The document provides instructions for a summative assessment math exam for Class 10 CBSE. It states that the exam will be 3 hours long and consist of 34 questions divided into 4 sections (A-D). Section A has 9 multiple choice questions worth 1 mark each. Section B has 6 questions worth 2 marks each. Section C has 10 questions worth 3 marks each. Section D has 10 questions worth 4 marks each. Calculators are not permitted and an extra 15 minutes is provided to read the paper only.

1. 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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2. 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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3. ( )( )
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 cm3
No. of cubes = = 40
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4. 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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5. 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
+ PQ2
102
= 62
+ PQ2
PQ2
= 102
– 62
= 100 – 36 = 64
PQ = √
PQ = 8 cm
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6. 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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7. 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 = m2
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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8. 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 = πr2
1 + πr2
2
π (r2
1+r2
2)
[( ) ( ) ]
[ ]
Area of shaded region = 192.5 cm2
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9. 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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10. ( )
( )
( )
( )
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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11. 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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12. 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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13. 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 P1
P1
QR1
is the required triangle.
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