SUMMATIVE ASSESSMENT MATHS

CLASS 10

CBSE Sample Paper-SA2-With-Solutions

Time: 3 Hrs Max Marks: 90

General Instructions:

A) All questions are compulsory.

B) The question paper consists of 34 questions divided into four sections A, B, C and D.

a. Section A comprises of 8 questions of 1 mark each

b. Section B comprises of 6 questions of 2 marks each

c. Section C comprises of 10 questions of 3 marks each

d. Section D comprises 10 questions of 4 marks each

C) Question numbers 1 to 8 in section A are multiple choice questions where you are to select one

correct option out of the given four.

D) Use of calculator is not permitted.

E) An additional 15 minutes time has been allotted to read this question paper only

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SECTION – A

1. Find the product of roots for the quadratic equation

a) -7

b) 10

c) -21

d) 21

= 0 = 0

Roots : x = -7, -3

Product of the roots = ( -7) (-3) = 21

2. A string of length 50√3 attached to a kite is temporarily tied to a point on the ground at an

inclination φ⁰. Find the angle of inclination of the string if the kite is flying at a height of 75 m

above the ground, assuming that there is no slack in the string.

a) ⁰b) ⁰c) 45⁰d) 90⁰According to the given figure,

The length of string = AC = √ m ....

AB = 75 m

In triangle ABC

sin φ =

sin φ =

√

sin φ =

√

√

√

√ =

√

⟹ Φ = 60

3. A circle inscribed in a ABC touches the sides AB, BC and CA at P, Q and R respectively,

Find BC given that AB = 14 cm, AR = 6 cm & RC = 7cm.

a) 7 cm

b) 15 cm

c) 13 cm

d) 12 cm

Since tangents drawn from an external point to a circle are equal,

AP = AR = 6 cm

CQ = CR= 7 cm

PB = AB -AP =14 - 6 = 8 cm

BQ = BP = 8 cm

BC = BQ + QC= 8 + 7 = 15 cm

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4. Tangents TA and TB are drawn from an external point T to a circle with center O. Another

tangent at point X on the circle is drawn cutting TA and TB at P and R respectively. Which of

the following is true?

a) PQ = TP – TQ

b) PQ = AP – BQ

c) PQ = AP + BQ

d) PQ = TP + TQ

Since tangents drawn from an external point to a circle are equal,

TA = TB ----- (1)

PX = PA ----- (2)

QX = QB ----- (3)

Adding (2) and (3)

PX + QX = PA + QB

PQ = AP + BQ

5. Two concentric circles with radii 10cm and 6 cm are given. The length of chord MB which

touches the inner circle at R is

a) 8 cm

b) 12 cm

c) 10 cm

d) 16 cm

In the right angled triangle OPA,

AP2 = OA

2 - OP

2

AP2 = 10

2 - 6

2 = 100 – 36 = 64; AP = 8 cm

AP = PB = 8 cm (Since radius OP is ⊥ bisector of the chord AB)

But BP = BR (Tangents from an exterior point are equal)

⟹ BR = 8cm

⟹ BM = 2(BR) = 2(8) = 16 cm

6. A metallic sphere of radius 4.5 cm is melted and recast into the shape of a cylinder of radius 4

cm. The height of the cylinder is ______ cm.

a) 11 cm

b) 7.6 cm

c) 6.7 cm

d) 76 cm

Radius of metallic sphere = r = 4.5 cm, Radius of cylinder(R) = 4 cm

Let height of cylinder = h cm

volume of metallic sphere = volume of cylinder

⟹ h

⟹

⟹ h =

⟹ h = 7.59 cm ≈ 7.6 cm

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7. Tickets numbered from 1 to 50 are mixed up together and then a ticket is drawn at random.

What is the probability that the ticket has a number which is a multiple of 3 or 7?

a)

b)

c)

d)

The multiples of 3 or 7 are : 3,6,7,9,12,14,15,18,21,24,27,28,30,33,35,36,39,42,45,48,49

Favorable number of events = 21

Total events = 50

Therefore, P(Getting a number which is a multiple of 3 or 7) =

P =

8. Find the values of K for which the equation 3 - 2kx + 3 = 0 has real roots.

a) k ≥ 3 or k ≤ -3

b) k ≥ -3 or k ≤ 3

c) k ≥ 2 or k ≤ -2

d) k ≥ 9 or k ≤ -9

For 3 - 2kx + 3 = 0 to have real roots, – 4ac ≥ 0

- 4(3) (3) ≥ 0 ⟹ - 36 ≥ 0

≥ 36 ⟹ ≥ 9 ⟹ k ≥ 3 or k ≤ -3

9. Given below is a triangle ABC and a triangle AB'C' similar to ABC. What is the ratio of the

sides of AB'C' to that of corresponding sides of the triangle ABC.

a) 4:3

b) 3:4

c) 4:7

d) 3:7

AB' is 3 parts: AA1 = AA2 = AA3

AB is 4 parts: AA1 = AA2 = AA3 = AA4

The ratio of the sides of AB'C' to that of corresponding sides of the triangle ABC is 3:4

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10. A wire is looped so as to form a circle of diameter 84 cm. It is rebent to form a square. The

length of the side of the square is ______ cm.

a) 42 cm

b) 66 cm

c) 132 cm

d) 76 cm

Radius =

= 42 cm

Perimeter of circle = 2 x x 42

Perimeter of square = 4 x side

Circumference of the circle = Perimeter of the square

2 x

x 42 = 4 x side

Side of the square = 66 cm

SECTION – B

11. If one root of the equation is –1, find the value of k and the other root.

since -1 is the root of this equation,

⟹

⟹

⟹

⟹

Putting this value of k in the equation

⟹

⟹

⟹

⟹

⟹ ⟹

⟹ ⟹

So the other root is 2

12. From the well shuffled pack of 52 cards few cards of same colour are missing. If probability of

picking a card is P (Red card) =

and P (Black card) =

then which colour of cards are

missing and how many?

Since the probability of drawing a black card is less than red cards, black cards are missing

Let x be the number of missing black cards.

P(black) =

=

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⟹

13 black cards are missing

13. Find the AP if the 6th

term of an AP is 19 and the 16th

term is 15 more than the 11th

term.

Let a be the 1st term and d be the common difference

⟹ ⟹

Therefore AP =

14. A van has two wipers which have a blade of length 25 cm sweeping through an angle of 110 .Find the total area cleaned at each sweep of the blades if the wipers do not overlap at all.

Area of the sector =

25 25 = 600 cm²

Area cleaned by the two blades = 600 2 = 1200 cm²

15. A well 30 m deep with diameter 6 m is dug and the earth from digging the well is evenly

spread out to form a platform 24 m by 12 m. What is the height of the platform?

Diameter of well = d = 6 m

Radius of well = r =

=3 m

Height of well = h = 30 m

Length of platform = l = 24 m

Breadth of platform = b = 12 m

Let height of platform = H

According to given condition we have:

Volume of Well = Volume of Earth Dug out

⟹

⟹

⟹

⟹ m

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16. If M and N are the points of trisection of the line segment joining the points A (3, -2) and

B (9, 6). If M is (5,

) near to A, find the coordinates of N.

Let the point N be (x, y)

Since N is other point of tri-section, it will be the mid-point between M(5,

) and B(9,6)

so,

=

= 7

=

=

(7,

) is the other point of tri-section.

17. In the given image, the tangent to a circle of radius 6 cm from an external point P, is 8 cm long.

What is the distance of P from the nearest point on the circle?

In △OPQ

QP ⊥ OQ

OP2 = OQ

2 + PQ2

= 62

+ 82

= 36 + 64 = 100

⟹ OP = √ = 10 cm

PR = OP – OR = 10 – 6 = 4 cm

The distance of P from the nearest point on the circle is 4 cm

18. The centre of a circle is O (2a, a - 7). If the circle passes through the point (11, -9) and has

diameter 10√ units, find the value of a.

Given : Co-ordinates of O are (2a, a - 7).

Let the co-ordinates of A and B be (11, -9) and (x, y) respectively.

AB = 10 √ units

0 is the mid-point of AB

Co-ordinates of O are

,

)

Given O = (2a, a - 7)

⟹

= 2a and

= a - 7

= 2a ⟹11 + x = 4a ⟹x = 4a - 11

= a - 7 ⟹ -9 + y = 2a - 14 ⟹y = 2a - 5

A = (11, -9) and B = (4a - 11, 2a - 5)

Length of AB = √

⟹√

= √

⟹√

Given AB = 10 √ units

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⟹√ = 10 √

Squaring both sides,

20a2 - 160a + 500 = 200

⟹20a2 - 160a + 300 = 0

⟹a2 - 8a + 15 = 0

⟹ (a - 5) (a - 3) = 0

⟹a = 5 or a = 3

19. If x = 2 and x = 3 are the roots of the equation , find the value of k and

m.

Substituting x = 2 in the equation, we get

⟹

⟹ ⟹ -----(1)

Substituting x = 3 in the equation, we get

⟹

⟹ (since m = )

⟹

⟹- ⟹

Substituting k in (1), we get

(

) -

20. A bag contains 12 balls out of which x are white.

i) If 4 non-white balls are taken out, then what is the probability that the next ball will be a

white ball?

ii) If 6 more balls are put in the bag, the probability of drawing a white ball will be double.

Then what is the number of white balls in the bag?

Total number of balls = 12, Let number of white balls be = x

i) Since 4 non-white balls are removed,

Total number of balls = 12 – 4 = 8 ,

White = x

P1(getting a white ball) =

ii) Since 6 more balls are put in the bag and the probability of drawing a white ball becomes

double => white balls are added.

Total number of white balls = x + 6

Total number of balls = 12 + 6 =18

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P2 (white ball) =

P2 = 2 P1

= 2 (

)

⟹ ⟹

21. In an A.P. the Sum of first n terms is

+

. Find its 25

th term.

Sn=

+

= +

S(n-1) = (n-1)2 +

Sn - S (n-1) = n2- (n-1)

2 +

-

an = 2n +

a25 = 2(25) +

= 50 +

=

22. Two poles of equal height are on either side of a road, which is 200 m wide. The angles of

elevation of the top of the pillars are 60 and 30 at a point on the road between the pillars. Find

the position of the point between the pillars and the height of each pillar.

Let AB = CD = h m

P is the point from where the elevation angles are measured.

In Δ PAB,

tan 60⁰ =

⟹ x√ = h ---(1)

In Δ PCD,

tan 30⁰ =

⟹

√ =

⟹

√ = h ---(2)

Since the poles are of equal heights, CD = AB

from (1) and (2)

x√ =

√

⟹ ⟹

= 50 m

Substituting x in (1)

h = 50√ = m

Therefore, the height of poles is 86.6 m and the point is 50 m and 150 m far from these poles.

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23. The inner edge of a circular track has the diameter of 70 cm. The track is 2.5 m wide and the

cost of laying the bricks is Rs. 12 per m2. Find the total cost of laying the bricks on the track?

Radius of inner edge =

= 35 cm

Radius of outer edge = 35 + 2.5 = 37.5 cm

Area of the inner circle =

35 35

= 3850 m²

Area of outer circle =

37.5 37.5

= 4419.6 m2

Area of track = 4419.6 – 3850 = 569.6

Total cost = 569.6 12 = Rs. 6835.2

24. Draw a circle of radius 6 cm. From a point P, 10 cm away from its centre, construct a pair of

tangents to the circle. Measure the lengths of the tangents.

OA = 6 cm OP = 10 cm ----- given

Join OP and bisect it. Let M be the midpoint of OP

Taking M as centre and MO as radius, draw a circle,

intersecting the given circle at A and B.

Join PA and PB

PA and PB are required tangents.

Lengths of tangents

= √ = √ = √ = 8 cm

25. A toy was made by scooping out a hemisphere from each end of a solid cylinder. If the height

of the cylinder is 12 cm and its base has a radius of 2.5 cm, find the volume of the toy.

(Assume that the base of the hemisphere is same as the base of the cylinder.)

Volume of toy = Volume of cylinder - Volume of 2 hemispheres

= π r2h – 2 (

π r

3)

= π r2 [ h -

r ]

=

2.5 2.5 [ 12 -

2.5 ]

=

2.5 2.5 [ 12 - 3.3 ]

= 19.64 8.7

= 170.868 cubic cm

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26. What is the length of the median through the vertex A given that A (-4, 2), B (-5, -3) and C (1,

-1) are the vertices of the ∆ ABC.

If AP is the median, then P(x, y) is the mid-point of BC

x =

=

= -2 ;

y =

=

= -2

Co - ordinates of

The length AP of the median is

√

= √

= √

= √ √ units

27. Find the coordinates of the points which divide the line segment joining (-8, 0) and (0, 6) into

four equal parts.

Let the points of division of line segment PQ be A, B, C such that PA = AB = BC = CQ

Since B is the mid-point of PQ,

B = (

,

) = (-4, 3)

Since A is the mid-point of PB,

A = (

,

) = (-6,

)

Since C is the mid-point of BQ,

C = (

,

) = (-2,

)

The points are : (-4, 3), (-6,

), (-2,

)

28. In the given figure n, l, m are the lengths of the sides LM, MN and NL respectively of a right

triangle LMN. If r is the radius of the in-circle, prove that 2r = (n + l - m).

Let the circle touch the sides

LM, MN and NL at E, F and G respectively.

Given that LM = n, MN = l, NL = m

we know that tangents drawn from an external point are equal in length

⟹ LE = LF, NG = NF and

ME = MG = r (Since MGOE is a square)

⟹ n - r = LF

⟹ l - r = NF

NL = NF + LF

NL = l - r + n - r

m = l + n -2r

2r = l + n - m (Hence proved)

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29. A train travels 418 km at a uniform speed. If the speed had been 16 km/h less, it would have

taken 8 hours more for the same journey. Find the speed of the train.

Let the original speed of the train = x km/h

The time taken will be

hrs

the decreased speed = x - 16 km/h.

then time taken will be =

Given that

–

= 8

= 8

= 8

8x2 - 128x - 6688 = 0

x2 - 16x - 836 = 0

x2 - 38x + 22x - 836 = 0

(x - 38 )(x + 22 )

As speed cannot be negative, x = -22 is not valid

x = 38 or speed = 38 km/h.

30. A container shaped like a right circular cylinder has a diameter (d) equal to 12 cm and height

(h) equal to 15 cm, is full of ice cream. The ice cream is to be filled into 10 equal cones having

a hemispherical shape on the top. If the height of the cone is 4 times its radius, find the height

of the cone.

r =

=

= 6 cm , h = 15 cm

Volume of the cylinder = πr2 h = π(6)

2 (15) = 540π

Let the radius of the cone be r cm

height of the cone = 4r cm

Volume of the cone with hemispherical part

=

πr

2 h +

πr

3

=

πr

2 (4r) +

π(r)

3

= 2πr3

Volume of cylinder = 10 (Volume of cone with hemisphere)

⟹ 540 π = 10 (2πr3)

⟹

= r

3

⟹ r3 = 27 ⟹ r = 3 cm

⟹ height = 4r = 12 cm

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31. If m times the mth

term of an AP is equal to t times the tth

term, show that it’s (m + t)th

term is

zero.

Given m( a + (m-1)d ) = t ( a + (t-1)d )

ma + md (m-1) = ta + td (t-1)

ma - ta = d( t (t-1) – m (m-1) )

a ( m – t) = d ( t2 - t – m

2 + m )

a ( m – t ) = d ( t2 – m

2- t + m )

a ( m – t ) = d ( (t - m) (t + m) -(t-m) )

a ( m – t ) = d ( t – m ) ( t + m – 1 )

a ( m – t ) = -d ( m – t ) ( t + m - 1)

a = -d ( t + m – 1 )

a + ( t + m – 1 ) d = 0

But a + ( t + m -1 ) d is the ( t + m)th

term

⟹ (t + m)th

term of AP = '0 '

32. In the given figure, AP and AQ are two tangents drawn to the circle with center O. B is a point

on the extended tangent QA and PAB = 135⁰, Find POQ.

PAB = 135

PAQ + PAB = 180 (Linear pair)

PAQ + 135 = 180

PAQ = 180 - 135

PAQ = 45

PA = QA

(Since lengths of tangent from an external point to a circle are equal.)

⟹APQ = AQP (angles opposite to equal sides are equal) ----(1)

In triangle APQ

APQ +AQP +PAQ = 180

APQ +AQP + 45 = 180

2(APQ) = 180 – 45 (from (1))

APQ =

⁰

APQ = AQP =

OQA = 90 & OPA = 90 (Since radius is perpendicular to tangent)

OPQ + QPA = OPA = 90 (Linear Pair)

OPQ +

=90

OPQ = 90 -

= (

) =

= OQP (Since angles opp to equal sides i.e radii are equal)

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In ∆ POQ

OQP + OPQ + POQ = 180 (angle sum property)

+

+ POQ = 180

POQ + 45 =180

POQ =180 - 45

POQ =135

33. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of

the cone is 6 cm and the diameter of the base is 10 cm. Find the volume of the toy. If a right

circular cylinder circumscribes the toy, what is the difference between the volume of the

cylinder and the toy. (Take π = 3.14 )

Volume of the toy = Volume of the cone + Volume of the hemisphere

Volume of the cone =

πr

2 h =

π(5)

2 (6) = 50π cm

3

Volume of the hemisphere =

π5

3 =

π cm

3

Total Volume of the toy = 50π +

π =

π cm

3 = 133.3π cm

3

Height of the cylinder = height of the cone + radius = 6 + 5 = 11 cm

Volume of the cylinder = πr2 h

= π 5 5 11

= 275π cm3

The difference of the volume of the cylinder and toy

= 275π - 133.3π cm3 = 141.7π cm

3 = 141.7(3.14) = 444.9 cm

3

Volume of the toy = 133.3π = 418.56 cm3

The difference in the volume of the cylinder and toy = 444.9 cm3

34. A man standing at the top of the tower observes a van at an angle of depression of 30 , which

is approaching the foot of the tower with a uniform speed. Eight seconds later, the angle of

depression of the van is found to be 60 . Find the time taken by the van to reach the foot of the

tower from this point.

Let AB = h m be the vertical tower and BD = x m .

Let D and C be the positions of the van when the angle

of depression from the top of the tower is 60 and 30 respectively.

Let the uniform speed of the van be y m/sec.

Time taken for the angle of depression to change from 30 to 60 = 8 sec (Given)

ADB = 30 (Alternate angles)

ACB = 60 (Alternate angles)

CD = Distance covered by van in 8 sec

= y m/sec 8 sec = 8y m

In ∆ ABC

tan 30 = AB/BC

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√ =

h√ = 8y + x --(1)

In ∆ ABD

tan 60 = AB/BD

√ =

⟹ x =

√ ⟹ h√ = 3x ---- (2)

From (1) and (2), we get

3x = 8y + x ⟹ 2x = 8y ⟹

= 4 sec

Time taken by the van to reach the foot of the tower from D =

Time taken by the van to reach the foot of the tower = 4 sec

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