CBSE Class 10 Mathematics Standard Question Paper 2020 Set 30-5 Solutions

Strictly Confidential - (For Internal and Restricted Use Only)
Secondary School Examination-2020
Marking Scheme - Mathematics 30/5/1, 30/5/2, 30/5/3
General instructions
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.30/5/1. 1

QUESTION PAPER CODE 30/5/1
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question numbers 1 to 20 are of 1 mark each.
Question numbers 1 to 10 are multiple choice questions.
You have to select the correct choice :
Q.No. Marks

1. On dividing a polynomial p(x) by x2 – 4, quotient and remainder are found
to be x and 3 respectively. The polynomial p(x) is
(a) 3x2 + x – 12 (b) x3 – 4x + 3 (c) x2 + 3x – 4 (d) x3 – 4x – 3
Ans: (b) x3 – 4x + 3 1
2. In Figure 1, ABC is an isosceles triangle, right-angled at C. Therefore
(a) AB2 = 2AC2 B

(b) BC2 = 2AB2
(c) AC2 = 2AB2
(d) AB2 = 4AC2 C A
2 2 Figure 1
Ans: (a) AB = 2AC 1
3. The point on the x-axis which is equidistant from (–4, 0) and (10, 0) is
(a) (7, 0) (b) (5, 0) (c) (0, 0) (d) (3, 0)
Ans: (d) (3, 0) 1
OR
The centre of a circle whose end points of a diameter are (– 6,3) and (6, 4) is
⎛ 7⎞ ⎛ 7⎞
(a) (8, –1) (b) (4, 7) (c) ⎜ 0, ⎟ (d) ⎜ 4, ⎟
⎝ 2⎠ ⎝ 2⎠
⎛ 7⎞
Ans: (c) ⎜ 0, ⎟ 1
⎝ 2⎠
4. The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has
equal roots, is
(a) 4 (b) ±4 (c) – 4 (d) 0
Ans: (b) ± 4 1
5. Which of the following is not an A.P.?
(a) –1.2, 0.8, 2.8, … (b) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , …
4 7 9 12 −1 −2 − 3
(c) , , , ,… (d) 5 , 5 , 5 , …
3 3 3 3
4 7 9 12
Ans: (c) , , , ,… 1
3 3 3 3

.30/5/1. 2

6. The pair of linear equations
3x 5y
+ = 7 and 9x + 10y = 14 is
2 3
(a) consistent (b) inconsistent
(c) consistent with one solution (d) consistent with many solutions
Ans: (b) inconsistent 1
7. In Figure 2, PQ is tangent to the circle with centre at O, at the point B.
If ∠AOB = 100°, then ∠ABP is equal to
(a) 50°
A O
(b) 40° 100°
(c) 60° Q
(d) 80° B
Ans: (a) 50° P Figure 2 1
3
8. The radius of a sphere (in cm) whose volume is 12π cm , is
(a) 3 (b) 3 3 (c) 32/3 (d) 31/3
Ans: (c) 32/3 1
9. The distance between the points (m, – n) and (–m, n) is
(a) m2 + n 2 (b) m + n

(c) 2 m 2 + n 2 (d) 2m 2 + 2n 2

Ans: (c) 2 m 2 + n 2 1
10. In Figure 3, from an external point P, two tangents PQ and PR are drawn to a
circle of radius 4 cm with centre O. If ∠QPR = 90°, then length of PQ is
(a) 3 cm Q

(b) 4 cm P
4 cm

O
(c) 2 cm R

(d) 2 2 cm
Ans: (b) 4 cm Figure 3 1
Fill in the blanks in question numbers 11 to 15.
11. The probability of an event that is sure to happen, is _________.
Ans: 1 1

1 + tan 2 A
12. Simplest form of is _________.
1 + cot 2 A
Ans: tan2A 1
13. AOBC is a rectangle whose three vertices are A(0, –3), O(0, 0) and B(4, 0).
The length of its diagonal is _________.
Ans: 5 units 1

.30/5/1. 3

⎛ ∑ fi u i ⎞
14. In the formula x = a + ⎜ ⎟ × h, ui = _________.
⎝ ∑ fi ⎠
xi − a
Ans: 1
h
15. All concentric circles are _________ to each other.
Ans: similar 1
Answer the following question numbers 16 to 20.
16. Find the sum of the first 100 natural numbers.
100
Ans: [2 + 99] = 5050 1/2+1/2
2 A
17. In Figure 4, the angle of elevation of the top of a
tower from a point C on the ground, which is
30 m away from the foot of the tower, is 30°.
Find the height of tower. 30°
C
AB 1 30 B 30 m
Ans: = ⇒ AB = m or 10 3 m Figure 4 1/2+1/2
30 3 3
18. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers
is 26, find the other.
182 × 13
Ans: = 91 1/2+1/2
26
19. Form a quadratic polynomial, the sum and product of whose zeros are
(–3) and 2 respectively.
Ans: x2 + 3x + 2 1
OR
Can (x2 – 1) be a remainder while dividing x4 – 3x2 + 5x – 9 by (x2 + 3)?
Justify your answer with reasons.
Ans: No, degree of remainder < degree of divisor 1
2 tan 45° × cos 60°
20. Evaluate :
sin 30°

1
2 × 1×
Ans: 2 =2 1/2+1/2
1
2
SECTION – B
Question numbers 21 to 26 carry 2 marks each.
21. In the given Figure 5, DE || AC and DF || AE.
BF BE
Prove that = .
EF EC

.30/5/1. 4

A

D

B F E C
Figure 5

BD BF
Ans: In ΔABE, DF || AE, ∴ = … (i) 1
AD FE
BD BE
In ΔABC, DE || AC, ∴ = … (ii) 1/2
AD EC
BF BE
From (i) and (ii) = 1/2
FE EC
22. Show that 5 + 2 7 is an irrational number, where 7 is given to be an
irrational number.
Ans: Let us assume that 5 + 2 7 is not an irrational number..

∴ 5 + 2 7 is a rational number p i.e. 5 + 2 7 = p 1
p−5
⇒ 7= 1/2
2
Which is a contradiction as RHS is a rational but LHS is irrational.
Hence 5 + 2 7 can not be rational, so irrational. 1/2

OR
Check whether 12n can end with the digit 0 for any natural number n.
Ans: Prime factors of 12 are 2 × 2 × 3 1
n
Since 5 is not a factor, so 12 can not end with 0. 1
23. If A, B and C are interior angles of a ΔABC, then show that
⎛ B+C ⎞ ⎛A⎞
cot ⎜ ⎟ = tan ⎜ ⎟ .
⎝ 2 ⎠ ⎝2⎠
B+C A
Ans: A + B + C = 180°, ∴ = 90° − 1
2 2
⎛ B+C ⎞ ⎛ A⎞ A
∴ cot ⎜ ⎟ = cot ⎜ 90° − ⎟ = tan 1
⎝ 2 ⎠ ⎝ 2⎠ 2

.30/5/1. 5

24. In Figure 6, a quadrilateral ABCD is drawn to circumscribe a circle.
Prove that AB + CD = BC + AD.
A
B

D
Figure 6 C

Ans: Let the circle touches the sides AB, BC, A P
B
CD and AD at P, Q, R
and S respectively. S 1/2
Q
∴ AP = AS ⎫
BP = BQ ⎪ D 1
DR = DS ⎬ R C
⎪
CR = CQ ⎭
adding, we get (AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ) 1/2
∴ AB + CD = BC + AD
OR
In Figure 7, find the perimeter of ΔABC, if AP = 12 cm.
A

B D
C
P Q

Figure 7

Ans: AP = AB + BP = AB + BD ⎫
⎬
AQ = AC + CQ = AC + CD ⎭ 1
⇒ AP + AQ = AB + AC + (BD + CD) = AB + AC + BC
But AP = AQ ∴ 2 AP = Perimeter of ABC 1/2
∴ Perimeter = 2(12) = 24 cm 1/2
25. Find the mode of the following distribution:
Marks: 0-10 10-20 20-30 30-40 40-50 50-60
Number of
Students: 4 6 7 12 5 6

Ans: Modal Group : 30 – 40 1/2

f1 − f 0 5
Mode = L + × h = 30 + × 10 1
2f1 − f 0 − f 2 12
= 34.17 1/2

.30/5/1. 6

26. 2 cubes, each of volume 125 cm3, are joined end to end. Find the surface
area of the resulting cuboid.
Ans: Side of cube = (125)1/3 = 5 cm 1/2
∴ Dimensions of cuboid : 10, 5, 5 1/2
S.A = 2(50 + 25 + 50) = 250 cm2 1
SECTION – C
Question numbers 27 to 34 carry 3 marks each.

1
27. A fraction becomes when 1 is subtracted from the numerator and it
3
1
becomes when 8 is added to its denominator. Find the fraction.
4
x
Ans: Let the fraction be 1/2
y
x −1 1 x 1
∴ = , = 1/2+1/2
y 3 y +8 4
⇒ 3x – y = 3 , 4x – y = 8 1/2

5
Solving to get x = 5, y = 12 ∴ Fraction is 1
12
OR
The present age of a father is three years more than three times the
age of his son. Three years hence the father’s age will be 10 years
more than twice the age of the son. Determine their present ages.
Ans: Let the present age of son be x years
∴ Father’s present age = (3x + 3) years. 1
3 years hence, Son’s age = (x + 3) years ⎫
and father’s age = (3x + 6) years ⎬ 1/2
⎭
∴ 3x + 6 = 2(x + 3) + 10 1
⇒ x = 10 ∴ Son’s age = 10 years,
Father’s age = 33 years 1/2
28. Use Euclid Division Lemma to show that the square of any positive integer
is either of the form 3q or 3q + 1 for some integer q.
Ans: Any positive integer ‘n’ can be of the form 3m, 3m + 1, 3m + 2 1
(for some integer m)
∴ n2 = (3m)2 = 9m2 = 3(3m2) = 3q, ⎫
2 2 2 2
or n = (3m + 1) = 9m + 6m + 1 = 3(3m + 2m) + 1 = 3q + 1, ⎪
⎪ 1
2 2 2 ⎬ 1
or n = (3m + 2) = 9m + 12m + 3 + 1 ⎪ 2
= 3(3m2 + 4m + 1) + 1 = 3q + 1 ⎪⎭
Hence square of any positive integer is either of the form
3q or 3q + 1 for some integer q. 1/2

.30/5/1. 7

29. Find the ratio in which the y-axis divides the line segment joining the
points (6, – 4) and (– 2, –7). Also find the point of intersection.
Ans: K:1 Let the point P(0, y) on y-axis
A B divides the line segment AB in K : 1 1
(6, -4) (0, y) (-2, -7)

−2K + 6
∴ 0= K +1 ⇒ K = 3 ∴ Ratio is 3 : 1 1

3(−7) + 1(− 4) −25 ⎛ −25 ⎞
Also, y = = ∴ Point of intersection is ⎜ 0, ⎟ 1
3+1 4 ⎝ 4 ⎠
OR
Show that the points (7, 10), (–2, 5) and (3, – 4) are vertices of an
isosceles right triangle.
Ans: Let the points be A(7, 10), B(–2, 5) and C(3, – 4)
AB = ( −2 − 7) 2 + (5 − 10) 2 = 106 1

BC = (3 + 2) 2 + ( −4 − 5) 2 = 106 1/2

AC = (3 − 7) 2 + ( −4 − 10) 2 = 212 1/2

AB = BC and AC2 = AB2 + BC2
Hence ABC is isosceles right triangle. 1
1 + sin A
30. Prove that: = sec A + tan A
1 − sin A

1 + sin A 1 + sin A
Ans: LHS = ⋅ 1
1 − sin A 1 + sin A

(1 + sin A) 2 1 + sin A
= = 1+
1
cos 2 A cos A 2

= sec A + tan A 1/2
31. For an A.P., it is given that the first term (a) = 5, common difference
(d) = 3, and the nth term (an) = 50. Find n and sum of first n terms (Sn)
of the A.P.
1
Ans: 50 = 5 + (n – 1)3 ⇒ n = 16 1+
2

16
S16 = [10 + 15 × 3] = 440 1+
1
2 2

32. Construct a ΔABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
3
Then construct a triangle whose sides are of the corresponding
4
sides of ΔABC.
Ans: Constructing ΔABC with given dimensions 1
Constructing the similar triangle. 2

.30/5/1. 8

OR
Draw a circle of radius 3.5 cm. Take a point P outside the circle at a distance of
7 cm from the centre of the circle and construct a pair of tangents to the circle
from that point.
Ans: Drawing a circle of radius 3.5 cm and centre O, and taking a point
P such that OP = 7 cm 1
Constructing two tangents. 2
33. Read the following passage and answer the questions given at the end:
Diwali Fair
A game in booth at Diwali fair involves using of spinner first. Then, if the
spinner stops at an even number, the player is allowed to pick a marble from
bag. The spinner and the marbles in the bag are represented in Figure-8
Prizes are given, when a black marble is picked. Shweta plays the game
once.

1 4
2 10
6 8
Figure 8

(i) What is the probability that she will be allowed to pick a marble from
the bag?
(ii) Suppose she is allowed to pick a marble from the bag, what is the
probability of getting a prize, when it is given that the bag contains 20
balls out of which 6 are black?
5 1
Ans: (i) P(she will be allowed to pick a marble) = 1
6 2

6 3 1
(ii) P(getting a prize) = or 1
20 10 2

6 0
Both answers or for part (ii) in Q33 are to be treated correct
20 20
as the bag contains marbles only.
34. In Figure-9, a square OPQR is inscribed in a quadrant OAQB of a circle.
If the radius of the circle is 6 2 cm, find the area of shaded region.
B
R Q

O P A
Figure 9

.30/5/1. 9

( ) ⇒ a = 6 cm
2
Ans: Let side of square be ‘a’ cm ∴ a2 + a2 = 6 2 1

90 22
( )
2 1
∴ Area of shaded region = π r − a2 = × 6 2 ⋅ − 36
2 1
1+
360 7 4 2

396 − 252 144 2
= = cm or 20.57 cm2 1/2
7 7
SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. Obtain other zeroes of the polynomial
P(x) = 2x4 – x3 – 11x2 + 5x + 5
If two of its zeroes are 5 and − 5 .

Ans: Since 5 and − 5 are zeroes of p(x), so x − 5 and x + 5 ( ) ( )
are factors of p(x). Thus (x2 – 5) is a factor of p(x). 1

( 2x − x − 11x + 5x + 5) ÷ ( x − 5) = 2x − x − 1
4 3 2 2 2
1
1
2

2x2 – x – 1 = (2x + 1) (x – 1) 1

1
∴ Other zeroes of p(x) are 1, − 1/2
2
OR
What minimum must be added to 2x – 3x2 + 6x + 7 so that the resulting
3

polynomial will be divisible by x2 – 4x + 8?
Ans:

2x + 5
x − 4x + 8 2x − 3x + 6x + 7
2 3 2

2x3 − 8x2 +16x
– + –
5x −10x + 7
2
3
5x − 20x + 40
2

– + –
10x − 33

∴ We have to add (33 – 10x) 1
36. Prove that the ratio of the areas of two similar triangles is equal to
the square of the ratio of their corresponding sides.
1
Ans: For correct Given, To Prove, Constructions and figure 2
×4 = 2

For correct proof 2

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37. Sum of the areas of 2 squares is 544 m2. If the difference of their perimeter
is 32 m, find the sides of two squares.
Ans: Let ‘a’ and ‘b’ be the sides of two squares, with a > b.
1
then a2 + b2 = 544 and 4a – 4b = 32 1
2
or a – b = 8 ∴ a = b + 8
∴ (b + 8) + b = 544 ⇒ 2b2 + 16b – 480 = 0
2 2
1
∴ b2 + 8b – 240 = 0 ⇒ (b + 20) (b – 12) = 0 ⇒ b = 12 1
b = 12 m ⇒ a = 12 + 8 = 20 m 1/2
OR
A motorboat whose speed is 18 km/hr in still water takes 1 hour more to
go 24 km upstream than to return downstream to the same spot. Find the
speed of the stream.
Ans: Let speed of the stream be x km/h
24 24
− =1 2
18 − x 18 + x
⇒ 24(2x) = 324 – x2 or x2 + 48x – 324 = 0 1
⇒ (x + 54) (x – 6) = 0 ⇒ x = 6 1
∴ Speed of the stream = 6 km/h
38. A solid toy in the form of a hemisphere surmounted by a right circular
cone of same radius. The height of the cone is 10 cm and the radius of
its base is 7 cm. Determine the volume of the toy. Also find the area of
the colored sheet required to cover the toy.
22
(Use π = and 149 = 12.2)
7
2 1
Ans: Volume of toy = π(7)3 + π(7) 2 × 10 cm 3 1
3 3
1 22
= × ×49 (14 + 10) = 1232 cm 3 1
10 3 7
7 Area of Sheet = Surface area = 2π(7)2 + π (7) 102 + 7 2 1
= 308 + 22 × 12.2 = 576.4cm2 1
39. A statue 1.6 m tall, stands on the top of a pedestal. From a point on the
ground, the angle of elevation of the top of the statute is 60° and from
the same point the angle of elevation of the top of pedestal id 45°. Find
the height of the pedestal. (Use 3 = 1.73) D

Ans: For correct figure. 1
Let h m be the height of pedestal 1.6 m
h
Then from figure, = tan 45° = 1 ⎫
x ⎪
⎬ 1+1
h+1.6 ⎪
and = tan 60° = 3 ⎭
x h

⇒
h+1.6
h
= 3 ⇒ ( 3 − 1) h = 1.6 60°
45°
A
1/2
C x

.30/5/1. 11

160
⇒ h= = 2.19 m (approx) 1/2
73
40. For the following data, draw a ‘less than’ ogive and hence find the
median of the distribution.
Age
(In years): 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Number of persons: 5 15 20 25 15 11 9

Ans: The points to be plotted for less than ogive are
(10, 5), (20, 20), (30, 40), (40, 65), (50 , 80), (60, 91), (70, 100) 2
1
Drawing the ogive 1
2
Getting median = 34 (approx) 1/2
OR
The distribution given below shows that the number of wickets taken by
bowler in one-day cricket matches. Find the mean and the median of the
number of wickets taken.
Number of wickets : 20-60 60-100 100-140 140-180 180-220 230-260
Number of bowlers : 7 5 16 12 2 3
Ans:
No. of wickets : 20-60 60-100 100-140 140-180 180-220 220-260 Sum
(fi) No. of bowlers :
( f 7 5 16 12 2 3 45
xi 40 80 120 160 200 240
ui –2 –1 0 1 2 3 1/2
fixi –14 –5 0 12 4 9 6 1/2
cf 7 12 28 40 42 45 1/2

∑ fi u i 6 × 40
Mean = a + ∑ f × h = 120 + 45 = 125.33
1
1
i 2

N
−c
22.5 − 12
Median = l + 2 × h = 100 + × 40 = 126.25 1
f 16

.30/5/1. 12

QUESTION PAPER CODE 30/5/2
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question numbers 1 to 10 are multiple choice questions of 1 mark each.
You have to select the correct choice :
Q.No. Marks

1. The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has
equal roots, is
(a) 4 (b) ±4 (c) – 4 (d) 0
Ans: (b) ± 4 1
2. Which of the following is not an A.P.?
(a) –1.2, 0.8, 2.8, … (b) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , …
4 7 9 12 −1 −2 − 3
(c) , , , ,… (d) 5 , 5 , 5 , …
3 3 3 3
4 7 9 12
Ans: (c) , , , ,… 1
3 3 3 3
3. In Figure 1, from an external point P, two tangents PQ and PR are drawn to a
circle of radius 4 cm with centre O. If ∠QPR = 90°, then length of PQ is
(a) 3 cm Q

(b) 4 cm P
4 cm

O
(c) 2 cm R

(d) 2 2 cm
Figure 1
Ans: (b) 4 cm 1
4. The distance between the points (m, – n) and (–m, n) is
(a) m2 + n 2 (b) m + n

(c) 2 m 2 + n 2 (d) 2m 2 + 2n 2

Ans: (c) 2 m 2 + n 2 1
5. The degree of the polynomial having zeroes –3 and 4 only is
(a) 2 (b) 1
(c) More than 3 (d) 3
Ans: All the three options (a), (c) and (d) are acceptable
1 mark for any of the option (a), (c) or (d) 1
6. In Figure 2, ABC is an isosceles triangle, right-angled at C. Therefore
(a) AB2 = 2AC2 B

(b) BC2 = 2AB2
(c) AC2 = 2AB2
(d) AB2 = 4AC2
C A
Ans: (a) AB2 = 2AC2 Figure 2 1

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7. The point on the x-axis which is equidistant from (–4, 0) and (10, 0) is
(a) (7, 0) (b) (5, 0) (c) (0, 0) (d) (3, 0)
Ans: (d) (3, 0) 1
OR
The centre of a circle whose end points of a diameter are (– 6,3) and (6, 4) is
⎛ 7⎞ ⎛ 7⎞
(a) (8, –1) (b) (4, 7) (c) ⎜ 0, ⎟ (d) ⎜ 4, ⎟
⎝ 2⎠ ⎝ 2⎠
⎛ 7⎞
Ans: (c) ⎜ 0, ⎟ 1
⎝ 2⎠
8. The pair of linear equations
3x 5y
+ = 7 and 9x + 10y = 14 is
2 3
(a) consistent (b) inconsistent
(c) consistent with one solution (d) consistent with many solutions
Ans: (b) inconsistent 1
9. In Figure 3, PQ is tangent to the circle with centre at O, at the point B.
If ∠AOB = 100°, then ∠ABP is equal to
(a) 50°
(b) 40° A O
100°
(c) 60° Q
(d) 80° B
P Figure 3
Ans: (a) 50° 1
3
10. The radius of a sphere (in cm) whose volume is 12π cm , is
(a) 3 (b) 3 3 (c) 32/3 (d) 31/3
Ans: (c) 32/3 1
Fill in the blanks in question numbers 11 to 15.
11. AOBC is a rectangle whose three vertices are A(0, –3), O(0, 0) and B(4, 0).
The length of its diagonal is _________.
Ans: 5 units 1
⎛ ∑ fi u i ⎞
12. In the formula x = a + ⎜ ⎟ × h, ui = _________.
⎝ ∑ fi ⎠
xi − a
Ans: 1
h
13. All concentric circles are _________ to each other.
Ans: similar 1
14. The probability of an event that is sure to happen, is _________.
Ans: 1 1

.30/5/2. 14

15. Simplest form of (1 – cos2 A) (1 + cot2 A) is _________.
Ans: 1 1
Answer the following question numbers 16 to 20.
16. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers
is 26, find the other.
182 × 13
Ans: = 91 1/2+1/2
26
17. Form a quadratic polynomial, the sum and product of whose zeros are
(–3) and 2 respectively.
Ans: x2 + 3x + 2 1
OR
Can (x2 – 1) be a remainder while dividing x4 – 3x2 + 5x – 9 by (x2 + 3)?
Justify your answer with reasons.
Ans: No, degree of remainder < degree of divisor 1
18. Find the sum of the first 100 natural numbers.
100
Ans: [2 + 99] = 5050 1/2+1/2
2
19. Evaluate:
2 sec 30° × tan 60°
2
Ans: 2 × × 3 1/2
3
=4 1/2
20. In Figure 4, the angle of elevation of the top of a A
tower from a point C on the ground, which is
30 m away from the foot of the tower, is 30°.
Find the height of the tower.
AB 1 30
Ans: = ⇒ AB = m or 10 3 m 30° C 1/2+1/2
30 3 3 B 30 m
Figure 4
SECTION – B
Question numbers 21 to 26 carry 2 marks each.
21. Find the mode of the following distribution:
Marks: 0-10 10-20 20-30 30-40 40-50 50-60
Number of
Students: 4 6 7 12 5 6

Ans: Modal class : 30 – 40 1/2

f1 − f 0 5
Mode = L + × h = 30 + × 10 1
2f1 − f 0 − f 2 12
= 34.17 1/2

.30/5/2. 15

22. In Figure 5, a quadrilateral ABCD is drawn to circumscribe a circle.
Prove that AB + CD = BC + AD.
A
B

D
Figure 5 C

Ans: Let the circle touches the sides AB, BC, A P
CD and AD at P, Q, R B
and S respectively. S 1/2
∴ AP = AS ⎫ Q
BP = BQ ⎪ ⎪ 1
⎬ D
DR = DS ⎪ R C
CR = CQ ⎪ ⎭
adding, we get (AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ) 1/2
∴ AB + CD = BC + AD
OR
In Figure 6, find the perimeter of ΔABC, if AP = 12 cm.
A

B D
C
P Q

Figure 6

Ans: AP = AB + BP = AB + BD ⎫
⎬
AQ = AC + CQ = AC + CD ⎭ 1
⇒ AP + AQ = AB + AC + (BD + CD) = AB + AC + BC
But AP = AQ ∴ 2 AP = Perimeter of ABC 1/2
∴ Perimeter = 2(12) = 24 cm 1/2
23. How many cubes of side 2 cm can be made from a solid cube of side 10 cm?
10 × 10 × 10
Ans: No. of cubes = 1
2× 2× 2
= 125 1

.30/5/2. 16

24. In the given Figure 7, DE || AC and DF || AE.
BF BE
Prove that = .
FE EC

A

D

B F E C
Figure 7

BD BF
Ans: In ΔABE, DF || AE, ∴ = … (i) 1
AD EF
BD BE
In ΔABC, DE || AC, ∴ = … (ii) 1/2
AD EC
BF BE
From (i) and (ii) = 1/2
FE EC
25. Show that 5 + 2 7 is an irrational number, where 7 is given to be an
irrational number.
Ans: Let us assume that 5 + 2 7 is not an irrational number..

∴ 5 + 2 7 is a rational number p i.e. 5 + 2 7 = p 1
p−5
⇒ 7= 1/2
2
Which is a contradiction as RHS is a rational but LHS is irrational.
Hence 5 + 2 7 can not be rational, so irrational. 1/2

OR
Check whether 12n can end with the digit 0 for any natural number n.
Ans: Prime factors of 12 are 2 × 2 × 3 1
n
Since 5 is not a factor, so 12 can not end with 0. 1
26. If A, B and C are interior angles of a ΔABC, then show that
⎛ B+C ⎞ ⎛A⎞
cot ⎜ ⎟ = tan ⎜ ⎟ .
⎝ 2 ⎠ ⎝2⎠
B+C A
Ans: A + B + C = 180°, ∴ = 90° − 1
2 2
⎛ B+C ⎞ ⎛ A⎞ A
∴ cot ⎜ ⎟ = cot ⎜ 90° − ⎟ = tan 1
⎝ 2 ⎠ ⎝ 2⎠ 2

.30/5/2. 17

SECTION – C
Question numbers 27 to 34 carry 3 marks each.
27. In Figure-8, a square OPQR is inscribed in a quadrant OAQB of a circle.
If the radius of the circle is 6 2 cm, find the area of shaded region.
B
R Q

O P A
Figure 8

( ) ⇒ a = 6 cm
2
Ans: Let side of square be ‘a’ cm ∴ a2 + a2 = 6 2 1

90 22
( )
2 1
∴ Area of shaded region = π r − a2 = × 6 2 ⋅ − 36
2 1
1+
360 7 4 2

396 − 252 144 2
= = cm or 20.57 cm2 1/2
7 7
28. Construct a ΔABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
3
Then construct a triangle whose sides are of the corresponding
4
sides of ΔABC.
Ans: Constructing ΔABC with given dimensions 1
Constructing the similar triangle. 2
OR
Draw a circle of radius 3.5 cm. Take a point P outside the circle at a
distance of 7 cm from the centre of the circle and construct a pair of
tangents to the circle from that point.
Ans: Drawing a circle of radius 3.5 cm and centre O, and taking
a point P such that OP = 7 cm 1
Constructing two tangents. 2
29. Prove that:
2 cos3 θ − cos θ
= cot θ
sin θ − 2sin 3 θ

cos θ ( 2 cos 2 θ − 1)
Ans: L.H.S. =
sin θ (1 − 2sin 2 θ )
1/2

cos θ ⎡⎣ 2 (1 − sin 2 θ ) − 1⎤⎦
sin θ (1 − 2sin 2 θ )
= 1

cos θ (1 − 2sin 2 θ )
sin θ (1 − 2sin 2 θ )
= 1

= cos θ = R.H.S. 1/2

.30/5/2. 18

1
30. A fraction becomes when 1 is subtracted from the numerator and it
3
1
becomes when 8 is added to its denominator. Find the fraction.
4
x
Ans: Let the fraction be 1/2
y
x −1 1 x 1
∴ = , = 1/2+1/2
y 3 y +8 4
⇒ 3x – y = 3 , 4x – y = 8 1/2

5
Solving to get x = 5, y = 12 ∴ Fraction is 1
12
OR
The present age of a father is three years more than three times the
age of his son. Three years hence the father’s age will be 10 years
more than twice the age of the son. Determine their present ages.
Ans: Let the present age of son be x years
∴ Father’s present age = (3x + 3) years. 1
3 years hence, Son’s age = (x + 3) years ⎫
and father’s age = (3x + 6) years ⎬ 1/2
⎭
∴ 3x + 6 = 2(x + 3) + 10 1
⇒ x = 10 ∴ Son’s age = 10 years,
Father’s age = 33 years 1/2
31. Using Euclid’s Algorithm, find the largest number which divides
870 and 258 leaving remainder 3 in each case.
Ans: HCF of (870 – 3) and (258 – 3) 1
= 867 and 255
867 = 255 × 3 + 102 ⎫
255 = 102 × 2 + 51 ⎪⎬
1
1
2
102 = 51 × 2 + 0 ⎪
⎭
∴ HCF = 51 1/2
32. Find the ratio in which the y-axis divides the line segment joining the
points (6, – 4) and (– 2, –7). Also find the point of intersection.
Ans: K:1 Let the point P(0, y) on y-axis
A B divides the line segment AB in K : 1 1
(6, -4) (0, y) (-2, -7)

−2K + 6
∴ 0= K +1 ⇒ K = 3 ∴ Ratio is 3 : 1 1

3(−7) + 1(− 4) −25 ⎛ −25 ⎞
Also, y = = ∴ Point of intersection is ⎜ 0, ⎟ 1
3+1 4 ⎝ 4 ⎠

.30/5/2. 19

OR
Show that the points A(7, 10), B(–2, 5) and C(3, – 4) are vertices of an
isosceles right triangle.
Ans: Let the points be A(7, 10), B(–2, 5) and C(3, – 4)
AB = ( −2 − 7) 2 + (5 − 10) 2 = 106 1

BC = (3 + 2) 2 + ( −4 − 5) 2 = 106 1/2

AC = (3 − 7) 2 + (−4 − 10) 2 = 212 1/2

AB = BC and AC2 = AB2 + BC2
Hence ABC is isosceles right triangle. 1
33. In an A.P. given that the first term (a) = 54, the common difference
(d) = –3, and the nth term (an) = 0. Find n and sum of first n terms (Sn)
of the A.P.
Ans: an = 0 1/2
54 + (n – 1) (–3) = 0 1
n = 19
19
S19 = (54 + 0) 1
2
= 19 × 27 = 513 1/2
34. Read the following passage and answer the questions given at the end:
Diwali Fair
A game in a booth at Diwali fair involves using of spinner first. Then, if the
spinner stops at an even number, the player is allowed to pick a marble from
bag. The spinner and the marbles in the bag are represented in Figure-9
Prizes are given, when a black marble is picked. Shweta plays the game
once.

1 4
2 10
6 8
Figure 9

(i) What is the probability that she will be allowed to pick a marble from
the bag?
(ii) Suppose she is allowed to pick a marble from the bag, what is the
probability of getting a prize, when it is given that the bag contains 20
balls out of which 6 are black?
5 1
Ans: (i) P(she will be allowed to pick a marble) = 1
6 2

6 3 1
(ii) P(getting a prize) = or 1
20 10 2

6 0
Both answers or for part (ii) in Q34 are to be treated correct
20 20
as the bag contains marbles only.

.30/5/2. 20

SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. Sum of the areas of 2 squares is 544 m2. If the difference of their perimeter
is 32 m, find the sides of two squares.
Ans: Let ‘a’ and ‘b’ be the sides of two squares, with a > b.
1
then a2 + b2 = 544 and 4a – 4b = 32 1+
2
or a – b = 8 ∴ a = b + 8 1
∴ (b + 8) + b = 544 ⇒ 2b + 16b – 480 = 0
2 2 2
1
∴ b2 + 8b – 240 = 0 ⇒ (b + 20) (b – 12) = 0 ⇒ b = 12
b = 12 m ⇒ a = 12 + 8 = 20 m 1/2
OR
A motorboat whose speed is 18 km/hr in still water takes 1 hour more to
go 24 km upstream than to return downstream to the same spot. Find the speed
of the stream.
Ans: Let speed of the stream be x km/h
24 24
− =1 2
18 − x 18 + x
⇒ 24(2x) = 324 – x2 or x2 + 48x – 324 = 0 1
⇒ (x + 54) (x – 6) = 0 ⇒ x = 6 1
∴ Speed of the stream = 6 km/h
36. A solid toy in the form of a hemisphere surmounted by a right circular
cone of same radius. The height of the cone is 10 cm and the radius of
its base is 7 cm. Determine the volume of the toy. Also find the area of
the coloured sheet required to cover the toy.
22
(Use π = and 149 = 12.2)
7
2 1
Ans: Volume of toy = π(7)3 + π(7) 2 × 10 cm 3 1
3 3
1 22
10 = × ×49 (14 + 10) = 1232 cm 3 1
3 7
7
Area of Sheet = Surface area = 2π(7)2 + π (7) 10 2 + 7 2 1
= 308 + 22 × 12.2 = 576.4cm2 1
37. For the following data, draw a ‘less than’ ogive and hence find the
median of the distribution.
Age
(In years): 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Number of persons: 5 15 20 25 15 11 9

.30/5/2. 21

Ans: The points to be plotted for less than ogive are
(10, 5), (20, 20), (30, 40), (40, 65), (50 , 80), (60, 91), (70, 100) 2
1
Drawing the ogive 1+
2
Getting median = 34 (approx) 1/2
OR
The distribution given below shows that the number of wickets taken by
bowler in one-day cricket matches. Find the mean and the median of the
number of wickets taken.
Number of wickets : 20-60 60-100 100-140 140-180 180-220 220-260
Number of bowlers : 7 5 16 12 2 3
Ans:
No. of wickets : 20-60 60-100 100-140 140-180 180-220 220-260 Sum
(fi) No. of bowlers : 7 5 16 12 2 3 45
xi 40 80 120 160 200 240
ui –2 –1 0 1 2 3 1/2
fixi –14 –5 0 12 4 9 6 1/2
cf 7 12 28 40 42 45 1/2

∑ fi u i 6 × 40
Mean = a + ∑ f × h = 120 + 45 = 125.33
1
1+
i 2

N
−c
2 22.5 − 12
Median = l + × h = 100 + × 40 = 126.25 1
f 16
38. From a point on the ground, the angles of elevation of the bottom and
the top of a transmission tower fixed at the top of a 20 m high building
are 45° and 60° respectively. Find the height of the

(
tower Use 3 = 1.73 )
Ans: Let h be the height of the tower cor. fig. 1
In right ΔABD C

20
= tan 45°⎫
x ⎬ h 1
x = 20 m ⎭
A
In right ΔCBD
h + 20
= tan 60° 1
20 20 m
h + 20 = 20 3 60°

( 3 − 1)
45°
h = 20 B x D 1/2
= 20 × 0.73 = 14.60 m 1/2

.30/5/2. 22

39. Prove that in a right-angled triangle, the square of the hypotenuse is
equal to the sum of the squares of the two sides.
1
Ans: For correct Given, To Prove, Construction and figure ×4 = 2
2
For correct proof 2
40. Obtain other zeroes of the polynomial
P(x) = 2x4 – x3 – 11x2 + 5x + 5
If two of its zeroes are 5 and − 5 .

Ans: Since 5 and − 5 are zeroes of p(x), so x − 5 and x + 5 ( ) ( )
are factors of p(x). Thus (x2 – 5) is a factor of p(x). 1

( 2x − x − 11x + 5x + 5) ÷ ( x − 5) = 2x − x − 1
4 3 2 2 2
1
1
2

2x2 – x – 1 = (2x + 1) (x – 1) 1
1
∴ Other zeroes of p(x) are 1, − 1/2
2
OR
What minimum must be added to 2x – 3x2 + 6x + 7 so that the resulting
3

polynomial will be divisible by x2 – 4x + 8?
Ans:

2x + 5
x − 4x + 8 2x − 3x + 6x + 7
2 3 2

2x3 − 8x2 +16x
– + –
5x 2 −10x + 7
3
5x − 20x + 40
2
– + –
10x − 33

∴ We have to add (33 – 10x) 1

.30/5/2. 23

QUESTION PAPER CODE 30/5/3
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question numbers 1 to 10 are multiple choice questions of 1 mark each.
You have to select the correct choice :
Q.No. Marks

1. The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has
equal roots, is
(a) 4 (b) ±4 (c) – 4 (d) 0
Ans: (b) ± 4 1
2. Which of the following is not an A.P.?
(a) –1.2, 0.8, 2.8, … (b) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , …
4 7 9 12 −1 −2 − 3
(c) , , , ,… (d) 5 , 5 , 5 , …
3 3 3 3
4 7 9 12
Ans: (c) , , , ,… 1
3 3 3 3
3. The radius of a sphere (in cm) whose volume is 12π cm3, is
(a) 3 (b) 3 3 (c) 32/3 (d) 31/3
Ans: (c) 32/3 1
4. The distance between the points (m, – n) and (–m, n) is
(a) m2 + n 2 (b) m + n

(c) 2 m 2 + n 2 (d) 2m 2 + 2n 2

Ans: (c) 2 m 2 + n 2 1
5. In Figure 1, from an external point P, two tangents PQ and PR are drawn to a
circle of radius 4 cm with centre O. If ∠QPR = 90°, then length of PQ is
(a) 3 cm
(b) 4 cm
A O
(c) 2 cm 100°
(d) 2 2 cm Q
B
Ans: (b) 4 cm 1
P Figure 1

6. On dividing a polynomial p(x) by x2 – 4, quotient and remainder are found
to be x and 3 respectively. The polynomial p(x) is
(a) 3x2 + x – 12 (b) x3 – 4x + 3 (c) x2 + 3x – 4 (d) x3 – 4x – 3
Ans: (b) x3 – 4x + 3 1

.30/5/3. 24

AD 3
7. In Figure 2, DE || BC. If = and AE = 2.7 cm, then EC is equal to
DB 2
A
(a) 2.0 cm
(b) 1.8 cm
(c) 4.0 cm D E
(d) 2.7 cm C
B
Ans: (b) 1.8 cm Figure 2 1
8. The point on the x-axis which is equidistant from (–4, 0) and (10, 0) is
(a) (7, 0) (b) (5, 0) (c) (0, 0) (d) (3, 0)
Ans: (d) (3, 0) 1
OR
The centre of a circle whose end points of a diameter are (– 6,3) and (6, 4) is
⎛ 7⎞ ⎛ 7⎞
(a) (8, –1) (b) (4, 7) (c) ⎜ 0, ⎟ (d) ⎜ 4, ⎟
⎝ 2⎠ ⎝ 2⎠
⎛ 7⎞
Ans: (c) ⎜ 0, ⎟ 1
⎝ 2⎠
9. The pair of linear equations
3x 5y
+ = 7 and 9x + 10y = 14 is
2 3
(a) consistent (b) inconsistent
(c) consistent with one solution (d) consistent with many solutions
Ans: (b) inconsistent 1
10. In Figure 3, PQ is tangent to the circle with centre at O, at the point B.
If ∠AOB = 100°, then ∠ABP is equal to
(a) 50°
(b) 40° A O
100°
(c) 60° Q
(d) 80° B
Ans: (a) 50° P Figure 3 1

Fill in the blanks in question numbers 11 to 15.
1 + tan 2 A
11. Simplest form of is _________.
1 + cot 2 A
Ans: tan2A 1
12. ( )
If the probability of an event E happening is 0.023, then P E = _________.
Ans: 0.977 1
13. All concentric circles are _________ to each other.
Ans: similar 1

.30/5/3. 25

14. The probability of an event that is sure to happen, is _________.
Ans: 1 1
15. AOBC is a rectangle whose three vertices are A(0, –3), O(0, 0) and B(4, 0).
The length of its diagonal is _________.
Ans: 5 units 1
Answer the following question numbers 16 to 20.
16. Write the value of sin2 30° + cos260°.
2 2
⎛1⎞ ⎛1⎞
Ans: ⎜ ⎟ + ⎜ ⎟ 1/2
⎝2⎠ ⎝2⎠
1
= 1/2
2
17. Form a quadratic polynomial, the sum and product of whose zeros are
(–3) and 2 respectively.
Ans: x2 + 3x + 2 1
OR
Can (x2 – 1) be a remainder while dividing x4 – 3x2 + 5x – 9 by (x2 + 3)?
Justify your answer with reasons.
Ans: No, degree of remainder < degree of divisor 1
18. Find the sum of the first 100 natural numbers.
100
Ans: [2 + 99] = 5050 1/2+1/2
2
19. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers
is 26, find the other.
182 × 13
Ans: = 91 1/2+1/2
26
A
20. In Figure 4, the angle of elevation of the top of a
tower from a point C on the ground, which is
30 m away from the foot of the tower, is 30°.
Find the height of tower.
AB 1 30
Ans: = ⇒ AB = m or 10 3 m 30° C 1/2+1/2
30 3 3 B 30 m
SECTION – B
Question numbers 21 to 26 carry 2 marks each.
21. A cone and a cylinder have the same radii but the height of the cone
is 3 times that of the cylinder. Find the ratio of their volumes.
1 2
πr (3h)
Ans: 3 1
πr 2 h
1
= =1:1 1
1

.30/5/3. 26

22. In Figure 5, a quadrilateral ABCD is drawn to circumscribe a circle.
Prove that AB + CD = BC + AD.
A
B

D
Figure 5 C

Ans: Let the circle touches the sides AB, BC, A P
CD and AD at P, Q, R B
and S respectively. S 1/2
∴ AP = AS ⎫ Q 1
BP = BQ ⎪
⎬ D
DR = DS ⎪ R C
CR = CQ ⎭
adding, we get (AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ) 1/2
∴ AB + CD = BC + AD
OR
In Figure 6, find the perimeter of ΔABC, if AP = 12 cm.
A

B D
C
P Q

Figure 6

Ans: AP = AB + BP = AB + BD ⎫
⎬
AQ = AC + CQ = AC + CD ⎭ 1
⇒ AP + AQ = AB + AC + (BD + CD) = AB + AC + BC
But AP = AQ ∴ 2 AP = Perimeter of ABC 1/2
∴ Perimeter = 2(12) = 24 cm 1/2
23. Find the mode of the following distribution:
Marks: 0-10 10-20 20-30 30-40 40-50 50-60
Number of
Students: 4 6 7 12 5 6

Ans: Modal Group : 30 – 40 1/2

f1 − f 0 5
Mode = L + 2f − f − f × h = 30 + 12 × 10 1
1 0 2

= 34.17 1/2

.30/5/3. 27

24. In the Figure 7, if PQ || BC and PR || CD, prove that QB = DR .
AQ AR

B

Q

A P
C
R
D
Figure 7

QB PC
Ans: = …(i) 1
AQ AP
PC DR
= …(ii) 1/2
AP AR
From (i) and (ii)
QB DR
= 1/2
AQ AR
25. Show that 5 + 2 7 is an irrational number, where 7 is given to be an
irrational number.
Ans: Let us assume that 5 + 2 7 is not an irrational number..

∴ 5 + 2 7 is a rational number p i.e. 5 + 2 7 = p 1
p−5
⇒ 7= 1/2
2
Which is a contradiction as RHS is a rational but LHS is irrational.
Hence 5 + 2 7 can not be rational, so irrational. 1/2

OR
Check whether 12n can end with the digit 0 for any natural number n.
Ans: Prime factors of 12 are 2 × 2 × 3 1
Since 5 is not a factor, so 12n cannot end with 0. 1
26. If A, B and C are interior angles of a ΔABC, then show that
⎛ B+C ⎞ ⎛A⎞
cot ⎜ ⎟ = tan ⎜ ⎟ .
⎝ 2 ⎠ ⎝2⎠
B+C A
Ans: A + B + C = 180°, ∴ = 90° − 1
2 2
⎛ B+C ⎞ ⎛ A⎞ A
∴ cot ⎜ ⎟ = cot ⎜ 90° − ⎟ = tan 1
⎝ 2 ⎠ ⎝ 2⎠ 2

.30/5/3. 28

SECTION – C
Question numbers 27 to 34 carry 3 marks each.
27. Prove that:

( sin θ − cos θ + 1) co sec θ = 2
4 4 2

Ans: L.H.S = ⎡⎣( sin θ + cos θ )( sin θ − cos θ ) + 1⎤⎦ co sec θ
2 2 2 2 2
1

= ⎡⎣1( sin θ − cos θ ) − 1⎤⎦ co sec θ
2 2 2
1
= 2 sin2θ × cosec2θ = 2 1
28. Find the sum:
(–5) + (–8) + (–11) + …+ (–230)
Ans: a = –5, d = –3, an = –230 1/2
⇒ – 5 + (n – 1) × (–3) = 230 1
225
(n – 1) = = 75
3
n = 76 1/2
76
S76 = [ −5 + (−230)] 1/2
2
= 38(–235) = –8930 1/2
29. Construct a ΔABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
3
Then construct a triangle whose sides are of the corresponding
4
sides of ΔABC.
Ans: Constructing ΔABC with given dimensions 1
Constructing the similar triangle. 2
OR
Draw a circle of radius 3.5 cm. Take a point P outside the circle at a distance of
7 cm from the centre of the circle and construct a pair of tangents to the circle
from that point.
Ans: Drawing a circle of radius 3.5 cm and centre O, and taking a point
P such that OP = 7 cm 1
Constructing two tangents. 2
30. In Figure-8, ABCD is a parallelogram. A semicircle with centre O and
the diameter AB has been drawn and it passes through D. If AB = 12 cm
and OD ⊥ AB, then find the area of the shaded region. (Use π = 3.14)

D
C

A O B
Figure 8

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Ans: Area of shaded portion = Ar of ||gm – Ar of Quadrant 1
1
= 12 × 6 – × 3.14 × 6 × 6 1
4
= 43.75 cm2 1
31. Read the following passage and answer the questions given at the end:
Diwali Fair
A game in a booth at Diwali fair involves using of spinner first. Then, if the
spinner stops at an even number, the player is allowed to pick a marble from
bag. The spinner and the marbles in the bag are represented in Figure-9
Prizes are given, when a black marble is picked. Shweta plays the game
once.

1 4
2 10
6 8
Figure 9

(i) What is the probability that she will be allowed to pick a marble from
the bag?
(ii) Suppose she is allowed to pick a marble from the bag, what is the
probability of getting a prize, when it is given that the bag contains 20
balls out of which 6 are black?
5 1
Ans: (i) P(she will be allowed to pick a marble) = 1
6 2

6 3 1
(ii) P(getting a prize) = or 1
20 10 2

6 0
Both answers or for part (ii) in Q31 are to be treated correct
20 20
as the bag contains marbles only.

1
32. A fraction becomes when 1 is subtracted from the numerator and it
3
1
becomes when 8 is added to its denominator. Find the fraction.
4

Ans: Let the fraction be x 1/2
y
x −1 1 x 1
∴ = , = 1/2+1/2
y 3 y +8 4
⇒ 3x – y = 3 , 4x – y = 8 1/2

5
Solving to get x = 5, y = 12 ∴ Fraction is 1
12

.30/5/3. 30

OR
The present age of a father is three years more than three times the
age of his son. Three years hence the father’s age will be 10 years
more than twice the age of the son. Determine their present ages.
Ans: Let the present age of son be x years
∴ Father’s present age = (3x + 3) years. 1
3 years hence, Son’s age = (x + 3) years ⎫
and father’s age = (3x + 6) years ⎬ 1/2
⎭
∴ 3x + 6 = 2(x + 3) + 10 1
⇒ x = 10 ∴ Son’s age = 10 years,
Father’s age = 33 years 1/2
33. Find the ratio in which the y-axis divides the line segment joining the
points (6, – 4) and (– 2, –7). Also find the point of intersection.
Ans: K:1 Let the point P(0, y) on y-axis
A B divides the line segment AB in K : 1 1
(6, -4) (0, y) (-2, -7)

−2K + 6
∴ 0= K +1 ⇒ K = 3 ∴ Ratio is 3 : 1 1

3(−7) + 1(− 4) −25 ⎛ −25 ⎞
Also, y = = ∴ Point of intersection is ⎜ 0, ⎟ 1
3+1 4 ⎝ 4 ⎠
OR
Show that the points (7, 10), (–2, 5) and (3, – 4) are vertices of an
isosceles right triangle.
Ans: Let the points be (7, 10), (–2, 5) and (3, – 4)
AB = ( −2 − 7) 2 + (5 − 10) 2 = 106 1

BC = (3 + 2) 2 + ( −4 − 5) 2 = 106 1/2

AC = (3 − 7) 2 + ( −4 − 10) 2 = 212 1/2

AB = BC and AC2 = AB2 + BC2
Hence ABC is isosceles right triangle. 1
34. Use Euclid Division Lemma to show that the square of any positive integer
is either of the form 3q or 3q + 1 for some integer q.
Ans: Any positive integer ‘n’ can be of the form 3m, 3m + 1, 3m + 2 1
∴ n2 = (3m)2 = 9m2 = 3(3m2) = 3q, ⎫
or n = (3m + 1) = 9m + 6m + 1 = 3(3m + 2m) + 1 = 3q + 1, ⎪⎪
2 2 2 2
1
1
⎬ 2
or n2 = (3m + 2)2 = 9m2 + 12m + 3 + 1 ⎪
= 3(3m2 + 4m + 1) + 1 = 3q + 1 ⎪⎭
Hence square of any positive integer is either of the form
3q or 3q + 1 for some integer q. 1/2

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SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. Sum of the areas of 2 squares is 544 m2. If the difference of their perimeter
is 32 m, find the sides of two squares.
Ans: Let ‘a’ and ‘b’ be the sides of two squares, with a > b.
1
then a2 + b2 = 544 and 4a – 4b = 32 1
2
or a – b = 8 ∴ a = b + 8 1
∴ (b + 8) + b = 544 ⇒ 2b + 16b – 480 = 0
2 2 2
1
∴ b2 + 8b – 240 = 0 ⇒ (b + 20) (b – 12) = 0 ⇒ b = 12
b = 12 m ⇒ a = 12 + 8 = 20 m 1/2
OR
A motorboat whose speed is 18 km/hr in still water takes 1 hour more to
go 24 km upstream than to return downstream to the same spot. Find the speed
of the stream.
Ans: Let speed of the stream be x km/h
24 24
− =1 2
18 − x 18 + x
⇒ 24(2x) = 324 – x2 or x2 + 48x – 324 = 0 1
⇒ (x + 54) (x – 6) = 0 ⇒ x = 6 1
∴ Speed of the stream = 6 km/h
36. For the following data, draw a ‘less than’ ogive and hence find the
median of the distribution.
Age
(In years): 0-10 10-20 20-30 0-40 40-50 50-60 60-70
Number of persons: 5 15 20 25 15 11 9

Ans: The points to be plotted for less than ogive are
(10, 5), (20, 20), (30, 40), (40, 65), (50 , 80), (60, 91), (70, 100) 2
1
Drawing the ogive 1
2
Getting median = 34 (approx) 1/2
OR
The distribution given below shows that the number of wickets taken by
bowler in one-day cricket matches. Find the mean and the median of the
number of wickets taken.
Number of wickets : 20-60 60-100 100-140 140-180 180-220 220-260
Number of bowlers : 7 5 16 12 2 3

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No. of wickets : 20-60 60-100 100-140 140-180 180-220 220-260 Sum
(fi) No. of bowlers :
( f 7 5 16 12 2 3 45
xi 40 80 120 160 200 240
ui –2 –1 0 1 2 3 1/2
fixi –14 –5 0 12 4 9 6 1/2
cf 7 12 28 40 42 45 1/2
∑ fi u i 6 × 40
Mean = a + ∑ f × h = 120 + 45 = 125.33
1
1
i
2

N
−c
22.5 − 12
Median = l + 2 × h = 100 + × 40 = 126.25 1
f 16
37. A statue 1.6 m tall, stands on the top of a pedestal. From a point on the
ground, the angle of elevation of the top of the statute is 60° and from
the same point the angle of elevation of the top of pedestal is 45°. Find
the height of the pedestal. (Use 3 = 1.73)
Ans: For correct figure. D 1
Let h m be the height of pedestal
h
Then from figure, = tan 45° = 1 ⎫⎪ 1.6 m
x
⎬ 1+1
h+1.6 ⎪
and = tan 60° = 3 ⎭
x

⇒ ( 3 − 1) h = 1.6 h 1/2

160 60°
⇒ h= = 2.19 m (approx) 45° 1/2
73 C x A
38. Obtain other zeroes of the polynomial
P(x) = 2x4 – x3 – 11x2 + 5x + 5
If two of its zeroes are 5 and − 5 .

Ans: Since (
5 and − 5 are zeroes of p(x), so x − 5 and x + 5 ) ( )
are factors of p(x). Thus (x2 – 5) is a factor of p(x). 1

( 2x − x − 11x + 5x + 5) ÷ ( x − 5) = 2x − x − 1
4 3 2 2 2
1
1
2

2x2 – x – 1 = (2x + 1) (x – 1) 1

1
∴ Other zeroes of p(x) are 1, − 1/2
2

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OR
What minimum must be added to 2x – 3x2 + 6x + 7 so that the resulting
3

polynomial will be divisible by x2 – 4x + 8?
Ans:

2x + 5
x 2 − 4x + 8 2x3 − 3x2 + 6x + 7
2x3 − 8x2 +16x
– + –
5x −10x + 7
2
3
5x − 20x + 40
2

– + –
10x − 33

∴ We have to add (33 – 10x) 1
39. In a cylindrical vessel of radius 10 cm, containing some water, 9000
small spherical balls are dropped which are completely immersed in
water which raises the water level. If each spherical ball is of radius
0.5 cm, then find the rise in the level of water in the vessel.
Ans: Volume of raised water in cylinder = Volume of 9000 spherical balls 1
4
π (10) 2H = 9000 × × π × (0.5)
3
2
3
∴ H = 15 cm 1
40. If a line is drawn parallel to one side of a triangle to intersect the other
two sides at distinct points, prove that the other two sides are divided in
the same ratio.
1
Ans: For correct Given, To Prove, Constructions and figure 2
×4 = 2

For correct proof 2

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