SUMMATIVE ASSESSMENT MATHS

CLASS 10

CBSE Sample Papers Maths SA II Solved

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 9 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. If a + b + c = 0, a ≠ 0, a, b, c ϵ R, then roots of the equation ax2 + bx + c = 0

a) 1, c

b) 1,

c) a, b

d) 1,

Given: ( )

Substituting in the equation

( ) ( )

( )( ) ( )

( )[ ( ) ]

( )

2. If the angles of elevation of the top of a tower from two distinct points at a distance of x and y

(x > y) from its foot are 30∘ and 60∘ respectively, then the height of the tower is

a) √

b) √

c) √

d) √

Tan

√ ( )

Tan

√ ( )

Multiplying (1) and (2)

√ √ √

3. The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear if

a) k =

, 2

b) k =

, 2

c) k = -1, 2

d) k = 2, 1

The points are collinear if area = 0

( )( ) ( )( ) ( )( )

( )( )

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( )( )

4. C (o, r1) and C (o, r2) are two concentric circles with r1 > r2. AB is a chord of C (o, r1) touching

C (o, r2) at C then

a) AB = r1 + r2

b) AC = BC

c) AB = r1

d) AB = r2

AO = BO, OC is common,

OCA = OCB = 90

AOC BOC

AC = BC - Corresponding parts of Congruent triangles

5. To divide a line segment AB in the ration 4 : 7, a ray AX is drawn first such that BAX is an

acute angle and then points A1, A2.... are located at equal distances on the ray AX and then the

point B is joined to

a) A10

b) A9

c) A11

d) A12

6. The perimeter of a quadrant of radius 'r' is

a) 2πr +

b)

c) 2πr

d)

r (π + 4)

Perimeter =

(2πr) + 2r

( )

7. A rectangular block of dimensions 6 cm x 12 cm x 15 cm is cut into exact number of equal

cubes. The least possible number of cubes will be

a) 11

b) 40

c) 6

d) 33

Volume of cuboid =

If side of cube = 3 cm then

V = 33 = 27 cm

3

No. of cubes =

= 40

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8. The probability of getting a number between 1 and 100 which is divisible by 1 and itself only

is

a)

b)

c)

d)

A number which is divisible by 1 and itself only is a Prime number.

Prime numbers between 1 and 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,

53, 59, 61, 67, 71, 73, 79, 83, 89, 97 i.e total 25 nos.

P =

9. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m

from the ground. The angle of elevation of the balloon from the eyes of the girl at that instant is

60 . After some time the angle of elevation reduces to 30 . Find the distance travelled by the

balloon during the interval.

a)

b) √

c) √

d)

h = 88.2 – 1.2 = 87

Tan

√

Tan

√

Distance travelled by the balloon

√

√

√

√

√

√

√

√

SECTION – B

10. Solve for x: ( )

( ) ( )

( )( )

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11. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that

the card drawn is

i) A card of spade or an ace

ii) A red king

iii) Neither a king nor a queen

iv) Either a king or a queen

i)

ii)

iii)

iv)

12. How many terms of the AP 3, 5, 7….. must be taken so that the sum is 120?

[ ( ) ]

[ ( ) ( ) ]

( )

( )

( )

( )( )

( ) ( )

No. of terms cannot be -ve n = 10

13. A tangent PQ at a point P to a circle of radius 6 cm meets a line through the centre 'O' at a

point Q so that OQ = 10 cm, find the length of PQ.

QP ⊥ PQ In △ OPQ

OQ2

= OP2 + PQ

2

102 = 6

2 + PQ

2

PQ2 = 10

2 – 6

2 = 100 – 36 = 64

PQ = √

PQ = 8 cm

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14. A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at

random from the bag. What is the probability that the ball drawn is

i) White or blue

ii) Red or black

iii) Not white

iv) Neither white nor black

White = 5,

Red = 7

Black = 4

Blue = 2

Total balls = 18

i)

ii)

iii)

iv)

15. For what values of P the equation ( ) ( ) ( ) has

coincident roots?

Condition for coincident roots is D = 0

[ ( )] ( )( )

( ) ( )

( )

( )

16. In an AP, if the 5th

and 12th

terms are 30 and 65 respectively. What is the sum of first 20 terms?

( ) ( )

Subtracting (1) from (2)

Substituting in 1 ( )

[ ( ) ]

[ ( ) ( )]

( ) ( )

.

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17. The angle of elevation of a cloud from a point 200 m above a lake is 30 and the angle of

depression of its reflection in the lake is 60 . Find the height of the cloud above the lake.

Let height of the cloud above the ground = H

But H = h + 200

Tan

√ ( )

Tan

√

√ ( )

Equating –(1) and (2)

√

√

18. The cost of fencing a circular field at the rate of Rs.24/m is Rs. 5280. The field is to be

ploughed at the rate of Rs. 0.5/m2. Find the cost of ploughing the field.

Perimeter of field =

m

Area of field =

m

2

Cost of ploughing =

19. Draw a circle of radius 3 cm. From a point P, 6 cm away from its centre, construct a pair of

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

OA = 3 cm. OP = 6 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 = 5.2 cm approximately.

20. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is

the rise in water level in the pond if the average displacement of the water by a person is 0.04

m3?

Rise in water level = Volume of 500 persons

m = 0.005 m = 0.5 cm

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21. Point P divides the line segment joining points A (2, 1) and B (5, -8) such that

. If P

lies on the line , find the value of k.

Co - ordinates of (

) ( )

∵ P lies on

Substituting x = 3 and y = -2

( ) ( )

22. In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will

appear.

Total outcomes are

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Doubles are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)

A total of 9 are (3, 6), (6, 3), (4, 5), (5, 4)

Possible outcomes = 36 - (6 + 4) = 26

P =

=

23. Find the area of the shaded region, if diameter of the circle with centre 'O' is 28 cm and

AB = 28 cm r = 14 cm

AQ = 7 cm r1 =

cm

QB = 21 cm r2 =

cm

Area of shaded region =

πr

21 +

πr

22

π (r

21+r

22)

[(

)

(

)

]

[

]

Area of shaded region = 192.5 cm2

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24. Find the roots of the equation by the method of completing the square :

Dividing the equation by 3

(

)

(

) (

)

= 0

(

) (

)

(

)

(

)

(

)

(

)

i) If (

)

ii) If (

)

25. A cone is cut into two parts by a horizontal plane passing through the midpoint of its axis. Find

the ratio of the volumes of the upper part and the cone.

Given: AD= ½ AB, DE ǁ BC

In △ ADE and △ABC:

DE ǁ BC

ADE = ABC and AED = ABC

△ ADE △ ABC−−−−AA similarity

=

Ratio of volumes of small cone and large cone =

(

) (

)

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( )

( ) (

)

(

)

26. The sum of first five terms of an AP and the sum of the first seven terms of the same AP is

167. If the sum of first ten terms of this AP is 235, find the sum of its first twenty terms.

( )

( )

( )

( )

Given:

( ) and

( )

( )

Multiplying by 6

( )

Subtracting (3) from (1)

Substituting in (2)

( )

[ ( ) ]

( )

( )

27. Median of a triangle divides it into two triangles of equal areas. Verify this result for △ ABC

whose vertices are A (4, -6), B (3, -2) and C (5, 2)

Co-ordinates of D =

( )

Area of △ ABD =

( ) ( ) ( )

Area of △ ADC =

( ) ( ) ( )

Area of ABD = Area of ADC = 3 units

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28. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction

and its reciprocal is

, find the fraction.

Let the numerator be x denominator = and fraction =

Given :

( ) ( )

( ) ( )

( )( )

( ) ( )

i) If , the fraction =

ii) If

; the fraction =

Since

is a mixed numeral rejected.

the fraction =

29. The length of the minute hand of a clock is 6.3 cm. Find the area swept by the minute hand

during the time period 5:45 am to 6:10 am.

The angle swept by minute hand during 5:45 to 6:10 am is 150

Area of sector =

cm2

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30. A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has been evenly spread

all around it in the shape of a circular ring of width 4 m to form an embankment. Find the

height of the embankment.

Volume of the earth taken out = Volume of the embankment

( )

r = 1.5 m, h = 14 m, R = 5.5 m

( )( )

( )( )

m

31. Prove that the tangents at the extremities of a chord of a circle make equal angles with the

chord.

To prove: PAB = PBA

Proof: PA = PB -- Tangents from the external point P

In △ PAB, PAB = PBA

Angles opposite to equal sides are equal

32. Prove that the parallelogram circumscribing a circle is a rhombus.

Given : Parallelogram ABCD circumscribes a circle

To Prove : ABCD is a rhombus

Proof : AP = AS---1 (Tangents from A)

BP = BQ---2 (Tangents from B)

CR = CQ---3 (Tangents from C)

DR = DS---4 (Tangents from D)

Adding --1, 2, 3 and 4

AP + BP + CR + DR = AS + BQ + CQ + DS

AB + CD = (AS + DS) + (BQ + CQ) = AD + BC

But AB = CD and AD = BC --Opposite sides of parallelogram

2 AB = 2 AD

AB = AD

AB = CD = AD = BC

ABCD is a Rhombus.

33. If AB is a chord of a circle with centre 'O', AOC is a diameter and AT is the tangent at A as

shown in the figure. Prove that BAT = ACB

Given: AT is tangent, AB is a chord, AOC is diameter

To prove : BAT = ACB

Proof: OA⊥AT OAT=90

ABC=90 −−−− angle in a semicircle

1+2+ABC=180 −−−− angle sum property

1+2+90 =180

1+2=90 −−−−1

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But 1+BAT=90 −−−−2

Equating 1 and 2

1+2=1+BAT

2=BAT

BAT = ACB

34. Construct a ∆PQR in which QR = 6 cm, Q = 60° and R = 45°. Construct another triangle

similar to ∆PQR such that its sides are 5/6 of the corresponding sides of ∆PQR.

Steps of construction:

Draw QR = 6 cm At Q draw an angle = 60

At R draw an angle = 45

The point of intersection of the arms of these angles is P.

PQR is the required triangle.

At Q draw an acute angle RQX.

Locate 6 points Q1,Q2.....Q6 on QX such that

QQ1=Q1Q2=Q2Q3=Q3Q4=Q4Q5=Q5Q6

Join RQ6 and draw a line through Q5 parallel to RQ6

intersecting QR at R1

Draw a line through R1 parallel to PR intersecting PQ at P

1

P1QR

1 is the required triangle.

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