CBSE SAMPLE PAPERS

SUMMATIVE ASSESSMENT-II (MATHS)CLASS 10

Max Marks: 90 Time: 3 Hrs

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

SECTION – A

1. A Line cuts X axis at (18, 0) and Y axis at (0, -8). The circum-centre of the triangle formed by

the line with the axis is

a) (-4, 9)

b) (9, - 4)

c) (0, 0)

d) (18, -8)

Midpoint of (18, 0) and (0, -8) is ( )

2. To construct the triangle similar to a given △ ABC with its sides of the corresponding

sides of △ ABC, draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side

of A with respect to BC. The minimum number of points to be located at equal distances on

ray BX is

a) 8

b) 13

c) 5

d) 3

3. If the circumference of a circle is equal to the perimeter of a square, then the ratio of their areas

is

a) 7 : 11

b) 14 : 11

c) 22 : 7

d) 7 : 22

4. A die is thrown twice. The probability that two will not come up either time is

a)

b)

c)

d) None of these

The probability that 2 will not come either time: events are

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

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

Total events are: 36

P =

5. The number of points on X axis which are at a distance of 2 units from (2,4) is

a) 0

b) 1

c) 2

d) 3

The ordinate of point (2, 4) is 4 because 4 > 2

There is no point that is on X axis at a distance of 2 units from (2, 4)

6. For to have equal roots, k can take the values

a) 2 and -2

b) 4 and -4

c) 8 and -8

d) 16 and -16

For a quadratic equation to have equal roots D = 0 i.e.,

7. The sum of first 11 terms of an AP whose middle term is 30 is

a) 320

b) 330

c) 340

d) None of these

The middle term [ ]

30 = 330

8. If AB = 4 m and AC = 8 m then angle of observation of point A as observed from C is

a) 60

b) 30

c) 45

d) Cannot be determined

Point on Y axis is (0, 1)

SECTION – B

9. Find the values of k for which the given equation has real and distinct roots :

_____________________

For a quadratic equation to have real and distinct root D > 0

10. If the 3rd

and 9th

terms of an AP are 4 and -8 respectively, which term of this AP is zero?

------- 1

----- 2

Subtracting 2 from 1

– – –

Substituting in 1

Let

11. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box,

what is the probability that it will be a black ball? If 6 more black balls are put in the box, the

probability of drawing a black ball is now double of what it was before. Find x.

i)

ii) ( )

12. Find a point on the Y axis which is equidistant from (-3,2) and (1,-2)

Let the point on Y axis be (0, y)

Distance between (-3, 2) & (0, y) = Distance between (1, -2) and (0, y)

√ √

Squaring both sides

( )

13. The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Find its area.

Perimeter of the quadrant =

cm

Area of the quadrant =

cm2

14. Cards marked with numbers 2 to 90 are placed in a box and mixed thoroughly. One card is

drawn at random from the box. Find the probability that the card drawn is

i) a two digit number________

ii) a number which is a perfect square_________

i) Two digit number from 10, 11…….90 are 81

ii) Perfect squares between 2 to 90 are 4, 9, 16, 25, 36, 49, 64, and 81

SECTION – C

15. In the given figure, O is the centre of the circle. Determine AQB and AMB if PA and PB

are tangents.

In quadrilateral PAOB, P = 75 , PAO = PBO = 90

AOB = 105 ------ angle sum property of quadrilateral

Q =

(Angle subtended by the arc AMB at centre = angle subtended at point Q on circle)

AMB = 180 – 52.5 = 127.5 (cyclic quadrilateral property)

16. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream

than to return down stream to the same spot. Find the speed of stream.

Let the speed of stream = x km/h

Speed of motor boat upstream = 18 - x km/h and downstream = 18 + x km/h

Time = Given:

( )

or 6

Since speed cannot be negative is rejected

km/h

17. Which term of the AP 3, 10, 17…..will be 84 more than its 13th

term?

given:

Ans: 25th

term

18. An airplane when 3000 m high, passes vertically above another plane at an instant, when the

angle of elevation of the two airplanes from the same point on the ground are 60 and 45

respectively. Find the vertical distance between airplanes.

√

√

√√

√ m

Height = 3000 – 1732 = 1268 m

19. ABCDE is a polygon are A(-1,0), B(4,0), C(4,4), D(0,7) and E(-6,2). Find the area of the

polygon.

Area whose vertices of polygon =

Area of ABC + Area of AEC + Area of EDC

[ ]

Area of polygon =

= 42 square units

20. ABC, an isosceles in which AB = AC, is circumscribed about a circle. Show that BC is

bisected at the point of contact.

AB = AC ----- given, also AF = AE (tangents from A)

AB - AF = AC – AE BF = CE

But BF = BD and CE = CD (tangents from external point)

BD = CD

BC is bisected at the point of contact D.

21. The co-efficient of x in the quadratic equation was taken as 17 in place of

13, its roots were found to be -2 and -15. Find the roots of the original equation.

Substituting for x = (- 2) in

In original equation

Original equation:

22. Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn

with vertex 'o' of an equilateral OAB of side 12 cm as centre ( )

AOB = 60

angle of major arc = 360 – 60 = 300

Area of major arc COD =

cm2

Area of equilateral = √

cm2

Area of the shaded region = 94.286 + 62.352 = 156.638 cm2

23. A sector of circle of radius 12 cm has the angle 120 . It is rolled up so that the two bounding

radius are joined together to form a cone. Find the volume of the cone.

Radius of the sector = 12 cm

Angle of sector = 120

Length of arc =

Circumference of the base of the cone = cm

Length of the cone = 12 cm

Volume of the cone =

√

√ √

√ cm3

cm3

24. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is

i) Black and a king _______

ii) Spade or an ace _______

iii) neither a heart nor a king _______

i)

ii)

iii)

SECTION D

25. Solve for x :

( )

( )

( ) ( )

or

26. Find the sum of the integers between 100 and 200 that are

i) divisible by 9 ii) not divisible by 9

i) Integers divisible by 9 are 108, 117 ….. 198

–

ii) Integers not divisible by 9 are: 101, 102 ….. 199 except those in (i)

Sum of integers between 100 and 200 that are not divisible by 9

= 14850 – 1683 = 13167

27. From a point on the ground, the angles of elevation of the bottom and top of a transmission

tower fixed at the top of a 20m high building are 45 and 60 respectively. Find the height of

the tower.

Tan m

Tan √

√ √

m

28. ABC is a right, right angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10

cm and O is the centre of the incircle of ABC. (Use = 3.14).

Let AP = AR = x; BP = BQ = y; CQ = CR = z ----- 1

AB2 + AC

2 = BC

2 AC

2 = BC

2 – AB

2 10

2 - 6

2 = 100 - 36

AC = √ = 8 cm

Given AB = 6 cm and BC = 10 cm

AB = AP + PB = x + y = 6 (From 1)

BC = BQ + QC = y + z = 10 ---- 2 (From 1)

AC = CR + RA = x + z = 8 (From 1)

Adding : 2(x + y + z) = 24 x + y + z = 12

Substituting --- 2 in the above , we get

x = 2 cm

AP ⊥ OP, AR ⊥ OR and A = 90

APOR is a square with side x = r = 2 cm

Area of shaded region = Area of ABC - Area of circle

Area of shaded region =

= 24 - 4 = 24 - 4 (3.14)

= 24 - (12.56) = 11.44 cm2

29. A metallic bucket is in the shape of a frustum of a cone mounted on a hollow cylindrical base

as shown in the figure. If the diameters of two circular ends of the bucket are 45 cm and 25 cm

respectively, the total vertical height is 30 cm and that of the cylindrical portion is 6 cm. Find

the area of the metallic sheet used to make the bucket.

√ √

√ √ √

Lateral Surface Area of frustum of cone

= ( ) ( )

SA of base of frustum =

SA of hollow cylinder at base =

Total SA = cm2

30. If A (5,-1), B (-3,-2) and C (-1,8) are the vertices of ABC, find the length of median through

A and the co-ordinates of the centroid.

By midpoint formula, co-ordinates of D are

( )

The centroid divides the median in the ration 2 : 1

Co-ordinates of G are ( ) ( )

Length of median AD = √ √

= √ √ units

31. Construct a ABC in which AB = 5 cm, B = 60 and altitude CD = 3 cm. Construct a

A1 BC1 and ABC such that each side is 1.5 times that of the corresponding sides of ABC.

Steps of construction :Draw AB = 5 cm.

Draw ABQ = 60

At A, draw AX ⊥ BA.

Mark AY = 3 cm on ray AX

At Y, draw a ⊥ that cuts ray BQ at C.

Join AC.

ABC is the required

At B draw acute angle ABP

Locate 3 points B1 , B2, B3 on BP

such that BB1 = B1 B2 = B2 B3

Join AB2 and draw a line parallel to AB2

through B3 intersecting the extended line segment BA at A1

Draw a line through A1 parallel to AC intersecting the extended line BC at C1

A1 BC1 is the required triangle.

32. QR is a tangent at Q. PR ǁ AQ, where AQ is a chord through A and P is a centre, the end point

of the diameter AB. Prove that BR is tangent at B.

AQ ǁ PR 2 = 3 and 1 = 4

But 1 = 2 (PA = PQ radius) 3 = 4

In PQR and PBR, PR is common

3 = 4 and PQ = PB radius

PQR PBR PQR = PBR = 90 ----- cpct.

BR is a tangent

33. A round table cover has six equal designs as shown in the figure. If the radius of the cover is

28 cm, find the cost of making the designs at the rate of 35 paise per square cm. (Use √ = 1.7)

Area of 6 segments

= 6 [Area of sector – Area of OAB]

= [

√

]

= [ ] [ ]

= [ ] [ ]

= cm2

Area of 6 designs = 464.8 cm2

Cost =

34. An iron pillar has some part in the form of a right circular cylinder and remaining in the form

of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The

cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar,

if one cubic cm of iron weighs 10 g. ( )

Volume of pillar = Vol. of cylinder + Vol. of cone.

( ) ( )

= cm2

Weight of the pillar = 50688 10 g = 506880 506.88 kg