SUMMATIVE ASSESSMENT - II MATHS

CBSE Samle Papers For Class 10 Maths - SAII - Solved - Paper 1

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. If - and 1 are the sum and product of roots of a quadratic equation respectively, then the

equation is _____.

a) = 0

b) = 0

c) = 0

d) = 0

= 0

= 0

= 0

2. The tops of two poles of height 25 m and 15 m are connected by a wire. If the wire makes an

angle of 30∘ with the horizontal, then find the length of wire.

a) 30 m

b) 20 m

c) 5 m

d) 10 m

According to the given figure,

AB = 25 – 15 = 10 m

The length of the string = AC

sin =

⇒

=

AC = 20 m

3. As shown in the given image , from an external point S, tangents SA and SB are drawn to

circle O. CD is tangent to the circle at E. If AS = 12 cm, find the perimeter of ∆SCD.

a) 20 cm

b) 24 cm

c) 25 cm

d) 12 cm

CA = CE and DE = DB (tangents from external points C and D) ------ (1)

Perimeter of ∆SCD

= SC + CD + SD

= SC + CE + ED + SD

= SC + CA + DB + SD (from (1))

= SA + SB

= 12 + 12 = 24 cm

Perimeter of ∆SCD = 24 cm

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4. TA and TB are tangents to circle O from point T. TA = 4x - 7 and TB = 3x + 5, find TA.

a) 12 cm

b) 24 cm

c) 41 cm

d) 36 cm

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

TA = TB -----(1)

4x - 7 = 3x + 5

x = 12 cm

TA = 4(12) – 7 = 48 – 7 = 41 cm

TA = 41 cm

5. If AB = 12 cm and CD = 9 cm are two chords of a circle, then which of the following is true?

a) AB is nearer to the centre of the circle

b) CD is nearer to the centre of the circle

c) AB and CD are equidistant from the centre of the circle

d) Can’t say

Of any two chords of a circle, the one which is larger is nearer to the centre of the circle

6. Which constant must be added or subtracted to solve the equation 5 - √ x + 3 = 0 by the

method of completing the square?

a)

b)

c) √

d)

5 - √ x + 3 = 0

⇒ - √

x + = 0

⇒ - 2( √

+

√

(

√

)

= 0

∴ √

should be added and subtracted

√

=

=

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∴ the number of red balls in the bag is 6

7. Given below are two similar triangles, ∆ABC and ∆AB'C' .Which of the following statements

are true?

a) AB' = AB

b) AB = AB'

c) AB = AB

d) None of the above

AB' is 5 parts: AA1 = AA2 = AA3 = AA4 = AA5

AB is 7 parts: AA1 = AA2 = AA3 = AA4 = AA5 = AA6 = AA7

The ratio of the sides of ABC to that of corresponding sides of the triangle AB'C' is 7:5

8. A rope that attaches a swing to a tree is 2.6 m long and the maximum difference between

trajectories is an angle of 150⁰. The maximum distance travelled by the swing when the swing

angle is maximum is

a) 4.2 m

b) 6.8 m

c) 13.2 m

d) 8.6 m

length of the arc =

2

r

=

× 2 ×

× 2.6 = 6.8 m

SECTION – B

9. If O (

, 2 ) is the midpoint of the line segment joining the points Q ( -7 , 3 ) and R (-5 , 1) ,

then find the value of m.

Applying the Mid-point Formula for the line joining Q (-7,3) and R(-5,1),we get

=

=

= -6

m = -6 x 5 = -30

10. A bag contains 12 green balls and some red balls. If the probability of drawing a red ball is half

that of a green ball, then the number of red balls in the bag is?

No. of green balls = 12

Let no. of red balls = x

∴ Total no. of balls = 12 + x

Given P (red ball) =

P(green ball)

=

⇒ x = 6

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11. Find the numerical difference of the roots of - 7x - 9 = 0.

Roots of the equation - 7x - 9 = 0 are

x = √

⇒ √

⇒x = √

, √

⇒ Numerical difference of the roots

= √ √

= √

= √

12. Two unbiased dice are thrown. Find the probability that

i) the total score is more than 5

ii) the total score is less than 5

i) Total outcomes are 36

(i) The favorable outcomes are:

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

∴ P =

=

ii) The favourable outcomes are:

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) = 6

∴ P =

=

13. Given below is the group-wise heights of children in cms in a sports academy:

Group A - 131, 141, 115, 161

Group B - 119, 125, 131, 137

Group C - 102, 130, 142, 150

Group D - 105, 104, 103, 101

Which of the following groups forms an AP?

Consider the group A 131,141,115,161

a2 – a1 = 10

a3 – a2 = -26

a4 – a3 = 46

as the difference is not same they do not form AP.

Consider the group B 119,125,131,137

a2 – a1 = 6

a3 – a2 = 6

a4 – a3 = 6

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as the difference is same they form an AP.

Consider the group C 102,130,142,150

a2 – a1 = 28

a3 – a2 = 12

a4 – a3 = 8

as the difference is not same it does not form an AP.

Consider the group D 105,104,103,101

a2 – a1 = -1

a3 – a2 = -1

a4 – a3 = -2

as the difference is not same they do not form A

14. A horse is tied to a pole fixed at the middle of the side of a square field of edge 65 m with a

rope of 18 m long. Find the area of the field in which the horse can graze?

As the pole is on side of square, θ = 180

Area of the sector =

× × r²

=

×

× 18 ×18

= 509.14 m²

Area of the field in which the horse can graze is 509.14 m²

SECTION – C

15. The surface area of a metallic sphere is 616 cm2. It is melted and recast into a cone of height 28

cm. Find diameter of the base of cone so formed.

TSA of sphere = 4 r2

4

r

2 = 616

r2 = 616

r2 = 49

r = 7 cm

Volume of sphere = Volume of cone

r

3 =

r

2 h

7 7 7 =

r

2 28

r2

= 49

R = 7 cm

Diameter of cone = 7 2 = 14 cm

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16. Find the length of wire of diameter cm that can be drawn from a solid sphere of radius 12 cm

(Write the answer in m)

Radius of wire (r) =

=

Volume of sphere =

× πr

3 =

× 3.14 × 12 × 12 × 12

Volume of wire = Volume of sphere

3.14 ×

×

× h =

× 3.14 × 12 × 12 × 12

h =

= 25600 cm = 256 m

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

⇒ = l × b × H

⇒

×3×3×30=24×12×H

⇒H =

× 3 × 3 × 30 ×

×

⇒H = 2.946 m

17.

If tangents PA and PB from a point P to a c ircle with centre O are inclined to each

other at angle of 70⁰, then find POA.

∠OAP = 90

∠OBP = 90

∠APB = 70

Sum of the angles of a Quadrilateral = 360⁰

∠AOB = 360 - (90 + 90 +70 )

∠AOB = 110

∠POA = (∠AOB)

∠POA = 55

18. Determine the ratio in which the line segment joining the points A (1, -1) and B (4, 9) is

divided by the line .

Let the line divide the line segment in the ratio k:1.

Then the co-ordinates of the point of division are : (

)

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Since this point also lies on the given line ,

3(

) – 2(

) + 4 = 0

= 0

-2k + 9 = 0 => k =

Therefore the ratio is 9: 2

19. Sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m, find the

sides of the two squares.

Let the sides of two squares be a and b meters.

Given a2+b

2 = 640 and 4a−4b = 64

⇒4(a−b) = 64

⇒a−b = 16 →(1)

(a−b2)=16

2

⇒a2 + b

2−2ab = 256

640 − 2ab = 256

⇒−2ab = 256 − 640 ⇒ 2ab = 384

(a + b)2

= a2

+ b2

+ 2ab = 640 + 384 = 1024

⇒a + b = √

⇒a + b = 32 →(2)

Solving equation (1) and (2), a = 24

⇒ a – b = 16 ⇒ 24 – b = 16 ⇒ −b = 16 – 24 = -8

b = 8 m, a = 24 m

20. A lot consists of 52 laptops of which 46 are good, 4 have only minor defects and 2 have major

defects. Smith will buy a phone if it is good but the trader will only buy it only if it has no

major defect. One phone is selected at random, from the lot. Find the probability that

i) it is acceptable to Smith

ii) it is acceptable to the trader

iii) it is acceptable neither to Smith nor to the trader

Total number of laptops = 52

No. of good laptops = 46

Minor defect laptops = 4 , Major defect laptops = 2

i) P(good laptops) =

=

ii) No. of good laptops + Laptops with minor defect = 46 + 4 = 50

∴ P(acceptable to the trader ) =

=

iii) P(acceptable neither to Smith nor to the trader) =

=

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21. In a nursery, 37 plants have been arranged in the first row, 35 in the second, 33 in the third and

so on. If there are 5 plants in last row, how many plants are there in the nursery?

The given sequence is: 37, 35, 33……, 5

Here a = 37 ; d= -2 and an = 5

an = a + (n-1)d

5 = 37 + (n-1)-2

5 = 37 +2 -2n

5 = 39 -2n

2n = 34

n = 17

Sn =

(a + an)

(37 + 5)

× 42

=17 x 21 = 357

There are 357 roses on the flower bed..

22. An electrician needs to reach a point 1.6 m below the top of the pole of height 8 m to undertake

the repair work. If he has to incline the ladder at 60 degrees angle to the horizontal, what

should be the length of the ladder to enable him to reach the required position?

Let BP be the required pole of height 8 m.

Let A be the point below the top P such that AP = 1.6 m

BA = 8 - 1.6 = 6.4 m

In ABC triangle,

Sin 60 =

√

=

AC =

√ =

√

4.26√

= 7.38 m

The length of the ladder should be 7.38 m

23. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m

and the slant height of the conical portion is 53 m, find the area of canvas needed to make the

tent.

Area of Canvas Required = CSA of Cylindrical part + CSA of Conical part

= 2

× (r) (h) +

× (r) (l)

=

(r) [2 (h) + (l)]

=

(52.5) [2 (3) + 53]

= 22 × 7.5 × 59

= 9735 m2

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24. Draw a line segment of 8 cm and then divide it internally in the ratio 3:2.

To divide a line segment of 8 cm internally in the ratio 3:2

a) Draw a line segment AB = 8 cm

b) Draw another ray Making an acute angle BAX with AB

c) Along AX mark off 5(2 +3 ) points A1,A2,A3,A4 and A5 such

that AA1 = A1A2 = A2A3 = A3A4 = A4A5

d) Join BA5

e) Through A3 draw a line A3P parallel to BA5

AP : PB = 3 : 2

Section - D

25. If the co-ordinates of the mid points of sides AB, BC and CA of ∆ABC are (1, 1), (2, -3) and

(3, 4) respectively, then find the co-ordinates of the centroid.

Let A = (x1, y1) B = (x2, y2) and C = (x3, y3)

Let P be the mid-point of AB.

⇒(

) = (1, 1)

⇒x1 + x2 = 2, y1 + y2 = 2 →(1)

Q is the mid-point of BC

⇒(

) =(2,−3)

⇒x2 + x3 = 4, y2 + y3 = −6 →(2)

R is the mid-point of CA

⇒ (

) = (3, 4)

⇒x3 + x1 = 6, y3 + y1 = 8 →(3)

From (1), (2) and (3),

x1 + x2 + x2 + x3 + x3 + x1 = 12

y1 + y2 + y2 + y3 + y3 + y1 = 2 – 6 + 8

⇒x1 + x2 + x3 = 6

y1 + y2 + y3 = 2→(4)

∴ Co-ordinates of the centroid of D ABC are

(

) = (

) = (

)

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26. A park’s jogging track has its left and right ends as semi-circles. The distance between the two

inner parallel line segments is 60 m and the length of the straight part of the track is 118 m. If

the track is 12 m wide, find

i) the length of the track along its inner edge.

ii) the area of the track.

i) Radius of inner semi-circle =

= 30 m

Circumference of inner semicircle = πr =

× 30

=

m

The length of the track along its inner edge

= length of two inner parallel lines

+ circumference of two inner semicircles.

= 118 + 118 +

+

= 236 + 188.57 = 424.57 m

ii) Area of the track

= area of two rectangles ABCD and EFGH + 2( area enclosed between outer and inner

semi-circles) ---- (1)

Area of rectangle ABCD = length x width of track = 118 x 12=1416 m

Area of two rectangles=2 x 1080=2832 sq m ----(2)

Diameter of outer semi-circle= 60 + width of track + width of track = 60 + 12 + 12=84 m

Radius of outer semi-circle =

= 42 m

Area enclosed between semi-circles

= Area of bigger semi-circle - area of smaller semi-circle

=

[(r1

2 - r2

2]

=

[(42

2 - 30

2]

=

(1764 - 900)

=

(864) = 1357.7 m

2

⇒Area enclosed by left and right semi-circles=2 x 1357.7 =2715.4 m2 -------- (3)

Putting (2) and (3) in (1), we get

Area of the track

= area of two rectangles formed in the track + area enclosed by the semi-circles

= 2832 + 2715.4 = 4988.56 m2

27. If TA and TB are two tangents drawn to a circle with centre O from an external point T, prove

that ATB = 2OAB.

OAT = OBT = 90 (radius is perpendicular to tangent)

In quadrilateral OATB

OAT + ATB + TBO + BOA = 360

90 +ATB + 90 +BOA = 360

ATB +BOA = 180 -----(1)

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If two tan gents are drawn from an external point to a circle then

the tangents are equal in length.

AF = AE , FB = BD , EC = CD

Perimeter of triangle ABC = AB + BC+ CA

= AF + FB + BD + DC + CE + EA

= AF + BD + BD + CE + CE + A F (Since AF = AE, FB = BD, EC =

CD) = AF + AF + BD + BD + CE + CE

= 2(AF + BD+ CE)

Perimeter of triangle ABC = 2(AF + BD+ CE)

In OAB,

OA = OB OAB = OBA (angles opp. to equal sides are equal)

OAB + OBA + BOA = 180

2(OAB) + BOA = 180

BOA = 180 - 2(OAB) ----(2)

from (1) and (2)

ATB = 2(OAB)

Hence proved.

28. In the given figure, D, E, and F are the points where the incircle of the triangle ABC touches

the sides BC, CA, and AB respectively.

Show that AF + BD + CE = AE + BF + CD =

(perimeter of triangle ABC)

AF + BD + CE =

(perimeter of ABC) ……..(1)

Perimeter of ABC = AB + BC + AC

= AF + BF + BD + CD + AE + EC

= AE + BF + BF + CD + AE + CD (Since AF = AE, BF = BD, EC = CD)

= AE + AE + BF + BF + CD + CD

= 2(AE + BF + CD)

Perimeter of ABC = 2(AE + BF + CD)

AE + BF + CD =

(perimeter of ABC) ……..(2)

From (1) and (2) ,we get

AF + BD + CE = AE + BF + CD =

(perimeter of ABC)

Hence proved.

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29. The sum of two numbers is 20 and the sum of their reciprocals is

. Find the numbers.

Let one number be x ∴ other number = 20 − x

Given (

) + (

) =

⇒

=

⇒ 480 = 100x − 5x2

⇒ 5x2 − 100x + 480 = 0

⇒ x2 − 20x + 96 = 0

⇒x2 − 12x − 8x + 96 = 0

⇒x(x − 12) − 8(x −1 2) = 0

⇒(x − 12) (x − 8) = 0

⇒x = 12 or x = 8

∴ Numbers are 8 and 12

30. From a solid cylinder of height h = 16 cm and diameter d = 8 cm, a conical cavity of same

height and same diameter is carved out. Find the volume and total surface area of the

remaining solid.

Height of cylinder (H) = 16 cm

Diameter of cylinder (D) = 8 cm

Radius of cylinder(R) =

= 4 cm

Height of cone (h) = 16 cm

Radius of cone = 4 cm

Vol. of remaining solid = vol. of cylinder - vol. of cone

=

× 4 × 4 × 16 –

×

× 4 × 4 × 16

=

× 4 × 4 × 16 [

]

= 536.38 cm3

l = √ = 16.5 cm

TSA of remaining solid = C.S.A. of cylinder + C.S.A. of cone + Area of the base

= 2 ×

× 4 × 16 +

× 4 × 16.5 +

× 4 × 4

=

+

+

=

= 660 cm2

Vol. of remaining solid = 536.38 cm3

TSA of remaining solid = 660 cm2

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=(TP + PB) (TP – PB) (from (1))

31. In which of the following situations, does the list of numbers involved make an arithmetic

progression, and why?

i) The amount of air present in a balloon when

th of the air remaining in the ballon is

removed by a vacuum pump at a time.

ii) The cost of digging a well after every meter of digging, if it costs Rs 125 for the first meter

and increases by Rs. 65 for every subsequent meter.

i) Let amount of air initially present in the balloon be u

vacuum pump removes 1/5th of the air remaining in the cylinder

So the amount of air left when vacuum pump pumps out air

= u, u-(1/5)u=4u/5,4/5u-1/5u=3/5u,....

= v,(4/5)u,(3/5)u,...

So, the sequence we get is u,(4/5)u,(3/5)u,...

The difference between consecutive terms is,

u − u = −

,

−

= -

Difference between consecutive terms is constant.

Therefore, it is an arithmetic progression.

ii) Cost of digging 1 meter of well = Rs 125

Cost of digging the well

= 125,125+65= Rs 190,190+65 = Rs. 255,.....

we get a sequence of the form 125, 190, 255 ....

The difference between consecutive terms is, 255-190 = 190-125 = 65

Difference between consecutive terms is constant.

Therefore, it is an arithmetic progression.

32. TBA is a secant to the circle with center O intersecting the circle at A and B. TC is a tangent

segment, prove that TA × TB = TC2

Given: A secant TBA to a circle C with center O intersecting the circle at A and B and TC is

a tangent segment.

To prove : TA × TB = TC2

Construction: Draw OP ⊥ AB. Join OP, OT and OA

Proof: Since OP ⊥ AB

⇒ PA = PB … (1) (Perpendicular from the centre to the chord bisects the chord)

TA × TB = (TP + PA) (TP – PB)

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= TP2 – PB

2

In right ΔOPT,

OT2 = OP

2 + PT

2

⇒ TP2 = OT

2 – OP

2

∴ TA × TB = OT2 – OP

2– PB

2

= OT2 – (OP

2 + PB

2)

In right ΔOBP

OB2 = OP

2 + PB

2

∴ TA × TB = OT2 – OB

2 = OT

2 – OC

2 ------- (2) (OB = OC = r)

In right ∆OCT,

OT2 = TC

2 + OC

2

⇒ TC2 = OT2 – OC

2 ------ (3)

From (2) and (3)

∴ TA × TB = PT2

33. Stancy's house has an overhead tank in the shape of a cylinder which is filled by pumping

water from a cuboidal sump. The sump has dimensions 1.57 m 1.44 m 95cm. the

overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the

sump after the overhead tank has been completely filled with water from the sump which had

been full. Compare the capacity of the tank with that of the sump. (Use pi as 3.14 )

Radius of Overhead tank = 60 cm = 0.6 m

Height of Overhead tank = 95 cm = 0.95 m

Volume of water in overhead tank = πr2 h

= 3.14 0.6 0.6 0.95 = 3.14 0.36 0.95 m3

Dimensions of the sump = 1.57 m 1.44 m 95 cm

Volume of water in the sump when full = lbh

=1.57 1.44 0.95 = 2 3.14 0.36 0.95 m3

Volume of water in the sump after filling the tank

= 2 3.14 0.36 0.95 - 3.14 0.36 0.95

= 3.14 0.36 0.95 (2 - 1)

= 3.14 0.36 0.95 m3

Area of the base of the sump = 1.57 1.44 = 1.57 4 0.36 = 2 3.14 0.36 m3

Height of water in sump =

=

= 47.5 cm

=

=

Height of water in sump = 47.5 cm

Capacity of the tank is half the capacity of sump

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34. From a point on the ground 60 m away from the foot of a tower, the angle of elevation of the

top of the tower is 30 . The angle of elevation to the top of a water tank (on the top of the

tower) is 45 . Find the (i) height of the tower (ii) the depth of the tank.

i) Let h be the height of the tower BM

In right triangle CBM, we have

= tan 30° =

√

h =

√ =

√

= 20√ = 34.34 m

Thus, the height of the tower is 20√ = 34.64 m approx.

ii) Let the depth AM of the water tank be x m

In right triangle ABC,

= tan 45° =

h + x = 60 m (Given)

x = 60 – 34.64 = 25.36 m

Hence, the depth of the water tank = 25.36 m Approx.

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