Class X Session 2024-25

Subject - Mathematics (Basic)

Sample Question Paper - 10

Time Allowed: 3 hours Maximum Marks: 80

General Instructions:

1. This Question Paper has 5 Sections A, B, C, D and E.

2. Section A has 20 MCQs carrying 1 mark each

3. Section B has 5 questions carrying 02 marks each.

4. Section C has 6 questions carrying 03 marks each.

5. Section D has 4 questions carrying 05 marks each.

6. Section E has 3 case based integrated units of assessment carrying 04 marks each.

7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of
2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E

8. Draw neat figures wherever required. Take π = wherever required if not stated.
22

Section A
1. Cards marked with numbers 1, 2, 3, ..., 25 are placed in a box and mixed thoroughly and one card is drawn at [1]
random from the box. The probability that the number on the card is a multiple of 3 and 5 is

a) 12

25
b) 4

25

c) 1

25
d) 8

25

2. The roots of a quadratic equation x2 - 4px + 4p2 - q2 = 0 are [1]

a) 2p + q, 2p - q b) p + 2q, p - 2q

c) 2p + q, 2p + q d) 2p - q, 2p - q
3. An icecream cone has hemispherical top. If the height of the cone is 9 cm and base radius is 2.5 cm, then the [1]
volume of icecream is

a) 91.67 cm3 b) 96.67 cm3

c) 90.67 cm3 d) 91.76 cm3

4. The sum of two numbers is 17 and the sum of their reciprocals is 17

62
. The quadratic representation of the above [1]
situation is

a) b)
1 1 17 1 17
+ = =
x x+17 62 x(17−x) 62

c) 1

x
+
1

17−x
=
17

62
d) 1

x
−
1

17−x
=
17

62

5. The 2nd term of an AP is 13 and its 5th term is 25. What is its 17th term? [1]

a) 69 b) 77

Page 1 of 17
 c) 81 d) 73
– –
6. The distance between the points (a, a) and (− √3a, √3a) is [1]
– –
a) 2√2 units b) 3√2a units
–
c) 2 units d) 2a√2 units

7. If α and β are zeros of x2 + 5x + 8, then the value of (α + β) is [1]

a) -8 b) 8

c) 5 d) -5
8. In a △ABC, it is given that AD is the internal bisector of ∠A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, the [1]
CD = ?

a) 3.5 cm b) 7 cm

c) 4.8 cm d) 10.5 cm
9. In the given figure, PA and PB are tangents to a circle from an external point P. If ∠AP B = 50
∘
andAC || PB , [1]
then the measures of angles of triangle ABC are

a) 65°, 50°, 65° b) 50°, 55°, 75°

c) 80°, 60°, 40 d) 50°, 50°, 80°

10. How many tangents can be drawn to a circle from a point on it? [1]

a) Two b) Zero

c) Infinite d) One
11. If secθ + tanθ = x, then tanθ = [1]
2 2

a) x +1

x
b) x +1

2x

2 2

c) x −1

x
d) x −1

2x

12. The prime factorisation of the number 5488 is [1]

a) 23 × 73 b) 24 × 73

c) d)

Page 2 of 17
 24 × 74 23 × 74
13. If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, [1]
then the angle of elevation of its top

a) decreases b) Falls

c) remains unchanged d) increases

14. Three horses are tethered with 7-meter-long ropes at the three corners of a triangular field having sides 20 m, 34 [1]
m, and 42 m. The area of the plot which can be grazed by the horses is

a) 77 m2 b) 80 m2

c) 100 m2 d) 30 m2

15. If the perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm, then the area of the sector is ________. [1]

a) 15.5 cm2 b) 15.6 cm2

c) 15.9 cm2 d) 15.1 cm2

16. The probability of guessing the correct answer to a certain test questions is . If the probability of not guessing [1]
x

12

the correct answer to this question is 2

3
, then x =

a) 6 b) 4

c) 2 d) 3
17. If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd and is a [1]
multiple of 3 is

a) b)
1 2

9 9

c) 2

3
d) 1

18. If the arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x = [1]

a) 2 b) 1

c) 6 d) 4
19. Assertion (A): In a solid hemisphere of radius 10 cm, a right cone of same radius is removed out. The volume of [1]

the remaining solid is 523.33 cm3 [Take π = 3.14 and √2 = 1.4]

–

Reason (R): Expression used here to calculate volume of remaining solid = Volume of hemisphere - Volume of
cone

a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.

c) A is true but R is false. d) A is false but R is true.

20. Assertion (A): Sum of natural number from 1 to 100 is 5050. [1]
n(n+1)
Reason (R): Sum of n natural number is 2
.

a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.

c) A is true but R is false. d) A is false but R is true.

Section B

Page 3 of 17
21. Prove that 1
is irrational. [2]
√2

22. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO

BO
=
CO

DO
. Show that [2]
ABCD is a trapezium.
23. In Fig., PQ is a chord of length 8 cm of a circle of radius 5 cm and centre O. The tangents at P and Q intersect at [2]
point T. Find the length of TP.

24. If sinθ + sin2θ = 1, find the value of cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ - 2 [2]

OR
Prove the trigonometric identity:
tan θ cot θ
+ = sin θ cos θ
2 2
2 2
(1+tan θ) (1+cot θ)

25. In Figure, two concentric circles with centre O, have radii 21 cm and 42 cm. If ∠ AOB = 60o, find the area of the [2]
shaded region.

OR

Find the area of the segment shown in Fig., if radius of the circle is 21 cm and ∠ AOB = 120o (Use π = 22

7
)

Section C
26. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi [3]
takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same
direction. After how many minutes will they meet again at the starting point?
27. Point A is on x-axis, point B is on y-axis and the point P lies on line segment AB, such that P (4, - 5) and AP : [3]
PB = 5 : 3. Find the coordinates of point A and B.
28. To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of [3]
smaller diameter for 9 hours, only half of the pool can be filled. Find how long it would take for each pipe to fill
the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the
pool.
OR
A 2-digit number is four times the sum of its digits and twice the product of its digits. Find the number.
29. A point P is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent [3]
drawn from P to the circle.
OR

Page 4 of 17
 Find the length of a tangent drawn to a circle with radius 5 cm, from a point 13 cm from the centre of the circle.
2 2

30. Prove the trigonometric identity:

cot θ(sec θ−1)
+
sec θ(sin θ−1)
= 0 [3]
(1+sin θ) (1+sec θ)

31. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are [3]

30o and 45o. If the bridge is at a height of 8 m from the banks, then find the width of the river.

Section D
32. A person invested some amount at the rate of 12% simple interest and the remaining at 10%. He received yearly [5]
interest of ₹ 130 but if he had interchanged the amount invested, he would have received ₹ 4 more as the
interest. How much money did he invest at different rates?
OR
Sangeeta went to a book-seller's shop and purchased 2 textbook of IX Mathematics and 3 textbook of X mathematics
for Rs.250. Her friend Meenu also bought 4 textbooks of IX Mathematics and 6 textbooks of X maths of same kind
for Rs.500. Represents this situation algebraically and graphically.
33. Show that the points A(3,1), B(0, -2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD. [5]
34. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the [5]
hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
OR
A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal
to 2

3
of the total height of the building. Find the height of the building, if it contains 67 1

21
3
m of air.

35. Let there be an A.P. with first term 'a', common difference 'd'. If an denotes its nth term and Sn the sum of first n [5]

terms, find. n and Sn, if a = 5, d = 3 and an = 50.

Section E
36. Read the following text carefully and answer the questions that follow: [4]
The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial.
A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of
load, and so can be found in bridges and in architecture in a variety of forms.

i. In the standard form of quadratic polynomial, ax2 + bx + c, what are a, b and c? (1)
ii. If the roots of the quadratic polynomial are equal, what is the discriminant D? (1)
iii. If α and 1

α
are the zeroes of the quadratic polynomial are 2x2 - x + 8k, then find the value of k? (2)
OR
What is the relation between zeros and coefficient for a quadratic polynomial? (2)
37. Read the following text carefully and answer the questions that follow: [4]

Page 5 of 17
 Heart Rate: The heart rate is one of the 'vital signs' of health in the human body. It measures the number of
times per minute that the heart contracts or beats. While a normal heart rate does not guarantee that a person is
free of health problems, it is a useful benchmark for identifying a range of health issues.

Thirty women were examined by doctors of AIIMS and the number of heart beats per minute were recorded and
summarized as follows:

Number of heart beats per minute Number of Women

65 - 68 2

68 - 71 4

71 - 74 3

74 - 77 8

77 - 80 7

80 - 83 4

83 - 86 2

Based on the above information, answer the following questions:

i. How many women are having heart beat in the range 68 - 77?
ii. What is the median class of heart beats per minute for these women?
iii. a. Find the modal value of heart beats per minute for these women.
OR
b. Find the median value of heart beats per minute for these women.
38. Read the following text carefully and answer the questions that follow: [4]
Swimmer in Distress: A lifeguard located 20 metre from the water spots a swimmer in distress. The swimmer is
30 metre from shore and 100 metre east of the lifeguard. Suppose the lifeguard runs and then swims to the
swimmer in a direct line, as shown in the figure.

i. How far east from his original position will he enter the water? (Hint: Find the value of x in the sketch.) (1)
ii. Which similarity criterion of triangle is used? (1)
iii. What is the distance of swimmer from the shore? (2)
OR
What is the length of AD? (2)

Page 6 of 17
 Solution
Section A
1.
(c) 25
1

Explanation: Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24

Multiples of 5 = 5, 10, 15, 20, 25
Number of possible outcomes (multiple of 3 and 5) = {15} = 1
Number of Total outcomes = 25
∴ Required Probability =
1

25

2. (a) 2p + q, 2p - q
Explanation: Given: x2 - 4px + 4p2 - q2 = 0
⇒ (x - 2p)2 - q2 = 0
Using a2 - b2 = (a + b) (a - b),
⇒ (x - 2p + q)(x - 2p - q) = 0

⇒ x - 2p + q = 0 and x - 2p - q = 0

⇒ x = 2p - q and x = 2p + q

3. (a) 91.67 cm3

Explanation: Height of ice-cream cone is 9 cm and radius of the hemispherical top is 2.5 cm.
Now, Volume of ice-cream cone = Volume of cone + volume of Hemispherical top
= 1

3
π r2h + 2

3
π r3
= 1

3
πr2(h + 2r)
= 1

3
×
22

7
× 2.5 × 2.5(9 + 5)
= 1

3
×
22

7
× 2.5 × 2.5 × 14
= 91.67 cm3
4.
(c) 1

x
+
17−x
1
=
17

62

Explanation: Let one number be x, As the sum of the numbers is 17, then the other number will be (17 - x). Their reciprocals
will be and 1

x
.
17−x
1

17
∴ According to question, 1

x
+
1

17−x
=
62

5.
(d) 73
Explanation: a + d = 13 ... (i) and
a + 4d = 25 ....(ii)
From (i) and (ii), we get a = 9 and d = 4
⇒ T17 = (a + 16d) = (9 +16 × 4) = 73

6.
–
(d) 2a√2 units
– –
Explanation: Let the points be A(a, a) and B(−√3a, √3a)
−−−−−−−−−−−−−−−−−−−−− −
– 2 – 2
∴ AB = √(−√3a − a) + (√3a − a)

−−−−−−−−−−−−
–−−−−−−− −−−−−−−−−
–−−
= √3a + a + 2√3aa + 3a
2 2 2 2
+ a − 2√3aa
−−−−−−−−
= √6a + 2a 2 2

−−
−
= √8a2
–
= 2a√2 units

Page 7 of 17
 7.
(d) -5
Explanation: x 2
+ 5x + 8
−Coefficient of x
α + β =
2
Coefficient of x
−5
=
1

= -5

8. (a) 3.5 cm
Explanation: By using angle bisector theore in △ABC, we have
AB BD
=
AC DC

10 6−x
⇒ =
14 x

⇒ 10x = 84 − 14x

⇒ 24x = 84

⇒ x = 3.5

Hence, the correct answer is 3.5.

9. (a) 65°, 50°, 65°
Explanation: Since PA = PB {Tangents from an external point to a circle]
∴∠PAB = ∠PBA

Let ∠PAB = ∠PBA = x

Now, in triangle APB, x + x + 50° = 180°
⇒ x = 65°

Since AC ∥ PB and AB is intersecting.

∴∠PBA = ∠BAC = 65 [Alternate angles] ∘

And ∠PAB = ∠CBA = 65 [Alternate angles] ∘

∴∠ACB = 180 − (65 + 65 ) = 50

∘
[Angle sum property of a triangle]
∘ ∘ ∘

∘ ∘ ∘
∴∠BAC = 65 ,∠ABC = 65 , ∠ACB = 50

10. (a) Two

Explanation: Two
11.
2
x −1
(d) 2x

Explanation: Given, secθ + tanθ = x ...(i)

We know that
2 2
sec θ − tan θ= 1

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ x(sec θ − tan θ) = 1

⇒ sec θ − tan θ =
1

x
...(ii)
Subtracting (ii) from (i)
2
1 x −1
2 tan θ = x − =
x x
2
x −1
tan θ =
2x

12.
(b) 24 × 73
Explanation: 24 × 73
13.
(c) remains unchanged

Explanation:

Page 8 of 17
 Let height of the tower be h meters and distance of the point of observation from its foot be x meters and angle of elevation be
θ ∴tan θ = ………(i) h

Now, new height = h + 10% of h = h + 10

100
h = 11h

10
And new distance = x + 10% of x = x + 10

100
x = 11x

10
∴

11h

h
……….(ii)
10
tan θ = =
11x x

10

From eq. (i) and (ii), it is clear that the angle of elevation is same i.e., angle of elevation remains unchanged.

14. (a) 77 m2
Explanation: 77 m2
15.
(b) 15.6 cm2
Explanation: Perimeter of a sector of circle = θ
∘
× 2πr + 2r
360

θ
⇒ (
360
∘ × 2π × 5.2) + (2 × 5.2) = 16.4
θ 16.4−10.4 6
⇒ ∘
π = =
360 10.4 10.4
θ
Area of sector of circle = 360
∘ × πr
2

= 6

10.4
× (5.2)
2
= 15.6 cm2

16.
(b) 4
Explanation: Probability of guessing the correct answer
x
=
12

and probability of not guessing the correct

2
answer = 3
x 2 ¯
+ = 1 ∵ (A + A = 1)
12 3
x 2 1 12
⇒ = 1 − = ⇒ x = = 4
12 3 3 3

∴ x=4
17.
(b) 2

Explanation: Total numbers of digits for 1 to 9(n) = 9

Number divisible by 3(m) = 3, 6, 9
Odd numbers out of 3, 6, 9 = 3, 9
∴ Probability =
m 2
=
n 9

18.
(d) 4
Explanation: Mean of x, x + 3, x + 6, x + 9, x + 12 = 10
x+x+3+x+6+x+9+x+12
⇒ = 10
5
5x+30
⇒ = 10
5

⇒ x + 6 = 10
⇒ x = 10 - 6 = 4

19.
(d) A is false but R is true.
Explanation: A is false but R is true.
20. (a) Both A and R are true and R is the correct explanation of A.
Explanation: Both A and R are true and R is the correct explanation of A.
Section B
21. Let us assume, to the contrary, that is 1
rational.
√2

So, we can find coprime integers a and b (≠ 0) such that

1 a – b
= ⇒ √2 =
√2 b a

–
Since, a and b are integers, b

a
is rational, and so is√2 rational.

Page 9 of 17
 –
But this contradicts the fact that is√2 irrational.
So, we conclude that is irrational. 1

√2

AO CO
22. Given: The diagonals of a quadrilateral ABCD intersect each other at the point O such that BO
=
DO

To prove: ABCD is trapezium.

Construction: Through O draw a line OE||BA intersecting AD at E.
Proof: In △DBA∵ OE||BA

DO DE CO DE
∴ = ⇒ =
BO AE AO AE
AO CO
∵ = [Given]
BO DO

⇒
DO

BO
=
CO

AO
⇒
AO

CO
=
AE

DE
.........[Taking reciprocals]
∴In △ADC
OE ∥ CD ...........[By converse basic proportionality theorem]
But OE ∥ BA
BA ∥ CD........[By construction]
The quadrilateral ABCD is a Trapezium.
23. Join OT and OQ.
TP = TQ

∴ TM ⊥ PQ and bisects PQ
Hence PM = 4 cm
−−−−− − –
Therefore OM = √25 − 16 = √9 = 3cm.
Let TM = x
From △PMT, PT2 = x2 + 16
From △POT, PT2 = (x + 3)2 - 25
Hence x2 + 16 = x2 + 9 + 6x - 25
⇒ 6x = 32 ⇒ x =
16

Hence PT2 = 256

9
+ 16 = 400

∴ PT = 20

3
cm .
24. We have,
sinθ + sin2θ = 1 ⇒ sinθ = 1 - sin2θ ⇒ sinθ = cos2θ
∴ cos12θ + 3 cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ - 2
= (cos12θ + 3cos10θ + 3cos8θ + cos6θ) + 2(cos4θ + cos2θ - 1)
= (cos4θ + cos2θ)3 + 2(cos4θ + cos2θ - 1)
= (sin2θ + cos2θ)3 + 2(sin2θ + cos2θ - 1) [∵ cos2θ = sinθ ∵ cos4θ = sin2θ]
= 1 + 2(1 - 1) = 1
OR
tan θ cot θ
+
2 2
(1+ tan2 θ) (1+ cot 2 θ)

tan θ cot θ
= +
2 2 2 2
(sec θ) (cos e c θ)

sin θ 1 cos θ 1
= × + ×
cos θ 4 sin θ 4
sec θ cose c θ
sin θ 4 cos θ 4
= × cos θ+ × sin θ
cos θ sin θ

= sinθcos3θ + cosθsin3θ

Page 10 of 17
 = sinθcosθ(cos2θ + sin2θ)
= sinθcosθ = RHS
Hence proved.
25. Area of shaded region
∘

= [π(42) − π(21) ]
2 2 300

360
∘

22 5
= 7
× 63 × 21 × 6
.
= 3465 cm2
OR
Draw OM ⊥ AB

∠ OAB = ∠ OBA = 30o

sin 30o = 1

2
=
OM

21
⇒ OM = 21

cos 30o =
√3 –
2
=
AM

21
⇒ AM = 21

2
√3
–
Area of △OAB = 1

2
× AB × OM = 1

2
× 21√3 ×
21

2
441 –
= 4
√3 cm
2

∴ Area of shaded region = Area (sector ΟACB) - Area (△OAB)

120 –
= 22

7
× 21 × 21 ×
360
−
441

4
√3

or 271.3 cm2 (approx.)

√3
= (462 − 441 4
) cm
2

Section C
26. By taking LCM of time taken (in minutes) by Sonia and Ravi, We can get the actual number of minutes after which they meet
again at the starting point after both start at the same point and at the same time, and go in the same direction.
2
18 = 2 × 3 × 3 = 2 × 3

12 = 2 × 2 × 3 = 22 × 3

LCM (18, 12) = 2 × 3 = 36 2 2

Therefore, both Sonia and Ravi will meet again at the starting point after 36 minutes.
27. Let coordinates of A are (x, 0) and coordinates of B are (0, y)

Using section formula, we get

5×0+3×x
4= 5+3

⇒ 32 = 3x
⇒ x =
32

3
5×y+3×0
Similarly, 5 = 5+3

⇒ 40 = 5y

Page 11 of 17
 ⇒ y=8
∴ Coordinate of A are ( 32

3
, 0) and coordinates of B are (0, 8).
28. According to question,two pipes are used to fill a swimming pool.
Pipe with larger diameter is used for 4 hours and pipe with smaller diameter is used for 9 hours.
Let x hours be the total time taken by the larger pipe to fill the tank
1
so in 1 hour it would fill part of the tank.
x

Similarly, y hours are needed for the smaller pipe,

1
then in 1 hour it would fill y
part.
So, y=10+x ...(1)
4 9

x
+ y
= 1/2
using (1), 4

x
+ 9

10+x
=1/2 ...(2)
⇒ x2 -16x - 80=0
⇒ (x-20)(x+4)=0

Since value of x cannot be negative

Therefore, x=20 and y=30
Hence, Larger diameter pipe fills in 20 hours, and Smaller diameter pipe fills in 30 hours.
OR
Let the ten's place digit be y and unit's place be x.
Therefore, number is 10y + x.
According to given condition,
10y + x = 4(x + y) and 10y + x = 2xy
⇒ x = 2y and 10y + x = 2xy
Putting x = 2y in 10y + x = 2xy
10y + 2y = 2.2y.y
12y = 4y2
4y2 - 12y = 0 ⇒ 4y(y - 3) = 0
⇒ y - 3 = 0 or y = 3
Hence, the ten's place digit is 3 and units digit is 6 (2y = x)
Hence the required number is 36.
29. PT is the tangent to the circle with centre O and radius OT = 20 cm.
P is a point 29 cm away from O.

OP = 29 cm, OT = 20 cm
OT is radius and PT is the tangent
OT ⊥ PT
Now, in right △OPT,
OP2 = OT2 + PT2 (Pythagoras Theorem)
⇒ (29)2 = (20)2 + PT2
⇒ 841 = 400 + PT2
⇒ PT2 = 841 – 400
⇒ PT2 = 441 = (21)2
⇒ PT = 21

Length of tangent, PT = 21 cm
OR
According to question draw a circle with centre O and radius 5cm also given that T is any point outside of the circle.

Page 12 of 17
 Now,Since tangent at a point on the circle is perpendicular to the radius through the point.

Therefore, OP is perpendicular to PT.

In right triangle OPT, we have
OT2 = OP2 + PT2
⇒ (13)2 = (5)2 + PT2
⇒ PT2 = 132 - 52
⇒ PT2 = 169 - 25
⇒ PT2 = 144
⇒ PT2 = 122
⇒ PT = 12 cm

Hence, the length of a tangent is 12 cm.

2 2
cot θ(sec θ−1) sec θ(sin θ−1)
30. LHS = (1+sin θ)
+
(1+sec θ)

2 2
cot θ(sec θ−1)(1+sec θ)+ sec θ(sin θ−1)(1+sin θ)
=
(1+sin θ)(1+sec θ)
2 2 2 2
cot θ(sec θ−1)+ sec θ(sin θ−1)
=
(1+sin θ)(1+sec θ)

2 2 2 2
cot θ tan θ+ sec θ(− cos θ)

=
(1+sin θ)(1+sec θ)

2 2 2 2
cot θ tan θ− sec θ cos θ
=
(1+sin θ)(1+sec θ)

2 1 2 1
cot θ× − sec θ×
c ot2 θ se c2 θ
=
(1+sin θ)(1+sec θ)

1−1
=
(1+sin θ)(1+sec θ)

=0
= RHS
31. If the line through A is the bridge
In △ACB, ∠B = 45o
AC = 8
– –
BC = AC, √3 = 8√3
In △ACD, ∠D = 30o
CD = AC = 8
– –
Hence width of the river 8 + 8√3 = 8(1 + √3) m
Section D
32. Suppose that he invested ₹ x at the rate of 12% simple interest and ₹ y at the rate 10% simple interest.
10y
Then, according to the question, 12x

100
+
100
= 130

⇒ 12x+10y=13000 ..........Dividing throughout by 2

⇒ 6x+5y=6500 .......(1)
12y
and, 100
+
10x

100
= 134

⇒ 12y+10x=13400
⇒ 6y+5x=6700 .............Dividing throughout by 2

⇒ 5x+6y=6700 ...........(2)

Multiplying equation (1) by 6 and equation (2) by 5, we get

36x+30y=39000 ...............(3)

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 25x+30y=33500 .......(4)
⇒ subtracting (3) and (4) we get x = 500

Substituting this value of x in equation (1), we get 6(500)+5y=6500

⇒ 3000+5y=6500
⇒ 5y=6500-3000

⇒ 5y=3500
3500
⇒ y = = 700
5

So, the solution of the equation (1) and (2) is x=500 and y=700
Hence, he invested ₹ 500 at the rate of 12% simple interest and ₹ 700 at the rate of 10% simple interest.
verification.Substituting x=500, y=700,
We find that both the equation (1) and (2) are satisfied as shown below:
6x + 5y=6(500)+5(700)=3000+3500=6500
5x+6y=5(500)+6(700)=2500+4200=6700
This verifies the solution.
OR
Let the cost of a IX Maths textbook be Rs.x and the cost of a X Maths textbook be Rs.y.
Then the algebraic representation is given by the following equations
2x + 3y = 250 ...(1)
and 4x + 6y = 500 ...(2)
To represent these equations graphically, we find two
solutions for each equation.
These solution are given below:
For equations (1) 2x + 3y = 250
250−2x
⇒ y =
3

Table 1 of solutions
x 50 125

y 50 0
For equation (2)
4x + 6y = 500
⇒ 6y = 500 - 4x
500−4x
⇒ y =
6

Table 2 of solutions
x 50 125

y 50 0
We plot these points on a graph paper, we find that both the lines coincide.
This is so, because, both the equations are equivalent, i.e., one can be derived from the other.

33. Let A(3, 1), B(0, -2), C(1, 1) and D(4, 4) be the vertices of quadrilateral
Join AC, BD. AC and BD, intersect other at the point O.

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 We know that the diagonals of a parallelogram bisect each other
Therefore, O is midpoint of AC as well as that of BD
3+1 1+1
Now midpoint of AC is ( 2
,
2
) ie., (2, 1)

0+4 −2+4
And midpoint of BD is ( 2
,
2
) ie., (2, 1)

Mid point of AC is the same as midpoint of BD

Hence, A, B, C, D, are the vertices of a parallelogram ABCD
34. According to the question,a hemispherical depression is cut from one face of the cubical block such that the diameter l of the
hemisphere is equal to the edge of the cube.
Let the radius of hemisphere = r
l
∴ r =
2

Now, the required surface area = Surface area of cubical block - Area of base of hemisphere + Curved surface area of hemisphere.
2 2 2
= 6( side ) − πr + 2πr
2 2
2 l l
= 6l − π( ) + 2π( )
2 2

2
2 πl π 2
= 6l − + l
4 2
2
2 πl
= 6l +
4

Surface area = 1

4
(24 + π)l
2
units.
1 22 2
= (24 + )l
4 7

OR

Let the radius of the hemispherical dome be r and the total height of the building be h.
Since, the base diameter of the dome is equal to of the total height 2

2
2r = h
3
h
⇒ r =
3

Let H be the height of the cylindrical position.

h 2h
⇒ H = h − r = h − =
3 3

Volume of air inside the building = Volume of air inside the dome + Volume of air inside the cylinder
1 2 3 2
⇒ 67 = πr + πr H
21 3

1408 2 2
⇒ = πr ( r + H)
21 3

2
1408 22 h 2 h 2h
⇒ = × ( ) ( × + )
21 7 3 3 3 3

2
1408×7 h 2h 2h
⇒ = × ( + )
22×21 9 9 3

2
64 h 8h
⇒ = × ( )
3 9 9

64×9×9 3
⇒ = h
3×8

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 3
⇒ h = 8 × 27

⇒ h = 6

Thus, the height of the building is 6 m.

35. Given,
First term(a) = 5
Common difference(d) = 3
and, nth term (an) = 50
⇒ a + (n - 1)d = 50
⇒ 5 + (n - 1)(3) = 50
⇒ 5 + 3n - 3 = 50

⇒ 3n = 50 - 5 + 3

⇒ 3n = 48
48
⇒ n= = 16
3

Therefore, Sn = n

2
[a + an ]

16
= [5 + 50]
2

= 8 × 55

= 440
Section E
36. i. a is a non zero real number and b and c are any real numbers.
ii. D = 0
iii. 2x2 - x + 8k
α × = 1

α
8k

1 = 4k
k= 1

OR
−b −coefficient of x
α + β = a
i.e., ( 2
)
coefficient of x

c constant term
αβ =
a
i.e., ( 2
)
coeff of x

37. i. Women having heart beat in range 68 - 77

= 4 + 3 + 8 = 15
ii. Median class = 74 - 77
f1 − f0
iii. a. Mode = l + ( 2f1 − f0 − f2
) × h

l = 74, f1 = 8, f0 = 3, f2 = 7, h = 3

8−3
∴ Modal value = 74 + ( 16−3−7
) × 3

= 76.5
OR

b. No. of heart beats f cf

65 - 68 2 2

68 - 71 4 6

71 - 74 3 9

74 - 77 8 17

77 - 80 7 24

80 - 83 4 28

83 - 86 2 30
N
−Cf

Median = I + 2

f
× h

(15−9)
= 74 + × 3
8

= 76.25

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38. i.

△ ABC ∼ △DEC
20 x
=
30 100−x

2000 - 20x = 30x

2000 = 50x
x = 40 m
ii. AA
iii. 60 metres
OR
AD = AC + CD
−−− −−−−− −−− −−−−2
2 2 2
= √20 + 40 + √60 + 30
−−−−−− −−− −−−−−−− −−
= √400 + 1600 + √3600 + 900
−−−− −−−−
= √2000 + √4500
– –
⇒ 20√5 + 30√5
–
⇒ 50√5 m

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