MATHS SAMPLE PAPER

PART-A
Section-I
Section I has 16 questions of 1 mark each.

3
1. The exponent form of √7 is _______.

2. If m and n are two natural numbers and 𝑚𝑛 = 32, then 𝑛𝑚𝑛 is _____.

3. If 𝑎 = 2 𝑎𝑛𝑑 𝑏 = 3, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 (𝑎𝑏 + 𝑏𝑎)−1 𝑖𝑠 _______.

OR
Find the value of the polynomial 6𝑡2 + 7𝑡 − 3 𝑤ℎ𝑒𝑛 𝑡 = −1 .

4. Write the degree of the polynomial √2.

5. In the adjoining figure, find the value of 𝑥

6. Find the semi perimeter of a triangle with sides 9 cm, 12 cm and 30

cm.

7. In the given figure, AD is the median, then find ∠𝐵𝐶𝐴.

OR
𝐼𝑛 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑓𝑖𝑔𝑢𝑟𝑒 ∆𝑃𝐿𝑂 ≅ ∆𝑃𝑀𝑂, Find the length of PM.

8. The area of a triangle is 48cm2. Its base is 12cm, Find its altitude.

9. In the given figure the congruency rule used in providing ∆𝐴𝐶𝐵 ≅

∆𝐴𝐷𝐵 is _______.
 10. There are 5 red and 3 black balls in a bag. Find the probability
of drawing a black ball.

11. The two angles measuring (300 − 𝑎) and (1250 + 2𝑎) are
supplementary to each other, then the value of ′𝑎 ′ is ________.
OR
What is the measure of an angle whose measure is 320 less than its
supplement?

12. In the figure l1 ∥ l2, what is the value of 𝑥?

OR
Angles of a triangle are in the ratio 2 : 4 : 3, Find the smallest angle
of the triangle.

13. In parallelogram ABCD, 𝑚∠A = (5𝑥 − 20)° and 𝑚∠𝐶 = (3𝑥 +

40)°. 𝐹𝑖𝑛𝑑 the value of 𝑥.
OR
If the degree measures of the angles of the quadrilateral are 4𝑥, 7𝑥,
9𝑥 and 10𝑥, What is the measure of the smallest angle and the
largest angle?

14. The sum of either pair of opposite angles of a cyclic

quadrilateral is _______.

15. The radius of the circle is 5 cm and the distance of the chord
from the Centre of the circle is 4 cm then, the length of the chord is
_______.

1
16. What will be the rationalising factor of ?
√2−5

Section II
Case-study based questions are compulsory. Attempt any four sub
parts of each question. Each subpart carries 1 mark
17. Case study based-1: The Greek mathematicians of Euclid’s time
thought of geometry as an abstract model of the world in which they
lived. The notions of point, line, plane (or surface) and so on were
derived from what was seen around them. From studies of the space
and solids in the space around them, an abstract geometrical notion
of a solid object was developed. A solid has shape, size, position, and
can be moved from one place to another. Its boundaries are called
surfaces. They separate one part of the space from another, and are
said to have no thickness. The boundaries of the surfaces are curves
or straight lines. These lines end in points. Consider the three steps
from solids to points (solids-surfaces-lines-points). In each step we
lose one extension, also called a dimension. So, a solid has three
dimensions, a surface has two, a line has one and a point has none.
Euclid summarised these statements as definitions. He began his
exposition by listing 23 definitions in Book 1 of the ‘Elements’
Now let us discuss Euclid’s fifth postulate. It is :
If a straight line falling on two straight lines makes the interior angles
on the same side of it taken together less than two right angles, then
the two straight lines, if produced indefinitely, meet on that side on
which the sum of angles is less than two right angles.

(a) According to the above postulate, if the lines are parallel, then
which of the following angles will be supplementary?
(i) corresponding (ii) Alternate interior (iii) Co-interior
(iv) Vertically opposite
(b) If two angles are complements of each other then each angle
is
(i) An acute angle
(ii) An obtuse angle
(iii) A right angle
(iv) A reflex angle
(c) The measure of an angle is 5 times its complement, the angle
measures
(i) 25° (ii) 35° (iii) 65° (iv) 75°
(d) In the given figure, AOB is a straight line. If ∠AOC + ∠BOD =
95°, then ∠COD = ?
 (i) 95° (ii) 85° (iii) 90° (iv) 55°
(e) In the given figure AB is a mirror, PQ is the incident ray and
QR is the reflected ray. If ∠PQR = 108°, then ∠AQP = ?

(i) 72°
(ii) 18°
(iii) 36°
(iv) 54°

18. Case study based – 2: You must have observed that two copies
of your photographs of the same size are identical. Similarly, two
bangles of the same size, two ATM cards issued by the same bank
are identical. You may recall that on placing a one rupee coin on
another minted in the same year, they cover each other completely.
Do you remember what such figures are called? Indeed they are
called congruent figures (‘congruent’ means equal in all respects or
figures whose shapes and sizes are both the same). Now, draw two
circles of the same radius and place one on the other. What do you
observe? They cover each other completely and we call them as
congruent circles.

(a) Which of the following is not a criterion for congruence of

triangles?
(i). SSA
(ii). SAS
(iii). ASA
(iv). SSS

(b) If AB=QR, BC=RP and CA=PQ, then which of the following holds?
(i). ∆ABC ≅ ∆PQR
(ii). ∆CBA ≅ ∆PQR
 (iii). ∆CAB ≅ ∆PQR
(iv). ∆BCA ≅ ∆PQR

(c) If ∆ABC≅∆PQR AND ∆ABC is not congruent to ∆RPQ, then which

of the following is not true?
(i). BC=PQ
(ii). AC=PR
(iii). BC=QR
(iv). AB=PQ

(d) It is given that ∆ABC≅∆FDE in which AB=5cm, ∠B=40°, ∠A=80°

and FD=5cm. Then, which of the following is true?
(i). ∠D=60°
(ii). ∠E=60°
(iii). ∠F=60°
(iv). ∠D=80°

(e) In the given figure, AE=DB, CB=EF And ∠ABC=∠FED. Then,

which of the following is true?

(i). ∆ABC ≅ ∆DEF

(ii). ∆ABC ≅ ∆EFD
(iii). ∆ABC ≅ ∆FED
(iv). ∆ABC ≅ ∆EDF

19. Case study based -3: let us mark four points and see what we
obtain on joining them in pairs in some order.
Note that if all the points are collinear (in the same line), we obtain a line
segment, if three out of four points are collinear, we get a triangle, and if
no three points out of four are collinear, we obtain a closed figure with four
sides
Such a figure formed by joining four points in an order is called a
quadrilateral.
A quadrilateral has four sides, four angles and four vertices. You may
wonder why should we study about quadrilaterals (or parallelograms) Look
around you and you will find so many objects which are of the shape of a
quadrilateral - the floor, walls, ceiling, windows of your classroom, the
blackboard, each face of the duster, each page of your book, the top of
your study table etc.

(a) Three angles of a quadrilateral are 80o, 95o and 112o. Its fourth
angle is
(i). 78o
(ii). 73o
(iii). 85o
(iv). 100o

(b) In the given figure, ABCD is a parallelogram in which ∠BAD =

75° and ∠CBD = 60°. Then, ∠BDC = ?

(i). 60o
(ii). 75o
(iii). 45o
(iv). 50o

(c) In which of the following figures are the diagonals equal?

(i). Parallelogram
(ii). Rhombus
(iii). Trapezium
(iv). Rectangle

(d) The lengths of the diagonals of a rhombus are 16 cm and 12 cm.

The length of each side of the rhombus is
 (i). 10 cm
(ii). 12 cm
(iii). 9 cm
(iv). 8 cm

(e) If ABCD is a parallelogram with two adjacent angles ∠A = ∠ B,

then the parallelogram is a
(i). rhombus
(ii). trapezium
(iii). rectangle
(iv). none of these

20. Case study based – 4: You may have come across many objects
in daily life, which are round in shape, such as wheels of a vehicle,
bangles, dials of many clocks, coins of denominations 50 p, Re 1 and
Rs 5, key rings, buttons of shirts, etc. (see Fig.10.1). In a clock, you
might have observed that the second’s hand goes round the dial of
the clock rapidly and its tip moves in a round path. This path traced
by the tip of the second’s hand is called a circle.

(a) The radius of a circle is 13 cm and the length of one of its chords
is 10 cm. The distance of the chord from the centre is
(i). 11.5 cm
(ii). 12 cm

(iii).
(iv). 23 cm

(b) In the given figure, BOC is a diameter of a circle and AB = AC.

Then, ∠ABC = ?

(i). 30o
(ii). 45o
(iii). 60o
(iv). 90o
 (c) In the given figure, O is the centre of a circle and ∠ACB = 30°.
Then, ∠AOB = ?

(i). 30o
(ii). 15o
(iii). 60o
(iv). 90o

(d) AB and CD are two equal chords of a circle with centre O such
that ∠AOB = 80°, then ∠COD = ?

(i). 100o
(ii). 80o
(iii). 120o
(iv). 40o

(e) The angle in a semicircle measures

(i). 45o
(ii). 60o
(iii). 90o
(iv). 36o

PART–B
Section III
21. If P(x) = x3 − 1, then find the value of P(1) + P(-1).
22. Express 0.4777… in the rational form.

23. Write the factors of the polynomial 4𝑥2 + 𝑦2 + 4 + 4𝑥𝑦 + 8𝑥 + 4𝑦

24. If the point (2𝑘 − 3, 𝑘 + 2) is a solution of linear equation 2𝑥 + 3𝑦 +

15 = 0, find the value of 𝑘.

25. ABC is an isosceles triangle in which altitudes BE and CF are drawn to

equal sides AC and AB respectively. Show that these altitudes are equal

26. In the figure, OA=OB, OD =OC, then choose the

congruence rule by which ΔAOD ≅ΔBOC.

Section IV

27. In the figure A,B,C,D are four points on a circle. AC and BD intersect
at a point E such that ∠𝐵𝐸𝐶 = 1300 and ∠𝐸𝐶𝐷 = 200. Find ∠𝐵𝐴𝐶.
28. Examine which of the numbers 1,−1, −3 are zeroes of the polynomial
𝑝(𝑥) = 2𝑥4 + 9𝑥3 + 11𝑥2 + 4𝑥 – 6.

29. Show that the bisectors of angles of a parallelogram form a rectangle.

30. Prove that: "The sum of the angles of a triangle is 1800"

31. Show that: "If the diagonals of a quadrilateral bisect each other at
right angles, then it is a rhombus."

32. Construct a right triangle ΔXYZ in which ∠𝑍 =900, YZ=3cm, XZ+𝑋𝑌 =

5𝑐𝑚.

33. Find the volume of a sphere whose surface area is 154cm2.

Section V

34. A conical tent is 10m high and the radius of its base is 24m. Find
(i) slant height of the tent.
(ii) Cost of the canvas required to make the tent, if the cost of
1m2 canvas is ₹ 70

35. The following table shows the number of people of different age
groups travelling in a metro during a day: Draw Histogram for the given
data:

36. Draw the graph of the equation 𝑥 − 𝑦 = 4

Answer the following using graph paper:
(i) Find the value of 𝑦, 𝑖𝑓 𝑥 = 7 from the graph.
(ii) Write the coordinate of the point where the graph intersects
on 𝑥 − 𝑎𝑥𝑖𝑠.

***
 HINTS & SOLUTIONS
Maths Sample paper

1. 71/3
2. 510
3. 1/17 OR -4
4. 0
5. x = 28
6. 51/2 cm
7. 35°
8. 8 cm
9. SAS
10. 3/8
11. a = 25° OR 106°
12. x = 85° OR 40°
13. x = 20° OR 24° and 60°
14. 180°
15. 6 cm

16. √2 + 5
17. (a) (iii) Co – interior angles
(b) (i) acute angle
(c) (iv) 75°
(d) (ii) 85°
(e) (iii) 36°
18. (a) (i) SSA
(b) (iii) ∆CAB ≅ ∆PQR
(c) (i) BC = PQ
(d) (ii) ∠E = 60°
(e) (i) ∆ABC ≅ ∆DEF
19. (a) (ii) 73°
(b) (iii) 45°
(c) (iv) Rectangle
(d) (i) 10 cm
(e) (iii) Rectangle
20. (a) (ii) 12 cm
(b) (ii) 45°
(c) (iii) 60°
(d) (ii) 80°
(e) (iii)90°

21. P(1) = x3-1

P (1) = 13-1

=1-1

=0

P (-1) = -13-1

= -1-1

= -2

P (1) + P (-1) = 0-2

= -2

22. Let x = 0.4777 ...

Multiply by 10, we get

10x = 4.777... … (I)
Again, multiply by 10, we get
100x = 47.77... … (II)
Subtracting (II) from (I), we get
90x = 43

x=
23.

24. Since (2k - 3, k + 2) lies on 2x + 3y + 15 = 0

2(2k - 3) + 3(k + 2) + 15 = 0
4k - 6 + 3k + 6 + 15 = 0
7k + 15 = 0

25. Given: BE and CF are altitudes, AC = AB

To show: BE = CF

Proof:
In

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

AB = AC (Given)

AAS Postulate (Angle-Angle-Side) If two angles and a non-included side of

one triangle are congruent to the corresponding parts of another triangle,
then the triangles are congruent.
Therefore, By AAS congruence axiom,
Thus,

BE = CF (By corresponding parts of congruent triangles)

Hence, Proved.

26. Given:

OA = OB

And OD = OC

To Prove:
ΔAOD ≅ ΔBOC

Proof :

Lines CD and AB intersect.

∠AOD = ∠BOC (Vertically Opposite Angles)

In ΔAOD and ΔBOC,

OA = OB (Given)
∠AOD = ∠BOC (Vertically Opposite Angles)

OD = OC (Given)
Therefore, ΔAOD ≅ ΔBOC (By SAS Congruence)----- (1)
27.

28. 𝑝(1) = 2(1)4 + 9(1)3 + 11(1)2 + 4(1) – 6 ≠ 0

𝑝(-1) = 2(-1)4 + 9(-1)3 + 11(-1)2 + 4(-1) – 6 ≠ 0

𝑝(-3) = 2(-3)4 + 9(-3)3 + 11(-3)2 + 4(-3) – 6 = 0

Hence -3 is a zero

29. P, Q, R and S are the points of intersection of bisectors of the angles

of the parallelogram.

In ADS,

DAS + ADS =
( A and D are interior angles on the same side of the transversal)
Also in ADS,
DAS + ADS + DSA = 180o (angle sum property)

Similarly, it can be shown that APB or SPQ = 90o

Also, SRQ = RQP = 90o
Hence, PQRS is a rectangle.

30.

Construct a line Parallel to BC, passing through A.

Label the angles as shown.
∠1 = ∠4 (Alternate interior angles)
∠3 = ∠5 (Alternate interior angles)
Also, ∠1 + ∠2 + ∠3 = 180° (Angles on a straight line)
Thus, ∠1 + ∠4 + ∠5 = 180°
Hence Proved.

31. To Prove: If diagonals of a quadrilateral bisect at 90º, it is a

rhombus.
Figure:
Definition of Rhombus: A parallelogram whose all sides are equal.
Given: Let ABCD be a quadrilateral whose diagonals bisect at 90º

In ΔAOD and ΔCOD,

OA = OC (Diagonals bisect each other)

∠AOD = ∠COD (Given)

OD = OD (Common)

ΔAOD ΔCOD (By SAS congruence rule)

AD = CD ..................(1)

Similarly,

AD = AB and CD = BC ..................(2)

From equations (1) and (2),

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, it can be said that
ABCD is a parallelogram. Since all sides of a parallelogram ABCD are
equal, it can be said that

ABCD is a rhombus

Hence, Proved.
32. Given below is the required construction.

∆XYZ is the required triangle.

33. Let the radius be "r".

Given, Surface area = 154 cm2

We know that Surface Area of a sphere is given by,S = 4πr2

Therefore,

4πr2 = 154

4× × r × r = 154

r = 3.5 cm
Now,

Volume = πr3

= × × (3.5)3

= cm3
= 179.67 cm3
34. (i) Let ABC be a conical tent.

Height (h) of conical tent = 10 m

Radius (r) of conical tent = 24 m

Let the slant height of the tent be l.

In ΔABO,
AB2 = AO2 + BO2
l2 = h2 + r2
= (10 m)2 + (24 m)2
= 676 m2
l = 26 m
(ii) CSA of tent =

= * 24 * 26

= m2
Cost of 1 m2 canvas = Rs 70

So, cost of m2 canvas = ( m2 * 70)

= Rs 137280
35.

Is the required Histogram

36.
Plot the required graph.
(i) y = 3, if x = 7
(ii) For x – axis, y = 0, Thus x = 4, the required point is (4, 0)

***