 π (c) 1200
(d) 30/ π2 (d) 1215
Q220. Q227.
The outer and inner diameter of a circular path be 728 If for an isosceles triangle the length of each equal side is
cm and 700 cm respectively. The breadth of the path is 'a' units and that of the third side is 'b' units, then its area
(a) 7 cm will be
(b) 14 cm (a)(a/4)√4a2- a2 sq. units
(c) 28 cm (b) (b/4)√4a2- b2 sq. units
(d) 20 cm (c) (a/2)√2a2- b2 sq. units
Q221. (d) (b/2)√ a2- 2b2 sq. units
Q228.

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What is the position of the Circumcentre of an obtuse- (b) 50 m2
angles triangle? (c) 45 m2
(a) It is the vertex opposite to the largest side यह सबसे (d) 55 m2
Q235.
बडे पाश्र्व के ववपरीतशिर्ष है , The perimeter of a rhombus is 40 cm. If the length of one
(b) It is the midpoint of the largest side यह सबसेऱंबे of its diagonals be 12 cm, the length of the other diagonal
is
पाश्र्व का मध्य बबंद ु है|
(a) 14 cm
(c)it lies outside the triangles ./यह बिभुज के बाहर होती (b) 15 cm
(c) 16 cm
है |
(d) 12 cm
(d) it lies inside the triangles./यह बिभज
ु के अंदर होती Q236.
है | The area of a rhombus is 150 cm2. The length of one of its
diagonals is 10 cm. The length of the other diagonal is:
Q229.
(a) 25 cm
The ratio of circumference and diameter of a circle is
(b) 30 cm
22:7. If the circumference be 11/7 m, then the rafius of
(c) 35 cm
the circle is :
(d) 40 cm
(a)1/ 4 m
Q237.
(b) 1/3 m
The area of a regular hexagon of side 2√3 cm is :
(c) 1/2 m
(a) 18√3 cm2
(d) 1 m
(b) 12√3 cm2
Q230.
(c) 36√3 cm2
The area of a circle whose radius is the diagonal of a
(d) 27 √3 cm2
square whose area is 4 is:
Q238.
(a) 4 π
Each side of a regular hexagon is 1 cm. The area of the
(b) 8 π
hexagon is
(c) 6 π
(a)3√3/2 cm2
(d) 16 π
(b)3√3/4 cm2
Q231.
(c)4√3cm2
The diagonals of a rhombus are 32 cm and 24 cm.
(d)3√2cm2
respectively. The perimeter of the rhombus is :
Q239.
(a) 80 cm The length of one side of a rhombus is 6.5 cm and its
(b)72 cm altitude is 10 cm. If the length of its diagonal be 26 cm,
(c) 68 cm the length of the other diagonal will be :
(d) 64 cm (a) 5 cm
Q232. (b) 10 cm
The diagonals of a rhombus are 24cm and 10 cm. The (c) 6.5 cm
perimeter of the rhombus (in cm) is : (d) 26 cm
(a) 68 Q240.
(b) 65 The measure of each of two opposite angles of a
(c)54 rhombus is 60° and the measure of one of its sides is10
(d) 52 cm, The length of its smaller diagonal is :
Q233. (a) 10cm
The perimeter of a rhombus is 40 cm, If one of the (b) 10√3 cm
diagonals be 12 сm long what is the length of the other (c) 10√2 cm
diagonal? (d) (5/2)√2 cm
(a)12cm Q241.
(b)√136cm The perimeter of a rhombus is 100 cm, If one of its
(c)16cm diagonals is 14 cm, Then the area of the rhombus is
(d)√44cm (a) 144 cm2
Q234. (b) 225 cm2
The perimeter of a rhombus is 40 m and its height is 5m (c) 336 cm2
its area is : (d) 400 cm2
(a) 60 m2 Q242.

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The ratio of the length of the parallel sides of a trapezium (b) 18 cm
is 3 : 2. The shortest distance between them is 15 cm. If (c) 16cm
the area of the trapezium is 450 cm2 the sum of the (d) 9cm
length of the parallel sides is Q248.
(a) 15 cm The adjacent sides of a parallelogram are 36 cm and
(b) 36 cm 27cm, if the distance between the shorter sides is 12 cm,
(c) 42 cm then the distance between the longer sides is:
(d) 60 cm (a) 10 cm
Q243. (b) 12 cm
A parallelogram has sides 15cm and 7 cm long . the (c) 16cm
length of one of the diagonals is 20 cm . the area of the (d) 9cm
parallelogram is Q249.
(a) 42 cm 2 If the diagonals of a rhombus are 8 cm and 6 cm ,then
the area of square having same side as that of rhombus
(b) 60cm 2
is
(c) 84cm2
(a) 25
(d) 96 cm2
(b) 55
Q244. (c) 64
(d) 36
Sides of a parallelogram are in the ratio 5:4. Its area is Q250.
1000 sq. units, altitude of the greater side is 20 units. Two circles with centres A and Band radius 2 units touch
Altitude on the smaller side is: each other externally at 'C' A third circle with centre 'C'
and radius '2' units meets other two at D and E. Then the
(a) 20 units area of the quadrilateral ABDE is
(a) 2√2 sq. units
(b) 25 units (b) 3√3 sq. units
(c) 3√2 sq. units
(c) 10 units (d) 2√3 sq. units
Q251.
(d)15 units The perimeter of a non-square rhombus is 20 cm. One its
diagonal is 8 cm, The area of the rhombus is
Q245. (a) 28 sq. cm
(b) 20 sq. cm
The perimeter of a rhombus is 40 cm and the measure of
(c) 22 sq. cm
an angle is 60°, then the area of it is:
(d) 24 sq. cm
Q252.
(a) 100√3cm2
The perimeter of a rhombus is 100 cm and one of its
(b) 50 √3cm2 diagonals is 40 cm. Its area (in cm) is
(a) 1200
(c) 160√3cm2 (b) 1000
(c) 600
(d) 100cm2 (d) 500
Q246. Q253.
Two adjacent sides of a parallelogram are of length 15 In ∆ABC, D and E are the points of sides AB and BC
cm and 18 cm, If the distance between two smaller sides respectively such that DE|| AC and AD : BD = 3:2 the
is 12 cm, then the distance between two bigger sides is ratio of area of trapezium ACED to that of ∆ BED is
(a) 8 cm (a) 4: 15
(b) 10 cm (b) 15: 4
(c) 12 cm (c) 4 :21
(d) 15 cm (d) 21 : 4
Q247. Q254.
A parallelogram ABCD has sides AB = 24 cm and AD = 16 ABCD is a trapezium in which AB||DC and AB = 2 CD, The
cm. The distance between the sides AB and DC ,is 10 cn, diagonals AC and BD meet at O. The ratio of area of
Find the distance between the sides AD and BC. triangles AOB and COD is
(a) 15 cm (a) 1 : 1 .

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(b) 1: √2 (c) 400√15 m2
(c) 4 : 1 (d)450√15 m2
(d) 1 : 4 Q262.
Q255. Perimeter of a rhombus is 2p unit and sum of length of
The length of each side of a rhombus is equal to the diagonals is m unit, then area of the rhombus is.
length of the side of a square whose diagonal is 40√2 cm. (a)1/4m2 p sq unit
If the length of the diagonals of the rhombus are in the (b) 1/4mp2 sq unit
ratio 3: 4, then its area (in cm2) is (c)1/4 (m2 – p2 ) sq unit
(a) 1550 (d) 1/4 (m2 – p2) sq unit
(b) 1600 Q263.
(c) 1535 I ∆ABC, D and E are two points on the sides AB so that DE
(d) 1536 ||BC and AD/BD = 2/3. Then the area of trapezium
Q256. DECB/the area of ∆ABC is equal to
ABCD is a parallelogram BC is produced to Q such that (a) 5/9
BC = CQ, Then (b) 21/25
(a) area (∆ABC ) = area (∆DCQ) (c) 9/5
(b) area (∆ABC) > area (∆DCQ) (d) 21/4
(c) area (∆ABC) < area (∆DCQ) Q264.
(d) area. (∆ABC) not equal area (∆DCQ) The sides of a rhombus are 10cm each and a diagonal
Q257. measures 16cm. Area of the rhombus is:
ABCD is parallelogram, P and Q are the mid- points of (a) 96sq. cm
sides BC and CD respectively, If the area of ∆ABC is 12 (b) 160sq. cm
cm2, then the area of ∆APQ is (c) 100 sq. cm
(a) 12 cm2 (d) 40 sq. cm
(b) 8 cm2 Q265.
(c) 9 cm2 The lengths of two parallel sides of a trapezium are 6 cm
(d) 10 cm2 and 8 cm, If the height of the trapezium be 4 cm, then its
Q258. area is
The area of a rhombus is 216 cm2 and the length of its (a) 28 cm2
one diagonal is 24 cm. The perimeter (in cm) of the (b) 56 cm2
rhombus is (c) 30 cm2
(a) 52 (d) 36 cm2
(b) 6021. Q266.
(c) 120 If diagonal of a rhombus are 24 cm and 22cm then
(d) 100 perimeter of that rhombus is
Q259. (a)80 cm
One of the four angles of a rhombus is 60°. If the length (b) 84 cm
of each side of the rhombus is 8 cm, then the length of (c)76cm
the longer diagonal is (d) 72 cm
(a) 8 √3 cm Q267.
(b) 8 cm The area of an isosceles trapezium is 176 cm 2 and the
(c) 4√3 cm height is 2/11th of the sum of its parallel sides. If the ratio
(d) 8/√3 cm of the length of the parallel sides is 4: 7, then the length
Q260. of a diagonal (in cm) is
The diagonals of a rhombus are 12 cm and 16 cm (a) 2√137
respectively. The length of one side is (b) 24
(a) 8 cm (c) √137
(b) 6 cm (d) 28
(c) 10 cm Q268.
(d) 12 cm The perimeter of a rhombus is 60 cm and one of its
Q261. diagonal is 24 cm. The area of the rhombus is
A parallelogram has sides 60 in and 40 m and one of its (a) 432 sq. cm
diagonals is 80 m long. Its area is
(b) 216 sq.cm
(a) 500√15 m2
(c) 108 sq.cm
(b) 600√15 m2
(d) 206 sq.cm

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Q269. (c) Rs.9047.50
The area of the parallelogram whose length is 30 cm, (d) Rs. 4186.50
width is 20 cm and one diagonal is 40 cm is Q276.
(a) 200 √15 cm2 The outer circumference of a Circular race-track is 528
(b) 300√15 cm2 meter. The track is everywhere 14 meter wide, Cost of
(c) 100√15 cm2 leveling the track at the rate of Rs. 10 per sq. meter is :
(d) 150√15 cm2 (a)Rs. 77660
Q270. (b) Rs.67760
The area of a rhombus is 256 sq. cm, and on of its (c)Rs. 66760
diagonal is twice the other in length Then length of its (d) Rs.76760
larger diagonal is Q277.
(a) 32 cm The length and breadth of a rectangular field are in the
(b) 48 cm ratio of 3 : 2. If the perimeter of the field is 80 m, its
(c) 36 cm breadth (in meters) is:
(d) 24 cm (a) 18
Q271.
(b) 16
The length of two parallel sides of a trapezium is 15 cm
(c) 10
and 20 cm. If its area is 175 sq.cm, then its height is:
(d)24
(a) 25 cm
Q278.
(b) 10 cm
The sides of a rectangular plot are in the ratio 5 : 4
(c) 20 cm
equal and its area is equal to 500 sq. m the perimeter of
(d) 15 cm
the plot is
Q272.
(a)80m
The cost of carpenting a room is Rs. 120 If the Width had
(b)100m
been 4 meters less, the cost of the Carpet would have
(c)90m
been Rs. 20 less. The width of the room is :
(d)95m
(a) 24 m
Q279.
(b) 20 m
ABC is a triangle with base AB, D is =5 DB=3, what is
(c) 25 m
the ratio of the area of ∆ADC to and its area the area of
(d) 18.4 m
∆ABC
Q273.
(a)2/5
The floor of a corridor is 100 m long and 3 m wide. Cost
(b)2/3
of covering the floor with carpet 50 cm wide at the ratio
(c)9/25
of Rs. 15 per m is
(d)4/25
(a) Rs.4500
Q280.
(b) Rs.9000
If the area of a triangle is 1176 cm2 and the ratio of base
(c) Rs.7500
and corresponding altitude is 3 : 4, then the altitude of
(d) Rs.1900
the triangle is :
Q274.
(a) 42 cm
A playground is in the shape of a rectangle. A sum of
(b) 52 cm
1,000 was spent to make the ground usable at the rate of
(c) 54 cm
25 paise per sq. m. The breadth of the ground is 50 m. if
(d) 56 cm
the length of the ground is increased by 20 m. what will
Q281.
be the expenditure (in rupees) at the same rate per sq.
The sides of a triangle are in the ratio . If the perimeter
m.?
of the triangle is 52 cm, the length of the smallest side is :
(a)1250
(a) 24 cm
(b) 1,000
(b) 10 cm
(c) 1,500
(c) 12 cm
(d) 2,250
(d) 9 cm
Q275.
Q282.
A hall 25 meters long and 15 meters broad is surrounded
If the diagonal of two squares are in the ratio of 2: 5.
by a veradah of uniform width of 3.5 meters. The cost of
Their area will be in the ratio of
flooring the varandah at Rs. 27.50 per square meter is
(a) √2: √5
(a) Rs. 9149.50
(b) 2:5
(b)Rs.8146.50
(c)4:25

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(d) 4: 5 Q290.
Q283. A square and an equilateral triangle are drawn on the
The ratio of base of two triangles is x:y and that of their same base. The ratio of their area is
areas is a : b, then the ratio of their corresponding (a) 2:1
altitudes will be : (b) 1 : 1
(a)a/x : b/y (c)√30:√4
(b)ax:by (d)4:√3
(c) ay : bx Q291.
(d) x/a : b/y If the area of a circle and a square are equal, then the
Q284. ratio of their perimeter is
The area of a field in the shape of a trapezium measures (a) 1 : 1
1440m2. The perpendicular distance between its parallel (b)2: π
sides is 24m. If the ratio of the parallel sides is 5 : 3, the (c)π:2
length of the longer parallel side is : (d)√π :2
(a) 75 m Q292.
(b) 45 m The area of two equilateral triangles are in the ratio 25 :
(c) 120 m 36. Their altitudes will be in the ratio
(d) 60 m (a) 36:25
Q285.
(b) 25: 36
If the ratio of areas of two squares is 225 : 256, then the
(c) 5: 6
ratio of their perimeter is :
(d) √5:√6
(a) 225 : 256
Q293.
(b) 256: 225
If the length and the perimeter of a rectangle are in the
(c) 15: 16
ratio 5 : 16. then its length and breadth will be in the
(d) 16: 15
ratio
Q286.
The area of a triangle is 216 cm2 and its sides are in the (a) 5: 11
ratio 3 : 4 : 5. The perimeter of the triangle is : (b) 5: 8
(a) 6 cm (c)5:4
(b) 12 cm (d) 5:3
(c) 36 cm Q294.
(d) 72 cm Through each vertex of a triangle, a line parallel to the
Q287. opposite side is drawn. the ratio of the perimeter the
A circular wire of radius 42 cm is bent in the form of a new triangle, thus formed, with that of the original
rectangle whose sides are in the ratio of 6 : 5. The triangle is
smaller side of the rectangle is ( Take pie = 22/7) (a) 3 : 2
(a) 60 cm (b) 1 : 2
(b) 30 cm (c) 2 : 1
(c) 25 cm (d) 2 : 3
(d) 36 cm Q295.
Q288. The ratio of the number giving the measure of the
The ratio of the Outer and the inner perimeter of a circumference and the area of a circle of radius 3 cm is:
circular path is 23:22, If the path is 5 meters wide the (a) 1 : 3
diameter of the inner circle is : (b) 2: 3
(a) 110 m (c) 2: 9
(b) 55m (d) 3 : 2
(c) 220 m Q296.
(d) 230 m The height of an equilateral triangle is 4√3 cm. the ratio
Q289. of the area of its circumcircle to that of its incircle is:
The ratio of the area of a square to that of the square (a) 2 : 1
drawn on its diagonal is: (b) 4: 1
(a) 1 : 1 (c) 4 : 3
(b) 1 : 2 (d) 3:2
(c) 1 : 3 Q297.
(d) 1 : 4

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The radius of circle A is twice that of circle B and the If in a triangle ABC, the medians CD and BE intersect
radius of circle B is twice of that of circle C. there area each other at O, then the ratio of the areas of triangle
will be in the ratio: ODE and triangle OBC is:
(a) 16:4:1 (a) 1 : 4
(b) 4:2:1 (b) 6 : 1
(c) 1:2:4 (c) 1 : 12
(d) 1:4:16 (d) 12 : 1
Q298. Q305.
A circle and a square have equal areas, the ratio of a side The ratio of the area of two isosceles triangles having the
of the square and the radius of the circle is: same vertical angle (i.e. angle between equal sides) is 1 :
(a) 1 : √π 4. The ratio of their heights is:
(b)√π : 1 (a) 1 : 4
(c) 1 : √π (b) 2 : 5
(d) π : 1 (c) 1 : 2
Q299. (d) 3 : 4
The sides of a triangle are in the ratio 1/3 : ¼ : 1/5 and Q306.
its perimeter is 94cm. the length of the smallest side of The ratio of length of each equal side and the third side
the triangle is: of an isosceles triangle is 3:4 . if the area is 8√5 units2.the
(a) 18 cm third side is
(b) 22.5 cm (a)3 units
(c) 24 cm (b)2√5 units
(d) 27m (c)8√2units
Q300. (d)12 units
The sides of an quadrilateral are in the ratio 3 : 4 : 5: 6 Q307.
and its perimeter is 72cm. the length of its greatest side The ratio of sides of a triangle is 3:4:5. If area of the
(in cm) is: triangle is 72 square unit then the length of the smallest
(a) 24 side is
(b) 27 (a)4√3 unit
(c) 30 (b)5√3units
(d) 36 (c)6√3 units
Q301. (d)3√3 units
The ratio of the radii of two wheels is 3 : 4. The ratio of Q308.
their circumference is: The ratio of sides of a triangle is 3:4:5 and area of the
(a) 4 : 3 triangle is 72 squares unit .then the area of an equilateral
(b) 3 : 4 triangle whose perimeter is same as that of the pervious
(c) 2 : 2 triangle is
(d) 3 : 2 (a)32√3 sq. units
Q302. (b)48√3 sq. units
The sides of a triangle are in the ratio 2 : 3 : 4. The (c)96 sq. units
perimeter of the triangle is 18cm. the area (in cm 2 ) of the (d)60√3 sq. units
triangle is: Q309.
(a) 2 The parallel sides of a trapezium are in a ratio 2:3 and
(b) 36 their shortest distance is 12 cm. if the area of the
(c) √42 trapezium is 480 sq. cm. the longer of the parallel sides
(d) 3√15 is of length :
Q303. (a) 56 cm
The ratio of the areas of the circumference circle and the (b) 36 cm
incircle of an equilateral triangle is: (c) 42 cm
(a) 2 : 1 (d) 48 cm
(b) 4 : 1 Q310.
(c) 8 : 1 An equilateral triangle is drawn on diagonal of a square .
(d) 3 : 2 the ratio of the area of the triangle to that of the square
Q304. is
(a)√3:2
(b)1:√3

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(c)2:√3 to the side of the triangle, The ratio of the area of the
(d)4:√3 rectangle and the triangle is
Q311. (a) √3:1
Two triangle ABC and DEF are similar to each other in (b) 1:√3
which AB =10cm, DE = 8cm.then the ratio of the are of (c)2: √3
triangle ABC and DEF is (d) 4:√3
(a) 4: 5 Q318.
(b) 25: 16 The radius of a circle is a side of a square. The ratio of the
(c) 64 : 125 area of the circle and the square is
(d) 4 : 7 (a) 1 : π
Q312. (b) π : 1
The ratio between the area of two circles is 4 : 7. What (c) π : 2
will be the ratio of their radii? (d) 2 : π
(а) 2: √7 Q319.
(b) 4: 7 ABC is an isosceles right angled triangle with ∠B 90°, On
(c) 16: 49 the sides AC and AB, two equilateral triangles ACD and
(d) 4: √7 ABE have been constructed, The ratio of area of ∆ABE
Q313. and ∆ACD is
The area of a circle is proportional to the square of its (a) 1 : 3
radius. A small circle of radius 3 cm is drawn within a (b)2:3
larger circle of radius 5 cm. Find the ratio of the area of (c) 1 : 2
the annualar zone to the area of the larger circle (Area of (d) 1: √2
the annular zone is the difference between the area of Q320.
the larger circle and that of the smaller circle ) Two triangles ABC and DEF are similar to each other in
(a) 9 : 16 which AB = 10 cm, DE=8 cm. Then the ratio of the area of
(b) 9:25 triangles ABC and DEF is
(c) 16:25 (a) 4 : 5
(d) 16 : 27. (b) 25: 16
Q314. (c) 64: 125
The diameter of two circles are the side of a square and (d) 4: 7
the diagonal of the square. The ratio of the area of the Q321.
smaller circle and larger circle is ABC is a right angled triangle, B being the right angle.
(a)1:2 Mid-points of BC and AC are respectively B' and A'. The
(b)1:4 ratio of the area of the quadrilateral AA'BB' to the area of
(c)√2:√3 the triangle ABC is
(d)1:√2 (а) 1 : 2
Q315. (b) 2: 3
The ratio of the area of an equilateral triangle and that (c) 3 : 4
of its circumcircle is (d) None of the above
(a)2√3:2π Q322.
(b)4:π The sides of a triangle are in the ratio 1/4 : 1/6: 1/8 and
(c)3√3:4π its perimeter is 91 cm. The difference of the length of
(d)7√2:2π longest side and that of shortest side is
Q316. (a) 19 cm
if the perimeters of a rectangle and a square are equal (b) 20 cm
and the ratio of two adjacent sides of the rectangle is 1 : (c) 28 cm
2 then the ratio of area of the rectangle and that of the (d) 21 cm
square is Q323.
(a) 1 : 1 If the arcs of unit length in two circles subtend angles of
(b)1:2 60° and 75° at their centers , the ratio of their radii is
(c) 2 : 3 (a)3:4
(d) 8: 9 (b)4:5
Q317. (c)5:4
The perimeter of a rectangle and an equilateral triangle (d)3:5
are same, Also, one of the sides of the rectangle is equal Q324.

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ABCD is a parallelogram in which diagonals AC and BD The length of a rectangle is decreased by 10% and its
intersect at O. If E,F,G and H are the mid point of breadth is increased by 10%. By what percent is its area
AO,DO,CO and BO respectively , then the ratio of the changed?
perimeter of the quadrilateral EFGH to the perimeter of (a)0%
parallelogram ABCD is (b)1%
(a)1:4 (c)5%
(b)2:3 (d)100%
(c)1:2 Q330.
(d)1:3 The percentage increase in the area of a rectangle. If each
Q325. of its sides is increased by 20% is:
If the circumference of a circle increases from 4π to 8π , (a)40%
what change occurs in its area ? (b)42%
(a)it doubles (c)44%
(b)it triples (d)46%
(c) it quadruples Q331.
(d) it is halved If the circumference of a circle is reduced by 50%, the
Q326. area will be reduced by?
If the length of a rectangle is increased by 25% and the (a)12.5%
width is decreased by 20%, then the area of the (b)25%
rectangle: (c)50%
(a)Increased by 5 % (d)75%
(b)decrease by 5% Q332.
(c)remains unchanged If the side of a square is increased by 25%, then its area
(d) Increased by 10 % is increased by:
Q327. (a)25%
If the circumference and area of a circle are numerically (b)55%
equal, then the diameter is equal to: (c)40.5%
(a) area of circle (d)56.25%
(b) π /2 Q333.
(c)2 π If the radius of a circle is increased by 50%, its area is
(d)4 increased by:
Q328. (a)125%
If D and E are the midpoints of the side AB and AC (b)100%
respectively of the triangle ABC in the figure given here, (c)75%
the shaded region of the triangle is what per cent of the (d)50%
whole triangular region? Q334.
If the length of a rectangle is increased by 20% and its
breadth is decreased by 20% then its area
(a) Increased by 4 %(
(b) decrease by 4%(
(c) decrease by 1%(
(d) none of these
Q335.
If each side of a rectangle is increased by 50%, its area
will be increased by
(a)50%
(b)125%
(c)100%
(d)250%
(a)50% Q336.
(b)25% If the altitude of a triangle is increased by 10% while its
(c)75% area remains same, its corresponding base will have to
(d)60% be decreased by
Q329. (a)10%
(b)9%

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(c)100/11% If the length of a rectangular plot of land is increased by
(d)100/9% 5% and the breadth is decreased by 10% how much will
Q337. its area increase or decrease
If the circumference of a circle is increased by 50%, then (a)6.5% increase
the area will be increased by (b)5.5 % decrease
(a)50% (c)5.5 increase
(b)75% (d) 6.5% decrease
(c)100% Q345.
(d)125% The radius of circle is increased by 1%. How much does
Q338. the area of the circle increase?
The length and breadth of rectangle are increased by (a)1%
12% and 15% respectively. Its area will be increased by (b)1.1%
(a)136/5% (c)2%
(b)144/5% (d)2.01%
(c)27% Q346.
(d)28% The length of a room floor exceeds its breadth by 20m.
Q339. the area of the floor remains unaltered when the length
If the sides of an equilateral triangle are increased by is decreased by 10m but breadth is increased by 5m. the
20%, 30% and 50% respectively to form a new triangle area of the floor (in square meters) is:
the increase in the perimeter of equilateral triangle is (a)280
(a)25% (b)325
(b)100/3% (c)300
(c)75% (d)420
(d)100% Q347.
Q340. In measuring the sides of a rectangle, there is an excess
Each side of a rectangular field is diminished by 40%. By of 5% on one side and 2% deficit on the other. Then the
how much percent is the area of the field diminished? error percent in the area is:
(a)32% (a)3.3%
(b)64% (b)3%
(c)25% (c)2.9%
(d)16% (d)2.7%
Q341. Q348.
The length of rectangle is increased by 60%. By what The length and breadth of a square are increased by 30%
percent would the breadth to be decreased to maintain and 20% respectively. The area of the rectangle so
the same area? formed exceeds the area of the square by:
(a)75/2% (a)46%
(b)60% (b)66%
(c)75% (c)42%
(d)120% (d)56%
Q342. Q349.
The length and breadth of rectangle are increased by If side of a square is increased by 40%, the percentage
20% and 25% respectively. The increased in the area of increase in its surface area is:
the resulting rectangle will be: (a)40%
(a)60% (b)60%
(b)50% (c)80%
(c)40% (d)96%
(d)30% Q350.
Q343. If the diameter of a circle is increased by 8%, then its
If each side of a square is increased by 10%, its area will area is increased by:
be increased by (a)16.64%
(a)10% (b)6.64%
(b)21% (c)16%
(c)44% (d)16.45%
(d)100% Q351.
Q344.

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One side of a rectangle is increased by 30%. To maintain (b)216 cm2
the same area, the other side will be have to be (c)432 cm2
decreased by: (d)108 cm2
(a)300/13% Q359.
(b)1000/13% The length, breadth and height of a room is 5m , 4 m and
(c)30% 3m respectively. Find the length of the largest bamboo
(d)15% that can be kept inside the room.
Q352. (а) 5 m
The length and breadth of a rectangle are doubled.
(b) 60 m
Percentage increase in area is:
(c) 7m
(a)150%
(d) 5√2 m
(b)200%
Q360.
(c)300%
A wooden box measures 20 cm by 12 cm by 10 cm.
(d)400%
Thickness of wood is 1 cm. volume of wood to make the
Q353.
box ( in cube cm) is
The length of a rectangle is increased by 10% and
breadth decreased by 10%. The area of the new (a) 960
rectangle: (b) 519
(a)neither increased nor decreased (c)2400
(b)increased by 1% (d) 1120
(c)increased by 2% Q361.
(d)decreased by 1% A cuboidal block of 6 cm x 9 cm x 12 cm is cut up into
Q354. exact number of equal cube, The least possible number
If diagonal of a cube is √12 cm, the its volume in cm3 is: of cubes will be
(a)8 (a)6
(b)12 (b) 9
(c)24 (c) 24
(d)3√2 (d) 30
Q355. Q362.
How many cubes, each of edge 3cm, can be cut from a A cistern of capacity 8000 liters measures externally
cube of edge 15cm? 3.3m by 2.6 m by 1.1 m and its walls are 5 cm thick. The
(a)25 thickness of the bottom is:
(b)27 (a)1m
(c)125 (b)10cm
(d)144 (c)1cm
Q356. (d)90cm
What is the volume of a cube in cubic cm whose diagonal Q363.
measures 4√3 cm? The area of three adjacent faces of a cuboid are c, y, z
(a)16 square units respectively If the volume of the cuboid by
(b)27 v, cube units, then the correct relation between x, y, z is
(c)64 (a) v2 = xyz
(d)8 (b) v3 =xyz
Q357. (c) v2 = x3 y3 z3
A cuboidal water tank has 216 liters of water. Its depth is (d) v3 =x2y2z2
1/3 of its length and breadth is ½ of 1/3 of the difference Q364.
of length and breadth. The length of the tank is: The largest sphere is carved out of side 7 cm.The volume
(a)72dm of the sphere (in cm3) will be
(b)18dm (a)718.66
(c)6dm (b)543.72
(d)2dm (c)481.34
Q358. (d)179.67
The volume of cuboid is twice the volume of a cube. If the Q365.
dimensions of the cuboid are 9 cm, 8 cm, 6 cm the total The length (in meters) of the longest rod that can be put
surface area of the cube is: in a room of (in m?) is dimensions 10 m x 10 m x 5 m
(a)72 cm2 is

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(a)15√3 (a) 192cm2
(b)15 (b) 512cm2
(c)10√2 (c) 768cm2
(d)5√3 (d)384cm2
Q366. Q373.
A rectangular sheet of metal is 40 cm by 15 cm. equal The maximum length of a pencil that can be kept in a
squares of side 4cm are cut off at the corners and the rectangular box of dimensions 8 cm × 6cm × 2cm
remainder is folded up to form an open rectangular box. (a) 2√13cm
The volume of the box is (b) 2√14cm
(a)896cm3 (c) 2√26cm
(b)986 m3 (d) 10√2cm
(c)600 m3 Q374.
(d)916 m3 The volume of cubical box is 3.375 cubic meters. The
Q367. length of edge of the box is:
The areas of three consecutive faces of a cuboid are 12 (a)75m
cm2, then the volume (in cm3) of the cuboid is (b) 1.5m
(a) 3600 (c) 1.125m
(b) 100 (d)2.5m
(c) 80 Q375.
(d) 24√3 Two cubes of sides 6 cm each are kept side to form a
Q368. rectangular parallelepiped. The area (in sq.cm) of the
The length of the longest rod that can be placed in a whole surface of the rectangular parallelepiped is:
room which is 12 m long. 9 m broad and 8 m high is: (a)432
(a) 27m (b) 360
(b) 19m (c) 396
(c) 17m (d) 340
(d) 13m Q376.
Q369. 2 cm of rains has fallen on a square km of land. Assuming
The floor of a room is of size 4m × 3m and its height is that 50% of the raindrops could have been collected and
3m. the walls and ceiling of the room require painting. contained in a pool having a 100m × 10m, by what level
The area to be painted is: would the water level in the pool have increased?
(a) 66m2 (a)1km
(b) 54m2 (b) 10m
(c) 42 m2 (c)10cm
(d) 33m2 (d)1m
Q370. Q377.
If the sum of three dimensions and the total surface area A parallelepiped whose side are in ratio 2 : 4: 8 have the
of a rectangular box are 12cm and 94 cm 2 respectively, same volume as a cube. The ratio of their surface area is:
then the maximum length of a stick that can be placed (a)7:5
inside the box is: (b) 4:3
(a) 5√2 cm (c)8:5
(b) 5 cm (d) 7:6
(c) 6 cm Q378.
(d) 2√5 cm If two adjacent sides of a rectangular parallelepiped are
Q371. 1cm and 2cm and the total surface area of the
The area of the four walls of a room is 660m 2 and its parallelepiped is 22 square cm, then the diagonal of the
length is twice its breadth, if the height of the room is 11 parallelepiped is:
m, then area of its floor in m. sq. is: (a)√10
(a) 120 (b) 2√3
(b) 150 (c) √14
(c) 200 (d) 4cm
(d)330 Q379.
Q372. If the sum of the length, breadth and height of a
If the length of the diagonal of a cube is 8√3 cm, then its rectangular parallelepiped is 24cm and the length of its
surface area is: diagonal is 15cm, then its total surface area is:

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(a) 256cm2 (a) 486cm2
(b) 265cm2 (b) 496cm2
(c) 315cm2 (c) 256cm2
(d) 351cm2 (d)658cm2
Q380. Q387.
If the total surface area of a cube is 96 cm square, its Some bricks are arranged in an area measuring 20m 3.. if
volume is: the length, breadth and height of each brick is 25cm,
(a) 56cm3 12.5cm and 8cm respectively, then the number of bricks
(b) 16cm3 are suppose there is no gap in between two bricks:
(c) 64cm3 (a) 6000
(d) 36cm3 (b) 8000
Q381. (c) 4000
The length of the largest possible road that can be placed (d)10000
in a cubical room is 35√3m. the surface area of the Q388.
largest possible sphere that fit within the cubical The whole surface of a cube is 150sq.cm. then the
room(assuming π = 22/7) in sq. m is: volume of the cube is:
(a)3500 (a) 125cm3
(b) 3850 (b) 216cm3
(c) 2450 (c) 343cm3
(d)4250 (d) 512cm3
Q382. Q389.
The volume of air in a room is 204 m3. The height of the The ratio of the length and breadth of a rectangular
room is 6m. what is the floor area of the room? parallelepiped is 5 : 3 and its height is 6cm. if the total
(a) 32m2 surface area of the parallelepiped be 558sq.cm, then its
(b) 46m2 length in dm is:
(c) 44m2 (a)9
(d)34m2 (b) 1.5
Q383. (c) 10
A square of side 3 cm is cut off from each corner of a (d) 15
rectangular sheet of length 24cm and breadth 18cm and Q390.
the remaining sheet is folded to form an open If the sum of the dimensions of a rectangular
rectangular box. The surface area of the box is: parallelepiped is 24cm and the length of the diagonal is
(a)468m2 15cm, then the total surface area of it is:
(b) 396m2 (a) 420cm2
(c) 615m2 (b) 275cm2
(d) 423m2 (c) 351cm2
Q384. (d) 378cm2
Three solid iron cubes of edges 4cm, 5cm, and 6 cm are Q391.
melted together to make a new cube. 62cm3of the melted The length, breadth and height of a cuboid are in the
material is lost due to improper handing. The area of the ratio 3 : 4 : 6 and its volume is 576cm3. The whole
whole surface of the newly formed cube is: surface area of the cuboid is:
(a) 294 (a)216cm2
(b) 343 (b) 324cm2
(c) 125 (c) 432cm2
(d) 216 (d) 460cm2
Q385. Q392.
Area of the floor of a cubical room is 48sq.m. the length If the number of vertices, edges and faces of a
of the longest rod that can be kept in that room is: rectangular parallelepiped are denoted by v, e and f
(a)9m respectively the value of (v- e + f) is:
(b) 12m (a) 4
(c) 18m (b) 1
(d) 6m (c) 0
Q386. (d) 2
Three cubes of sides 6cm, 8cm, 1cm are melted to form a Q393.
new cube. The surface area of the new cube is:

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A low land, 48m long and 31.5m broad is raised to (d)125
6.5dm. for this, earth is removed from a cuboidal hole, Q400.
27m long and 18.2m broad, dug by the side of the land. A cylindrical tank of diameter 35 cm is full of water. If 11
The depth of the hole will be: litres of water is drawn off, the water level in the tank
(a) 3m will drop by:
(b) 2m (a)21/2cm
(c) 2.2m (b)9/7cm
(d) 2.5m (c)14cm
Q394. (d)80/7 cm
A cuboidal shaped water tank 2.1m long and 1.5m broad Q401.
is half filled with water. If 630 litres more water is The volume of a right circular cylinder whose height is
poured into tank, the water level will rise. 40cm, and circumference of its base is 66 cm is:
(a)2cm (a)55440cm2
(b)0.15cm (b)3465cm2
(c) 0.20 m (c)7720cm2
(d)0.18cm (d)13860cm2
Q395. Q402.
A solid cuboid of dimensions 8 cm× 4cm× 2cm is melted The circumference of the base of a circular cylinder is 6π
and cast into identical cubes of edge 2cm. number of cm. the height of the cylinder is equal to the diameter of
such identical cubes is. the base. How many litres of water can it hold?
(a)16 (a)54 π cc
(b)4 (b)36 π cc
(c)10 (c)0.054π cc
(d)8 (d)0.54πcc
Q396. Q403.
A metallic hemisphere is melted and recast in the shape The volume of a right circular cylinder is equal to the
of cone with the same base radius (R) as that of the volume of that right circular cone whose height is 108cm
hemisphere. If H is the height of the cone, then: and diameter of base is 30cm. if the height of the cylinder
(a)H=2R is 9cm, the diameter of its base is:
(b)H=2/3R (a)30cm
(c) H=√3R (b)60cm
(d)B=3R (c)50cm
Q397. (d)40cm
If the radius of a sphere is increased by 2cm, its surface Q404.
area increased by 352cm3 the radius of sphere before Three solid metallic spheres of diameter 6cm, 8 cm, and
change is: 10cm are melted and recast into a new solid sphere. The
(a)3cm diameter of the new sphere is:
(b) 4cm (a)4cm
(c)5cm (b)6cm
(d)6cm (c)8cm
Q398. (d)12cm
The height of a conical tank is 60cm and the diameter of Q405.
its base is 64cm. the cost of painting it from outside at Three slid metallic balls of radii 3cm, 4cm, 5cm, are
the rate of Rs. 35 per sq. m is: melted and molded into a single solid ball. The radius of
(a)Rs. 52 approx the new balls is:
(b)Rs. 39.20 approx (a)2cm
(c)Rs. 35.20 approx (b)3cm
(d)Rs. 23.94 approx (c)4cm
Q399. (d)6cm
A solid metallic cone of height 10cm, radius of base 20cm Q406.
is melted to make spherical balls each of 4 cm diameter, Three solid spheres of a metal whose radii are 1cm, 6cm
how many such balls can be made? and 8 cm are melted to form another solid sphere. The
(a)25 radius of this new sphere is:
(b)75 (a)10.5cm
(c)50 (b)9.5cm

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(c)10cm (a)2424cm2
(d)9cm (b)2446cm2
Q407. (c)2484cm2
The slant height of a conical mountain is 2.5 km and the (d)2464cm2
area of its base is 1.54 km2. Taking π = 22/7, the height Q414.
of the mountain is: The surface area of a sphere is 64π cm2. Its diameter is
(a)2.2km equal to:
(b)2.4km (a) 16 cm
(c)3km (b) 8 cm
(d)3.11km (c) 4 cm
Q408. (d) 2 cm
The base of a conical a tent is 19.2 metres in diameter Q415.
and the height is 2.8metrres. the area of the canvas The diameter of the base of a cylindrical drum is 35dm.
required to put up such a tent ( in square meters) (taking and the height is 24 dm. it is a full of kerosene. How
π = 22/7) is nearly: many tins each of size 25 cm × 22cm × 35cm can be filled
(a)3017.1 with kerosene from the drum?
(b)3170 (a) 1200
(c)301.7 (b) 1020
(d)30.17 (c) 600
Q409. (d) 120
A hollow cylindrical tube 20cm long. Is made of iron and Q416.
its external and internal diameters are 8cm and 6cm A hollow iron pipe is 21cm long and its exterior diameter
respectively. The volume of iron used in making the tube is 8cm. if the thickness of the pipe is 1cm and iron
is (π = 22/7 ) weights 8g/cm3, then the weight of the pipe is:
(a)1760cu.cm (a) 3.696 kg
(b)880 cu.cm (b) 3.6 kg
(c)440 cu.cm (c) 36 kg
(d)220 cu.cm (d) 36.9 kg
Q410. Q417.
A sphere of radius 2cm is put into water contained in a The volume of a right circular cylinder, 14 cm in height,
cylinder of base radius 4cm. if the sphere is completely is equal to that of a cube whose edge is 11cm, the radius
immersed in the water, the water level in the cylinder of the base of the cylinder is:
rise by: (a) 5.2 cm
(a)1/3cm (b) 5.5 cm
(b)1/2cm (c) 11.0 cm
(c)2/3cm (d) 22.0 cm
(d)2cm Q418.
Q411. If the volume of a right circular cylinder is 9πh m 3, where
A solid metallic spherical ball of diameter 6cm is melted h is its height (in meters) then the diameter of the base
and recast into a cone with diameter of the base as 12cm. of the cylinder is equal to:
the height of the cone is: (a) 3 m
(a)6cm (b) 6 m
(b)2cm (c) 9 m
(c)4cm (d) 12 m
(d)3cm Q419.
Q412. Each of the measure of the radius of base of a cone and
The volume of a right circular cone is 1232cm3 and its that of a sphere is 8cm. also, the volume of these two
vertical height is 24cm. its curved surface area is: solids are equal. The slant height of the cone is:
(a)154cm2 (a)8√17cm
(b)550cm2 (b) 4√17cm
(c)604cm2 (c) 34√2cm
(d)704cm2 (d)3cm
Q413. Q420.
The volume of a sphere is 88/21 × (14)3 cm3 the curved
surface of the sphere is assuming π = 22/7.

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A well 20m in diameter is dug 14 m deep and the earth Q427.
taken out is spread all around it to a width of 5m to form If the volume and surface area of a sphere are
an embankment. The height of the embankment is: numerically the same, then its radius is:
(a) 10 m (a) 1 unit
(b) 11 m (b) 2 units
(c) 11.2 m (c) 3 units
(d) 11.5 m (d) 4 units
Q421. Q428.
The diameter of the iron ball used for the shot put game In a right circular cone, the radius of its base is 7cm and
is 14cm. It is melted and then a solid cylinder of height its height 24cm. a cross section is made through the
7/3 cm is made. What will be the diameter of the base of midpoint of the height parallel to the base.
the cylinder? (a)169 cm3
(a)14cm (b)154cm3
(b)28cm (c)1078 cm3
(c)14/3cm (d)800 cm3
(d)28/3cm Q429.
Q422. Some solid metallic right circular cones, each with radius
The sum of radii of two spheres is 10cm and the sum of of the base 3cm and height 4cm, are melted to form a
their volume is 880cm3. What will be the product of their solid sphere of radius 6cm. the number of right circular
radii? cones is:
(a)21 (a) 12
(b)79/3 (b) 24
(c)100/3 (c) 48
(d)70 (d) 6
Q423. Q430.
A rectangular paper sheet of dimensions 22cm × 12cm is A right circular cylinder of height 16 cm is covered by a
folder in the form of a cylinder along its length. What will rectangular tin foil of size 16 cm x 22 cm, The volume of
be the volume of this cylinder? the cylinder is
(a)460cm2 (a) 352 cm3
(b)462cm2 (b) 308 cm3
(c)624cm2 (c) 616 cm3
(d)400cm2 (d) 176 cm3
Q424. Q431.
A copper rod of 1cm diameter and 8cm length is drawn If the area of the base of a core is 770 cm2 and the area of
into a wire of uniform diameter and 18m length. The the its curved surface is 814 cm2. then find its volume.
radius (in cm) of the wire is: (a) 213√5 cm2
(a)1/15 (b) 392√5 cm2
(b)1/30 (c) 550 √5 cm2
(c)2/15 (d) 616√5 сm2
(d)15 Q432.
Q425. The size of a rectangular piece of paper is 100 cm x 44
12 spheres of the same size are made by melting a solid cm. A cylinder is formed by rolling the paper along its
cylinder of 16cm diameter and 2cm height. The diameter breadth. The volume of the cylinder is (use pie =22/7)
of each sphere is: (a) 4400 cm3
(a) 2 cm (b) 15400 cm3
(b) 4 cm (c) 35000 cm3
(c) 3 cm (d) 144 cm3
(d) √3 cm Q433.
Q426. The radius of the base and height of a metallic solid
When the circumference of a toy balloon is increased cylinder are r cm and 6 cm respectively. It is melted and
from 20cm to 25cm its radius (in cm) is increased by: recast into a solid cone of the same radius of base, The
(a)5 height of the cone is :
(b)5/π (a) 54 cm
(c)5/2π (b) 27 cm
(d)π/5 (c) 18 cm

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(d) 9 cm (a)72 π cm3
Q434. (b)144 π cm3
The total surface area of a metallic hemisphere is 1848 (c)108√cm3
cm2. The hemisphere is melted to form a solid right (d)54√6cm2
circular cone. If the radius of the base of the cone is the Q441.
same as the radius of the hemisphere its height is A solid metallic sphere of radius 3 dm is melted to form a
(a) 42 cm circular sheet 1 mm thickness. The diameter of the sheet
(b) 26 cm so formed is:
(c) 28 cm (a)26 m
(d) 30 cm (b)24m
Q435. (c)12m
A right circular cylinder is formed by rolling a (d)6m
rectangular paper 12 cm long and 3 cm wide along its Q442.
length. The radius of the base of the cylinder will be Water flows through a cylindrical pipe, whose radius is
(a)3/2 π 7cm, at 5 meter per second. The time, it takes to fill an
(b)6/ π empty water tank with height 1.54 metres and area of
(c)9/2 π the base ( 3×5) square metres is:
(d)2 π (a)6 minutes
Q436. (b)5 minutes
What part of a ditch, 48 metres long. 16.5 metres broad (c)10 minutes
and 4 meters deep can be filled by the earth got by (d)9 minutes
digging a cylindrical tunnel of diameter 4 meters and Q443.
length 56 metres? If S denotes the area of the curved surface of a right
(a)1/9 circular cone of height h and semi-vertical angle A then S
(b)2/9 equals
(c)7/9 (a)π h2tan2 α
(d)8/9 (b)1/3 π h2tan2 α
Q437. (c) π h2 sec α tan2 α
The volume of the metal of cylindrical pipe is 748 cm 3. (d) 1/3 π sec α tan2 α
The length of the pipe is 14 cm and its external radius is Q444.
9 cm. its thickness is: The height and the radius of the base of a right circular
(a) 1 cm cone are 12 cm and 6 cm respectively. The radius of the
(b) 5.2 cm circular cross section of the cone cut by a plane parallel
(c) 2.3 cm to base at a distance of 3cm from the base is:
(d) 3.7 cm (a) 4 cm
Q438. (b) 5.5 cm
Two iron sphere each of diameter 6 cm are immersed in (c)4.5cm
the water contained in a cylindrical vessel of radius 6 cm. (d)3.5 cm
the level of the water in the vessel will be raised by: Q445.
(a) 1 cm If S1 and S2 be the surface areas of a sphere and the
(b) 2 cm curved surface area of the circumscribed cylinder
(c) 3 cm respectively, then S1 is equal to
(d) 6 cm (a)3/4 S2
Q439. (b)1/2 S2
The height of the cone is 30cm. a small cone is cut off at (c)2/3 S2
the top by a plane parallel to its base. If its volume is (d)S2
1/27 of the volume of the cone. At what height above the Q446.
base, is the section made? The volume of a right circular cylinder and that of a
(a)6cm sphere are equal and their radii are also equal. If the
(b)8cm eight of the cylinder be h and the diameter of the sphere
(c)10cm d. then which of the following relation is correct?
(d)20cm (a)h=d
Q440. (b) 2h=d
The total surface area of a solid hemisphere is 108π cm 2. (c) 2h=3d
The volume of the hemisphere is: (d) 3h=2d

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Q447. Q454.
Water is being pumped out through a circular pipe Two solid right cones of equal height and of radii r1 and
whose internal diameter is 7cm. if the flow of water is r2 are melted and made to form a solid sphere of radius
12cm. if the flow of water is 12cm per second, how many R. then the height of the cone is:
litres of water is being pumped out in one hour? (a)4R2/ r12r22
(a) 1663.2 (b)4R/ r1 r2
(b) 1500 (c) 4R3/ r12r22
(c) 1747.6 (d) R2/ r12r22
(d) 2000 Q455.
Q448. The ratio of height and the diameter of a right circular
The lateral surface area of a cylinder is 1056cm2 and its cone is 3 : 2 and its volume is 1078cc, then its height is:
height is 16cm. find its volume. (a)7
(a) 4545 cm3 (b)14
(b) 4455 cm3 (c)21
(c) 5445 cm3 (d)28
(d) 5544 cm3 Q456.
Q449. From right circular cylinder of radius 10cm and height
A solid metallic cone is melted and recast into a solid 21 cm a right circular cone of the same base radius
cylinder of the same base as that of the cone. If the height removed. If the volume of the remaining portion is
of the cylinder is 7cm, the height of the cone was 4400cm3 then the height of the removed cone is:
(a) 20 cm (a) 15 cm
(b) 21 cm (b) 18 cm
(c) 28 cm (c) 21 cm
(d) 24 cm (d) 24 cm
Q450. Q457.
A copper wire of length 26m and diameter 2mm is A child reshapes a cone made up of a clay of height 24cm
melted to form a sphere. The radius of the sphere (in cm) and radius 6 cm into a sphere. The radius (in cm) of the
is: sphere is:
(a) 2.5 (a)6
(b) 3 (b)12
(c) 3.5 (c)24
(d) 4 (d)48
Q451. Q458.
The diameter of the base of a right circular cone is 4cm A solid cylinder has total surface area of 462sq.cm. its
and its height 2√3 cm. the slant height of the cone is: curved surface area is one third of the total surface area.
(a)5cm Then the radius of the cylinder is:
(b)4cm (a)7cm
(c)2√3cm (b)3.5cm
(d)3cm (c)9cm
Q452. (d)11cm
The rain water from a roof 22m × 20m drains into a Q459.
cylindrical vessel having a diameter of 2m and height The diameter of a cylinder is 7cm and its height is 16cm.
3.5m, if the vessel is just full, then the rainfall (in cm ) is: using the value of π=22/7 the lateral surface area of the
(a)2 cylinder is::
(b)2.5 (a)352cm2
(c)3 (b)350cm2
(d)4.5 (c)355cm2
Q453. (d)348cm2
From a solid cylinder of height 10cm and radius of the Q460.
base 6cm, a cone of same height and same base is The height of a solid right circular cylinder is 6 metres
removed. The volume of the remaining solid is: and three times the sum of the area of its two end faces is
(a)240 π cu. cm twice the area of its curved surface, the radius of its
(b)5280 π cu. cm base( in meters) is:
(c)620 π cu. cm (a) 4
(d)360 π cu. cm (b) 2

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(c) 8 A cylinder has 'r' as the radius of the base and 'h' as the
(d) 10 height. The radius of base 'of another cylinder, having
Q461. double the volume but the same height as that of the first
A semi- circular sheet of metal of diameter 28 cm is bent cylinder must be equal to
into an open conical cup. The depth of the cup is (a)r/√2
approximately (b)2r
(a) 11 cm (c)r√2
(d)√2r
(b) 12 cm
Q468.
(c) 13 cm
From a solid cylinder whose height is 12 cm and
(d) 14 cm
diameter 10cm a conical cavity of same height and same
Q462.
diameter of the base is hollowed out. The volume of the
A right angled sector of radius r cm is rolled up into a
remaining solid is approximately (pie =22/7)
cone in such a way that the two binding radii are joined
(a) 942. 86 cm3
together. Then the curved surface area of the cone is:
(b) 314.29 cm3
(a) π r2 cm2
(c) 628. 57 cm3
(b) π r2 /4 cm2
(d) 450.76 cm3
(c) π r2 /2 cm2
Q469.
(d) 2π r2 cm2
The radius of a cylinder is 10 cm and height is 4 cm. The
Q463.
number of centimeters that may be added either to the
The radius of the base of a conical tent is 16 meter. if
radius or to the height to get the same increase in the
2992/7 sq. meter canvas is required to construct the
volume of the cylinder is
tent , then the slant height of tent is (take pie =22/7)
(a) 5 cm
(a) 17 metres
(b) 4 cm
(b) 15 metres
(c) 25 cm
(c) 19 metres
(d) 16 cm
(d) 8.5 metres
Q470.
Q464.
The radius of the base of a right circular cone is doubled
A circus tent is cylindrical up to a height of 3 m and
keeping its height fixed. The volume of the cone will be :
conical above it. If its diameter is 105 m and the slant
(a) Three times of the previous volume
height of the conical part is 63 m, then the total area of
(b) four times of the previous volume
the canvas required to make the tent is (take pie =22/7 )
(c) √2 times of the previous volume
(a) 11385 m2
(d)double of the pervious volume
(b) 10395 m2
Q471.
(c) 9900 m2
The base of a right circular cone has the same radius a as
(d) 990 m2
that of a sphere, Both the sphere and the cone have the
Q465.
same volume. Height of the cone is
A toy is in the form of a cone mounted on a hemisphere,
(a) 3a
The radius of the hemisphere and that of the cone is 3 cm
(b) 4a
and height of the cone is 4 cm. The total surface area of
(c)7/4 a
the toy (take pie =22/7 )n is
(d) 7/3 a
(a)75.43 sq.cm
Q472.
(b)103.71 sq.cm
The circumference of the base of a 16 cm high solid cone
(c)85.35 sq.cm
is 33 cm What is the volume of the cone in cm3 ?
(d)120.71 sq.cm
(a)1028
Q466.
(b) 616
A cylindrical rod of iron whose height is eight times its
(c)462
radius is melted and cast into spherical balls each of half
(d) 828
the radius of the cylinder. The number of such spherical
Q473.
balls is
A solid sphere of 6 cm diameter is melted and recast into
(a) 12
8 solid spheres of equal volume. The radius(in cm) of
(b) 16
each small sphere is
(c)24
(a) 1.5
(d) 48
(b) 3
Q467.
(c) 2

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(d) 2.5 The area of the curved surface and the area of the base of
Q474. a right circular cylinder is a square cm b by square cm
In a cylindrical vessel of diameter 24 cm filled up with respectively. The height of the cylinder is
sufficient quantity of water, a solid spherical ball of (a)2a√πb cm
radius 6 cm is completely immersed. Then the increase (b)a √b/2√π cm
in height of water level is : (c)a/2√πb cm
(a) 1.5 cm (d)a √π /2 √b
(b) 2 cm Q481.
(c) 3 cm The volume of a solid hemisphere is 19404 cm 3. Its total
(d)4.2 cm surface area is
Q475. (a) 4158 cm2
A solid wooden toy is in the shape of a right circular cone (b) 2858 cm2
mounted on a hemisphere. If the radius of the (c) 1738 cm2
hemisphere is 4.2 cm and the total height of the toy is (d) 2038 cm2
10.2 cm find the volume of wooden toy (nearly), Q482.
(a) 104 cm3 A solid hemisphere is of radius 11 cm. the curved surface
(b) 162 cm3 area in sq.
(c) 421 cm3 (a) 1140.85
(d) 266 cm3 (b) 1386.00
Q476. (c) 760.57
If a solid cone of volume 27π cm3 is kept inside a hollow (d) 860.57
cylinder whose radius and height are that of the cone, Q483.
then the volume of water needed to fill the empty space The base of a cone and a cylinder have the same radius 6
is cm. They have also the same height 8 cII. The ratio of the
(a) 3 π cm3 curved surface of the cylinder to that of the cone is
(b) 18 π cm3 (a) 8: 5
(c) 54 π cm3 (b) 8 : 3
(d) 81 π cm3 (c) 4:3
Q477. (d) 5 : 3
A cylindrical can whose base is horizontal and is of Q484.
internal radius 3.5 cm contains sufficient water so that A right cylindrical vessel is full with water. How many
when a solid sphere is placed inside, water just covers right cones having the same diameter and height as that
the sphere. The sphere fits in the can exactly. The depth of the right cylinder will be needed to store that water
of water in the can before the sphere was put, is (take 22/7 )
(a)35/3cm (a)4
(b)17/3cm (b)2
(c)7/3 cm (c)3
(d)14/3 cm (d)6
Q478. Q485.
The base of a right circular come has the same radius 'a' A spherical lead hall of radius 10 cm is melted and small
as that of a sphere, Both the sphere and the cone have lead balls of radius 5 mm are made the total number of
the same volume. Height of the come is possible small lead balls is (take pie =22/7)
(a) 3a (a)8000
(b) 4a (b) 400
(c) 7/4a (c) 800
(d) 7/3a (d) 125
Q479. Q486.
The radius and height of a cylinder are in the ratio 5: 7 The number of spherical bullets that can be made out of
and its volume is 550 cm3. Calculate its curved surface solid cube of lead whose edge measures 44 cm each
area in sq. cm. bullet being of 4 cm diameter is (take pie =22/7)
(a) 110 (a) 2541
(b) 444 (b) 2451
(c) 220 (c) 2514
(d) 616 (d) 2415
Q480. Q487.

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The radius of a metallic cylinder is 3 cm and its height is The volume of a conical tent is 1232 cu. m and the area of
5 cm. It is melted and molded into small cones, each of its base is 154 sq. m Find the length of the canvas
height 1 cm and base radius 1 mm. The number of such required to build the tent, if the canvas is 2 m in width.
cones formed is (Take pie =22/7)
(a) 450 (a) 270 m
(b) 1350 (b) 272 m
(c) 8500 (c) 276 m
(d) 13500 (d) 275 m
Q488. Q495.
A sector is formed by opening out a cone of base radius If the ratio of the diameters of two right circular cones
8 cm and height 6 cm. Then the radius of the sector is (in of equal height be 3: 4 then the ratio of their volume will
cm) be
(a) 4 (a) 3: 4
(b) 8 (b) 9 : 16
(c) 10 (c) 16 : 9
(d) 6 (d) 27 : 64
Q489. Q496.
A solid cone of height 9 cm with diameter of is base 18 The surface area of two spheres are in the ratio 4 : 9.
cm is cut out from a wooden solid sphere of radius 9cm Their volumes will be in the ratio
the percentage of wood wasted is : (a) 2: 3
(a)25% (b) 4: 9
(b) 30% (c) 8:27
(c)50% (d) 64 : 729
(d) 75% Q497.
Q490. A semicircular sheet of metal of diameter 28 cm is bent
The perimeter of the base of a right circular cylinder is ‘a’ in an open conical cup. The capacity of the cup ( take pie
unit. If the volume of the cylinder is V cubic unit, then the =22/7 )
height of the cylinder is (a) 624.26 cm3
(a)4a2V/π unit
(b) 622.36 cm3
(b) 4 π a2 /V unit
(c) 622.56 cm3
(c) π a2 V/4 unit
(d) 623.20 cm3
(d) 4 π V/a2 unit
Q498.
Q491.
A conical flask is full of water. The flask has base radius r
What is the height of a cylinder that has the same volume
and height h, This water is poured into a cylindrical flask
and radius as a sphere of diameter 12 cm ?
of base radius m, height of cylindrical flask is
(a) 7 cm
(a)m/2h
(b) 10 cm.
(b)h/2 m2
(c) 9 cm
(c)2h/m
(d) 8 cm
(d) r2h/3m2
Q492.
Q499.
The perimeter of the base of a right circular cone is 8
A solid spherical copper ball whose diameter is 14 cm is
cm. If the height of the cone is 21 cm, then its volume is :
melted and converted into a wire having diameter equal
(a) 108 π cm3
to 14 cm. The length of the Wire is
(b) 112/ π cm3
(a) 27 cm
(c) 112 π cm3
(b) 16/3 cm
(d) cm 108/ π cm3
(c) 15 cm
Q493.
(d)28/3 cm
If the volume of two right circular cones are in the ratio 4
Q500.
: 1 and their diameter are in the ratio 5 : 4, then the ratio
A sphere of diameter 6cm is dropped in a right circular
of their height is
cylindrical vessel partly filled with water. The diameter
(a) 25 : 16
of the cylindrical vessel is 12 cm. If the sphere is just
(b) 25 : 64
completely submerged in water, then the rise of water
(c) 64 : 25
level in the cylindrical vessel is
(d) 16:25
(a)2cm
Q494.
(b)1cm

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(c)3cm (b) less than 8800
(d)4cm (c) equal to 8400
Q501. (d) greater than 9000
A copper sphere of diameter 18 cm is drawn into a wire Q508.
of diameter 4 mm The length of the wire in meter is : If a metallic cone of radius 30 cm and height 45 cm is
(a) 2.43 m melted and recast into metallic spheres of radius 5 cm,
(b) 243 m find the number of spheres,
(c) 2430 m (a) 81
(d) 24.3 m (b) 41
Q502. (c) 80
In A rectangular block of metal has a dimension 21 cm, (d) 40
77 cm and 24 cm, The block has been melted into a Q509.
sphere. The radius of the sphere .is (take pie 22/7) A metallic sphere of radius 10.5 cm is melted and then
(a) 21 cm recast into small cones each of radius 3.5 cm and height
(b) 7 cm 3 cm. The number of cones thus formed is
(c) 14 cm (a) 140
(d) 28 cm (b) 132
Q503. (c) 112
The radius of cross-section of a solid cylindrical rod of (d) 126
iron is 50 cm. the cylinder is melted down and formed Q510.
into 6 solid spherical balls of the same radius as that of A right circular cone is 3.6 cm high and radius of its base
the cylinder. The length of the rod (in metres) is is 1.6 cm. It is melted and recast into a right circular cone
(a)0.08 with radius of its base as 1.2 cm. Then the height of the
(b)2 cone (in cm) is
(c)3 (a) 3.6 cm
(d)4 (b) 4.8 cm
Q504. (c) 6.4 cm
Two right circular cones of equal height of radii of base (d) 7.2 cm
3 cm and 4cm melted together and made a solid sphere Q511.
of radius 5 cm, the height of a cone is If surface area and volume of a sphere are S and V
(a) 10 cm respectively ,then value of S3/V2 is
(b) 20 cm (a)36 π units
(c) 30 cm (b)9 π units
(d) 40 cm (c)18π units
Q505. (d) 27 π units
The radius of the base and the height of a right circular Q512.
cone are doubled, The volume of the cone will be Assume that a drop of water is spherical and its diameter
(a) 8 times of the previous volume is one tenth of 1 cm. a conical glass has a height equal to
(b) three times of the previous volume the diameter of its rim. If 32,000 drops of water fill the
(c) 3√2 times of the previous volume glass completely, Then the height of the glass (in cm) is
(d) 6 times of the previous volume (a) 1
Q506. (b) 2
If h, c, v are respectively the height, curved surface area (c) 3
and volume of a light circular cone then the value of 3 (d) 4
πvh3- c2h2 + 9v2 is Q513.
(a) 2 A tank 40 m long, 30 m broad and 12 m deep is dug in a
(b) - 1 field 1000 m long and 30 m wide. By how much will the
(c) 1 level of the field rise if the earth dug out of the tank is
(d) 0 evenly spread over the field?
Q507. (a) 2 meter
The total number of spherical bullets, each of diameter 5 (b) 1.2 meter
decimeter , that can be made by utilizing the maxim In of (c) 0.5 meter
a rectangular block of lead with 11 meter length, 10 (d) 5 meter
meter breadth and 5 meter width is (assume pie =3) Q514.
(a) equal to 8800

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 A sphere is cut into two hemispheres. One of them is A solid metallic spherical ball of diameter 6 cm is melted
used as bowl. It takes 8 bowlfuls of this to fill a conical and recast into a cone with diameter of the base as 12
vessel of height 12 cm and radius 6 cm. The radius of the cm. The height of the cone is
sphere (in centimeter) (a) 2 cm
(a)3 (b) 3 cm
(b)2 (c) 4 cm
(c)4 (d) 6 cm
(d)6 Q521.
Q515. A hemispherical bowl of internal radius 15 cm contains a
The height of a cone is 30 cm, A small cone is cut off at liquid. The liquid is to be filled into cylindrical shaped
the top by a plane parallel to the base . if its volumebe bottles of diameter 5 cm and height 6 cm. The number of
1/27th of the volume of the given cone, at what height bottles required to empty the bowl is
above the base is the section made? (a) 30
(a) 19 cm (b) 40
(b) 20 cm (c) 50
(c) 12 cm (d) 60
(d) 15 cm Q522.
Q516. If V1 , V2 and V3, be the volumes of a right circular cone, a
A ball of lead 4 cm in diameter is covered with gold. If the sphere and aright circular cylinder having the same
volume of the gold and lead are equal then the thickness radius and same height then
of gold [given 3√2 =1.259) is approximately (a)V1 = V2/4 =V3/3
(a) 5.038 cm (b) V1/2 =V2/3 = V3
(b) 5.190 cm (c) V1/3 =V2/2 = V3
(c) 1.038 cm (d) V1/3=V2 =V3 =V3/2
(d) 0.518 cm Q523.
Q517. the surface area of a sphere is 346.5 cm2, then its radius
A conical cup is filled with ice cream. The ice-cream [taking Pie =22/7]
forms a hemispherical shape on its open top. The height (a)7cm
of the hemispherical part is 7 cm. the radius of the (b)3.25cm
hemispherical part equals the height of the cone. Then (c)5.25 cm
the volume of the ice cream is (d) 9 cm
(a)1078 cubic cm Q524.
(b)1708 cubic cm Deepali makes a model of a cylindrical kaleidoscope for
(c)7108 cubic cm her science project. she uses a chart paper to make it . if
(d)7180 cubic cm the length of the kaleidoscope is 25 cm and radius 35
Q518. cm , the area of the paper she used , in sq. cm. is [ take
A hollow sphere of internal and external diameter 6 cm pie = 22/7]
and 10 cm respectively is melted into a right circular (a) 1100
cone of diameter 8 cm. The height of the cone is (b) 5500
(a) 22.5 cm (c) 500
(b) 23.5 cm (d) 450
(c) 24.5 cm Q525.
(d) 25.5 cm If the volume of a sphere is numerically equal to its
Q519. surface area then its diameter is;
A flask in the shape of a right circular cone of height 24 (а) 4 cm
cm is filled with water. The water is poured in right
(b) 6 cm
circular cylindrical flask whose radius is 1/3 rd of radius
(c) 3 cm
of the base of the circular cone. Then the height of the
(d) 2 cm
water in the cylindrical flask is
Q526.
(a) 32 cm
5 persons live in a tent. If each person requires 16 m 2 of
(b) 24 cm
floor area and 100 m3 space for air then the height of
(c) 48 cm
the cone of smallest size to accommodate these persons
(d) 72 cm
would be?
Q520.
(a) 16m
(b) 18.75 m

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(c) 10.25 m Q533.
(d) 20 m Assume that a drop of water is spherical and its
Q527. diameter is one tenth of a cm. A conical glass has a height
The numerical values of the volume and the area of the equal to the diameter of its rim. If 32000 drops of water
lateral surface of a right circular cone are equal. If the fill the glass completely, then the height of the glass is
height of the cone be h and radius be r, the value of 1/h 2+ (a) 3
1/r2 is (b) 4
(a) 9/1 (c) 1
(b) 3/1 (d) 2
(c) 1/3 Q534.
(d) 1/9 If the height of a cylinder is 4 times its circumference, the
Q528. volume of the cylinder in terms of its circumference, c is
Then is wooden sphere of radius 6√3 cm. the surface (a) 2c3/ π
area of the largest possible cube cut out from the sphere (b)c3 / π
will be (c) 4 πc3
(a) 464 √3 cm2 (d) 2 π c3
(b) 646√3 cm2 Q535.
(c) 864 cm2 The radii of a sphere and a right circular cylinder are 3
(d) 462 cm2 cm each. If their volumes are equal, then curved surface
Q529. area of the cylinder is (Assume π =22/7)
If a hemisphere is melted and four spheres of equal (a) 528/7 cm2
volume are made, the radius of each sphere will be equal (b)458/ 7 cm2
to (c)521/7 cm2
(a) 1/4th of the hemisphere (d)507/3 cm2
(b) radius of the hemisphere Q536.
(c) 1/2 of the radius of the hemisphere The radius of a hemispherical bowl is 6 cm. The capacity
of the bowel is: (take π =22/7)
(d) 1/6 th of the radius of the hemisphere
(a) 452.57 cm3
Q530.
(b) 452 cm3
The portion of a ditch 48 m long. 16.5 m wide and 4 m
(c) 345.53 cm3
deep that can be filled with stones and earth available
(d) 405.51 cm3
during excavation of a tunnel, cylindrical in shape, of
Q537.
diameter 4 m and length 56 m is[take pie =22/7]
The total surface area of a right circular cylinder with
(a) 1/9 भाग radius of the base 7 cm and height 20cm is :
(b)1/2 भाग (a)140 cm2
(b)1000 cm2
(c)1/4 भाग
(c)900 cm2
(d) 2/9 भाग (d) 1188 cm2
Q531. Q538.
From a solid right circular cylinder of length 4 cm and The radius of base and curved surface area of a right
diameter 6 cm, a conical cavity of the same height and cylinder 'r' units and 4 πrh square units respectively.
base is hollowed out. The whole surface of the remaining The height of the cylinder is:
solidi [in square cm.) is (a)4h units
(a) 48 π (b)h/2 units
(b) 63 π (c)h units
(c) 15 π (d) 2h units
(d) 24 π Q539.
Q532. A hemi-spherical bowl has 3.5 cm radius. It is to be
A spherical ball of radius 1 cm is dropped into a conical painted inside as well as outside. The cost of painting it
vessel of radius 3 cm and slant height 6 cm, The volume at the rate of Rs. 5 per 10sq. cm. will be:
of water (in cm3), that can just immerse the ball is (a) Rs. 77
(a)5 π /3 (b) Rs. 175
(b)3 π (c)Rs. 50
(c) π /3 (d) Rs. 100
(d)4 π/3 Q540.

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 The volume of a right circular cone which is obtained Water is flowing at the rate of 3 km/ Hr. through a
from a wooden cube of edge 4.2 dm wasting minimum circular pipe of 20 cm internal diameter into a circular
amount of wood is: cistern of diameter 10 m and depth 2m. In how much
(a) 194.04 cu.dm time will the cistern be filled?
(a) 1 hour
(b) 19.404 cu. dm
(b) 1 hour 40 minutes
(c) 1940.4 cu. dm
(c) 1 hour 20 minutes
(d) 1940. 4cu. dm
(d) 2 hours 40 minutes
Q541.
Q548.
If the radius of a sphere is increased by 2 cm, then its
Water flows at the rate of 10 metres per minute from
surface area increase by 352cm2. The radius of the
cylindrical pipe 5 mm in diameter, How long it will take
sphere initially was : [use π=22/7]
to fill up a conical vessel whose diameter, at the base is
(a) 3 cm
30 cm and depth 24 cm?
(b) 5 cm
(a) 28 minutes 48 seconds
(c) 4 cm
(b) 51 minutes 12 seconds
(d) 6 cm
(c) 51 minutes 24 seconds
Q542.
(d) 28 minutes 36 seconds
A right triangle with sides 9 cm, 12cm and 15 cm is
Q549.
rotated about the side of 9 cm to form a cone. The
The radius of the base of conical tent is 12 m, The tent is
volume of the cone so formed is:
9 m high. Find the cost of canvas required to make the
(a) 432 π cm2
tent, if one square meter of canvas costs Rs. 120 (Take
(b) 327 π cm3
pie = 3.14)
(c) 334 cm2
(a) Rs.67,830
(d) 324 cm2
(b) Rs.67,800
Q543.
(c) Rs.67,820
The volume of the largest right circular cone that can be
(d) Rs.67,824
cut Out of a cube of edge 7 cm? (use π= 22/7)
Q550.
(a) 13.6 cm3
A plate of square base made of brass is of length x cm
(b) 147.68 cm3 and thickness 1 mm. The plate weighs 4725 gm., If 1
(c) 89.9 cm3 cubic cm of brass weighs 8.4 gram, then the value of c is:
(d) 121 cm3 (a) 76
Q544. (b) 72
By melting two solid metallic spheres of radii 1 cm and 6 (c) 74
cm, a hollow sphere of thickness 1 cm is made. The (d) 75
external radius of the hollow sphere will be Q551.
(a)8cm The diameter of a 120 cm long roller is 84 cm . it takes
(b)9cm 500 complete revolution of the roller to level a ground.
(c)6 cm The cost of travelling the ground at Rs. 1.50 sq.cm. is
(d)7 cm (a) Rs. 5750
Q545. (b) Rs. 6000
Water is flowing at the rate of 5 km/ h through a pipe of (c) Rs. 3760
diameter 14 cm into a rectangular tank which is 50 m (d) Rs. 2376
long 44m wide, The time taken (in hours ) for the rise in Q552.
the level of water in the tank to be 7 cm is Two right circular cylinders of equal volume have their
(a) 2 heights in the ratio 1 : 2. The ratio of their radii is
(b)3/2 (a)√2:1
(c) 3 (b) 2 : 1
(d) 5/2 (c) 1 : 2
Q546. (d) 1 : 4
The volume (in m3) of rain water that can be collected Q553.
from 1.5 hectares of ground in a rainfall of 5 cm is If the volume of two cubes are in the ratio 27: 1, the ratio
(a) 75 of their edge is :
(b) 750 (a) 3 : 1
(c) 7500 (b) 27: 1
(d) 75000 (c) 1 : 3
Q547.

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(d) 1:27 Spheres A and B have their radii 40 cm and 10 cm
Q554. respectively, Ratio of surface area of A to the surface area
The edges of a cuboid are in the ratio 1 : 2 : 3 and its of B is :
surface area. is 88cm. The volume of the cuboid is : (a) 1 : 16
(a) 48 cm3 (b) 4: 1
(b) 64 cm3 (c) 1 : 4
(c) 16 cm3 (d) 16: 1
(d) 100 cm3 Q562.
Q555. If the radius of the base of a cone be doubled and height
The volume of two spheres is in the ratio 8: 27. The ratio is left unchanged, then ratio of the volume of new cone to
of their surface area is : that of the original cone Will be :
(a) 4: 9 (a) 1 : 4
(b) 9:3 (b)2 : 1
(c)4:5 (c)1 : 2
(d)5:6 (d) 4: 1
Q556. Q563.
The curved surface area of a cylindrical pillar is 264 m 2 A cube of edge 5 cm is cut into cubes each of edge of 1
and its volume is 924 m2 (take π =22/7) find the ratio cm. The ratio of the total surface area of one of the small
of its diameter to its height . cubes to that of the large cube is equal to :
(a) 7 : 6 (а) 1 : 125
(b) 6: 7 (b) 1 : 5
(c) 3: 7 (c) 1 : 625
(d) 7 : 3 (d) 1 : 25
Q557. Q564.
The ratio of the volume of two cones is 2 : 3 and the ratio The diameter of two hollow spheres made from the same
of radii of their base is 1 : 2 The ratio of their height is metal sheet is 21 cm and 17.5 cm respectively. The ratio
(a) 3: 8 of the area of metal sheets required for making the two
(b) 8: 3 spheres is
(c) 4 : 3 (a) 6: 5
(d) 3 : 4 (b) 36:25
Q558. (c) 3 : 2
If the volume of two cubes are in the ratio 27 : 64, then (d) 18: 25
the ratio of their total surface area is: Q565.
(a) 27 : 64 By melting a solid lead sphere of diameter 12 cm, three
small spheres are made whose diameters are in the ratio
(b) 3: 4
3 : 4 : 5. The radius (in cm) of the smallest sphere is
(c) 9 : 16
(a) 3
(d) 3: 8
(b) 6
Q559.
(c) 1.5
A hemisphere and a cone have equal base. If their heights
(d) 4
are also equal, the ratio of their curved surface will be:
Q566.
(a)1:√2
A cone is cut at mid-point of its height by a frustum
(b) √2:1
parallel to its base. The ratio between the volumes of two
(c)1:2
parts of cone would be
(d) 2 : 1
(a)1 : 1
Q560.
(b) 1: 8
If the height of a given cone be doubled and radius of the
(c) 1 : 4
base remains the same the ratio of the volume of the
(d) 1 : 7
given cone to that of the second cone will be
Q567.
(a) 2 : 1
The ratio of the area of the in-circle and the circum-circle
(b) 1 : 8
of a square is
(c) 1 : 2
(a) 1 : 1
(d) 8 : 1
(b) √2:1
Q561.
(c) 1: √2
(d) 2 : 1

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Q568. Q575.
The ratio of the surface area of a sphere and the curved A solid metallic sphere of radius 8 cm is melted to form
surface area of the cylinder circumscribing the sphere 64 equal small solid spheres, The ratio of the surface
is area of this sphere to that of a small sphere is
(a) 1 : 2 (a) 4: 1
(b) 1 : 1 (b) 1:16
(c) 2 : 1 (c) 16:1
(d) 2: 3 (d) 1 : 4
Q569. Q576.
The radii of two spheres are in the ratio 3 : 2. Their The diameter of two cylinders, whose volumes are
volume will be in the ratio: equal, are in the ratio 3 : 2, Their heights will be in the
(a)9:4 ratio
(b)3:2 (a) 4: 9
(c)8:27 (b) 5: 6
(d)27:8 (c) 1 :2
Q570. (d) 8: 9
The volume of a sphere and a right circular cylinder Q577.
having the same radius are equal the ratio of the The radius of base and slant height of a cone are in the
diameter of the sphere to the height of the cylinder is ratio 4 : 7. If slant height is 14 cm then the radius (in cm)
(a) 3: 2 of its base is 2 ( use pie = 22/7)
(b) 2: 3 (a) 8
(c)1:2 (b) 12
(d)2:1 (c) 14
Q571. (d) 16
A cone, a hemisphere and a cylinder stand on equal Q578.
bases and have the same height. The ratio of their A right circular cylinder just encloses a sphere of radius
respective volume is r. The ratio of the surface area of the sphere and the
(a) 1 : 2: 3 curved surface area of the cylinder is
(b) 2 : 1 : 3 (a)2:1
(c) 1 : 3 : 2 (b)1:2
(d) 3 : 1 : 2 (c)1:3
Q572. (d)1:1
dii of the base of two cylinders are in the ratio 3 : 5 and Q579.
their heights in the ratio 2 : 3. The ratio of their curved The ratio of radii of two cone is 3:4 and the ratio of their
surface will be : height is 4: 3. Then the ratio of their volume will be
(a) 2: 5 (a) 3: 4
(b) 2: 3 (b) 4:3
(c) 3 : 5 (c) 9 : 16
(d) 5:3 (d) 16:9
Q573. Q580.
If the radii of two spheres are in the ratio 1 : 4, then their If a right circular cone is separated into solids of volumes
surface area are in the ratio : V1, V2, V3 by two planes parallel to the base which also
(a) 1: 2 trisect the altitude, then V1: V2:V3 is
(b) 1:4 (a) 1 : 2 : 3
(c) 1 : 8 (b) 1: 4: 6
(d) 1 :16 (c) 1: 6:9
Q574. (d) 1 : 7 : 19
The radii of the base of two cylinders A and B are in the Q581.
ratio 3:2 and their height in the ratio x:1 If the volume of The total surface area of a solid right circular cylinder is
cylinder A is 3 times that of the cylinder B, the value of x twice that of a solid sphere. If they have the same radii,
is the ratio of the volume of the cylinder to that of the
(a)4/3 sphere is given by
(b)2/3 (a) 9 : 4
(c)3/4 (b) 2 : 1
(d)3/2 (c) 3 : 4

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(d) 4:9 The radii of the base of cylinder and cone are in ratio
Q582. √3:√2 and their heights are in the ratio √2:√3 . their
The respective height and volume of a hemisphere and a volumes are in the ratio of
right circular cylinder are equal, then the ratio of their (a)√3:√2
radii is - (b)3√3:√2
(a)√2:√3 (c)√3:2√2
(b)√3:1 (d)√2:√6
(c)√3:2 Q589.
(d)2:√3 The heights of two cones are in the ratio 1 : 3 and the
Q583. diameters of their base are in the ratio 3 : 5, The ratio of
The ratio of the volume of a cube and of a solid sphere is their volume is
363 : 49. The ratio of an edge of the cube and the radius (a) 3 : 25
of the sphere is ( take pie =22/7) (b) 4: 25
(a) 7: 11 (с) 6: 25
(b) 22: 7 (d) 7 : 25
(c) 11:7 Q590.
(d) 7 : 22 A sphere and a hemisphere have the same volume. The
Q584. ratio of their radii is
The radius and the height of a cone are in the ratio 4 : 3. (a) 1 : 2
The ratio of the curved surface area and total surface (b) 1 : 8
area of the cone is (c) 1 :√ 2
(a) 5: 9 (d) 1 : 3√2
(b) 3: 7 Q591.
(c) 5 :4 The diameter of the moon is assumed to be one fourth of
(d) 16:9 the diameter of the earth. Then the ratio of the volume of
Q585. the earth to that of the moon is
A sphere and a cylinder have equal volume and equal (a)64:1
radius. The ratio of the curved surface area of the (b)1:64
cylinder to that of the sphere is (c)60:7
(a) 4: 3 (d)7:60
(b) 2: 3 Q592.
(c) 3 : 2 If A denotes the volume of a right circular cylinder of
(d) 3: 4 same height as its diameter and B is thevolume of a
Q586. sphere of same radius then A/B is :
A right circular cylinder and a cone have equal base (a)4/3
radius and equal height. If their curved surfaces are in (b)3/2
the ratio 8: 5, then the radius of the base to the height (c)2/3
are in the ratio : (d)3/4
(a) 2: 3 Q593.
(b) 4: 3 The radii of the base of cylinder and cone are in ratio √3:
(c) 3 : 4 √2 and their heights are in the ratio √2: √3 . their
(d) 3: 2 volumes are in the ratio of
(a) √3: √2
Q587. (b)3√3: √2
(c) √3:2√2
The edges of a rectangular box area in the ratio 1 : 2: 3 (d)√2:√6
and its surface area is 88 cm2. The volume of the box is Q594.
Diagonal of a cube is 6√3 cm. ratio of its total surface
(a) 24 cm3 area and volume (numerically) is
(a)2:1
(b)48cm3 (b)1:6
(c)1:1
(c) 64 cm3
(d)1:2
Q595.
(d)120 cm3
Q588.

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 A sphere and a hemisphere have the same volume. The (d) 3 : 4
ratio of their curved surface area is : Q602.
(a)23/2: 2 The volumes of a right circular cylinder and a sphere are
(b)22/3:1 equal. The radius of the cylinder and the diameter of the
(c)42/3:1 sphere are equal, The ratio of height and radius of the
(d)21/3:1 cylinder is
Q596. (а) 3 : 1
The volume of a cylinder and a cone are in the ratio 3 : 1.
(b) 1 : 3
Find their diameters and then compare them when their
(с) 6: 1
height is equal.
(d) 1 : 6
(a) diameter of cylinder = 2 times of diameter of cone
Q603.
(b) diameter of cylinder = diameter of cone
A large solid sphere is melted and molded to form
(c) diameter of cylinder > diameter of cone
identical right circular cones with base radius and height
(d) diameter of cylinder < diameter of cone
same as the radius of the sphere. One of these cones is
Q597.
melted and molded to form a smaller solid sphere. Then
A solid sphere is melted and recast into a right circular
the ratio of the surface area of the smaller to the surface
cone with a base radius equal to the radius of sphere.
area of the larger sphere is
What is the ratio of the height and radius of the cone so
(a) 1: 34/3
formed
(b) 1: 23/2
(a) 4 : 3
(c) 1:32/3
(b) 2 : 3 (1)
(d) 1:34/3
(c) 3 : 4
Q604.
(d) 4: 1
A plane divided a right circular cone into two parts of
Q598.
equal volume . if the plane is parallel to the base, then
Two cubes have their volumes in the ratio 27 : 64, The
the ratio in which the height of the cone is divided, is
ratio of their surface areas is
(a)1:√2
(a) 9 : 25
(b) 1:3√2
(b) 16: 25
(c) 1:3√2-1
(c) 9 : 16
(d) 1:3√2+1
(d) 4: 9
Q605.
Q599.
On increasing each side of a squire by 50%, the ratio of
The ratio of weights of two spheres ratio of different
the area of new formed square and the given square will
materials is 8 : 17 and the ratio of weights per 1 cc of
be
materials of each is 289 : 64. The ratio of radii of the two
(a) 9 : 5
spheres is
(b) 9 :7
(a) 8:17
(c) 9: 3
(b)4:17
(d) 9 : 4
(c)17:4
Q606.
(d)17:8
A cone of height 7 cm and base radius 1 cm is carved
Q600.
from a cuboidal block of wood 10 cm x 5 cm x 2 cm[
The total number of spherical bullets each of diameter 5
assume π=22/7]
decimeter that can be made by utilizing the maximum of
(a)278/3%
a rectangular block of lead with 11 meter length , 10
(b)139/3%
meter breadth and 5 meter is (assume that π =3 )
(c)127/2%
(a) 8800
(d)124/3%
(b) 8000
Q607.
(c) 7800
If radius of a cylinder is decreased by 50% and the
(d) 7790
height is increased by 50% to from a new cylinder , the
Q601.
volume will be decreased by
If the ratio of volumes of two cones is 2 : 3 and the ratio
(a)6%
of the radii of their bases is 1 : 2, then the ratio of their
(b)25%
heights will be
(c)62.5%
(a) 8 : 3
(d)75%
(b) 3 : 8
Q608.
(c) 4: 3

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Each of the height and base radius of a cone is increased (a) 10%
by 100%. The percentage increase in the volume of the (b) 60%
cone is (c) 40%
(a) 700% (d) 20%
Q616.
(b) 400%
The length, breadth and height of a cuboid are in the
(c) 300%
ratio 1 : 2: 3. If they are increased by 100%, 200% and
(d) 100%
200% respectively. then compared to the original
Q609.
volume the increase in the volume of the cuboid will be
If both the radius and height of a right circular cone are
(a) 5 times
increased by 20%, its volume will be increased by
(b) 18 times
(a) 20%
(c) 12 times
(b) 40%
(d) 17 times
(c) 60%
Q617.
(d) 72.8%
Each of the radius of the base and the height of a right
Q610.
circular cylinder is increased by 10%. The volume of the
A cone of height 15 cm and base diameter 30 cm is
cylinder is increased by
carved out of a wooden sphere of radius 15 cm. The
(a) 3.31%
percentage of used wood is :
(b) 14.5%
(a) 75%
(c) 33.1%
(b) 50%
(d) 19.5%
(c) 40%
Q618.
(d) 25%
If the height of a cone is increase by 100% then its
Q611.
volume is increased by :
If the height of a right circular cone is increased by
(a) 100%
200% and the radius of the base is reduced by 50%, the
(b) 200%
volume of the come
(c) 300%
(a) increases by 25%
(d) 400%
(b) increases by 50%
Q619.
(c) remains unaltered
A hemispherical cup of radius 4 cm is filled to the brim
(d) decreases by 25%
with coffee, The coffee is then poured into a vertical cone
Q612.
of radius 8 cm and height 16 cm. The percentage of the
If the height and the radius of the base of a cone are each
volume of the cone that remains empty is :
increased by 100%, then the volume of the cone become
(a) 87.5%
(a) double that of the original
(b) 80.5%
(b) three times that of the original
(c) 81.6%
(c) six times that of the original
(d) 88.2%
(d) eight times that of the original
Q620.
Q613.
The height of a circular cylinder is increased six times
If the height of a cylinder is increased by 15 per cent and
and the base area is decreased to one ninth of its value.
the radius of its base is decreased by 10 percent then by
The factor by which the lateral surface of the cylinder
what percent will its curved surface area change?
increases is
(a) 3.5 percent decrease
(a) 2
(b) 3.5 percent increase
(b)1/2
(c) 5 percent increase
(c) 2/3
(d) 5 percent decrease
Q614. (d) 3/2
If the radius of a sphere is doubled, its volume becomes Q621.
(a) double If the radius of a sphere be doubled the area of its
(b) four times surface will become
(c) six times (a) Double
(d) eight times
(b) Three times
Q615.
(c) Four times
If the radius of a right circular cylinder is decreased by
(d) None of mentioned
50% and its height is increased by 60% its volume will
Q622.
be decreased by

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If each edge of a cube is increased by 50 % the between the parallel sides is 8 cm , if the volume of the
percentage increase in its surface area is prism is 1056 cm3, then the height of the prism is
(a) 150% (a) 44cm
(b) 75% (b) 16.5 cm
(c) 100% (c)12 cm
(d) 125 % (d) 10.56 cm
Q623. Q630.
If the radius of a sphere be doubled, then the percentage Each edge of a regular tetrahedron is 3cm, then its
increase in volume is volume is
(a) 500% (a)9√2/4 c.c.
(b) 700% (b)27√3 c.c.
(c) 600% (c)4√2/9 c.c.
(d)800% (d)9√3 c.c.
Q624. Q631.
If radius of a circle is increased by 5%, then the The perimeter of the triangular base of a right prism is
increment in its area is 15 cm and radius of the incircle of the triangular base is
(a) 10.25% 3 cm. If the volume of the prism be 270 cm3 then the
height of the prism is
(b) 5.75%
(a) 6 cm
(c) 10%
(b) 7.5 cm
(d) 5%
(c)10cm
Q625.
(d)12 cm
If the length of each side of a regular tetrahedron is 12
Q632.
cm, then the volume of the tetrahedron is
The base of a solid right prism is a triangle whose sides
(a) 144√2 cu. Cm
are 9 cm, 12 cm and 15 cm, The height of the prism is 5
(b) 72√2 cu. Cm
cm. Then the total surface area of the prism is
(c) 8√2 cu. Cm
(d) 12√2 Cu. cm (a) 180 cm2
Q626. (b) 234 cm2
If the radii of the circular ends of a truncated conical (c) 288 cm2
bucket which is 45cm high be 28 cm and 7cm then the (d) 270 cm2
capacity of the bucket in cubic centimeter is[take Q633.
π=22/7] The base of a right prism is an equilateral triangle of
(a)48510 area 173 cm2 and the volume of the prism is10380 cm2.
(b)45810 The area of the lateral surface of the prism is
(c)48150 (a) 1200 cm2
(d)48051 (b)2400 cm2
Q627. (c)3600 cm2
There is a pyramid on a base which is a regular hexagon (d) 4380 cm2
of Side 2a cm. if every slant edge of this pyramid is of Q634.
length 5a/2 cm then the value of the pyramid is The base of a right pyramid is a square of side 16 cm
(a) 3a3cm3 long. If its height be 15 cm , then the area of the lateral
(b)3√2 a3cm3 surface in square cm is:
(c)3√3 a3cm3 (a) 136
(d) 6 a3cm3 (b) 544
Q628. (c) 800
The base of a right pyramid is a surface of the prism is (d) 1280
square of side 40 cm long. If the volume of the pyramid Q635.
is 8000 cm3 , then its height is Area of the base of a pyramid is 57 sq. cm and height is
(a) 5 cm 10 cm, then its volume (in cm3), is
(b) 10 cm (a) 570
(c) 15 cm (b) 390
(d)20 cm (c) 190
Q629. (d) 590
The base of a right prism is a trapezium . the length of Q636.
the parallel sides are 8 cm and 14 cm and the distance

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The height of a right prism with a square base is 15 cm, If (b) 10 cm
the area of the total surface of the prism is 608 sq. cm, its (c) 10√3cm
volume is (d)12 cm
(a) 910 cm3 Q643.
(b) 920 cm3 A right prism stands on a base of 6 cm side equilateral
(c) 960 cm3 triangle and its volume is 8 1√3 cm3. the height (in cm )
(d) 980 cm3 of the prism is
Q637. (a) 9
The base of a right prism is an equilateral triangle of side (b) 10
8 cm and height of the prism is 10 cm. Then the volume (c) 12
of the prism is (d) 15
(a) 320 √3 cubic cm Q644.
(b) 160√3 cubic cm A right pyramid stands on a square base of diagonal
(c) 150 √3 cubic cm, 10√2 cm. If the height of the pyramid is 12 cm, the area
(d) 300 √3 cubic cm (in cm2) of its slant surface is
Q638. (a) 520
A prism has as the base a right angled triangle whose
(b) 420
sides adjacent to the right angles are 10 cm and 12 cm
(c) 360
long. The height of the prism is 20 cm. the density of the
(d) 260
material of the prism is 6 gm /cubic cm. the weight of the
Q645.
prism is
If the altitude of a right prism is 10 cm and its base is an
(a) 6.4 kg
equilateral triangle of side 12 cm, then its total surface
(b)7.2 kg
area (in cm2) is
(c) 3.4 kg
(a)(5+ 3√3)
(d) 4.8 kg
(b)36√3
Q639.
(c)360
If the slant height of a right pyramid with square base is
(d)72(5+√3)
4 meter and the total slant surface of the pyramid is 12
Q646.
sq. m. then the ratio of total slant surface and area of the
A right pyramid stands on a square base of side 16 cm,
base is
and its height is 15 cm. The area (in cm2) of its slant
(a) 16: 3
surface is
(b) 24 : 5
(a) 514
(c) 32 : 9
(b) 544
(d) 12: 3
(c) 344
Q640.
(d) 444
The length of each edge of a regular tetrahedron is 12
Q647.
cm. The area (in sq. cm) of the total surface of the
The base of a right prism is a right angled triangle whose
tetrahedron is
sides are 5 cm, 12 cm and 13 cm. If the total surface area
(a) 288√3
of the prism is 360 cm2, then its height (in cm) is
(b) 144√2
(a) 10
(c) 108√3
(b) 12
(d) 144√3
(c) 9
Q641.
(d) 11
The base of right prism is a triangle whose perimeter is
Q648.
28 cm and the in radius of the triangle is 4 cm. If the
A right pyramid 6 m high has a square base of which the
volume of the prism is 366 cc, then its height is
diagonal is √1152 m. Volume of the pyramid is
(a) 6 cm
(a) 144 m3
(b) 8 cm
(b) 288 m3
(c) 4 cm
(c) 576 m3
(d) None of these
(d) 1152 m3
Q642.
Q649.
The base of a right pyramid is equilateral triangle of side
The height of the right pyramid whose area of the base
10√3 cm. if the total surface are of the pyramid is 270√3
is 30 m and volume is 500 m3 is
cm2. its height is
(a) 50 m
(a) 12√3 cm
(b) 60 m

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(c) 40 m (d)4√3 cm2
(d) 20 m Q657.
Q650. Base of a right pyramid is a square of side 10 cm . if the
The base of a right prism is an equilateral triangle . if the height of the pyramids 12 cm, then its total surface area
lateral surface area and volume is 120 cm2 , 40√3 cm3 is
respectively then the side of base of the prism is (a)360 cm2
(a) 4 cm (b)400cm2
(b)5 cm (c)460cm2
(c) 7 cm (d)260cm2
(d) 40 cm Q658.
Q651. A right prism has a triangular base whose sides are 13
Each edge of a regular tetrahedron is 14cm. its volume ( cm, 20 cm and 21cm, if the altitude of the prism is 9 cm,
in cubic cm ) is then its volume is
(a)16√3/3 (a)1143 cm3
(b)16√3 (b)1314 cm3
(c)16√2/3 (c)1413 cm3
(d)16√2 (d)1134 cm3
Q652. Q659.
The base of a prism is a right angled triangle with two Base of a prism of height 10 cm square. Total surface
side 5 cm and 12 cm. the height of the prism is 10 cm. area of the prism is 192 sq. cm . the volume of the prism
the total surface area of the prism is is
(a)360 sq. cm (a)120cm3
(b)300 sq. cm (b)640 cm3
(c)330 sq. cm (c) 90 cm3
(d)325 sq. cm (d)160 cm3
Q653. Q660.
The base of a right prism is a quadrilateral ABCD, given A right prism has triangular base. If v be the number of
that AB = 9cm , BC=14 cm, CD =13 cm, DA=12 cm and vertices, e be the number of faces of the prism. The value
∠DAB =90° . if the volume of the prism be 2070 cm3then of v + e –f/2 is
the area of the lateral surface is (a)2
(a)720cm2 (b)4
(b)810 cm2 (c)5
(c)1260cm2 (d)10
(d)2070 cm2 Q661.
Q654. The base of a right prism is a trapezium whose lengths
If the area of the base , height and volume of a right of two parallel sides are 10 cm and 6 cm and distance
prism be {3√3/2} p2c m2, 100√3 cm and 7200 cm3 between them is 8 cm , its volume is
respectively, then the value of P ( in cm) will be ? (a)300cm3
(a)4 (b)300.5 cm3
(b)2√3 (c)320 cm3
(c)√3 (d)310 cm3
(d)3/2 Q662.
Q655. Base of a right prism is a rectangle, the ratio of whose
If the base of right prism remains same and the lateral length and breadth is 3:2 if the height of the prism is 12
edges are halved then it s volume will be reduced by cm and total surface area is 288 sq. cm, the volume of the
(a) 33.33% prism is :
(b)50% (a)288 cm3
(c)25% (b)290 cm3
(d)66% (c)286 cm3
Q656. (d)291 cm3
The total surface area of a regular triangular pyramid Q663.
with each edges of length 1 cm is? Height of a prism shaped part of a machine is 8 cm and
(a)4/2(√2)cm2 its base is an isosceles triangle, whose each of the equal
(b)√3 cm2 sides is 5 cm and remaining side is 6 cm. the volume of
(c)4 cm2 part is

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(a) 90 cm3 (b) 1 times
(c) 9 times
(b) 96 cm3
(d) 6 times
(c) 120 cm3
Q671.
(d) 86 cm3
Let ABCDEF be a prism whose base is a right angled
Q664.
triangle, where sides adjacent to 90° are 9 cm and 12
The sides of a triangle are 7 cm,8 cm, 9 cm, then the
cm, If the cost of painting the prism is Rs. 151.20 at the
area of the triangle (in cm2) is
rate of 20 paisa per sq. cm then the height of the prism is:
(a) 12√5
(a)16 cm
(b) 6√5
(b) 17 cm
(c) 24√5
(c) 18 cm
(d) 30√5
(d) 15 cm
Q665.
Q672.
A cylindrical rod of radius 30 cm and length 40 cm is
The total surface area of a right pyramid on a square
melted and made into spherical balls of radius 1 cm. The
base of side 10 cm with height 12 cm is
number of spherical balls is.
(a) 260 square cm
(a) 40000
(b) 300 square cm
(b) 90000
(c) 330 square cm.
(c) 27000
(d) 360 square cm
(d) 36000
Q673.
Q666.
If the area of a square is increased by 44%, retaining its
The ratio of the radii of two cylinders is 2: 1 and their
shape as a square, each of its sides increases by :
heights are in the ratio 3: 2. Then their volumes are in
(a) 20%
the ratio.
(b) 19%
(a) 4:3
(c) 22%
(b) 6:5
(d) 21%
(c) 3 : 1
(d) 6:1
Q667.
The radii of the base of a cylinder and a cone are equal
and their volumes are also equal. Then the ratio of their
heights is
(a) 1 :2
(b) 2: 1
(c) 1: 4
(d) 1 : 3
Q668.
The curved surface area of a cylinder with its height
equal to the radius, is equal to the curved surface area of
a sphere. The ratio of volume of the cylinder to that
sphere is
(a) 3:2 √2
(b) √2:3
(c)2 √2:3
(d) 3: √2
Q669.
The base of a right prism whose height is 2cm is a
square. If he total surface area of the prism is 10 cm 2
(a)2 cm3
(b)1 cm3
(c)4 cm3
(d)3 cm3
Q670.
The radius of a wire is decreased to one third. If volume
remains the same, length will increase by:
(a) 3 times

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ANSWER :
1b 2c 3a 4c 5b 6b 7b 8c 9d 10 d 11 a 12 d 13 a 14 b
15 d 16 d 17 b 18 d 19 d 20 b 21 a 22 a 23 c 24 d 25 d 26 a 27 d 28 b
29 c 30 d 31 a 32 d 33 a 34 c 35 a 36 a 37 c 38 b 39 d 40 a 41 a 42 a
43 d 44 d 45 b 46 a 47 b 48 d 49 b 50 b 51 b 52 a 53 c 54 c 55 d 56 b
57 c 58 d 59 d 60 b 61 b 62 c 63 b 64 b 65 b 66 c 67 b 68 a 69 b 70 c
71 a 72 c 73 c 74 b 75 b 76 a 77 b 78 c 79 b 80 c 81 b 82 b 83 a 84 b
85 a 86 a 87 c 88 b 89 c 90 c 91 a 92 a 93 b 94 b 95 c 96 c 97 b 98 b
99 c 100 a 101 a 102 c 103 a 104 c 105 c 106 d 107 b 108 c 109 a 110 c 111 a 112 c
113 d 114 b 115 d 116 c 117 a 118 c 119 b 120 b 121 c 122 d 123 c 124 b 125 b 126 c
127 a 128 b 129 b 130 a 131 a 132 b 133 b 134 b 135 c 136 b 137 c 138 c 139 b 140 c
141 a 142 b 143 a 144 a 145 c 146 c 147 a 148 b 149 a 150 a 151 d 152 b 153 d 154 b
155 a 156 c 157 a 158 a 159 b 160 d 161 a 162 c 163 c 164 c 165 b 166 c 167 c 168 d
169 d 170 a 171 d 172 c 173 c 174 c 175 b 176 c 177 b 178 a 179 a 180 b 181 a 182 a
183 c 184 a 185 c 186 a 187 d 188 c 189 b 190 d 191 a 192 a 193 d 194 c 195 a 196 a
197 a 198 d 199 b 200 d 201 b 202 b 203 a 204 c 205 a 206 a 207 b 208 b 209 d 210 c
211 b 212 b 213 d 214 d 215 d 216 b 217 b 218 d 219 d 220 b 221 c 222 a 223 c 224 b
225 a 226 b 227 b 228 c 229 a 230 b 231 a 232 d 233 c 234 b 235 c 236 b 237 a 238 a
239 a 240 a 241 c 242 d 243 c 244 b 245 b 246 b 247 a 248 d 249 a 250 b 251 d 252 c
253 d 254 c 255 d 256 a 257 c 258 b 259 a 260 c 261 b 262 c 263 b 264 a 265 a 266 a
267 a 268 b 269 d 270 a 271 b 272 a 273 b 274 a 275 d 276 b 277 b 278 c 279 d 280 d
281 c 282 c 283 c 284 a 285 c 286 d 287 a 288 c 289 b 290 d 291 d 292 c 293 d 294 c
295 b 296 b 297 a 298 b 299 c 300 a 301 b 302 d 303 b 304 a 305 c 306 a 307 c 308 b
309 d 310 a 311 b 312 a 313 c 314 a 315 c 316 d 317 c 318 b 319 c 320 b 321 c 322 d
323 c 324 c 325 c 326 c 327 d 328 c 329 b 330 c 331 d 332 d 333 a 334 b 335 b 336 c
337 d 338 b 339 b 340 b 341 a 342 b 343 b 344 b 345 d 346 c 347 c 348 d 349 d 350 a
351 a 352 c 353 d 354 a 355 c 356 c 357 b 358 b 359 d 360 a 361 c 362 b 363 a 364 d
365 b 366 a 367 d 368 . c 369 b 370 a 371 c 372 d 373 c 374 b 375 b 376 b 377 d 378 c
379 d 380 c 381 b 382 d 383 b 384 a 385 b 386 a 387 b 388 a 389 b 390 c 391 c 392 d
393 b 394 c 395 d 396 a 397 d 398 d 399 d 400 d 401 d 402 a 403 b 404 d 405 d 406 d
407 b 408 c 409 c 410 c 411 d 412 b 413 d 414 b 415 a 416 a 417 b 418 b 419 a 420 c
421 b 422 b 423 b 424 b 425 b} 426 c 427 c 428 b 429 b 430 c 431 d 432 b 433 c 434 c
435 b 436 b 437 a 438 b 439 d 440 b 441 d 442 b 443 c 444 c 445 d 446 d 447 a 448 d
449 b 450 b 451 b 452 b 453 a 454 c 455 c 456 c 457 a 458 a 459 a 460 a 461 b 462 b
463 d 464 a 465 b 466 d 467 c 468 c 469 a 470 b 471 b 472 c 473 a 474 b 475 c 476 c
477 e 478 b 479 c 480 c 481 a 482 c 483 a 484 c 485 a 486 a 487 d 488 c 489 a 490 d
491 d 492 b 493 c 494 d 495 b 496 c 497 b 498 d 499 d 500 b 501 b 502 a 503 d 504 b
505 a 506 d 507 a 508 a 509 d 510 c 511 a 512 d 513 c 514 a 515 b 516 d 517 a 518 c
519 d 520 b 521 d 522 a 523 c 524 b 525 b 526 b 527 d 528 c 529 c 530 d 531 a 532 a
533 b 534 b 535 a 536 a 537 d 538 d 539 a 540 b 541 d 542 a 543 c} 544 b 545 a 546 b
547 b 548 a 549 d 550 d 551 d 552 a 553 a 554 a 555 a 556 d 557 b 558 c 559 b 560 c
561 d 562 d 563 d 564 b 565 a 566 d 567 a 568 b 569 d 570 a 571 a 572 a 573 d 574 a
575 c 576 a 577 a 578 d 579 a 580 d 581 c 582 c 583 b 584 a 585 b 586 c 587 b 588 b
589 a 590 d 591 ' a 592 b 593 b 594 c 595 d 596 b 597 d 598 c 599 a 600 a 601 a 602 d
603 d 604 c 605 d 606 a 607 c 608 a 609 d 610 d 611 d 612 d 613 b 614 d 615 b 616 d

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 617 c 618 a 619 a 620 a 621 c 622 d 623 b 624 a 625 a 626 a 627 c 628 c 629 c 630 a
631 d 632 b 633 c 634 b 635 c 636 c 637 b 638 b 639 a 640 d 641 d 642 d 643 a 644 d
645 d 646 b 647 a 648 d 649 a 650 a 651 c 652 a 653 a 654 a 655 b 656 b 657 a 658 d
659 d 660 c 661 c 662 a 663 d 664 a 665 c 666 d 667 d 668 d 669 a 670 c 671 c 672 d
673 a

=√
= √( ) =
5. (b) Perimeter of square = 44 cm
4 × side = 44
1. (b) Side of square = side = 11 cm
√
area of square = side2 = 112 = 121 cm2
Area of square =
√ Circumference of circle = 44cm
2 π(radius) = 44
= radius =
area of circle = cm2
= 2.6 × 5.2 = 13.52 cm2
Option (b) is the answer. (circle, 33 cm2)
2. (c) Area of square 6. (b) Let the side of square = a
and the radius of circle = r
= =
perimeter of square = circumference of circle = 4a
= 2πr
r =
3. (a) Let the length of rectangular hall = x
Thus, Breadth of rectangular hall area of circle = 3850 m2
= π×
According to Question, 16a2 = , a2 = 3025m2
Area = 768 m2 7. (b) 2(l + b) = 28
x× l + b = 14
and l × b = 48
(l + b)2 = l2 + b2 + 2lb
(14)2 = l2 + b2 + 48 × 2
196 – 96 = l2 + b2
= 256 × 4 l2 + b2 = 100
x =√ √( l2 + b2) = 10
= 32m Diagonal = 10m
Difference of length and breadth 8. (c) Let the length of rectangular hall = x
=x– Thus, Breadth of rectangular hall
=
According to question,
4. (c) Since the room is in cuboid shape = Area = 192m2
Length of largest rod = = Diagonal of cuboid x ×

= √( )
= √( ) x2 - = 64 × 4
√( ) x = √(64 × 4) = 16 cm

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 Difference of length and breadth a =
=x–
Length = a + 5
= 16/4
= 4 cm
9. (d) Side of the square Breadth = a – 3 =
√
= =4 =
√ √
Area of the square = 16 perimeter of the rectangular = 2(l + b)
Area of new square =32 =2( )
Side of new square = √32
13. (a) According to question,
= √ 2(l + b) = 160
Diagonal of new square = √ √ (l + b) = 80 ………. (i)
= 8 cm (l - b) = 48 ……… (ii)
10. (d) on solving (i) and (ii)
Diagonal of square A = (a + b) 1 = 64, b = 16
Side of square = =* +2 Area of square = Area of rectangle
√ √
= (side)2= 64 × 16
side = √
Area of square = B = 2 × Area of square = 32 m
=
= 2× (a + b)2 14. (b) Side of square, whose perimeter = 24 cm
side of square B = √ = cm
= (a + b)b So area of square = 62 = 36 cm2
Diagonal of square B =√2(a + b) Again, side of square, whose perimeter is 32 cm
11. (a) = 32/4 = 8 cm
So area of this square = 82
= 64 cm2
5m According to the question,
Area of new square = 64 + 36 = 100 cm2
Side of the new square
12 m = √100 = 10 cm
Area of the rectangular garden = 12 × 5 = 60 m2 Hence perimeter of new square = 10 × 4 = 40 cm
Thus, Area of square = 60 15. (d) (side)2 = 484 cm
(side)2 = 60 side = 22 cm
Side = √60 Perimeter of square = 4 × 22 = 88 cm
Diagonal of the square = √2 = √2 × √60 = √120 According to question, 2πr = 88 cm
= 2√30 m r =
12. (c)
Area of circle = π r2 =
= 616 cm2
(a - 3) 16. (d) l = 10 m, b = 6 m, h = 4 m
Length of diagonal (longest + rod)
a (a + 5) =√ =√ m
According to question, 17. (b) Let the length of smaller line segment = x cm
a2 = (a + 3 (a + 5) The length of larger line segment = (x + 2) cm
a2 = a2 + 5a – 3a – 15 According to question,
2a = 15 (x + 2)2 – x2 = 32
x2 + 4x + 4 - x2 = 32

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 x = =7 Perimeter of third square = 4 × 6
The required length = x + 2 = 24 cm
= 7 + 2 = 9 cm 22. (a) Side of the square=
18. (d) A D thus, Sides of all five squares are
=
b
ATQ
Area of another square = 62 + 82, 102, 192, 202
Side2= 36 + 64 + 100 + 361 + 400
B C
side = √961 = 31
BD = Length of diagonal = Speed ×Time
23. (c)
Area of the tank = Length × breadth
= 180 × 120 = 21600 m2
BD = √(l2 + b2)
Total area of the circular plot
→ (l2 + b2) = 132 = 169
= 40000 + 21600 = 61600
Again, (l2 + b2) =
π (radius)2 =
(l + b) = l2 + b2+ 21b
radius = √
172 = 169 + 21B
=√
1B = m2
= 7 × 20
19. (d) Let the breadth be = x m = 140
Thus, Length = 206
2 (x + 23 + x) = 206 24. (d) Let the breadth of rectangle = x m
4x = 206 – 46 Thus, Length = (x + 5) m
x = = 40 m Thus, Area of hall = Length × breadth
Thus, Length = 40 + 23 = 63 m (x + 5)x
Thus, Required area= 63 × 40 = 2520 m2 = 750
20. (b) Length of rectangle = 48 m = 30 × 25
Breadth of rectangle = 16m (clearly 750 = 30 × 25)
According to question, Thus, x = 25, Breadth = 25m
Perimeter of square = Length = 25 + 5 = 30 m
Perimeter of rectangle = 2 (48 + 16) 25. (d) Required total area
4 × side = 2 × 64 = Area of four walls + Area of base
side = = 2 × 1.25(6 + 4) + 6 × 4
= 49 m2
Thus, Area of the square = (side)2 = (32)2
26. (a)
= 1024
2x
21. (a)
3x
Ratio of length and breadth = 3 2
2(l + b) = 20 cm
2 (3x + 2x) = 20 cm
2 × 5x = 20 cm
a
10x = 20
Area of third square= a2 – b2
x =2
= 102 – 82
Thus, Length = 3 × 2 = 6 cm
= 100 – 64
breadth= 2 × 2 = 4 cm
= 36 cm2
area = Length × breadth
Side of third square= √36 = 6 cm

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 = 6 × 4 = 24 cm2 r =
27. (d)
31. (a) a2 = 21 a = 11
b a → Perimeter of square
= 11 × 4
l a = 44 cm
2(l + b) = 160 m 4a = 160 → Cricumference of circle
(l + b) = 80 m …….. (i) = 44
a = 40 m 2π r = 44
ATQ 2 × × r = 44
a2 – lb = 100 → r = 7cm
(40)2 - lb = 100 32. (d)
1600 – lb = 100 Let the no. of hours be x
lb = 1500 → (0.3 × 0.2 × 20000) × x = 200 × 150 × 8
Clearly , 50 + 30 = 80 →x = = 200 hrs
and 50 × 30 = 1500
33. (a)
length = 50 m
28. (b)
220

32 m
10 180 200

200
38 m
2
area of path = 600 m
Area of path = 200 × 220 – 200 × 180
(l + b – 2x) = 600
= 44000 – 36000 = 8000 m2
(38 + 32 - 2x)2x = 600
34. (c) Diagonal of square = diaeter of circle
(70 + 2x)2x = 60
=8×2
(70 – 2x)x = 600/2 = 300
= 16 cm
70x – 2x2 = 300
2x2 - 70x + 300 = 0 Thus, Side of square = 16/√
x2 – 35x + 150 = 0 → Area of square √ 2
x2 - 30x + 5x + 150 = 0 = 128 cm2
x (x – 30) – 5 (x - 30) = 0 35. √
(a) Side of square = = 8 cm
√
(x – 30 )(x - 5) = 0
Thus, Area of square = 8 × 8 = 64 cm2
x = 30 not possible
36. (a) x2 + 7x + 10 = x2 + 5x + 2x + 10
x = 5 (right)
= x(x + 5) + 2 (x + 5)
29. (c) Area of walls = Perimeter of base × height = 18
= (x + 2) (x + 5)
× 3 = 54 m2
Thus, Two sides of rectangle
a =9
= (x + 2) (x+ 5)
30. (d) a2 = 81, a = 9
Thus, Perimeter = 2(x + 2 + x + 5)
→ Perimeter of square
= 2 (2x + 7) = 4x + 14
=9×4
37. (c) Let the sides of rectangle be 6 cm and 2 cm (or
= 36 cm
any other number)
→ 2r + π r = 36
→ Area of rectangel Q = 6 × 2
r (r + π ) = 36
= 12 cm2

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 Side of square = 4 cm Breadth =
→ Side of square (P) = 4 × 4 = 16 cm2 = 25
→P> Q 42. (a) Let the side of squqre = a cm
38. (b) No. of cubes with no side painted = (n - 2)8 ATQ
Where n is the side of the bigger cube l × 2 = 3x
Requried number = (6 - 2)8 = 64
20 × = 3a2
39. (d) Side of squre =
√ a = 10
√
= 43. (d)
√
2
Area of square= (side) = (15) 2 4m
= 225 cm2
40. (a)

4m

Area of path = (l + b + 2x)2x

Thus, Where x = thinkness of path
A Let 1 = 7p,b = 4p
7p + 4p + 2(4) 2(4) = 416
(11p + 8)8 = 416
B D 35 cm 11p + 8 = 52
11p = 44
p = = 4, p = 4
Thus, breadth = 4 × 4 = 16 m
C 8 cm 44. (d) Area of the floor= 8 × 6 = 48 m2
E M = 4800 dm2 (1m = 10 dm)
Area of square = 2 = 4 × 4 = 16 dm2
2 No. of tiles =
=
45. (b)
=
= 512 cm2
√
Area of triangel = (8)2 =
= 1.732 × 2 × 8
= 27.712 cm2 15 m
Required are = (512 + 27.712) cm2 15 m
= 539. 712 cm2
41. Shape of godown is cuboidal
(a) Area of the lawn hectare
(1) Length = 15 m , breadth = 12 m height
Length × breadth = 1/12 × 10000 m2 Area of four walls = 2 (l + b) × h
Length × breadth = m2 area of floor = (l + b)
= 4x × 3x = m2 ATQ
(l + b) + (l + b)= 2(l + b) × h
= 12x2 = 2(l + b)= 2(l + b) × h
= x2 = 2(15 × 12) = 2(15 + 12) × h
= 2 × 27 × h
x=
2 × 180 = 2 × 27 × h

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 h =
Volume of the cuboid = 1 × b × h 35 35
= 15 × 12 ×
= 60 × 20 = 1200 m3
46. (a) D 35 C
According to the question,
D C
Radius of circle = =
Required area of shaded portion
O = (35)2 -
= 1225 – 962.5 = 262.5 m2
A B 51. (b) Diagonal of square = √2 side
side of a square = AB of square
= √2 a units Here a = (x + 1) and d =
√
Thus, AC = Diagonal = √2 × √2 a
Thus, d = √2a
Thus, Diameter = 2 a units
Circumference = π × diameter → = √2 * ]
√
= π × 2a = 2πa units Thus, x = 1 unit
47. (b) Perimeter of rectangle = 40m 52. (a)
Length = 12 meter A B
Thus, 2(12 + b) = 40
12 + b = = 20 b
b = 20 – 12 = 8 m
D c
48. (d) Percentage increase in area
l
= ( ) Let ABCD is a rectangular carpet having length 1
Here, x = 100%, y = 100% meter and breadth b meter and
= % BD is a diagonal
= 300% → As we know
49. → l × b = 120 ……. (i)
D C → Area
→
16 m 2(l × b) = 46
→ Perimeter
Using formula
A B → (l × b)2 = l2 × b2 + 2lb
30 m → (23)2 = l2 × b2 + 2 × 120
AC =√(AB2 + BC2) = √(302 + 162) → 529 = l2 × b2 + 240
= √(900 + 256) l2 × b2 = 529 – 240
= √1156 = 34 meter → l2 × b2 = 529 – 240
Distance travelled by elephant → l2 × b2 = 289
= 34 – 4 = 30 meter → √(l2 × b2) = √289
Diagonal = 17
Speed of elephant =
Diagonal of carpet is17 meters
50. (b)
A 35 B 53. (c)
Diagonal of a square = 6√2 cm

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 √ 59. (d)
Side of a square = cm
√
Area of a square = = 6 × 6 = 36 cm2
28 cm
54. (c) 14 cm
Let the breadth; of floor = x m
Then the length of floor = (x + 3)m Radius to the largest circle
ATQ
=
x2 + (x + 3) = 70
x2 + 3x - 70 = 0 = × 28 = 14 cm
x2 + 10x + 7x – 70 = 0 Area of the circle = π (radius)2
(x + 10) (x - 7) = 0
x = 7 , x = - 10 = × 14 ×4 = 616 cm2
Breadth = 7m
60. (b)
Length = 10m
Perimeter of floor = 2 (l + b)
= 2 (10 + 7)
= 34m
55. (d) Let the breadth of rectangle = x m
then the length of rectangle = 2x m
ATQ
x × 2x = 417.605
R
2x2 = 417.605
Thus: 2 πr = 88
x2 =
r = = 14 cm
x = √(83521/400)
2 πR = 88
x =
R =
Breadth = m The area between two circle
Length = π (21 + 14) (21 – 14)
56. (b) Radius of circle = 5 cm = 770 cm2
Length of arc = l = 3.5 cm 61. (b)
Thus, Area of sector =
cm2
7 cm
57. (c)
Radius of circular wheel = 1.75m
Circumference of circular wheel = 2πr = 2 Circumference of wheel = 2 π r
=2×
No. of revolutions = = 44 cm
= Thus, total distnace travelled by wheel in 15
revolutions = 15 × 44 cm = 660 cm
58. Circumference of wheel = 2 π r\ 62. (c)
= 2 × × 21 cm = 132 cm
No. of revolutions =
= = 700
r

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Height =

Circumference = 2 π r 66. (c)

Distance covered in 1 min A
= 52 ×
new circumference = 2 × π × r × 10
Time taken = = 50 min D E
63. (b)

12 cm B F C
2 2 2
Thus, 3 + 4 = 5
ABC is a right angled triangle
ar(ABC) =
15 cm cm2
Areao fthe triangel = thus, Requred Area of (Δ DEF)
= +1/2 × 15 × 12 = cm2
= 90 cm2
Area of another triangle = 2 × 90 = 180 cm2 = cm2

67.

Height = A 3.5 3.5 C

64. (b)
3.5 3.5

3.5 3.5 7 cm

Area of the square = 81 cm2 B

Side of the square = √81 = 9 cm
Perimeter of the squre = 4 × 9 = 36 cm AB = BC = AC = 7 cm
Now, According to the question, Area enclosed =
π r + 2r = 36 Area of equiliateral Δ ABC -
r (π + 2) = 36
r = √
= - * +
2
= = 1.967 cm
Area of the semi circle with radius 7 68. (a) Thus, πr2 = 2464 cm2
√
= 72/ 2 = 77 cm2 → r =
65. (b) Area of square = (12)2 = √794 = 28 m
= 144 cm2 Thus, Diameter = 2r = 2 × 28 = 56 cm
=
Area of triangel =
69. (b) Requried area = Area o0f square Area of circle
= × 12 × height = (2a)2 - π (a)2
= = 4a - a2

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 = = =
√ √
70. (c) Diameter of the circle = Side of square Perimeter = 3 × sides
2r = 21 = 3 × 16
r = = 48 cm
73. (c) Perimeterof Δ = 30 cm
Area = π r2 = π ( )2 Area = 30 cm2
=
cm2 Check the triplet =
{(5, 12, 13), (3, 4, 5)}
= cm2
Whose largest side is 13,
71. (a) Also, 52 + 122 = 132
A And perimeter = 5 + 12 + 13
= 30 cm
Smallest side = 5 cm
74. (b) Diameter of the wheel = 3m
Circumference = (π × diameter = )
B C Since a wheel covers a distance equal to its
Thus, Area of an equilateral triangle = 400 √3 circumfrerence in one revolution therefore
√ distance covered in 28 revolutions
(side)2 = 400√3
= 28 ×
√
(side)2= = 1600 264 meters covered = 1 minute
√
side = 40m 1 meter covered =
perimeter = 3 × side
5280 meters covered =
= 3 × 40
= 120 m = 20 minutes
72. (c) 75. (b) Distance covered = 2 km 26 decameters
A = (2 × 1000 + 26 × 10)
(1 decameter = 10 meter)
= 2260 m
D F Distance coverd in 1 revolution =
P
= 20 m
B C Now, π × dameter = 20
E
diameter =
Let P be the point inside the equilateral Δ ABC
Let, PD = √3, PE = 2√3 = =6 m
PF = 5√3 76. (a) Distance coverd in 1 revolution
and = AB = BC = AC = x = Circumference of wheel
ar (ABC) =2× × 1.75 m
√ √ √ No. of revolution
√ = √ √ √ = 16 =
Thus, Perimeter of triangel = 3x = 3 × 16 = 48 cm 77. (b) Radius of circle = =
Alternative :
Side of equilatera Δ When a square is inscribed in the circle, diagonal
of the squre is equal to diameter of the circle
= (√ √ √ )
√ Thus, Diagonal of square =

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 √ area enclosed by conins = (Area of equilateral
Thus, side of square=
√ √ triangle – 3x (area of sector of angle 60°)
78. (c) Let outer Radius =R √
and inner Radius = r =
2πR - 2πr = 132 √
=
2π(R - r) = 132
= (√ ) cm2
R–r =
82. (b) A
Hence, width of path = 21 meters
79. (b)

B C
28 cm 2r

side of squre papersheep

M is the centre , BM = CM = r
= √794 = 28 cm
AM | BC1 (AM = r)
=
area of Δ ABC = r2
Thus, Circumference of each circular plate
=2πr
83. (a)
= 2 × 22/7 × 7
= 44 cm
80. (c) 15 30

Circumm radius

In radius Side of the square = = 120/4 = 30 cm

radius of the circle =
2
Area of the circle =
= (15)2
84.
Circum radius of equilateral triangle
=
√
In radius of equilatera tringle
=
√
=8
√
Side = 8√3
Thus, In radius of equilateral triangle radius of in circle
= =
√
= 4 cm √ cm
√ √
√ √
81. (b) radius of each circle = 1 cm area of incricle
with all the three centres an equilateral triangle of = ( )2
√
side 1 cm is formed.

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 =( )
8
= 22 × 21 = 462 cm2
Diagonal of square = 8√2 cm
85.
88. (b)A

r = 20
cm
x 5 cm
whell of radius 20 cm no. of revolutions
B C
= isoceleus right traingle
Thus, x2 + x2 = 52 = 25
86. (a)
Area of trinagle
A
= x2 = cm2
89. (c) A
P1 P3

P2 6 cm 8 cm
B C
7cm
In an equilatera triangel
side = (P1+ P2+ P3) B C
√
87. (c) Length of side = (P1 + P2 + P3 )
√
=
√
=
√
√ √
=
√ √
= 14√3 cm
90. (c) Remember : area of isoceleus triangle
= a2 sin θ (θ is angle between equal sides)
4√3 = (10)2 × sin 45°
√
√ √ √
= 25√2 cm2
side of equilateral triangle 91. (a)
= 4√3
√
Ccircumradius of traingle = =4
√ √
See the figure o A E C
side of square = 2 × circum radius
=2×4 =8
Radius of circule = 6 cm
Area of smallest circle = = 12 π
8√2 8
Radius of smallest circle
√
= = 2√3 cm

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 A
92. a

2a a
a

B C
Area of shaded region
√ 2 = area of semicircle - area of triangle
a = 4√3
2
a =4×4 =
a = 4 cm = –a2 = a2 ( )
Circum radius = 96. (c) According to question
√ √
Area of circle = πr2 π (R + 1)2 – π R2 = 22
= π ( )2 π ,(R + 1)2 - R2} = 22
√
= cm2 (R + 1 + R) (R + 1 - R) =
93. (b) 2R + 1 = 7
R = 3 cm
r 97. (b)

7cm 14 cm
Circumference – diameter = 30 cm
2πr – 2r = 30 148 cm
2r (π - 1) = 30 radius of lartgest circle
r=
94. (b) Area = cm2
98.

12 cm
13 cm 8 cm

In radius of circle (r) =

B 5 cm C side/√3 =
2 2 √ √
AC = √*(12) + (5) = √
Area of circle = π ( )2
√ √
Thus, Length of perpendicular to hypotenuse to =

=
= = 16.76
95. (c) Required area

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 √
( ) 2
-
√
=( )
√ 5 3 5

cm2
99. (c) 30 : 72 : 78 C B C
5 : 12 : 13 4 4
So the trinagle is right tirangle The cone so formed after rotating about Side AB.
So, slant height of cone
= 5 cm
radius = 4 cm
30 78 Height = 3 cm
Thus, Volume of cone
= r2h
72 r = radius
= h = height
altitude = 30 m Thus, Volume of cone
100. (a) =
= 16 π cm3
102. (c)

x √2x
1 1

x
Perimeter of triangle 1 1
= √
x + x + √2x = 4√2 + 4 1 1
2x + √2x = 4√2 + 4 Area of bounded region
x(2 + √2 ) = 4(√2 + 1) √
22 - π (1)2
x =
√ =(√ )cm2
Hypotenuse = √2x
103. (a)
=√
2πr = 11
=√
√ →r =
101. (a)
Area of sector
A 2
=

3 5 = cm2
104. (c) Let the side of the triangle be ‘a’ cm
→ Circumradius =
B 4 C √

cone is rotated about and In radius =

√
A

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 → π( = m
111. (a)
The angles of the given tinagle are 90° , 30°
and 60°
→ = 14
2
→ a = 56
a = 2√14
√
→ Area = √ √ P 10
2
= 14√3 cm
105. (c) Side of square = Daimeter of the circle
πr2 = 9π 30°
→ r = 3cm B
→ Side of squre = 3 × 2 = 6 cm P = 10/2 = 5
→ Area = 6 × 6 = 36 cm2 B = 5√3
→ Area = √
106. (d) The given triangle is a right angled triangle √ 2
=
→ Side of the square =
112. (c) Let the altitude = x cm
→ Area of square= cm2 → 2

107. (b) Radius of circumcirclel =

√ →x =
Radius of incricle = → x = 16π
113. (d) The sides of the given tiangle are 3,4 and 5 cm
√ √ Area = 2
→ Required area = π (R2- r2)
114. (b)
= ( )
= cm2
108. (c) Side of square= √121 = 11 cm 16 √ 2
Perimeter of squre = Circumference of circle = 44
cm
→ 2πr = 44
√
→r = Other sides = = 16 cm
√
→ Area =π r2 (as the Δ issosceles)
= → Area = 128 cm2
= 154 cm2 115. (d)
√
a2 = 4√3
109. (a) 2r + πr = 36
→ a2 = 16
→ r(2 + π) = 36
→ a = 4 cm
→ r (2 + ) = 36 116. (c)
→r = =7m Side of hexagon
110. (c) 2 + π r = π r = = 2cm
√ √
→ r(2 + π) = πr Area of hexagon = a2 =
→ 4 + 2π = πr = 6√3 cm2
→r= +2 117. (a) The radius of park =
→ Diameter = 2 ( ) = 28m

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 → Area of road = π(28 + 7)2 - π(28)2 = 28 m → a = 2 cm
→ Area of road = π (35 + 28)(38 - 28) →3a = 66 cm
= m2 Circumference of circle = 66 cm
118. (c) 2π r = 66
A M B r =
12 Area = πr2 =
13
o = 346.5 cm2
124. (b) Area grazed by the cow
2
=
In Δ OMB MB = √(132 – 122) = 5
→ AB = 5 × 2 = 10 cm =
119. (b) 4 4 = 77 m2
S = = 42
4 4
Area of filed = √
4 4 =√
= 836m2
4 4 → Remaining area
Area of shaded portion = 336 – 77
= 8 × 8 – π × 42 = 259 m2
= 64 – 16π 125. (b) A
= 16(4 – π) cm2
120.
3 3 P Q

3 3
3 3 B C
As P and Q are mid-point and PA
3 3 → Δ APQ – ABC
Area orf shaded portion → AP/AB = PA/BC =
= 6 × 6 – π 32 → PQ =
= 36 – 3π
→ BC = 2PQ = 2 × 5
= 9 (4 - π) cm2
= 10 cm
121. (c) Radius of incircle
√
126. (c)
= A
√
= 7 cm
→ Area = πr2 =
P Q
= 154 cm2
122. (d)
Radius of incircle
B C
√ As PQ || BC
√ → Δ APQ – ΔABC
Area = π r2
→ APQ is also an equilatera Δ
= 3π cm2
√
123. √ → Δ APQ = (5)2
(c) a2 = 121 √3

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 √ 134. (b)
→ m2
Length of rubber band
127. √
(a) 2 π r = (5)2 = 3d + 2πr
→r = = 30 + 10π
135. (c)
→ πr2 = cm A M B
8
128. A
15
O
8
E F
C N D

In Δ OMB
B D C OB = √(152 + 82) = 17 cm
ar Δ AOE = 15 cm2 OB = OD = radius
ar BOFD = 2 × ar Δ AOE = 30 cm2 In Δ OND
129. (b) ND = √(172 - 82)
= 15 cm
CD = 15 × 2
= 30 cm
136. (b) Perimeter = 2r + πr
A = 63 +
= 63 + 99 = 162 cm
137. (c) A
D E

B C
ar ABE = ar Δ ACD = 36 cm2
130. (a) The third side will be either 15 or
B D F E C
→ Possible perimeter
In triangle AFB
= 15 × 2 + 22
AF | BC
= 52
AF2 = AB2 – FB2 = 100 = 25
and 22 × 2 + 15
AF = 5√3
= 59
In triangle ADF
131. (a)
AD2 = AF2 + BF2
No. of revolutions
AD2 = 75 + 2

√
AD =
= 500
138. (c) Let sides of trangel are, a,b, and c respectively
132. (b) 2πr = Thus, Largest side given = 17 cm
= Perimeter = a + b + c = 40 cm (given)
→r = = 0.7
area = 60 cm2 (given)
→ Diameter = .14 cm In such questions take the help of triplets which
133. (b) 2π r = from right angle trinangle
→r = = 35 cm A

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 3, 4, 5 (a2 + 4 + 4a – a2) = 1 + √3
6, 8, 10 8 17
= = 1 + √3
8, 15, 17
etc. 1 + a = 1 + √3
B C a = √3 units
15 145. (c) S = = 15
So, here we have a side 17 cm Area = √
→ by triplet we git sides 8 and 15
√
→ check the sides periemter = 8 + 15 + 17 = 40
=√ √ cm2
Area = 146. (C) Let the length of each equal of each side be a
Hence sides are 15,8 unit
smaller side = 8 cm → √(4a2 - 4) = 4
139. (b) 2 2
√(4a2 - 4) 8
4a2 – 4 = 64
2 2
a2 – 1 = 16
a2 - 1 = 16
2 2
a2 = 17
a = √17 units
2 2
147. (a)
Area of shaded region
Sum of other two sides
= (4)2 – π(2)2 = 16 – 4π cm2
(a + b) = 32 – 11 = 21
140. (c) Let the side of the triangle be a
and a- b = 5
→ Perimeter = 3a
√ → a=
3a = a2) √3
b =
3 = a
Sides of the Δ = 11, 8 , 13 cm
a = 4 units
S =
141. (a)
√ √ → Area
Area of Δ = a2 = √ cm2 =
142. (b) √ (16 - 11)
Area of Δ = = √(16 × 3 × 8 × 5)
= 8√30 cm2
148. (b)
( ) √ √
Area of Δ a2 = 2
(Thus, 9, 12, 15 from triplet
√3 cm2
= cm2 149. (a) Let the original radius
143. (a) 3x + 2y = 6 = r cm
→ π (r + 1)2 – r2 = 22
(Make R.H.S. equal to one) r2 + 1 + 2r - r2 = =7
→ Coordinates of Δ 2r + 1 = 7
= (0,3), (2,0), (0,0). r = 3 cm
2
Area of Δ = 150. (a) Area of two circles = π(52 + 122)
144. (a) = 169π
Let each side of the triangle be a units → πr3 = 169π
√ r2 = 169
→ (a + 22 - a2) = 3 + √3

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 r = 13 cm area = 2

Thus, Radius of thir circlel = 13 cm Perimeter

151. (d) Let the radius of the semicircle be = r 156. √
→ 2r + πr = 36 (c) Length of median = a
r (2 + π) = 36 = 6√3
= a = 12 cm
r (2 + ) = 36 Thus, Perimeter = 12 × 3 = 36 cm
157. (a) Area of equilateral Δ
r( ) √
= a2 = 4√3
→ Area = = a2 = 16
= = 77 m2 a =4
152. (b) Side of squqre = √2 m Thus, Perimeter = 4 × 3 = 12 cm
= Diameter of circle
√ 158. (a) Distance covered by small gear = 2 πr × 42
→ Radius of circle = = m
√ = 84 π × = 504π
Thus, Area = 22/7 × ( ) 2 = m2 = No. of revolution by big gear
√
153. (d) k =
159. (b) Perimeter of semi-circle = 2r + πr = r(2 + r)
A 4 6 C
= 18
4 6
=
8 8 160. (d) Perimeter of circle = 2πr
B = 2(18 + 26)
k = 88 cm
Side of Δ ABC = 10, 14, 12 = π r = 44 cm
S = = 18 r = 14 cm
2
Thus, Area of circle =
Area = √
161. (a)
=√
Area of circle = 38.5 cm2
= 24√6 cm2
πr2 = 38.5
154. D E
r2 =
r = cm
B B Cricumference of a circle = 2πr
C 2× = 22 cm
162. (c) Diameter of circle
√
Area □ ABDE = 3 × ar Δ ADC = 12 cm
√ √
√ Radius of circle =
=3× (2)2 (ADC is equinteral triangel)
=3√3 units2 Radius of circumcircle of equilaternal Δ
= √3
155. (a) Check tripletes → a = Radius × √3 = 6√3
3, 4, 5 cm
6, 8, 10 163. (c)
7, 24, 25 (0,3) A
→ 7,24,25 ful fill the given conditions 3x + 4y = 12

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 B √
=

(4, 0) x = = =
Note: The ratio of area of two similar triangles is
equal to the ratio fo square of their corresponding
sides.
168. (d)
3x + 4y – 12 A

Thus, Divide by 12 on both sides make

R.H.S. = 1

B D C
Thus, Cordinates of point A = (0,3)
Point B = (4,0) AB = AC = BC
Area of Δ OAB = × 4 × 3 = 6 sq. units AB + BC + AC = 544
164. (c) hight of equilateral Δ = 15 cm BC BC + BC = 544
√ = 544
= 544
side =
√ BC=
√ 2
Area = = 204
→ AB = AC =
√ √
( )2 = × Area of Δ ABC = b/4 √(4a2 - b2)
√
= 75√3 cm2 Thus, Where = equal side
165. (b) √ (side)2 = 9√4 b = base
2
(side) = 9 × 4 = 36 = √ 2
- (204)]2
side = √36 = 6 cm = 51 √(115650 - 41616)
Length of median = √ (side) = 51 √73984
√ = 51 × 272
= √ cm
= 13872 cm2
Note: In an equilatera triangle, length of median,
169. (d) A
angle bisector, altitude is equal to √3/2 sides
166. (c) Clearly,
a a
12 cm, 16 cm and 20 cm from a triplet
8
3 4 5 → triplet
B b b C
×4 ×4 ×4
D
12 16 20 → triplet
AB = AC = a cm
BD = DC = b cm
They from a right triangle
Altitude of isoscles triangel is also median
area of triangle = × 16 × 12 = 96 cm2 In right Δ ADC
AD2 = a2 - b2
167. (c) = = 64 = a2 - b2 …… (i)
Perimeter = 64

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 a + a + 2b = 64 Thus, Diagonal of square = =
√ √
2a + 2b = 64
a + b = 32 ……….. (ii) Now, b = =
√ √ √
√
On dividing ( ) Required ratio =
√
a3 × 6/a2
Thus, = (a + b) (a - b) √
√
a–b =2 → 3√3 : 2
a + b = 32 173. (c)
On solving a = 17, b = 15 Let the side of equilateral triangle
area of Δ ABC = = 120 cm2 =s
170. (a) √
Area of equilateral = s2
A
A c a C
c a

b b
B 3/2 s s
x = AB = b + c
y = BC = a + b
z = AC = a + c B C
Thus, semi-perimeter,s
= =
= Height of equilateral triangle
√ ] √
=
=√
√
171. (d) π R2 = π (10)2 + π (24)2
R2 = 102 + 242 = 100 + 576 = = √3
√ √ √
R = √676
= 26 cm 174. (c)
172. (c) A A D
.

a M a B 3 C E F
3 3 3 4

o Δ ABC - ΔDEF
]
]
B C =
a 2
ar(ΔDEF =
Let the side of equilateral triangle = ‘a’
175. ]
and the side of square = ‘b’ (b) ]
in circle radius of equilateral Δ
=
√

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 DE2 = = 2 × ×PQ × OQ
DE =
√
= = 12 × 5 = 60 cm2
178.
176. (c)

B C

o A
Width
C2 12

R
C

AB = AC triangle drawn from the same point equal Let radius of outer circle = R
OB = OC = 3cm and radius of inner circle = r
OA = 12 cm 2 π R 2 π r = 66
∠ABO = ∠ACO 90°
2 R - r) = 66
In ight ΔABO
AB √ 2 - 32 √ R -r = =
√ √ width = 10.5
ar ABOC = 2 × ar (ABO) = 2 × × AB × OB 179. (a)
Perimeter of the circle =
= 3√ √ cm2
Circumference of circle =
Let ‘R’ be the radius
177.
ATQ
Q
2π R – 2R = x
2R (π - 1) = x
5 2R =
13 Diameter =
o P Thus, 2R = Diameter of the circle
180.

R
√
∠OQP = ∠ORP °
(radius is | tangent )
and PQ = PR (tangent drawn from same point
are equal)
PQ √ OP 2 - (OQ)2 √ 3 - 52) = 12

ar of (PQOR) = 2 × ar (PQO)

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Let the side of equilateral triangle = ‘a’ B C
Thus, Circumcircle radius = 20 m
√
Thus 15, 20, 25 from a triplet
Area of circumcircle = π ( )2 = πa2/3
√ Clearly, 252 = 152 + 202
πa2/3 = 3π ABC is a right triangle
a2 = 9. a = 3
Area of Right ΔABC =
Perimeter = 3 × a = 3 × 3 = 9 cm
181. (a) = × 15 × 20
= 150
Cost of sowing seeds = 150 × 5 = 750
184. (a)
r 72° r

o
r 8

15
88 A M15 B
AB = 30 cm
Length of are = OM | AB and OM = 8
Thus, AM = BM = 15 cm
In right Δ OMA
OA2 = OM2 + AM2
r = 70 m
OA2 = 289
182. (A) AO = √
OA = 17 cm
Radius of circle = 17cm
4, 5 3, 5 185. (c)
C B A
4, 5 4,5

5, 5 5, 5 r 15 cm
A 9 cm r
r
B C
In Δ ABC 12 cm
Perimeter of ΔABC = (AB + BC + AC) Since, 9 , 12, 15 forms a triplet
= 2 (3.5 + 4.5 + 5.5) 2
area of Δ ABC =
= (13.5 × 2) = 27
In circle radius of triangle
183. (c) A =
=
Alternate:
In a right triangle, with , P, B and H incircle radius
15 m 25 m =
Hence, r =

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Also circum circle radius =
186.

D C B B C
In Δ ABC and Δ A’B’C’
∠ C = ∠C common
∠B ∠B Thus AB || AB
a/2 a
Δ ABC - Δ A’B’C
√2a
( ) ( )2
2 =¼
A B
a Δ A’B’C
189. (b) Perimeter of square = 44 cm
Let the side of square = a Area of square = ( )2 = 121 cm2
in circle radius of square = Circumference of circle = 2 π r = 44
√ r =
Circumcircle radius of square =
→ Area of circle = π r2 = 2
Thus, = √ √ = 154 cm,2
= 1 : √2 Required difference = 154 – 121 = 33 cm
187. (d) 190. (d) C
a
A a a C

2a 2a B
B
A
Hence, ABC is the equilateral triangle
AB= BC = AC = ‘2a’ cm
√ √
area of Δ ABC = (2a)2 = = 2

= √3 a2 - ∠ ACB = 90° (angle in semi-circle)

area of 3 sectors of θ °
°
AC:BC = 3 : 4
= a 2
AB2 √ AC2 + BC2 √ 2+ 42) = 5 units
°
a /2
2 5 units = 5 cm
A ea of shaded egion a ea of Δ ABC - Area Thus a Δ ABC 2

of 3 sector 191. (a) A

Δ a2 - a2/ 2
√
= a2 cm2
188. (c) A
N

C B
A’ M

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 ar Δ ABC = cm2 B 24 C
Δ ABC - Δ MCN ∠ABC = 90°
∠C = ∠C AC √ AB2) + (BC2 √ 2) + (242)
∠M ∠B Thus MN | | AB √
2 √ m
Thus, ( = 2 =¼
Now a ea of Δ ABC
ar□MNAB = 2
= ½ × AB × BC
192. (a) = ½ × 32 × 24 = 384 cm2
Now, in Δ ADC
s =
a ea of Δ ADC √
24 cm 24 cm =√
=√ = 20 × 3 × 5
= 300 m 2

Area of the plot

= 3841 + 300 = 684 m2
24 cm 194. (c)
Inradius of an equilateral triangle A
= = = 4√3 cm
√ √
Area of equilatera triangle
√ 2
= = √3/4 × 24 × 24 a M H
= √3 × 6 × 24 P
= 144 √3 cm2
= 144 × 1.732
= 249.408 cm2 B b c
Now, Area of incircle
2
= Lengthof perpendicular drawn from the right angle
hypotenuse,
= √ √
=
= ( )
2
= 150.86 cm 2 P =
Area of remaining part = area of Δ - area of P2 = (Thus, H2 = )
incircle
= 249.408 – 150.86 195. (a)
= 98.548 cm2
193. (d)

A 25
A 8 4 B
D

32 25 Diameter of the circle = AB = 8 + 4 = 12 units

Raduys =
Thus, Area of circle

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 = π r2 = π × (6)2 R
2πR - 2πr = 33
196. √ √
(a) = (side)2 (R - r) = =
side = 2 units thickness = 5.25 m
197. (a) Let the ;emgtj pf sode pf sqiare= a 202. (b) Ratio = 5 : 6 : 7
Let the diameter of circle = d Sum of sides = perimeter
According to the question, = 18
a =d sides,
Thus, =
( )
=S=
= =
√
= 2
=√ √
→ 14 : 11 203. (a) Circumference of circle
198. (d) Length of median of an equilateral triangle = π × diameter
√
= (side) = cm
Length of median, altitude, and angle bisector is 2 (l + b) = 352
= √3/2 (side) l+b =
√
Thus, = 6√3 Thus, smaller side =
√
a = = 12 cm = 77 cm
√
√ 204. (c) Perimeter of equilateral triangle
Thus, Area of Δ ABC a2 = 18
√ 3 × side = 18 cm
=
√ √ cm2 side = 6 cm
2 √
199. (b) πr = 2πr Length of median = side
r = 2 units
Thus, Area of circle = π(2)2 √
= 4 π sq. units
= 3√3 cm
200. (d) Area of equilateral triangle
√
205. (a)
= (side)2
√ .
= (side)2 = 48
(side)2 = √ = 64√3 r r
side = ]1/2

201. (b)

r Circumference of paper sheet

r = 352
2πR = 352
R =

r = =

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 . Thus, Circumference of circular plate = 2 π r r
=2× 210. (c) Circumference of front wheel
= 176 cm × no. of its revolutions= circumference of rear
206. (a) Inradius of equilatera triangle wheal
= × no. of its revolutions
√
2πx × n = 2πx y × m (let ‘m’ is the revolution of
√3 = rear wheel)
√
side = 6 cm m =
perimeter of of equilateral triangle = 3 × 6 = 18 cm
211. (b) Distance to be covered in one revolution =
207. (b) Circumference of circle = πd
Circumference of wheel = π × diameter
Thus, πd - d = 150
d (π - 1) = 150 =
d( ) = 150 Total distance = 2.2 km
= (2.2 × 1000 × 100)cm
d× = 150 = 220000 cm
d= .Thus, Number of revolutions = = 1250
Radius = = = 35 m 212. (b) A
208. (b)
Let radius of circle = r
Side of square = a
Side of equilateral Δ = b
According to question, a a
2πR = 4a = 3b
G
Thus, a =
Ratio of their areas
√
πR2 : a2 b2 B C
√
πR2 : )2 : 2
a
√ We know that in an equ7ilateral trinagle a median
1 : : π
also be a altitude
C : S : T
→Altitude of an equilateral triangle
Here, we can see that C>S>T
√
Quicker approach: When primeter of two or more =
figures are same then the figure who has more √
→ a = 12√3 (given)
vertex is greater in the area. Since, here, circle has
infinite vertex. → a = 24 cm
→ Then ara of an equilatera triangle
√
Therefore, C>S>T = a2
209. (d) √
Distance covered in 1 revolution =
= Circumference of circular field = 2πr = 144√3 cm2
Distance = Speed × Time 213. (d) Let a triangle ABC has sides of measurement
3 cm, 4 cm, and 5 cm
= 66 m/s ×
→ Δ ABC will be a right angled tri-angle
Thus, 2 πr = 165 → Inner radius of circle C1
2 × × r = 165 =

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 = r = 1 cm 11R2 - 200 R - 400 = 0
By option (d), In such type of equation go through
A the option to save your valuable time
R = 20
11 × 202 - 200 × 20 - 400 = 0
4400 - 4000 - 400 = 0
3 0 =0
Therefore, radius of pool R = 20 cm
5 215. (d) A
C1
B C
4
3x 5x
→Circum-radius of cirlce = C2
R =
In a right angled triangle half of hypotenuse is
circum B C
radius 4x
Area of right angled triangle = 7776
R =
→
→ πr2/πr2 = l2/( )2 = → 6x2 = 7776
214. (d) Let the radius of swinmming pool → x2 = 1296
=R → x = 36
→ Perimeter of triangle =
3x + 4x + 5x = 12x
= 12 × 36
= 432 cm Ans.

216. (b)

R
R+4
l

r
90°

r
Outer radius of Pool with concrete wall = (R + 4)
According to question , According to the firgure,
2
πR × - π R2 → Perimeter = r +r+ l
3→ 75 cm = 2r + Length of arc
R2 × R2 + 16 + 8R – R2
2
→ 75 cm = 2r +
= = 16 + 8R
→ r = 21 cm

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 → Its area = ¼ = 2π - π =
= 346.5 cm2 = π= m2
217. 219. (d) According to the question,
A Circumference of a circle= 2 π sr
2π r =
r = 15/π2
8 8 D = 2 = 30/ π2
220. (b) According to the question
O
D = 728 m
SD = 728
B C
D
8
According to the question,
Here OD = radius,
Thus, r = R = 364 cm
√ √
r = 4/√3 r = 350
Required area of slided portion
√ 2 2
= –π
√
√
=
= √3 × 16 –
= 10.95 m2
= m2
218. (d) D

4 A
d = 700 m
45° The breadth of the path
= (R - r)
3 B D = (364 – 350) cm
= 14 cm
221. (c)
According to the questions
According to the question
Area of sector OED
= πr2 ×
r = 84 cm
=π×4×4×
= 2r m2
Area of the sector OAB
= π r2 × =π×3×3×
= m2
So, Area of shaded protion = Area of OED – Area of
OAB

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 15°
B C
Sin 15° = =
AB = Sin15°
a =? Cos15°
Let the length of side of the square be a cm BC = Cos15°
Circumference c = 4a (Peimeter square) = 2πr = Area of ΔABC =
4a = ° °
= 2 × × 84 = 4a
= *Thus, sin2 θ = 2 sinθ cosθ+
132 cm = a
222. (a) Area of circle= 324 π cm2 = m2
π r2 = 324 π = 1250 cm2
r = 18 cm 227.
Longest chord = diameter = 2r
= 2 × 18 A
= 36 cm
223. (c) Circumference of a Δ = 24 cm a a
a + b + c = 24 cm
or S = 12 cm
Cricumference of incircle
2πr (Inner) = 14 cm
Area of Δ = S × r (inner) B C
2
= b
224. (b) Area of Δ = AB = AC = a
BC = b
= °
2 Thus, S =
= 25 √2 cm
225. (a) Accroding to the question S =a+
r = Area = √
Area =
Semiperimeter =
Inner radius = √( )( )( )( )
(Semiperimeter)
Area = √( )( )( )( )
6 =
Area = b/2 √(a2 - b2/ 4)
Area = 150 cm2
226. (b) According to the question Area = √ 2 – b2) sq. units
A 228. (c) As we know circum centre always made by the
intersection of half altitude
→ In obtuse angle it will always be out.
75°

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 → side =
B D C
6cm
D → Circum centre 10 cm 10 cm
229. (a) According to the question, 6 cm
2πr → Circumference
2r → Diameter A B
=

→
In Δ AOB,
→ OB = √(102 - 62)
→ = √(100 - 36)
= √64 = 8 cm
Diagonal BD = 8 × 2 = 16 cm
230. (b) Given 234. (b) A D
→ Area of square = 4
Side2 = 4
side = 2 5m
→ Diagonal of square = radius of circle
√2 side = r
→ r = 2√2
→ Area of circle = π r2 B C
→ π × (2√2)2 3 4 × side of rhombus = 40m
231. (a) side of rhombus = 10 m
We know that rhombus is parallegogram Since rhombus is also a parallelogram
Therefore its area = base × height
whose all four sides area equal and its diagonals
= 10 × 5 = 50 m2
bisect each other at 90°
235. D C
. A B
12 16
6
10
16 12 o
6
D C
2 2 A B
Thus, AB = √(16) + (12) = √400 = 20 cm
= side of rhombus 10
Thus, Perimeter of the rhombus = 20 × 4 = 80 cm Perimeter of Rhombus = 40 cm
232. (d) If d1 and d2 are the lengths of diagonals of 4 × side = 40
rhombus. side = 10 cm
then, We know that diagonals of rhombus bisect each
other at right angle,
Perimeter = 2√(d12 + d22 )
Therefore, In right Δ OAB
= 2√(242 + 102)
=2 √676 OB2 = AB2 - OA2
= 2 × 26 = 52 cm = 102 - 62 = 100 – 36 = 64
233. (c) 4 × side = 40 cm OB = √64 = 8 cm
(given) Diagonal BD = 2 × OB = 2 × 8

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 = 16 cm B C
Alternative: 6.5
Side of rhombus
= √(d12 + d22) (Thus, Rhombus is a | | gm, Thus, Area of
Rhombus = Base × Heinght)
10 = √(12)2 - d2
Area of Rhombus = Base × Height
20 = √144 + d22 = 6.5 × 10 = 65 cm2
144 + d22 = 400
Also area of Rhombus = d 1 × d2
d22 = 400 – 144 = 256
d2 = √256 = 16 cm = 26 × d2 = 65
236. (b) diagonal = d1 = 13 × d2 = 65
Area of Rhombus = 150 d2 = 5 cm
d1 × d2 = 150 240. A B

d2 =
237. (a)
60°

D C
In the above figure Δ ADC is equilateral triangel (as
AC is angle bisector)
→ AC = 10 cm (smaller diagonal)
241. (c) Side of rhombus
=
1 We know that in a rhhombus = 4a2 = d12 + d22
A regular hexagon consists of 6 equilateral triangle →
d22 = 4 × 252 - 142 = 2500 – 196
area of regular hexagon = 2304
√
=6× (side)2 → d2 = √2304 = 48 cm
√ → Area = d 1 × d2
6× a2
√ 2
cm2
=6× × (2√3)
242. (d) Let the parallel sides be 3x and 2x
√
=6× × 12 = 18√3 cm2 →
238. √ → 5x = 60
(a) Area of Hexagon = = 6 × (side)2
√
x = 12
=6× (1)2 → Sum of length of parallel sides
√ √ = (3 + 2)× 12 = 60 cm
=6× cm2
243. (c) A B
239. (a)
15
A D
20 7

D C
10
Using Hero’s formula

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 S = = 21 cm
Area of Δ ABC Area of parallegogramm
= BC × FC = 15 × 12 = 180 cm2
=√ ]
Area of parallegogram =
√ DC × AE = 180
= 42 cm2 18 × AE = 180
→ Area of □ ABCD = 42 1 2 = 84 cm2 AE = 10 cm
244. (b) Area of parallelogram = 1000 units 2 Thus, Distance between bigger sides
Let the altitude on smaller side = x units = 10 cm
→ 5 × 20 = 1000 247. A 24 B
100 units → 1000
1 unit 10
→ Greater side = 10 × 5 16
= 50 units
and smaller side = 10 × 4 = 40 units F
→ 40 × x = 1000 units
→ x = 25 units
245. (b) ) A B D E C
10 60°
AB = 24 cm
10 AD = 16 cm
AE = 10 cm (given)
Area of Parallelogram = AE × DC = 10 × 24 = 240
D C cm2
Also, area of Parallelogram = FC × AD = 10 × 24
as □ ABCD is a rhombus = 240 cm2
Thus, Δ is equilateral FC × 16 = 240
√ FC = 15
→ ar Δ ABC = × 10 × 10
= 25√3 cm 2 Thus, Distance between AD and BC = 15 cm
→ ar □ ABCD = 25√3 × 2 248. cm
= 50√3 cm2 A 36 B
246. (b)

27

F 12

A B
D E C

15 Area of parallegogram
= AE × DC
F 12 = CF × AD
AE × 36 = 12 × 27
= AE = 9 cm
D C Thus, Distance between bigger sides = 9 cm
E 18 249. (a) In a rhombus

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 4a2 = d12 + d22 Side of Rhombus =
4a2 = 82 + 62 Let BD = 40 cm
a2 = OB = 20 cm
→ Side of square = 5 cm In right Δ OBC
Thus, Area of square = 25 cm2 OC2 = BC2 - OB2
250. (b) OC = √252 - 202 = 15 cm
D E Thus, AC = 2 × OC = 2 × 15 = 30 cm
Area =
=
A B
2
C =
253.

Area □ ABDE = 3 × ar Δ ADC

(ADC is equilateral triangle)
√
3× 22 = 3√3 unit2
251. (d) Side of rhombus =
= 5 cm
A D
B
o
4 AC = 8 cm 2

D E
B 5 C 5

OC = 4 cm 3
In Right Δ OBC
OB2 = BC2 – OC2 A C
= 5 2 - 42 = 9
OB = √9 = 3 Thus, DE | | AC
BD = 2 × OB = 2 × 3 = 6 cm Thus, BDE – ΔBAC
Area of Rbombus
2
= 22/52 =

=
ar (trap ACED ) = ar(BAC) - ar(BDE)
Note : in the question do not get confused with the = 25 – 4 = 21
words non – square its simply to clear that it is
Rhombus. Thus, = = 21: 4
252. (c) 254. (c)
A D D C

B C
A B

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Thus, AB = 2 CD

= ( )2 = =4:1 N o P 12
12
Thus, ΔAOB - Δ COD
255. (d) A 40 D
A M B A
o
B
40 40 40 ara of ABCD = 24 ar (ABCD) = 24
Draw QM and PN and intersect them a O
ar□ POQC =
B 40 C 40
Thus, area PQC =
Let AC =4x and BD = 3x PQC = 3
Thus, OA = 2x and OB = QMAD =
In Right Δ OAB QAD =
√*(2x)2 + ( )2] = 40 ABP = 6
4x2 + 9x2/ 2 = 402 = 1600 ar(PQC) + ar(QAD) ar(ABP) = 15
16x2 + 9x2 = 1600 × 4 ar(APQ) = 24 – 15 = 9 cm2
25x2 = 6400 Also
x2 = =
√ Thus, Always it will be 3 : 8
x = =
258. (b) A
Thus, Ac = 4x = 4 × 16 = 64
BD = 3x = 3 × 16 = 48
area = 12
=
= 1536 cm2 B D
256. (a) o 9

A D

C
Let d1 = 24 cm
B C Q area of Rhombus = 216
d1 × d2 = 216
in Δ ABC & Δ DCQ d2 = = 216
2
∠ABC ∠ DCQ
∠ACB ∠DQC d2 =
BC = CQ OA = × d1 =
ΔAABC ΔDCQ
Thus,
a Δ ABC a ΔDCQ
Diagonals of Rhombus bisect each other at right
257. (c) D Q C D
angle
C

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 OD = × d2 = 82 + 62 = 100
Now, In right ΔAOB AB = ∠100 = 10 cm
AD2 = AO2 + OD2 = 122 + 92 261. (b)
= 144 + 81 = 225 A 60 B
AD = √225 = 15 cm
Thus, Perimeter of Rhombus
= 4 × AD 40 80 40
= 4 × 15 = 60 cm
259. (a) A D
D 60 C
o
S (Δ ABD) = = 90
arΔ ABD
90° = √
30° = √ √ m2
ar□ABCD = 2 × arΔABD = 600√ m2
B 8 C 262. (c) A D

Let ∠ ABC = 60° O

∠OBC °
Thus, Diagonals of Rhombus area the angle
bisectors
In ight Δ BOC b a
= Cos 30°
√

OB √
Thus, BD = 2 × OB
√ side of Rhombus = =
√ cm Let, AC = 2a
260. (c) ) A D thus, OA =OC = a
CD = 2b
OB = OD = b
o In right Δ OBC,
a2 + b2 =
8 6
4a2 + 4b2 = p2 ………. (i)
Also, 2a + 2b = m
On squareing, 4a2 + 4b2 + 2ab = m
B 10 C 4a2 + 4b2 =m2 - 8ab
AC = 16, BD = 12 cm from (i) and (ii)
Thus, OA = 8 cm, OB = 6 cm m2 - 8ab = P2
Thus, Diagonals of rhombus bisect each other at 8ab = m2 - P2
90° 4 × (2ab) = m2 - p2
In right Δ OAB (m2 - p2)
AB2 = OA2 + OB2 Area of Rhombus

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 ×s1 × d2 = A D
12
= 2ab = (m2 - P2)
263. (b) A 16
o
12
2 16

D E B C
Thus, OB = OD = 16 and OA = OC = 12
3 (Diaggonals of Rhombus bisect each other at 90°)
In Δ OBC,
BC2 = OB2 + OC2 = 162 + 122 = 400
BC = √400 = 20 cm
B C Perimeter = 20 1 4 = 80 cm
267. (a) D 4x C
Thus, DE || BC
Thus, ∠ADE = ∠ABC and ∠AED ∠AEB
Thus ΔADE - ΔABC
22/52 = 2x
Thus, Area DECB = Area(ABC) - area (ADE)
= 25 – 4 = 21
Thus, =
A N 4x M B
264. (a)
A B 7x
area =

11x2 = 176 → x2 = 16
x =4
D C
AB = 7 × 4 = 28 cm
AB = BC = CD = DA = 10 cm
CD = 4 × 4 = 16 cm
BD = 16 cm
CM = 2 × 4 = 8 cm
In Δ ODC
AM = AN + NM
OD = 8, CD = 10, ∠DOC = 90°
→ AN + 16
Thus, OC = √(CD2 - OD2) = √ 2 - 82)
→ 6 + 16 = 22 (AN = BM =
= 6 cm
Thus, Now, Area of Rhombus ABCD = d1 × d2 AC2 = CM2 + AM2
AC2 = 82 + 222
2
AC = √ √ √
265. (a) 268. (c)
Area if trapezium A
= =
2
= 28 cm
266. (a)

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 271. (b) A 15 cm B
B D
P

C h
ABCD is a rhombus
AB = 6
(Perimeter = 60 cm) D 20 cm C
AC = 24 As we know
AP = 12 → Area of trapezium
[Diagnonals of rhombus bisect perpendicularly] =
In Δ APB → 175 = ×h
AB = 15, AP = 12
Thus, BP = 9 → height = 10 cm
(By pythagoras theorm) 272. (a) Let the rate of carpenting = Rs. x meter2
Bd = 9 × 2 = 18 Thus, Length × breadth × x = Rs. 120 ……… (i)
Area of rhombus Length × breadth – 4 × x = Rs. 100 ………. (ii)
Diagonal1 × diagonal2
=
→ 216 sq cm Breadth = 24 m
269. (d) Let ABCD is || gm
|| Gm area of ABCD = 1 ÷ area of ADC 273. (b) Area of Room = 100 × 3 = 300 m2
Fro area of (ADC) Carpet Length =
A 30 cm B
274. (a) Old expenditure = 1000
40cm Increase in area = 50 × 20 m2
20cm 20 cm Increase in expenditure = 50 × 20 × 25 = 250
→ New expenditure = 1000 + 250 = 1250
D 20 cm C 275. (d) Area of veramdhah =
(25 + 3.5) × (15 + 3.5) – 25 × 15
Let a = 20 cm, b = 30 cm, c = 40cm = 527.25 – 375 = 152.25 m2
S= = = 45 cm cost of flooring = 152.25 × 27.5 = Rs. 4186.50
area = ADC (app.)
276. (b) 2 πR1 = 528
=√
→2× R1 = 528
=√
→ R1 = 84 m
=√
→ New Radius = R1 - K = R2
= 75√15 cm2
→ R2 = 84 – 14 = 70
270. (a)
New Radius R2 = 84 – 14 = 70
Let the diagonal of rhombus
Area of Road = π(R12 - R22)
d1 = x & d2 = 2x
→ π × 14(154)
Area of rhombus = d1 d2 → Total expenditure
→
277. (b) Since the ratio of length and breadth = 3 : 2
Let length of rectangular field = 3x

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Breadth of rectangular field = 2x x2 = = 196
Perimeter of the filed = 80 m
x = 14
2 (l + b) = 80
Thus, Altitude = 4 × 14 = 56 cm
2 (2x + 3x) = 80
281. (c) According to question,
2 × 5x = 80
Ratio of sides of triangle are =
x =
then breath = 2x
= 2 × 8 = 16 cm (Take LC M of 2, 3, and 4 which is 12)
278. (c) Since the sides of a rectangular plot are in the Now, 6x + 4x + 3x = 52
ratio = 5 : 4 13x = 52
Let the length of rectangular field = 5x x =4
and the breadth of rectangualr field = 4x Thus, Length of smallest side = 3x = 3 × 4 = 12 cm
According to tquestion = 282. (c) Let diagonals the 2x and 5x
Area = 500 m2 * +
5x × 4x = 500 m2
x2 =
x =5 → 4 : 25
Length = 5x = 5 × 5 = 25 m 283. (c) a b
Breadth = 4x = 4 × 5 = 20 m
Perimeter of the rectangle = 2 (25 + 20)
= 2 × 45 = 90 m
279. (d) C h1 h2

x y

* +

×
D
A D ay : bx
2 3 284. (a) Ratio of parallel sides = 5 : 3
5 Let sides are 5x and 3x
AB = 5cm (sum of parallel sides) × Perpendicular
DB = 3cm distance = 1440 m2
Thus, AD = 2 cm
2
=
4x × 24 = 1440
2
= x =
280. (d) Base = Corresponding altitude Thus, Length of longer side = 5x
= 3 :4 = 5 × 15
Let the base = 3x = 75 m
Altitude = 4x 285. (c)
Thus, Area of triangle = 1176

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 ( )

a1 a2
Let R = 23x , s = 22x
Thus, R – r = 5
a12 /a22 =
23x – 22x = 5
√
a1 /a2 = → r = 22 × 5 = 110
Ratiio of their perimeters Diameter of inner circle = 2r
= 2 × 110
= 4a1/4a2 = a1/a2 =
= 220 m
→ 15 : 16 289. (b) a
286. (d) Clearly, 3, 4, 5, from a triplet therefore,
consider the triangle, a right triangle
Let the sides are 3, 4, 5 triplet a√2
Perimeter = 3x + 4x + 5x = 12x a a a√2
area of triangle =
= 216 a

x2 =
x = √36 = 6 a√2
thus, Perimeter = 12 × 6 = 72 cm a√2
287. (a)

Ket tge sude if sqyare = a

r 42 cm Thus, Diagonal = a√2
5x ,√(a2 + a2) / a√2-

6x
Perimeter of rectangle = circumference of circlare
wire √
2 (6x + 5x) = 2 290.
22x = 2 × 22 × 6
x = 12
Clearly,
Smaller side of rectangle = 5 × 12 = 60 cm a a
288. (c)
a a a a
5m

r
R a

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 √
= 4 : √3
√
291. a

When we draw such figures as mentioned in the

r a a question the vertex of the old triangle are the mid
points of the sides of new traingle and the sides of
the old triangle are half of the opposite side.
Thus, Required ratio = 2 : 1
a 295. (b)
2 2
πr = a2 = =
r2 = 296.
(b) Ratio of area = (Ratio of radius)2 = . √ /2
r = √
√
= 4:1
Rato of perimeter =
297. (a)
Ratio of area = (Ratio of Radius)2
A B C
√ √ Radius 4 : 2 : 1
= √ Area 16 : 4 : 1
292. 298. (b) π r2 = a2

√
h1 h2 299. (c)
Ratio of sides =
= 20 : 15 : 12
= 20 + 15 + 12 = 47
→ 47 → 94
a1 a2 →1 →2
√ → Smallest side = 12 × 2 = 24 cm
300. (a) Let the sides be = 3x, 4x, 5x, and 6x
√ → 18x → 72, x → 4
→ Greatest side = 6 × 4 = 24 cm
√
301. (b) Ratio of circumference = Ratio of radius
√ =3:4
√ √
Ratio of altitudes = * + ] 302. (b)
Let the sides be = 2x, 3x and 4x
= =
→ 9x = 18 = x = 12
293. (d) Let length = 5x → Sides are 4, 6 , and 8 cm respectively
→ Breadth = Using hero’s formula
Thus, Required ratio = S =
=5:3 → Area = (√ ]
294. (c) =√ ]
2
√

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

303. (b) Ratio of area = (Ratio of radius) 2
2
= = 4:1
√ √
304. (a) A 306. (a)
Let the sies be 3x, .3x, and 4x
→ Area = √
E E = 4x √(36x - 16x2)
2 2

o = 4x2√20x2
= 8x3√5 = 8√5
= x3 = 1
B C x=1
As D and E are mid-points Thus, 3rd side = 3 × 1 = 3 units
→ DE || BC 307. (c) 3,4 and from treplet
→ Δ ODE – ΔBOC Let the sides be 3x, 4x = 72
and also → 6x2 = 72
(as D and E are mid-points) → x2 = 12
→ x = 2√3
( )2 =
Thus, Smallest side = 3 × 2√3 = 6√3
308. (b)
Let the sides be 3x, 4x and 5x
As D and E are mid-points
→ area =
→ DE || BC
→ Δ ODE - ΔBOC → 6x2 = 72
x2= 12
and also
X = 2√3
(as D and E are mid-points) → Perimeter of equilateral Δ
→ =( ) =¼ = 12 × 2√3 = 24 √ 3 units
305. √
Side of Δ = 8√3 units
A D √
Area of Δ = × (8√3)2
√
= √
θ θ
309. (d) Let the parallel sides be 2x and 3x
→ Area =
5x = 80
θ x = 16
B C E F → Longer parallel side = 16 1 3 = 48 cm
310. (a)
The given angle is same let verticle angle = θ Let the side of square = a
(Thus,Δ ABC and ΔDEF are isoceleus triangles) Thus, Side of equilateral Δ = √2 a
→ When two angles are equal then third angle is
also equal √ √
Thus, ΔABC – ΔDEF Required ratio =
ΔABC is similaer to Δ DEF
Thus, = √ ×2 = √3 : 2
√ 311. (b) Ratio of area = (Ratio of side)2
→
√
( ) = 25 : 16

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 a =

Required ratio =
( )
=8:9
317. (c) 2(l + b) = 3a
312. Thus, (a = side of equilateral traingle)
(a) = Let (b = a)
→ 2(l + a) = 3a
= 2 (l + a) = 3a
2l + 2a = 3a
√ 2l = a
√
= 2 : √7 Required ratiio =
√ √
313. (c)
– –
√

= = 2 : √3
√
→ 16 : 25 318.
(b) Required ratio =
314. (a) Let side of square = a
319. (c) Let AB = 1, BC = 1
radius of smaller circle = A √2
D
√
Radius of larger circle = 1
( )
E 1 √2 √2
ratio = =
√
( )
B 1 C
315. (c) A
Thus, AC =
√ √
a √
a
√
√3 √
B C 320.
(b) ( ) ( )
Circum radius = 321. (c) A
√ √
Equilateral Δ
√
√
Requird ratio =
( )
√
= 3√3 : 4π
316. (d) 2(l + b) = 4a (a = side of square)
2(2 + 1) = 4a
B 1 B 1 C
2 × 3 = 4a

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Old area = π
2 New circumference = 8 π
2πR = 8 π
AB || AB
R = = 4cm
Thus, A’ and B’ are the mid-point By mid point
theorem New area = 16 π
Thus, ΔA’B’C - ΔABC Option (c) is the answer
Let BB’ = B’C = 1 (Thus, area is quadruples)
Thus, BC = 2 (B is the mid point of BC) 326. (c) Length 4 → 5
Breadth 5 → 4
= Area 20 → 20
ar(AA’B’B) = ar(ABC) – ar(A’B’C) Area remains unchanged.
327. (d)
= 3/4 = 3 : 4 According to question,
Circumference of a circle = Area of circle
2πr = πr2
322. r =2
(d) Ratio of sides = Thus, Diameter of circle = 2r
= =2×2 =4
328. (c)
Thus, Take LCM = 24
ATQ perimeter = 91
6 + 4 + 3 = 91
13 unit = =7
Diff. between long and shor side = 6 – 3 = 3 unit
→ 3 unit = 7 × 3 = 21 cm
323. (c) By using result,
R1θ1 = R2 θ2
°
°

324. (c) In Δ OBC,

H and G are the midpoints of OB and OC
Dand E are the mid points of sides AB and AC
Thus, HG = ½ BC
Thus, DE||BC
Similarly, FG = ½ CD and EF = ½ AD,
(By mid Point theorem)
HE = ½ AB
Δ ADE - Δ ABC
On adding,
HE + HG + FG + EF
=
Perimeter of EFGH. ( ) ( )
=½
Thus,
Percentage of ar(DECB)
325. (c) Old circumfference = 4π = ¾ × 100 = 75%
2πr = 4π 329. (b) Increment in breadth = 10%
r = = 2 cm = (1 = Increment, 10→ Breadth)

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Length Breadth Area -1 Area of no change
Original 10 10 100 11 10
New 9 11 99 -1 Decrease in base
% change = - = 1% =
Alternate: 337. (d) Increase in circumference = Increase in radius
using x = 10% (Breadth), = 50% = ½ (1 → Increment, 2 → Original)
y = - 10% (length)
are x + y Radius Area
2 4
= 10 – 10 + +5
= - 1% 3 9

330. (c) Use x +y + 338. (b) Use of x + y +

20 + 20 + =44% Percentage change
331. (d) If circumference of circle is reduced by 50%
= 12 + 15 +
then radius is reduced by 50%.
50% = (1 → Decrement, 2→ Original) =27+
Radius Area = 144/5
Original 2 4 339. (b)
New 1 1 -3
(π is constant)
Reduction in area = =
332. (d) Increase in area 100 100 120 150

= 25 + 25 +

use formula (x + y + ) = 50 + 6.25

100 130
= 56.25%
333. (a) Increase in area Perimeter of equilateral triangle
= 50 + 50 + = 100 + 25 = 125% = 100 + 100 + 100 = 300
= 120 + 150 + 130 = 400
334. (b) Using x + y + % = Increase =
= 20 – 20 + =
= - 4% 340. (b) Length 5 → 3
(decrease by 4%) Breadth 5 →3
335. (b) Increase in area Area 25→9
% decrease =
= 50% + 50% +
341. (A) Length 5 8
= 100 + 25 = 125%
Breadth 8 5
336. (c) Increase in altitude = 10%
Area 40 → 40
(1 → Increament, 10 → Original)
→ % Decrease × 100
Altitude Base
10 11 =
342. (b) Length 5 → 6

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Breadth 4 → 5 = 80 + 16 = 96%
Area 20 → 30 % effect using x + y + xy/100
350. (a) Percentage increase in area
% increase = ×100 =50%
343. (c) Side 10 → 11 ( )
Area 100 → 121
% increase = ×100 =21% = 16 + 0.64 = 16.64%
351. (a) Side of square is increased by 30%
344. (b) Length 20 → 21
Breadth 10 → 9 =
Area 200 → 189
Initial +3 Final
% Decrease = × 100 = 5.5%

10 13
345. (d) Radius 100 → 101
Area 10000 π → 10201 π
% increase = × 100 = 2.01% -3
Other side will have to be decreased by
346. (c) Let the breadth = x cm
→ Length = (x + 10) cm →
According to the question,
x (x + 20) = (x + 10)(x + 5) 352. (c) Percentage increase in area
→ x2 + 20x = x2 + 15x + 50
= 100 + 100 +
→ 5x = 50
→ x = 10 = 300%
→ Area = 10 (10 + 20) = 300 m2 Alternate :
347. (c) Length 20 → 21 L B Area
Breadth 50 → 49 1 1 1
Area 1000 → 1029 +3
2 2 4
% error = × 100 = 2.9%
Percentage increase = ×100 =300%
348. (d) Length 10 → 13
Breadth 10 → 12 353. (d) x + y +
Area 100 → 156
% increase in area = 10 – 10 + = - 1%
= ×100 =56% (Negative sign shows decrease)
354. (a) Let the side of cube = a
349.
(d) 40% =
Side Surface area
5 (5)2 = 25
40% 24
2
7 (7) = 49
% = Increase =
Alternative:
Percentage increase in surface are
= 40 + 40 + %

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Diagonal of cube = a√3 cm = 960 cm2
a√3 = √12 361. (c) The number of cubes will be least if each cube
on squaring, a2(3) = 12 will be of maximum edge
a2 = 4  Maximum possible length
a = 2 cm = HCF of 6, 9, 12 = 3
Volume of cube = a3 = 23 = 8 cm Volume of cube = 3 × 3 × 3 cm3
355. Number of cubes
(c) Number of cubes = =125
=24
362. (b) Volume of the cistern = (330 – 10 ) × (260 –
356.
(c) Side of the cube = 10) × (110 - x) = 8000 × 1000
√
(where x = thickness of bottom)
√
= x = 110 – 100 = 10 cm
√
363. (a) Let the length, breadth and height be l, b, h
Volue of the cube (side)3
respectively.
= (4)3 = 4 × 4 × 4 = 64 cm3
→ lb = x
bh = y
lh = z
→ l2b2h2 = xyz
357. Let l = 9x, h = 3x, b=x
→ v2 = xyz
l × b × h = 216 × 1000
364. (d) The diameter of sphere = side of cube = 7 cm
(1 litre = 1000 cm2)
9x × 3x × x = 216000 Radius (r) = cm
27x3 = 216000
x3 = 8000 Volume of sphere = r3
x = 20
= cm3
l = 180 cm = 18 cm
358. Volume of cuboid 365. (b) Length of rod = √(102 + 102 + 52)
= 2 × volume of cube = √225
l × b × h = 2 × (side)3 = 15 cm
366. (a) Volume of the box = l × b × h
= (side)3 = (40 - 8) × (15 - 8) × 4
side = ∛(6 × 6 × 6) = 6 cm = 32 × 7 × 4
Total surface area of cube= 6 (side)2 = 6 (6)2 = 6 = 896 cm3
× 36 = 216 cm2 367. (d) Let the three sides of the cuboid be l, b and h
359. (d) Length of largest bomboo = √*(5)2 + (4)2 + (3)2 → lb = bh = hl = 12
=√(25 + 16 + 9) l2 b2 h2 = 12 × 12 × 12 = 1728
= √ √ → lbh = √1728 = 12√12
360. (a) The external dimensions of the box are = 1 = 24√3 cm3
1 = 20 cm, b = 12 cm, h = 10 cm 368. dimension’s of room
External volume of the box = 20 × 12 × 10 = 2400 Length(l) = 12 cm
cm2 Breadth(b) = 9 cm
Thickness of the wood = 1 cm Height (h) = 8 cm
Internal length = 20 – 2 = 18 cm Thus, diagonal of cuboid = √(l2 + b2 + h2)
Internal breadth = 12 – 2 = 10 cm Thus, Length of longest rod
Internal height = 10 – 2 = 8 cm → Length of diagonal of cuboid
Internal volume of the box = 18 × 10 × 8 = 1440 → √*(12)2 + (9)2 + (8)2]
cm3 → √ = √289 = 17 cm
Volue of wood = (2400 - 1440) cm2 369. (b) Area of floor → 3 × 4 = 12 m2

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 height = 3 m 375. (b) Whole surface area of cuboid = 2(Whole
 Area of walls of room surface area of cube)
→ (Perimeter of floor) × height of room - 2 × Area of one face
→ 2(l + b) × h ( two facesa of the two cibes area not visible)
l = length = 4 cm → Required area = 12a - 2a2 = 10a2
b = breadth = 3 cm = 10×62 = 360 cm2
h = height = 3 cm 376. (b) Let the increase in level = x m
 Area of walls → 2 (4 + 3) × 3 = 42 m2 →(1000 × 1000 × ) × ½ = 100 × 10 × x
Area of painted part = 42 m2 + 12 m2 = 54 m2 → x = 10 m
370. (a) Let length = l, breadth= b 377. (d) Sides of parallelopiped area in ratio = 2 : 4 : 8
Height = h Let length = 2 units
Given that Height = 8 units
(l + b + h) = 12 cm Let the side of cube = a unit
= total surface area of box According to question,
= 2 (lb + bh + hl) Volume of cube = volume of parallelopiped
(12)2 = l2 + b2 + h2 + 94 a3 = 2 × 4 × 8
144 – 94 = l2 + b2 + h2 a3 = 64
→ 50 = l2 + b2 + h2 a = ∛64 = 4 units
Diagonal of box = √(l2 + b2 + h2)
 Length of longest rod that
can be put iniside the box
√(l2 + b2 + h2) = 5√2 cm

371. (c) Let breadth = b m = =7:6

Thus, Length of room = 2b m 378. (c) Let length = 1 cm
(1 = 2b) Breadth = 2 cm, Height = h cm
height = 11m 2(lb + bh + hl) = 22
Area of four walls of room 2 + 3h = 11
→ 660 m2 (given) 3h = 9
2(l + b) × h = 660 h = 3 cm
2 (2b + b) × 11 = 660 √(12 + 22 + 32) = √(1 + 4 + 9) = √14 cm
3b × 22 = 660 379. (d √(l2 + b2 + h2) = 15
b = 10 (l2 + b2 + h2) = 225 …… (i)
Thus, breadth = 10 m  l + b + h = 24
Length =20 m (l + b + h) = 576
area of floor = l × b → 225 + 2(lb + bh + hl) = 576
length × breadth = 351 cm2
20 × 10 = 220 m2 380. (c) Total surface area of cube = 6(side)2
372. √ 6(side)2 = 96
(d) Side of cube (a) = = 8 cm
√
→ Total surface area = 6(a) = 6 × 82
2 (side)2 = =16
= 384 cm2 side = √16 = 4 cm
373. (c) Length of pencil = √(a2 + b2 + c2) Volume of the cube = (side)3
= √(82 + 62 + 22) = √(64 + 36 + 4) = (4)3
= √104 = 2√26 cm = 64 cm2
381. (b)
374. (b) Edge of box = ∛(3.375) = 1.5 M

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 3 18 3
The box will be of cuboid shape
Length of the box, l = 24 – (2 × 3) = 24 – 6 = 18 cm
Breadth of the box, b = 18 – (2 ×3) = 18 - 6 = 12
cm
Height of the box, h = 3 cm
Surface area of the box = 2 (l + b) × h + l × b
= 2 (18 + 12) × 3 + 18 × 12
= 2 × 30 × 3 + 18 × 12
Diagonal = 35√3 = 180 + 216 = 396 cm2
 The length of largest rod = Diagonal = side √3 384. (a) Volume of all three cube = 43 + 53 + 63
side = 35 = 64 + 125 + 216 cm3
Side √3 = 35√3 = 405 cm2
√ Now,, 62 cm2 is
side = =35 Thus, Volume of new cube = 405 – 62
√
side of cube = 35 = 343
(side of new cube) = 343
side of new cube - 3√343 = 7
Total surface area of new cube = 6 × (side) 2
= 6 × (7)2
= 6 × 49
= 294 cm2
385. (b) Area of cubical floor = 48
Diameter of the sphere = side of the cube Side2 = 48
2 × radius = side side = √48 = 4√3
radius = Diagonal of cube = side √3
= 4√3 × 53 = 12 cm
Surface area of the sphere
Length of longest rod = 12 m
= 4 π r2
386. (a)
= 4× =3850m2 Let side of new cube = a
According to question,
382. (d) Volume of air in room = 204 m3 a3 = 63 + 83 + 13 =
(area of floor) ×height = 204 216 + 512 + 1
Thus, Volume = area of base × height = 729
(area of floor) × 6 = 204 a = ∛729 = 9
then surface area = 6(a)2
Area of floor = = 34 cm2 6 × 92 = 6 × 81 = 486 cm2
383. (b) 387. (b) Volume = 20m3 = 20 × (100)3 cm3
Volume of one brick = (25 × 12.5 × 8) cm3
 Required number of bricks
=8000

12 388. (a) The toal surface area of cube

= 150 cm2
6 (side)2 = 150 cm2
(side)2 = = 25

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 side = √25 = 5 cm = 2 (6 × 8 + 8 × 12 + 12 × 6)
 Volume of cube = (side)3 = 2 (48 + 96 + 72)
= (5)3 = 125 cm3 = 2 × 216 = 432 cm2
389. (b) Ratio of length : breadth = 5 : 3 392. (d) We know that
Total surface area of parallelopiped = 558 cm 2 A parallelopiped has vertices (v)= 8
2 (lb + bh + hl) = 558 cm2 edge (e) = 12
2 (5x × 3x + 3x × 6 + 6 × 5x) = 558 face (f) = 6
2 (15x2 + 18x + 30x) = 558 Put into equation (v- e + f)
15x2 + 48x = 279 → 8 – 12 + 6 → 2
15x2 + 48x – 279 = 0 393. (b) According to the question,
On Solving, 1 dm = m
 Length = 5 × 3 = 15 cm = Let depth of the hole = d
= 1.5 cm Thus, 48m m × 31.5 × m
Alternatively:
Take help from the options = 27 × 18.2 × d
Covert alloptions m cm, 90, 15, 100, 150 d = 2m
then divide all by 5(because we have to find length) 394. (c)
18, 3 , 20, 30. Put all these values one by one. 2.1m × 1.5 m × h = 630 It
390. (c) l + b + h = 24 cm = 3

Length of diagonal = 15 cm
[1 m3 = 100 It
√(l2 + b2 + h2) = = 15
1000 cm3 = 1 It]
(l2 + b2 + h2) = = 225 cm
(l + b + h)2 - 2(lb + bh + hl) = 225 h = m = 0.20 meter
(24)2 - 2(lb + bh + hl) = 225 395. (d)
576 – 225 = 2(lb + bh + hl) Number of cubes = =8
Thus, Total surface area = 351 cm2
396. (a) When we change shape of a solid figure,
volume remains constant
 Volume of hemisphere = volume of cone
R3 = π R3 h
Thus, 2 R = h
397. (d)
According to the question,
Let the radius of sphere = r cm
4 π (r + 2)2 – 4πr2 = 352
In such type of questions take help from the
options to save your valuable time
391. (c) Let length = 3x, breadth = 4x
4π,(r + 2)2 - r2} = 352
Height = 6x
4π,r2 + 4 + 4r – r2} = 352
3x × 4x × 6x = 576
π (1 + r) = = 22
x3 = =8
3 Take = r = 6
x = √8 = 2cm
Thus, Length = 3 × 2 = 6 cm
breadth = 4 × 2 = 8 cm, = 22
height = 6 × 2 = 12 cm Then option (d) is the right answer.
Total surface area = 2 (lb + bh + hl) 398. (d)

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 400. (d) Radius of tank , r =
Let initial height = H
H = 60 cm
Final height = h
According to question,
( ) ( ) 3

R = 32 cm
We have to find the slant height H -h=
Take ratio of H and R = cm
= 60 : 32 401. (d)
15 : 8
L = √*(15)2 + 82 = 17
= 17 × 4 = 68 cm h = 40 cm
Cost of painting = Surface area of cone × 35
= π r L × 35
× 35
= Rs. 23.94 (approx)
399.
 Circumference of its base = 66 cm
2 π r = 66
r =
Thus, Volume = πr2h
H = 10 cm
r = 2 cm
= 13860 cm3
402. (a)

[Spherical balls]

R = 20 cm
[Solid Cone] h
Let the spherical balls made
= ‘x’
According to question, r
Volume of cone = x × volume of sphere
r
R2 H = x × π r3
According to question,
(20)2 × 10 = x × 4 × (2)3 2πr = 6π
x = 125 r = 3π
height of cylinder = diameter
= 2 ×r = 2 × 3 = 6 cm
Volume of water = π r2 h
= (3)2 × 6 = 54π cm3
403. Volume of the cone = π (15)2 × 108 cm3
Volume of the cylinder = π × r2 × 9 cm3
According to the question,

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 π × r2 × 9 = = π h (R + r) (R - r)

r2 = =900
r = √900 = 30 =
404. (d) Volume of new sold sphere
= ( ) ( ) ( )
410. (c) Volume of Sphere = Volume of displaced water
π r3 = ]
3
r = 216, r = 6 cm
 Diameter of the new sphere = 2 × 6 = 12 cm h =
405. (d) Let the radius of new ball 411. (d)
= R cm
Then, πR3 = π (33 + 43 + 53)
R3 = 27 + 64 + 125 = 216
R = ∛(6 × 6 × 6) = 6 cm
406. (d) Volume of the new sphere
= [r13 + r23 + r33] r = 3cm =

R3
[r13 + r23 + r33]
R3 = r13 + r23 + r33 R = 6 cm
R 3 = 13 + 63 + 83 Volume of cone = Volume of sphere
= 1 + 216 + 512 = 729 π R2 h = πr3
R = 729 = 9 cm π×6×6×h = π ×3×3×3
407. (b) 1 = 2.5 km
h =3
Area of base = 1.54 km2
412. (b) Volume of a cone = π r2h
πr2 = 1.54
r2 = π r2(24) = 1232 cm3
√ r2 =
r = =0.7 km
r2 = 7 × 7
We know that, l = r2 + h2
2
r = = 7 cm
h2 = √(l2 – r2) l = √r2 + h2 = √(72 + 242) = √625 = 25
= √( – Curved surface area = πrl
= √5.76 = 2.4 km = cm2
408. 413. (d)
(c) Radius = 9.6 m
Volume of a sphere
height = 2.8
l2 = r2 + h2 = (14)3
= = (14)3 , -
1 = √100 = 10 m Radius = 14
Area of the canvas = πrl Curved surface area of sphere = 4π(radius)2
= =4× × 14 × 14 = 2464 cm2
409. (c) External radius R = 4 cm 414. (b)
Internal Radius = 3 cm
Volume of iron used = πR2 h – πr2 h
= πh (R2 + r2)

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 r Now 1 cm3 = 8g
462 cm3 = 8 × 462 g
= 3696 g
= 3.696 kg
417. (b)

Surface area of sphere = 64 π

4 π (Radius)2 = 64π
(radius)2 = h = 14 cm
Radius = √16 = 4 cm 11 cm
Diameter = 8 cm

415. (a) 1 dm = 10 cm

h = 35 cm Volume of the cylinder = volume of cube

b = 22 cm π r2 h = (side)3
240 1 = 25 cm r2 × 14 = 11 × 11 × 11
cm r3 =
=
r = cm = 5.5 cm

x × volume of 1 tin = volume of cylinder 418. (b) Let the radius = r

π r2h = 9 π h
→ x × (25 × 22 × 35) = × 240
r2 = 9
x = 1200 r = √9 = 3 m
416. (a) r= diameter = 3 × 2 = 6 m
4 – 1 = 3 cm 419. (a)

8 cm
1m

r
= 21 cm
8 cm

volume of cone = value of sphere

π (8)2 × h = π (8)3
R = 4 cm 8×8×h =4×8×8×8
Volume of hollow iron pipe h = 32
= π ,R2 - r2} × h slant height = (l)
= π , 42 - 32} × 21 √(r2 + h2) = √(82 + 322) = √(64 + 1024) = 8 √17
420. (c) r = 10 m
=
= 462 cm2

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 100 – 3Rr = 21
h = 14 m 3Rr = 100 – 21 = 79
r = 10
Rr = =
423. (b)
2πR = 22 cm
22 cm

R = 15 12 cm
Volume of well = volume of embankment 12 cm = 4
π(10)2 × 14 = π (152 - 102) × H
H = = 11.2 m
421. (b)
R
Thus, Cylinder is folded along the length of
rectangle
 2 π R = 22
r = 7 cm
H = R = =
Volume of the cylinder = πR2 H

= 22 × 7 × 3 = 462 cm3
Volume of sphere = volume of cylinder 424. (b) R= r
(7)3 = π (R)2 ×
R2 = 4 × 7 × 7 = 2 × 2 × 7 × 7
R = √(2 × 2 × 7 × 7) = 2 × 7 = 14 cm
Diameter of base of cylinder = 2R = 2 × 14 = 28 H = 8 cm
cm h = 1800 cm
422. (b)

R r (Rod)
wire
Volume of wire = Volume of Rod
π r2 h = π R2 h

ATQ R + r = 10 r2 =
(R + r)2 = 100
2 2
R + r + 2Rr = 100 r =√
R2 + r2 + 100 + 2Rr …… (i)
3 3
πR + r = 880
π (R3 + r3) = 880
R3 + r3 = =
10 × (100 – 3Rr + Rr) = 210 425. (b)

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 8 cm Thus, When the cone is cut in between
then the ratio of volume of smaller cone to the
bigger one is always equal to the ratio of the cubes
1 cm = 12 × r of their heights.
429. (b)

h
6 cm
Volume of cylinder = 12 ×
volume of sphere
π (8)2 × 2 = 12 × πr3
r3=
3 cm
r =√ = 2 cm
r = 2 cm
n =
d = 4 cm
426. (c) 2 π r - 2 π r = 5
(R - r) =
427. (c)
430. (c) Height of cylinder = Breadth of tin foil
r → Circumference of the base of cylinder
= Length of the foil = 22 cm
→ 2 π r = 22
r = cm

π r3 = 4 π r2 Volume = π r2 h = × = 616 cm3

Radius (r) = 3 units 431. (d) π r2 = 770
428. (b) → r2 =
→ r = 7√5 cm
and πrl = 814
12 = h
→l = =
√
12 H 24 cm 2 2
l =h +r 2

= h2 + r2
R = 7 cm
h2 + 245
Volume of bigger cone
= × π (7)2 × 24 → h2
= → h2 =
= 22 × 7 × 8 = 1232 cm3 √
Volume = π r2h
=
√ √ √
=
= 616√5 cm2
2
Volume of smaller cone = 154 cm 432. In this case the breadth becomes the

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 circumference of the cylinder
h =
→ 2πr = 44
→r= = 7 cm
New volume = πr h 2 439. (d) Ratio of height = ∛(Ratio of volume)
→
3 units = → 30
= 1540 cm3
2 units → 20
→ The cut is made 20 cm above the base

440. (b) 3 π r2 = 108 π

433. 2
(b) π r H = r2h
→ r2 = 36
→ r = 6 cm
→H = h
Volume = π r3 = × 216 × π
→ h = 3H = 3 × 6 = 18 cm = 144 π
441. Radius = 3 decimeters = 30 cm
434. (c) 3 π r2 = 1848 Height of circular sheet = 1 mm = .1 cm
r2= =196 → (30)3 = πr2
→ r = 14 cm → r2 = √
According to the Question, → r = 600 cm = 6 meters
π r3 = π r2h 442. (b) Let no. of seconds rquired to fill the tank = x
→ 2r = h → (πr2h)× x = 3 × 5 × 1.54
→ h = 2 × 14 = 28 cm →x =
435. (b) The length of the paper becomes the
= 300 seconds
circumference of the base of cylinder when it is
→ Time requried = 5 minutes
rolled along its length
443. (c)
→ 2πr = 12
→r = cm
436. (b) Volume of tunnel = π × r2 × H
m3 h l
Volume of ditch
= 48 × 16.5 × 4
= 3168 m3
r1
Required part =
= tan
437. According to the question,
π 4(r2 – r2) = 748 → r = h tan
and = sec
R2 – r2 =
→ l = h sec
= 92 - r2 = 17 →S = π × tan × h sec
→ r2 = 81 – 17 = 64 → π h2 (tan sec )
→ r =8 444. (c)
→ Tickness = 9 – 8 = 1 cm A
438. (b) 2 × ( ) R2 h
→2×

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 D E 12 cm = 21 cm
450. (b) π × r2 × H = π r3
3 cm
= π r3
C r3 = 27
B r = 3 cm
As DE || BC, Δ ADE - Δ ABC 451. (b)

→
→ DE = 4.5 cm
2√3

Slant height = √( √ )
445. Height ofcylinder = Diameter of sphere
= √
→ 452. (b) Volume of vessel = Volume of roof
π × r2 × h = 22 × 20 × x
(Where is rain in cm)
446. → =
(d) =1
→ x = 2.5 cm
453. (a)
Volume of remaining solid
= π r2h
cm3
3h = 2d 454. (c) Let the height be H3→
447. (a) R3
Volume of water pumped out in one hour
→ H =
= ×12×3600
455. (c) Let height and diameter be 3x and 2x
= 1663200 cm3 → x3 × 3x = 1078
= 1663.2 ltr.
448. (d) 2 π rh = 1056 → x3 =49×7
r = →x =7
→ Height = 7 × 3 = 21 cm
Volume = π r2h = ×16 456. (c) Radius of cylinder r = 10 cm
= 5544 cm 3 height of cylinder h = 2 cm
449. Volume of cylinder = πr2h
(b) π r2 H = r2 h
radius of cone → radius of cylinder = 10 cm
H = Let height of cone = h1

→ h = 3H = 3 × 7 Thus, Volume of cone = π r2h1

=3×7  Volume of shaded portion

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 → 4400 cm3 (given)  Radius of sphere = 6 cm
(after removing cone) 458. Total surface area of cylinder
→ π r2h - π r2 h1 = 4400 → 462 (given)
→ (2 π rh + 2πr2 ) = 462 cm2
πr2 (h – ) = 14 2πr2 = 154 × 2 = 308
π r2 = 154
21 – 14 = r2 = × 7 = 49
h1 = 21 r = 7 cm
459. (a) Diameter of cylinder = 7 cm
radius = cm, height = 16 cm
Thus, Lateral or curved surface area
→ 2πrh
→ r = radius
h = height
Thus, 2 × → 352 cm2
460.

457. (a) .
6 cm

6 cm

24

Height of cylinder h = 6 meters

Let radius of cyliner = r meter curved surface
area = 2 π rh
area of end face = π r2
radius of cone = 6 cm → Total area of two end faces
height of cone = 24 cm
 Volume of cone π (6)2 × 24 cm3
Cone is converted to sphere
Let radius of sphere = r r
 Volume of sphere → π r3
Volume of sphere = Volume of cone
→ 2πr2
 r3
given that 3 × 2πr2 = 12 × 2πrh
→ 3r = 2h
→ r3 → ×3 r = 4 cm
→ r3 = 3 × 3 × 24  radius of base = 4 cm
=3×3×3×8
r3 = (3)3 × (2)3
r = 3 × 2 = 6 cm

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 Sheet is folded to form a cone
Let radius of cone = r1

r r

r1
 The circumference of base of cone
→ Circumference of sheet
2 π r1 = 14 π
r1 = 7 cm
 radius of cone = 7 cm
Slant height = radius of semi-circular sheet
r = 14 cm

 Height = √ –

=√ = 12 cm (approx)

462.

(b)

461. (b)
r r
r

r r r1
r →
r1 r

Folded Cone → Circumference of sectors

Radius of semi-circular sheet → Circumference of base of cone of radius = r1 = 2

π r1
=r →
r1
r = 14 cm
Circumference of sheet = πr r1 =
= 14 π cm

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 
Radius of cone =
curved surface area of cone = πr1l
l = slant height
l =r h = 3m

 surface area of cone → π× × r →

463. (d) radius of cone = r = 16 meter (given)
Let slant height = 1 meter
curved surface area = π rl Thus, radius of cone = m
2
= 427 m (given) slant height of cone = 263 m
→ curved surface area of cone
×16×l =
= (πrl)
= =8.5 meter 63
2
= 10395 m
→ radius of cylinder = m
height = 3 m(given)
 Curved surface are of cylinder = 2 π rh
=2× = 990 m2
Thus, Total curved area of structure
→ curved area of structure
→ curved area of cone + curved area of cylinder =
10395 + 390
= 11385 m2
465.

4 cm

3 cm

Surface area of hemisphere = 3πr2 = 2 × =

56.57 cm2
Height of cone = 4 cm
radius = 3 cm
Thus, slant height = √
464. Thus, surface area of cone = π rl
→ 47.14 cm2
1 = 63 m Thus, Total surface area of the toy
area of cone + area of hemisphere
→ 47.14 + 56.57 → 103.71 cm2
466. (d) Let radius of iron rod = r
 Height = 8r

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

  volume of iron rod = r = cm
= π× (r)2 × 8r → 8πr3
→ radius of spherical ball Thus, Volume of cone = π × r2 × h
= × 16
Volume of sphnerical ball = 462 cm 3

( ) 473. (a) Let the radius of small spheres be = r cm

Let n balls area cast → π r3) × 8 = (3)3
→ 8r3 = 33
 n × × π ( ) = 8 π r3
r = = 1.5 cm
→ → 8 → n = 48
474. (b) 12
467. (c) Let the radius of base of second cylinder = R
→ 2 (πr2h) = πR2h
→ 2r2 =R2
----------
√ - -- - - - - - - - r = 6 cm
468. (c) Volume of remaining solid 6 cm
r2h ----------
- - - - - - - - - --
----------
= 628.57 cm2
469. (a) Let the requried increase Let the increase in height = h cm
= x cm
→ πR2h = πr3
→ π (10 + x)2 × 4 = π × 102 × (4 + x)
100 + x2 + 20x = 25 (4 + x) (12)2 × h = × 63
x2 + 20x + 100 = 100 + 25x
x2 - 5x = 0 h = = 2 cm
x -5 =0
475. (c) Height of the cone = 10.2 – 4.2
x=5
= 6 cm
 Required increase =5 cm
470. (b) Let the sold volume → Volume of the toy = r2 h + π r3
π r2h → π r3 (h + 4r)
→ New volume = π(2r)2 h = 4πr2 h
= × (4.2)2 (4 × 4.2 + 6)
→ New volume is four times the old volume
471. (b) Let the height of cone be ‘h’ cm (4.2)2 × 22.8
π r2 h = π r3 = 421 cm3- (opp.)
476. (c) Volume of water = Volume of cylinder -
a2 h = 4a2 Volume of cone
h = 4a
π r2h

( )
→ 2 × 27π = 54π cm2
477. (c)

472. (c)
Radius of base = = ----------
- -- - - - - - - -

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 - - - - - - - - - -- 481. (a) r3 = 19404
----------
r3 =
- - - - - - - - - -- → r = 21 cm
---------- → Total surface area = 3πr3
---------
= = 4158 cm2
Height of water after ball is immersed = 3.5 × 2 = 7
cm
482. (c) Curved surface area = 2πr2
2 3
→ Volume of water = π r h - π r 2× = 760.57 cm2
π r2( ) 483. (a) Slant height of the cone
( ) (l) = √
→ Requried ratio =
= 11 × 3.5( ) = cm3
Volume of water before ball was imersed =
= π (3.5)2 × h = =8:5

=h = 484. (c) The volume of cone having same height &

diameter as that of a cylinder
= cm
= ×volume of cylinder
478. (b)
Let the height of cone = h cm No. of cones requried = 3
→ π r2h = π r3 485. (a) Let the no. of small balls = x
h = 4r → (10)3 = x ( )3
h = 4a (Thus, r = a cm)
→ 1000 = x ×
479. (c) Let height and radius be = 7x and 5x
respectively → x = 8000
→ π r2 h = 550 486. (a)
π (5x)2 × 7x = 550 Let the no. of balls = x
x2 × 7x = 550 → 44 × 44 × 44 = x × π( )
x3 = 1
→
x =1
Thus, height = 7 cm → x = 2541
radius = 5 cm 487. (d) Let the no. of cones = x
→ Curved surface are = 2 π rh
→ π 32 × 5= x × ×π×( ) ×1
= = 220 cm2
480. (c) Let the height of the cylinder be ‘h’ cm and → x = 9 × 5 × 3 × 100 = 13500
the radius be r cm
488. (c) Slant height of cone = √ cm
→ π r2 = b Slant height of cone = Radius of sector = 10 cm
489. (a)
→r=√ Volume of sphere
2 π rh = a = π r3= × π ×(9)3

2π√ ×h = a = 972 π cm3

Volume of cone = π r2 h = π× (9)3
h = cm
√

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 → 972 π cm3 r =
2 2
Volume of cone πr h = π 9 × 9 h = 21 cm
= 729 π cm3 Volume of cone
→ % of wasted wood % = ×π× = cm2

490. (d) r

r 493. (c)

2 πr = a,
r =
Volume of cylinder = V
h1 h2
π r2h = V
π( )2 ×h = V

×h =V r1 r2.
h = = =
491. (d) Radius of sphere =
Let the height of the cylinder = h ( ) × =
ATQ
Volume and radius are same Thus,
π (6)2 ×h = (6)3
( ) =
h = = 8 cm
492. (b)

=
h = 21 cm
494. (d)

Perimeter of base = 8 cm
h
2πr =8
2m

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 radius will be l
become slant height
l
In this question just cut the semi-circular paper and
told it to form cone
r Circumference of cone
2
π r = 154
r2= =49 
r = √49 7 m 2π r = π × 14
also volume = 1232 r = 7 cm
r2 × h = 1232 slant height, (l) of cone = radius of semicircular
plate
h = =8 l = 14 cm
=8m h2 = l2 – r2
Area of canvas required 142 - 72
→ π rl = 196 – 49
= πr√(r2 + h2) = 147
× 7 × √(242 + 72) h = √147 = 4√3
Volume of cone = √
= m2
= 622.36 cm3
Length (l) = = 275 m 498. (d)
495. (b) Ratio of the volume of cones r πrl

= ( ) =( ) =

= 9 : 16
496. (c) Ratio of surface area of sphere
h H
=

( )
Volume of water in conical flask = π r2 h
If the height of water level in cylindrical flask = H
Ratio of their volume units
Thus, π m2 H = π r2h
( ) =( )
H =
= 8 : 27 499. (d) R = 7 cm
497.
r

28
Volume of the solid sphere

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 (radius)3 = ∛(7 × 7 × 7 × 3 × 3 × 3)
r3
radius = 7 × 3 = 21 cm
= π × 7 × 7 × 7 cm3
Let the length of wire = h cm
π R2h = ×7×7×7

7×7×h= ×7×7×7
503. (d) r = 50 cm
h = cm
500. (b)

h =6× r = 50
r=3
H

Volume of cylinder
R
= 6 × volume of a sphere
Volume of sphere = = π r3 π 502h = 6 × π 503
= π × 3 × 3 × 3 = 36 π cm3 h =6×
If the water level rises by H cm = 400 cm
π R2 H = 36 π =4m
6 × 6 × h = 36 504. (b)
h =1
501. (b)

h + h =
R=9 4 5 cm

Volume of sphere = π R3
Volume of both the cones will be equal to the
= 972 π cm3 volume of sphere
Let the length of wire = h cm
π32 h + π42h = π53
π (0.2)2 ×h = 972 π
h = cm h 3 2 + 42 = × 5 × 5 × 5
= 243 meters × 25 = × 5 × 5 ×5
502. (a) Volume of sphere = Volume of rectangular h = × 3 = 20 cm
block
505. (a) Volume of cone = πr2h
3
π (radius) = Length × Breadth × height
Now, r1 = 2r, h1
π (radius)3 = 21 × 77 × 24 = 2h
Thus, Volume of second cone

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 = π r12h1
= π 2r2 2h = π r2 h × 8
= 8 times of the previous volume
Alternate:
In the formula of volume of cone, there is power 2
on radius and power 1 on height = 126
 (2)2 × 2 = 8 times
506. (d)

h l

r 510. (c)
C = π rl
C2 = πr2r2l2
V = π r2h
V3 = π2r4h2
3 πvh3 - c2h2 + 9v 3.6 cm h h
3π × π r2h h3 - π2r2l2h2 + 9 × π2 r4 h2
= π2r2h4 - π2r2h2 (r2 + h2) + π2r4h2
= π2r2h4 - π2r4h2 – π2r2h4 + π2r4h2 = 0 1.6 cm 1.2
507. (a) Volume of rectangular block = 11 ×10 × 5 cm
= 550 m3
= 550000 dm3 (1 m = 10 dm) According to question,
Volume of a sphere
× π × 1.6 × 1.6 × 3.6
π× dm3
→
= dm3
h =
ATQ n× = 550000
n = = 8800 = 6.4 cm
508. (a) Required number of spheres 511. (a)
× 32
( )
= 4×π×9 = =36π units
= 81
512. (d) Radius of sphere = cm
509.
(d) Volume of a sphere = π ×
Let the radius of cone = R

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 height = 2R We are given that:
According to the question
= π × R × R × 2R

π× ×32000 ( )

R3 =

R= But, h2 = 30
Height of glass = 2R × 2 × 2 = 4 cm  3 = 30 cm
513. Volume of earth taken out = 40 × 30 × 12 = 14400 1 = 10 cm→ h1 = 10
m3 height from base = 30 – 10 = 20 cm
Area of rectangular field = 1000 × 30 = 30000 m2
Area of region of tank = 40 × 30 = 1200 m2
Remaining area = 30000 – 1200 = 28800 m2
Increase in height =
= 0.5 m
514. (a)
R = 6 cm

r r

H = 12 cm

According to question,
8 × π r3 = π (6)2× 12 516. (d)

r3 =
=3×3×3
r = ∛(3 × 3 × 3) = 3 cm
515. . r
R

h1
Height = 30 cm
h2
Volume of lead = π r3
Volume of Gold = π r3 - r3
According to question,
R
π R3 – π r3 = π r3
π R3 = π r3
r3 = 2 r3

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 R3 = 2(2)3
π ( ) = 8π r2
R = 3√2×2 = 1.259 × 2
= 2.518 H = 9 × 8 = 72 cm
Thus, Thickness = R – r 520. (b) Volume of metallic sphere = volume of cone
= 2.518 – 2 = ×π3×3×3
= 0.518 cm = π R2 h
517. (a)
π×3×3×3

7
7 h = =3 cm
7 521. (d)
Number of bottle
=

In the question, =60

Radius of hemisphere = Radius of cone = height of
522. (a) Volume of cone V1
cone = 7 cm
 height of hemisphere = radius of hemisphere π r2 h
Volume of ice cream
= Volume of hemisphere part + volume of conical = r3 ( h = r)
part Volume of sphere V2 = π r3
= × (7)3 + × 73
Volume of cylinder V3
3 2 3
= × 7 = 22 × 7 = 1078 cm = πr2h = π r3
518. 4 cm  V1 : V2 : R3 = 1
= 1 :4 :3

r V1 = =
2
R 523. 4 π (Side) = 346.5
(Side)2 = = 5.25 cm

Volume of material of hollow sphere = Volume of

cone
π (52 - 32) = π (4)2 1 h
98 = 4 h
h= = 24.5 cm
519. (d) Radius of the base of conical shape = r cm 524. (b)
 Radius of base of cylinder =
Volume of water = Volume of cone
π r2 h = π r2 × 24
= 8 π r2 cm3
Volume of cylinder = volume of water

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 √3a = 2 × 6√3
a = 12
Surface area = 6a2 = 6 × 12 × 12 → 864 cm2
529. (c) Let hemisphere radius be = R
& Sphere radius be = r
ATQ,
Height of kaleidoscope = 25 cm 2R3 = 16 r3
Radius of kaledoscope = 35 cm
Paper used = curved surface area of cylinder = 2
× =
= 2 × 22 × 5 × 25 So option, ‘C’ is answer
= 5500 cm2 530. (d) Let part filled be =x
525. (b) According to the question, ATQ,
→ Volume of sphere = surfacea area of sphere x × (48 m × 16.5 m × 4m) = π (2) 2 × 56
→ π r3 = 4 π r2 → = 1 → r = 3 units x=
→ then diameter of sphere will be = 2r x = Ans.
= 2 × 3 = 6 units 531. (a) According to the question,
526. (b) Let the height of cone h meter
→ Total area of ground will be requried = 5 × 16 5
m2 = 80 m2 3
→ Total volume of air is needed = 100 × 5 m3 =
500 m3
According to the question,
Volume of cone = 500 m3
→ area if ground height = 500 Whole surface of remaining solid = π r l + 2πrh +
2
→ × π r × h = 500 πr2
Hence l = √
Height =
l =√
→ height of cone = 18.75
l =5
meters
Thus, π r *l + 2h + r+
527. (d)
Volume of cone = Lateral × 3 [5 + 2 × 4 + 3]
Surface Area
π r2h = π rl *l = √ ]

=√
Squaring both sides
=

= +
528. (c) Diagonal of cube will be equal to diameter of
sphere
√3a = 2 × r

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 ( rs = 1 cm)
π cm3
Vol. of water that can immerse the ball
= cm3
533. (b) According to the question,
Radius of the frop of water = cm

 Volume of drop of water = ×π

Volume of 32000 drops
532. (a)
× 32000 = π
A P B Thus, Volume of cone =
× π r2× h
2r = h
= π× ×h
3
Y ---------------------- X 16 × 4 = h
------------------- h = 4 cm
D ---------------- 534. (b)
--------------
O Here h = 4c
----------- Volume of cylinder = πr2h
---------
------
--- (Miltiply 4 π both in Numerator & denominator)

C
Δ ABC = equilateral Δ (Multiply 4π both in Numberator & denominator)
 ACB = 60°
=
BCP = 30°
ΔCDO, LDO = 90° (Angle b/w radius and tangent
is 90) 535. (a) According to the question,
OD = 1P = 1 cm Volume of sphere = π r3
OC = 2P = 2(1) = 2 cm Volume of cylinder = π r2h
then, CZ = OC + OZ
= 2 + 1 = 3 cm r=h = = 4 cm
Δ CZY CZY = 90 °
C.S.A of cylinder = 2 π rh
CZ = √3P = 3 cm
YZ = 1P = √3 cm =2× ×3×4
Now, In cone XYC
r = ZY = 3 cm = = cm2
h = CZ = 3 cm 536. According to question,
Vol. cone = π r2h = π (√3)2 (3) R = 6 cm
= 3 π cm2 → The capacity of the hemispherical bowl
Vol. of sphere = π r s3 = π r3 = × × 63

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 = 452.57 cm3 4 π r h = 2 π × r × Height
→ Height = 2h = units
539. (a) According to the question,
Radius = 3.5 cm
→ In the question it is given that A hemi-sherical
bowl is to be painted insided as well as outside
Total area that is to be painted = Inside area of
bowl + outside area of bowl
= 2 π r2 + 2 π r2
= 4 π r2
=4× cm2
→ Painting Rate = 10 cm2 in 5 Rs.
1 cm2 will be painted = = Rs.
So total cost will be painted in
=4× = Rs. 77
540. (b) 4.2 cm
537. (d)

r = 2.1 cm

20 cm

r = 2.1 dm
h = 4.2 dm
(for max)
According to the question,
Volume of cone = π r2h
→ r = 7 cm
→ h = 20 cm = × 2.1 × 2.1 × 4.2
→ Total surface Area of cylinder = curved surface = 19.404 dm3
Area + 2 × area of base 541. (d) Let the initial radius = r
2 π rh + 2 πr2 According to the question,
= 2πr (r + h) 4π(r + 2)2 - 4πr2 = 352
r2 + 4 + 4r – r2 =
= 44 × 27
4r + 4 = 28
→ TSA of cylinder = 1188 cm2
r =6
538. (d) According to question,
542. (a)
Given,
→ Radius of cylinder
=r
→ CSA of cylinder = 4 π r h
→ As we know
9 15
→ Curved surface area of cylinder
=2πRH

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 → 15000 m2
12 12 1 hectare = 10000 m2
Volume × 2
πr h → Level of rainfall
→ height of rainfall
→ π 12 × 12 × 9
= height of water level
→ 144 × 3π → 432 π = 5 cm = m
543. (c) According to the question,
Volume of cone Thus, Volume of collected water
π r2 h → 15000 × = 750 m2
Height = 7 cm 547. (b) Let the no. of hours be ‘x’
x (πR2H)= π r2h
Radius =
Thus, Volume of cone → 3000 × π × ×x
= = 89.8 cm3 =π× ×2

→ ×x =1

x =
= 1 hours 10 minutes
548. (a) Diameter = 5 mm = 0.5 cm
radius = 0.25 cm
Volume of water flowing from the pipe in 1 minute
= π × 0.25 × 0.25 × 1000 m3
Volume of conical vessel = π × 15 × 15 × 24 cm3

Thus, Time =
544. (c) Radius of 1st. solid metallic spheres = R = 6 cm =
Radius of 2nd. solid metallic spheres = r = 1 cm
= 28 minutes 48 seconds
Internal Radius of hollow spheres= x
549. (d) r = 12 m, h = 9 m
External Radius of hollow sphere = x + 1
l = √
So, π (R3 + r3) = π *(x + 1)3 – x3]
=√ = 15 m
216 + 1 = x3 + 1 + 3x (x + 1) – x3 Cost of canvas = curved surface area = × cost of 1
216 = 3x (x + 1) m2
72 = x2 + x Rs. 67824
→ x2 + x – 72 550. (d)
=0
After solving,
x = 8 cm 0.1 cm
So, the external radius of the hollow sphere
= x + 1 = 8 + 1 = 9 cm
545. (a) Let the time taken to fill the tank = x hrs x
→ (π r2 h) × x = 50 × 44 ×
8.4 gm = 1 cm3
→x=
4725 gm = cm3
= 2 hrs
Volume = x × x × 0.1
546. (b) → The area of ground

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 cm3 Thus, Volume = l b h
x2 × 0.1 = 2×4×6
x = 75 cm = 48 cm3
551. (d) According to the question 555.
(a) =
diameter = 84 cm
radius = 42 cm = 0.42 m
Height = 120 cm = 1.2 cm =
Thus, Circumference of cylinder = 2 π r h
( ) =
= Rs. 2376 Ratio of surface area
552. (a) Since the volume of the two cylinders is same

 =1
( )
( ) ( ) =
( ) =
556. (d) 2 πr h = 264 …. (i)
π r2h = 924 ……… (ii)
√
On diving =
√ r = 7 cm
=
Diameter = 2r = 2 × 7 = 14 cm
Thus, Ratio of their ratio
putting, r = 7 in (i)
= √2 : 1
2π r h = 264
h = = 6 cm
Required ratio =
553. (a) We are given that volume of two cube are in
the ratio= 27 : 1 =
557.
( )
(b) =
3
√
=
=
554. (a)
Ratio of edges of cuboid = 1 : 2 : 3 558.
(c) =
Let, l = x , b = 2x, h = 3x
Surface area = 88 cm2 ( ) =
2 (lb + bh + hl) = 88
2 (2x2 + 6x2 + 3x2) = 88 Ratio of their total surface area
11x2 = 44
x2 = 4 = = ( )
x =2
Thus, l = 2 cm, b = 4 cm, = ( ) =
h = 6 cm = 9 : 16

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559. (b) Radius of both hemisphere and cone = R
Also heightof hemisphere is equal to its Radius = R
 Height of both hemisphere and cone = R
Now, In cone
Slant height, l = √ =
= √2R →4:1
563. (d)

√
=
√
= √2 : 1
560. (c) Let height of cone = h 1
radius of cone = r 5
Volume of cone = π r2h
Now height is doubled Ratio of total surface area
2
Volume of new cone = (2h) = π r2 h =
Required ratio = 1 : 2 → 1 : 25
561. (d) 564.
(b) Let r1 = cm

r1 = 40 cm r2 = r2= cm
10 cm Required ratio =
=

A
B =
= 36 : 25
= ( ) 565.
( ) → 16 : 1 6 cm
= +
562. (d) +

3x
h1 4x 5x
h2 π ,(3x)3 + (4x)3 + (5x)3 = π (6)3
x3 (27 + 64 + 125) = 216
x3 - × 216 = 216
r1 x3 = =1
3
2r1 x = √1 = 1
R2 = 2r Radius of smallest sphere = 3x = 3 × 1 = 3 cm
h 2 = h1 x3 × 216 = 216
x3 = =1

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 x = 3√1 = 1
Radius of smallest sphere = 3x = 3 × 1 = 3 cm =
566. (d) M
= = 1:1
569. Ratio of Volume = (ratio of radius)3
h1 =
A D ( )
2 = h2
570.
(a) πr3 = πr2h

H= r
B C
R=
= = + =
Area of part (ABCD) (i.e. frustrum) = 8 – 1 2r (diameter ) × 2h = h
Required ratio = 1 : 7
567. (a)
571. (a) In this case height of cylinder and cone is equal
to the radius od hemisphere
H= r
√ Ratio of volumes

πr2 × r : πr3 : πr2 × r

1 : 2: 3
572. (A) Ratio of curved surface area = ratio of product
of height and radius
( )
= =  required ratio =
√
( ) 573. Ratio of surface area = (Ratio of radius)2
→1:2
=( )

574.
=

568. (b) R

H = 2R X=
R
575.
(c) R
(Height of cylinder = 2 × R)
=

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 equal to height of cylinder
=64 Ratio of volume of cylinder and
sphere
( ) = πr2 × r : =3:4
 r = 2 cm 582. (c)
Ratio of area
= (Ratio of radius)2
= (8 : 2)2
(H = R)
= 16 : 1
576. R : r = √3 : √4 = √3 : 2
(a) =1 583.
(b)
=1

 ( )

577. (a) Let the radius and slant height be 4x and 7x

 y = 1 cm
x = 2 cm
 Radius = 4×2 = 8 cm
578. (d) Height of cylinder = Diameter of sphere
 2πrh = 4πr2
 4πr2 = 4πr2 (h = 2r)
 Required ratio = 1 : 1
579. 584. (a) cone  radius : height
(a) 4 : 3
580. (d) Ratio of volume of bigger cone Let 4x : 3x
and smaller cones Curved surface area of cone
= (Ration of altitude)3  πrl
= (1 : 2 : 3)3 r = radius
Ratio of parts = 1 : 8 – 1 : 27 – 8 l = slant height
= 1 : 7 : 19 l=√
581. (c) Let radii of cylinder and sphere be r =√ = 5x
Volume of cylinder of height (h) curved surface area
= πr2h  π × 4x × 5x
Total surface area of cylinder  20πx2
= 2πrh + 2πr2 total surface area  πrl + πr2
Total surface area of sphere  πr (l+r)
= 4πr2  π × 4x (5x + 4x)
given that  π × 4x × 9x
4πr2 = 2πrh + 2 πr2  36 πr2
= 4 πr2 Curved area : total area
= 2πrh (h+r) 20πx2 : 36πx2
 2r = h+r 5 : 9
r =h 585. (b) Let radius of sphere =
radius of sphere or cylinder’s

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 radius of cylinder = r = 3 : 25
let height of cylinder = r 590. (d) Let the radius of sphere and hemisphere
given that volume of sphere be = R and r
= volume of cylinder

=
2R3 = r3

curved surface area
cylinder : sphere  R : r = 1 : ∛2
591. (a) Ratio of radius of earth and moon
2×π×r× : 4πr2
=4:1
 : 4  Ratio of volume = 43 : 13
2 : 3 = 64 : 1
592. (b) Let the radius of cylinder and sphere be = r cm
586. (c) radius of cone = radius of cylinder  height of cylinder = 2r cm
height of cone = height of cylinder = h  A = πr2 × 2r = 2πr3
B=

=
 =3 : 2
=
593. (√ ) √
(b) Ratio of volume =
 (√ ) √
= √ √

594. √
 l2 = 25 √ (c) Side of cube = 6 cm
√
 h2 = 16
Required rate = 1 : 1
√ = 25
√ = 25 595. (d) Let the radius of hemisphere and sphere be ‘r’
r=3 and ‘R’
radius : height 
3 : 4
587. (b) Let the sides of the rectangular box he
x, 2x, 3x
 2(2x2+6x2+3x2) = 88 √
 11x2 = 44  Ratio of curved surface area
x2 = 4
=
x=2 (√ )
 Sides are 2, 4, 6 cm 596.
Volume = 2×4×6 = 48 cm3 (b)
588. (b) Ratio of volume = r2 r2
√ √ √
(√ ) √ √ h
√ √ h h
589.
(a) Ratio of volume =

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 = 11×10×5 = 550 m3
= 550000 dm3. (1 m = 10 dm)
 Volume of a sphere
Diameter of cylinder = Diameter of cone = dm3
597. (d) rr
dm3
r h
A.T.Q n × =550000

n= = 8800
volume remains same 601.
(a)
volume of sphere
= volume of cone

4r = h
=4 : 1

598. (c) ( ) ( )
Let the side of first cube = a1
and the side of second cube = a2 ( )

√
Ratio of their surface area =  8 : 3

= = 9 : 16
599. (a)
602. (d)
r1 r2
R
Ratio of volume of sphere
x ratio of weight per 1 cc. of material of R/2
each
= Ratio of weight of two sphere

Let the Radius of cylinder = R

 Therefore, Radius of sphere =
Volume of right circular cylinder = πr2H
volume of sphere

 8 : 17 = ( )
600. (a)
volume of rectangular block =

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 According to question, = If a cone is cut in any parts parallel to
volume of cylinder = Volume of sphere its base then the ratio of volume of smaller
cone to the volume of larger cone is equal to
πr2H = the ratio of the cubes of the corresponding
heights/radii/slant a height(it is proved
by similarity)
 1 : 6
( )
603. (d) = h 1 : h2 – h1 = 1 : √
Radius of longer sphere = R units √
1 : (√ )
Its volume =
Now cones are formed with base radius 605. (d)
and height same as the radius of larger Let side of square be = x
sphere Area of square = x2
volume of smaller cone = Side of new formed square
= x + 50% of x
and one of the cone is converted into
= 1.5 x
smaller sphere
Area of new formed square
Therefore, volume of smaller sphere = = (1.5x)2
= 2.25x2
Ratio of the area (new square) : area of
(original square)
= 2.25x2 : x2
9 : 4
√ Quicker approach
Let side of square = 100%
(100% + 50%)2
9 : 4

( ) ( )
606. (a)
1 :
604. (c) A volume of cone =
h1 = cm3
B C Volume of cubical block = 10×5×2 cm3 = 100 cm3
h2 Wastage of wood
( ) cm3
D E
Cone =( ) cm3

= BC = ED (Volume are equal) % wastage = ×100=278/3

= %
=

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607. (c) Decrease in radius = 50%
= ×100
increase in height = 50%
Inc ement
= = ×100
Radius Height Volume = 25%
original 2 2 (2) 2×(2) = 8 611. (d) Height 1 → 3
50% 50% –5 Radius 2 → 1
decrease Increase Volume 4 → 3
New 1 3 (1)2×(3) = 3 % decrease = = 25%
Reduction in volume 612. (d)
= height = 100% Radius = 100%
Inc ement
608. (a) O igian
Increase in radius = 100% height Radius volume
= original 1 1 (1) 21 = 1 is
2
Radius Height Volume New 2 2 (2) (2) = 8 constant)
original 1 1 (1) 2×(1) = 1 = eight times that of original
7
613. (b) use x + y+
2
New 2 2 (2) ×(2) = 8
percentage change in area
% increase = = 700%
–
= 15 – 10 +
609. = 5 – 1.5 = 3.5 %
(d) 20% =
o igina (3.5% increase)
Radius light volume Remember : When change in area is
5 5 (5)25 = 125 asked in the question, then use this formula
91 To save your valuable time.
2
6 6 (6) (6) = 216
×100= 72.8%

610. (d)
614. (d)
r = 15 cm h = 15 cm
Let old radius = r
 volume =
R = 15 cm
New radius = 2r
Volume of cone
 New volume =

Volume of sphere =
 volume becomes eight times.
Percentage wanted 615. (b) Radius 2 → 1
Height 5 → 8
×100 Volume 20 → 8

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  volume decreases = 6 πrh
% decrease = = 60% New lateral surface area
= 2 × π × r × 6h
616. (d) Length 1 → 2 = 12 πrh
Breadth 2 → 6 Required factor = =2
Height 3 → 9
Volume 6 → 108
621. (c)
 New volume = 18 times the
Let the original radius be ‘r’
original volume
 Area = 4πr2
 Increase in volume
New area = 4π(2r2) = 16πr2
= 18 – 1 = 17 times
 New area is 4 times the old area
617. (c) Radius 10 → 11
622. (d)
Height 10 → 11
Edge is increased by 50%
Volume 1000 → 1331
 % Increase = =
= 33.1% =
o igina
618. (a) Let original edge = 2
% change in height = % Increased edge = 3
change in volume = 100% edge surface area
619. (a) 2 4
Volume of coffee +5[ 6 is constant]
3 9
=
% increase = = 125%
= cm3

Volume of cone =

= (8)2×16

=
Required percentage

= ×100

= 87.5%
620. (a) 623. Let the initial radius = 1 unit
Decrease in these radius New radius = 2 unit (radius is
doubled)
= (decrease in base area)1/2 = ( ) Radius : Volume
(1)3
=
(2)3
Let initial radius and height be 3r and h
New radius and height are r and 6h ( is constant)
Old lateral surface area 1 1
= 2 × π × 3r × h 7

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 2 8 G a F
Percentage increase = ×100  HF = (slant height)
= 700 %  HG = slant height(1)
 GF = base
624. (a) Quicker approach
slant height  √( )
in A = a + b +
√
Here a = b = 5% =√
in A = ( ) 2a
= 10.25 % G
A F
625. (a)
Volume of tetrahedron
2a 2a
=
√ √
= = √ cm3 o
√
626. (a) AOF is equilateral triangle of side 2a (AOF)
√
volume of bucket = (R2+r2+Rr) Altitude GO = ×2a
=√ a
= ×45(282+72+28×7) √
Slant height  a
3
= ×15×1029 = 48510 cm altitude = √ a
height of the pyramid = h
627. (c) √
 √( ) (√ )
Side of regular hexagon = 2a cm
area of hexagon
√ =√ √
= 6× ×(2a)2
 √ cm2 = a
slant edge of pyramid Volume of pyramid
 cm = area of base × height
H = √
= √ cm3
G
A a a F
2a 2a
B E
O
C D 628. (c)
40 cm
slant edge 
(Given)
H 40 cm 40 cm

40 cm
 area of base = 40×40

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 = 1600 cm2
3
Let height of pyramid = h
Volume = × h × area of base
= area of triangle
= × h × 1600
 cm = area of triangle
 8000 (given)
= h = 15 cm volume of prism  270 cm3
629. (c) area of trapezium (given)
= × h (AB + CD) 270 = h ×
= × 8 × (8 + 14)  h = 12 cm
= 4 × 22 = 88 cm2
= volume of prism 632. (b)
= Height of prism
area of base 15 cm
 height × 88 = 1056 (given) 9 cm
 height × 88 =
12 cm
 12 cm 9, 12, 15 is a triplet which forms a right
630. Edge of regular tetra hadron = 3 cm Angle triangle.
a = 3 cm area of base of prism
 ×9×12 = 54 cm3
3 cm 3 cm # Perimeter of triangle = 9 + 12 + 15
3cm = 36
Total surface area of prism = perimeter × height +
area of base
 height of prism = 5 cm given
√
Volume  cm3 total surface area = 36 × 5 + 54
 180 + 54 = 234 cm3
√
 × (3)3 = √ cm3

631. (d) A

a a
a

B C
r – in radius of in circle of triangle
perimeter = 15 cm (given)
Semi-perimeter (S) =
In radius of any triangle
633. (c)
r a
a ea
r=
where Δ is the area of triangle
r = 3 cm given

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 a a 1 = 17
b=8

Let side equilateral triangle b = a

√  height of pyramid  15 cm
area = a2  base = 8 cm
√  slant height of pyramid
= = 173 cm2
l =√  17 cm
 a2 = ×4  curved surface area of pyramid
√
(√3 = 1.73)  ×64×17  544 cm2
a2 = ×4
635. (c)
×4×100 Volume of pyramid
a2 = 400
= × Area of base × height
a = 20 cm
Perimeter of base = 20×3 = 60 = ×57×10 = 190 cm3
Volume of prism × 10380 cm3
(given) 636. (c)
area of base × height Let the side of square base = a cm
 2a2 + 4a × h = 608
height = = 60  2a + 4a × 15 = 608
LSA = 60×60 = 3600 cm2  a2 + 30a = 304
 a2 + 38a – 8a – 304 = 0
634. (b) 16  a(a + 38) – 8 (a + 38) = 0
 a = – 38,8
 a = 8 cm
16 16 Volume of prism = 8 × 8 × 15 = 960 cm3

16 637. (b)
Perimeter of the base = 4×16 = 64 cm √
volume of prism = a2 × h
# Curved or lateral surface area of pyramid
√
= × (perimeter of base) × height = × (8)2 × 10
= 160√ cm3
638. (b)
1 Volume of prism
×10×12×20 = 1200 cm3
8cm 8 cm
 Weight of prism = 1200×6 = 7200 gm
16 cm
= 7.2 kg

639. (a) Total slant surface area

= 4 × × 4 × a = 12

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 (Where a is the side of the square base) (perimeter of base × slant height + Base area =
a= cm 270√
√
 area of base = cm2 , √ √ -
(10√ )2 = 270√
Required ratio = = 16 : 3 15√ √ √ = 270√
√ = 13
2
640. (d) h + 25 = 169
Total surface area of tetrahedron h2 + 169 – 25 = 144
= √3 a2 h=√ = 12
=√3 × 122 643. (a)
= 144√3 cm2
641. (d) A 6 cm 6 cm

6 cm
r r Volume of prism = area of base × height
r √
= (6)2 × height
B C
In radius of triangle √
× 6 × 6 height = 81√
= √
pe imete height = = 9 cm
ar (Δ ABC) = Inradius × semi-perimeter √
= 4× = 4×14 = 56 cm
Volume of the prism = 366 cm3
(area of base)× height = 366 cm3
56 × height = 366 cm 644. (d)
height = = 6.535 cm
10√ 12 slant height
642. (d) A 5
10√
√ O M M Side of square = √ = 10 cm
√
slant height
(h) slant height = √ = 13 cm
E
lateral surface area = ×perimeter of base × slant
B 10√ C O E

Base is equilateral triangle height

In radius of equilateral triangle = × 40 × 13
= OE = = 260 cm2
√
√ 645. (d)
= = 5 cm Total surface area of prism = perimeter of base ×
√
slant length, 1 = √ height
+ 2 × base area
=√
√
Total surface area = (3 × 12 × 10) + 2 × × 122
= 270√ = 360 + 72√

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

 = 72(5+√ ) cm2 650. (a)
Lateral surface area of prism = 120
base perimeter × height = 120
L.S.A of prism = (base perimeter × height)
3 × (side) × height = 120
(perimeter of eq. Δ = 3 × side )
646. (b) M side × height = = 40 ……..(i)
16 cm
slant height volume of prism = 40√
o M 15cm
area of base × height = 40√
√
E O 8 cm E (side)2×height = 40√
16 √
(side)2 × height = = 160 ….(ii)
Slant height of pyramid = √ = 17 √
Dividing (ii) by (i)
(8, 15, 17) is triplet
lateral surface area
= × perimeter of base × slant height side = 4 cm
= × 64 × 17
651. (c) Volume of tetrahedron
= 32×17 = 544 cm2
√ √
647. (a) = (side)3 = (4)3
Perimeter of right Δ = (5+12+13) = 30
√ √
total surface area = lateral surface area + 2 × area  cm3
of base
= (perimeter of base × height + 2 × area of base
= (30 × height) + 2 × × 5 × 12
= (30 × height) + 60
ATQ 30 × height + 60 = 360
30 × height = 360 – 60 = 300
height = 10 cm
648. (d)
Height of pyramid = 6 cm
Diagonal of square base = 24√ m 652. (a)
Side of square = 24 m Area of the base of prism ( a right triangle)
Area of square = (24)2 = ×5×12 = 30 cm2
= 576
Volume of the pyramid Third side of the triangle
=√ = 13cm
= × area of base × height
Perimeter of the triangle
= × 576 × 6 = 5+12+13 = 30 cm
= 576 × 2 = 1152 m3 Total surface area
649. (a) = lateral surface + 2 × (base area)
Volume of pyramid = (perimeter of base × height + 2 × (base area
= (30×10)+2×30
= × area of base × height
= 300+60 = 360 cm2
500 = × 30 × height 653. (a)
13 cm
D C
height = = 50 m 15 cm

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 12 cm 14 cm 10

A 9 cm B
In Δ ABD, Area of base = 10×10 = 100 cm2
BD = √AB AD = √ Area of 4 phase
=√ =√ = 15 cm = × Base × slant height) × 4
Area of Δ ABD = × AB × AD  × 10×13×4
2
= × 9 × 12 = 54 cm = 65×4 = 260
In Δ BCD [slant height = √ =√ = 13]
semi-perimeter = Total Surface area
 260+100
= = 21  360 m2
Area of Δ BCD = 658. (d)
√ volume of prism = (area of base × height)
=√ Area of base (i.e. area of triangle)
=√ = 21×4 = 84 cm2  Area of base
=
area ABCD = 84+54 = 138 cm2
 Area of base
height of prism =
=√
= 15 cm = (By Hero’s formula)
ase
perimeter of base =
So, S = = 27
9+14+13+12 = 48 cm
Area of lateral surface = perimeter × height  = 27
= 48 × 15 = 720 cm2 √ ( – )
654. (a) As we know,
volume of Right Prism = Area of the base × Height √
√ √
 7200 = P2 × √  9×7×2
2
 72×2 = 9 P Volume of Prism
 P2 = 16 =(9×7×2)×9 = 1134 cm3
P=4
655. (b) 659. (d)
Half of its lateral edges Let the side of the square = a cm
 Half of its edges ATQ
 Half of its volume T.S.A = C.S.A + 2 base area
Then, volume reduced by = 50 % C.S.A = base perimeter × h
656. (b) T.S.A = base perimeter × h + 2 base area
Total surface area 192 = 4a × 10 + 2a2
√ 2a2 + 40a – 192 = 0
= 4×* +
a2 + 20a – 96 = 0
2
= √ cm a2 + 24a – 4a – 96 = 0
657. (a) a(a+24)–4(a+24) = 0
(a+24)(a–4) = 0
a = 4, (–24)
a = 4 (side can never be in –ve)
10
Volume = base area × h
10 10

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 Volume = 16×10 = √ = 12 cm2
Volume = 160 cm3 Volume of prism = 12×8 = 96 cm3
660. (c ) According to the question, 664. (a) Semi-perimeter of
V = number of vertices of prism = 6
e = edges of prism = 9 Δ = = 12 cm
f = faces of the prism = 5 Area of Δ = √
=√
=√
= =5
= √
661. (c ) 665. (c) According to the question
ATQ Volume of cylinder = πr2h
Volume of prism = Area of base × height
= trapezium area × height Volume of Sphere = πr3
The number of spherical balls
= (10+6)×5×8
= 16×5×4 = 320 cm3 =
662. (a)
= = 27000
666. (d) According to the question
2x Volume of the cylinder = πr2h

3x
Base of prism
667. (d) According to the question
 length : breadth
Volume of cylinder = Volume of cone
3x : 2x
Perimeter of base
= 2(3x + 2x) = 10x
area of base
 2x × 3x = 6x2
height of Prism = 12 cm (given)
total surface area of prism
= Perimeter of base × height + 2 × area of base
288 = 10x × 12 +12 x2 668. (d) According to the question C.S.A of cylinder
12x2 + 120x – 288 = 0 = 2 πrh = 2 π 2
x2 + 10x – 24 = 0 C.S.A of sphere = 4 π
x=2 2π =4π
area of base  6×4 √
 24 cm2
volume of prism
 24×12
 288 cm3
√
663. (b) =
√
Volume of the part (prism) = 669. (a) Total surface area of prism
Area of base × height
= perimeter of Base × Height
Area of base (Isosceles Δ) + 2 × Base Area
= √ 10 = 4a × 2 + 2 × a2
10 = 8a + 2 a2

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 a2+ 4a – 5 = 0
(a + 5)(a – 1) = 0
a = 1, a = – 5
volume of Prism = Area of base × height
= 1×1×2 = 2 cm3
670. (c ) Let the radius of wire = 1 cm
Volume of cone =

= (1)2h = πh
New radius of wire = cm
volume of new cone
= ( )

volume of old cone = Volume

of new cone

H = 9h
Height of new cone is increased by 9 times.

671. ( c) Painted Area of Prism

= 151.20×5 = 756.00 m2
AC = 15
[By using Pythagoras theorem]
Total surface Area = Perimeter
of base × height + 2 × Area of base
756 = 15+9+12×h+2× ×9×12
756 = 36×h+108
36 h = 756 – 108
h= = 18 cm
672. (d) Total surface Area
×(perimeter of base) ×
Slant height + Area of base
= (4×10)×13+10×10
= 260+100
= 360 cm2

[Slant height = √( )
=√
= 13 cm
673. (a) By option (a)
Are Increment = 20+20+
= 44%

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