CBSE Sample Papers with Solutions SA2 Maths for 2014 - 15 From www.LearnCbse.in

1. SUMMATIVE ASSESSMENT MATHS
CLASS 10 CBSE Sample Papers - Solutions - SA2
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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

2. 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)
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

3. 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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

4. 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 are4 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 whichx 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. Findx.
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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

5. ( )
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:
( )
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

6. 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.
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

7. 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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

8. 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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

9. 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
+ AC2
= BC2
AC2
= BC2
– AB2
102
- 62
= 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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

10. 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 
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

11. 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 =
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in

12. 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
www.LearnCBSE.in
www.LearnCBSE.in
w
w
w
.LearnCBSE.in