CBSE Class 10 Mathematics Standard Question Paper 2020 Set 30-1 Solutions

Strictly Confidential - (For Internal and Restricted Use Only)
Secondary School Examination-2020
Marking Scheme - Mathematics 30/1/1, 30/1/2, 30/1/3
General instructions
1. You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A
small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education
system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and
understand the spot evaluation guidelines carefully. Evaluation is a 10-12 days mission for all of us. Hence, it is
necessary that you put in your best effortsin this process.
2. Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one's
own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed.
However, while evaluating, answers which are based on latest information or knowledge and/or are innovative,
they may be assessed for their correctness otherwise and marks be awarded to them. In class-X, while evaluating
two competency based questions, please try to understand given answer and even if reply is not from marking
scheme but correct competency is enumerated by the candidate, marks should be awarded.
3. The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to
ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. The remaining
answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the
marking of individual evaluators.
4. Evaluators will mark( √ ) wherever answer is correct. For wrong answer 'X"be marked. Evaluators will not put right
kind of mark while evaluating which gives an impression that answer is correct and no marks are awarded. This is
most common mistake which evaluators are committing.
5. If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of
the question should then be totaled up and written in the left-hand margin and encircled. This may be followed
strictly.
6. If a question does not have any parts, marks must be awarded in the left-hand margin and encircled. This may also be
followed strictly.
7. If a student has attempted an extra question, answer of the question deserving more marks should be retained and the
other answer scored out.
8. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
9. A full scale of marks __________(example 0-100 marks as given in Question Paper) has to be used. Please do not
hesitate to award full marks if the answer deserves it.
10. Every examiner has to necessarily do evaluation work for full working hours i.e. 8 hours every day and evaluate 20
answer books per day in main subjects and 25 answer books per day in other subjects (Details are given in Spot
Guidelines).
11. Ensure that you do not make the following common types of errors committed by the Examiner in the past:-
• Leaving answer or part thereof unassessed in an answer book.
• Giving more marks for an answer than assigned to it.
• Wrong totaling of marks awarded on a reply.
• Wrong transfer of marks from the inside pages of the answer book to the title page.
• Wrong question wise totaling on the title page.
• Wrong totaling of marks of the two columns on the title page.
• Wrong grand total.
• Marks in words and figures not tallying.
• Wrong transfer of marks from the answer book to online award list.
• Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly
indicated. It should merely be a line. Same is with the X for incorrect answer.)
• Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
12. While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as cross (X) and
awarded zero (0)Marks.
13. Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall
damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to
uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously.
14. The Examiners should acquaint themselves with the guidelines given in the Guidelines for spot Evaluation before
starting the actual evaluation.
15. Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly
totaled and written in figures and words.
16. The Board permits candidates to obtain photocopy of the Answer Book on request in an RTI application and also
separately as a part of the re-evaluation process on payment of the processing charges.

.30/1/1. 1

QUESTION PAPER CODE 30/1/1
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Q. No. 1 to 10 are multiple choice type question of 1 mark each.
Select the correct option.
Q.No. Marks

1. If one of the zeroes of the quadratic polynomial x2 + 3x + k is 2, then the value
of k is
(a) 10 (b) –10 (c) –7 (d) –2
Ans: (b) – 10 1
2. The total number of factors of a prime number is
(a) 1 (b) 0 (c) 2 (d) 3
Ans: (c) 2 1
3. The quadratic polynomial, the sum of whose zeroes is –5 and their product is
6, is
(a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2 – 5x – 6 (d) –x2 + 5x + 6
Ans: (a) x2 + 5x + 6 1
4. The value of k for which the system of equations x + y – 4 = 0 and
2x + ky =3 has no solution, is
(a) –2 (b) ≠2 (c) 3 (d) 2
Ans: (d) 2 1
5. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3,140 (b) 12,420 (c) 3,420 (d) 420,3
Ans: (c) 3,420 1
6. The value of x for which 2x,(x + 10) and ( 3x + 2 ) are the three
consecutive terms of an AP, is
(a) 6 (b) –6 (c) 18 (d) –18
Ans: (a) 6 1
7. The first term of an AP is p and the common difference is q, then its 10th term is
(a) q + 9p (b) p – 9q (c) p + 9q (d) 2p + 9q
Ans: (c) p + 9q 1
8. The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ – b cos θ), is
(a) a2 + b2 (b) a2 – b2 (c) a 2 +b 2 (d) a 2 − b2
Ans: (c) a 2 + b2 1
9. If the point P(k, 0) divides the line segment joining the points A(2, –2) and
B(–7, 4) in the ratio 1 : 2, then the value of k is,
(a) 1 (b) 2 (c) –2 (d) –1
Ans: (d) –1 1
10. The value of p, for which the points A(3, 1), B(5, p) and C(7, –5) are collinear, is
(a) –2 (b) 2 (c) –1 (d) 1
Ans: (a) –2 1

.30/1/1. 2

In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.
11. In Fig. 1, Δ ABC is circumscribing a circle, the length of BC is _______cm.

A

4 cm
11 cm

P R
3 cm

B Q C
Fig. 1

Ans: 10 1

AB 1 ar ( ΔABC )
12. Given Δ ABC ~ Δ PQR, if = , then = __________.
PQ 3 ar ( ΔPQR )
1
Ans: 1
9
13. ABC is an equilateral triangle of side 2a, then length of one of its
altitude is _______.
Ans: 3a 1
cos 80°
14. + cos 59° cosec 31° = _______.
sin 10°
Ans: 2 1
⎛ 1 ⎞
15. The value of ⎜ sin 2 θ + ⎟ = _________.
⎝ 1 + tan θ ⎠
2

Ans: 1 1
OR
The value of (1 + tan2θ) (1 – sin θ) (1 + sin θ) =_______.
Ans: 1 1

Q. Nos. 16 to 20 are short answer type questions of 1 mark each.
16. The ratio of the length of a vertical rod and the length of its shadow
is 1: 3 . Find the angle of elevation of the sun at that moment?
1
Ans: tan θ = ⇒ θ = 30° 1/2+1/2
3
17. Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1.
What is the ratio of their volumes?
r1 3 h1 1
Ans: r = 1 , h = 3 1/2
2 2

.30/1/1. 3

1 2
πr1 h1
3 = 3 :1
∴ Ratio of volumes = 1 2 1/2
πr2 h 2
3
18. A letter of English alphabet is chosen at random. What is the probability
that the chosen letter is a consonant.
21
Ans: P (consonant) = 1
26
19. A die is thrown once. What is the probability of getting a number less than 3?
2 1
Ans: P (number less than 3) = or 1
6 3
OR
If the probability of winning a game is 0.07, what is the probability of losing it?
Ans: P (losing) = 1 – 0.07 1/2
= 0.93 1/2
20. If the mean of first n natural number is 15, then find n.
n(n + 1)
Ans: 2 = 15 1/2
n
∴ n = 29 1/2
SECTION – B
Q. Nos. 21 to 26 carry 2 marks each.
21. Show that (a – b)2, (a2 + b2) and (a + b)2 are in AP.

Ans: ( a + b ) − ( a − b ) = 2ab
2 2 2
1

( a + b )2 − (a 2 + b2 ) = 2ab 1
Common difference is same. ∴ given terms are in AP
BE BC
22. In Fig. 2 DE ||AC and DC ||AP. Prove that = .
EC CP
A

D

B E C P
Fig 2

BD BE
Ans: In ΔABC, DE || AC, ∴ = ... (i) 1
DA EC
BD BC
In ΔABP, DC || AP, ∴ = ... (ii) 1/2
DA CP
BE BC
From (i) & (ii), = 1/2
EC CP

.30/1/1. 4

OR
In Fig. 3, two tangents TP and TQ are drawn to a circle with centre O from
an external point T. Prove that ∠PTQ = 2∠OPQ.

P

T O

Fig 3 Q

Ans: Let ∠OPQ = θ
∴ ∠TPQ = ∠TQP = 90° – θ 1/2
In ΔTPQ, 2(90° – θ) + ∠PTQ = 180° 1
∴ ∠PTQ = 2θ 1/2
= 2∠OPQ
23. The rod AC of a TV disc antenna is fixed at right
angle to the wall AB and a rod CD is supporting A
C
the disc as shown in Fig. 4. If AC = 1.5m long
and CD = 3m, find (i) tanθ (ii) secθ + cosecθ. θ
AC 1
Ans: = sin θ ⇒ sin θ = ⇒ θ = 30° D 1/2
CD 2 B
1 Fig 4
(i) tan θ = tan 30° = Wall
1/2
3
(ii) sec θ + cosecθ = sec30° + cos ec30°
2 2(3 + 3)
= + 2 or 1
3 3
24. A piece of wire 22 cm long is bent into the form of an arc of circle
subtending an angle of 60° at its centre. Find the radius of the circle.
⎛ 22 ⎞
⎜ Use π = ⎟
⎝ 7 ⎠

22 60°
Ans: 2 × ×r× = 22 1
7 360°
∴ r = 21 cm 1
25. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3.
What is the probability that x2 ≤ 4?
Ans: Total number of outcomes = 7
Favourable outcomes are –2, –1, 0, 1, 2, i.e., 5 1

5
∴ P(x 2 ≤ 4) = 1
7

.30/1/1. 5

26. Find the mean of the following distribution:
Class: 3-5 5-7 7-9 9-11 11-13
Frequency: 5 10 10 7 8

Ans: Classes xi fi fxxi 1½
3–5 4 5 20
5–7 6 10 60
7–9 8 10 80
9 – 11 10 7 70
11 – 13 12 8 96
Total 40 326

x=
∑ fi xi = 326 = 8.15
∑ fi 40 1/2
OR
Find the mode of the following data:
Class: 0-20 20-40 40-60 60-80 80-100 110-120 120-140
Frequency: 6 8 10 12 6 5 3
Ans: Modal class : 60 – 80 1/2
f1 − f 0 12 − 10
Mode = l + 2f − f − f × h = 60 + 24 − 10 − 6 × 20 1
1 0 2
= 60 + 5 = 65 1/2
SECTION – C
Question numbers 27 to 34 carry 3 marks each.
27. Find the quadratic polynomial whose zeroes are reciprocal of the zeroes
of the polynomial f (x) = ax2 + bx + c, a ≠ 0, c ≠ 0.
Ans: f (x) = ax 2 + bx + c
b c
α + β = − , αβ = 1/2
a a
1 1 b
New sum of zeroes = + =− 1
α β c
1 1 a
New product of zeroes = × = 1
α β c

∴ Required quardratic polynomial = x +
2 b
c
a
(
x + or cx 2 + bx + a
c
) 1/2

.30/1/1. 6

OR
Divide the polynomial f(x) = 3x2 – x3 – 3x + 5 by the polynomial
g(x) = x – 1 – x2 and verify the division algorithm.
Ans: − x 2 + x − 1 − x 3 + 3x 2 − 3x + 5 x−2 2
−x 3 + x 2 − x
+ – +
2x 2 − 2x + 5
2x 2 − 2x + 2
– + –
3
Divisor × Quotient + Remainder
= ( − x 2 + x − 1)(x − 2) + 3
= − x 3 + 3x 2 − 3x + 5 = Dividend 1
28. Determine graphically the coordinates of the vertices of a triangle,
the equations of whose sides are given by 2y – x = 8, 5y – x = 14
and y – 2x = 1.

Ans:
y
2y–x=8
6
5 C(2,5) 5y–x=14
(0,4) 4
A(–4,2) 3 (6,4)
B(1,3)
2
1 (0,1)
x′ x 2y – x = 8
–6 –5 –4 –3 –2 –1 O1 2 3 4 5 6
–1
–2
x 0 2 –4
y 4 5 2

y–2x=1 5y – x = 14
x 1 6 –4
y′ y 3 4 2
y – 2x = 1
x 1 2 0
y 3 5 1
1
Drawing 3 lines 1
2

Coordinates of the vertices of the triangle are A(– 4, 2),
1
B(1, 3) and C(2, 5) 1
2

.30/1/1. 7

OR
If 4 is the zero of the cubic polynomial x3 – 3x2 – 10x + 24,
find its other two zeroes.
Ans: x – 4 is a factor of given polynomial.

x−4 x 3 − 3x 2 − 10x + 24 x 2 + x − 6
x 3 – 4x 2
– +
x 2 − 10x + 24
x 2 – 4x 2
– +
−6x + 24
−6x + 24
+ –
0

x 2 + x − 6 = (x + 3)(x − 2)
∴ Other than zeroes are –3 and 2. 1
29. In a flight of 600 km, an aircraft was slowed due to bad weather.
Its average speed for the trip was reduced to 200 km/hr and time
of flight increased by 30 minutes. Find the original duration of flight.
Ans: Let the speed of aircraft be x km/hr
600 600 30
∴ − = 1
x − 200 x 60

2
⇒ x − 200x − 240000 = 0 1
(x – 600) (x + 400) = 0
x = 600, – 400 (Rejected) 1/2
Speed of aircraft = 600 km/hr
∴ Duration of flight = 1 hr 1/2
30. Find the area of triangle PQR formed by the points
P(–5, 7), Q(– 4, –5) and R(4, 5).
1
Ans: ar(PQR) = ⎡⎣ −5 ( −5 − 5 ) − 4 ( 5 − 7 ) + 4 ( 7 + 5 ) ⎤⎦ sq. units 2
2
1
= [50 + 8 + 48] sq. units
2

= 53 sq. units 1
OR
If the point C(–1, 2) divides internally the line segment joining A(2, 5)
and B(x, y) in the ratio 3 : 4, find the coordinates of B.
⎛ 3x + 8 3y + 20 ⎞
⎟ = ( −1, 2 )
1
Ans: Coordinates of C arc ⎜ , 1
⎝ 7 7 ⎠
3:4
2
⇒ x = –5, y = –2 A B 1
(2, 5) (-1, 2) (x, y)
∴ Coordinates of B are (–5, –2) 1/2

.30/1/1. 8

A
AD AE
31. In Fig.5, ∠ D = ∠ E and = ,
DB EC
prove that BAC is an isosceles triangle. D E

Ans: ∠D = ∠E ⇒ AE = AD B C 1
Fig 5
AD AE
∴ = ⇒ DB = EC 1/2
DB EC
⇒ AD + DB = AE + EC 1
∴ AB = AC 1/2
Hence ΔBAC is an isosceles triangle.
32. In a triangle, if square of one side is equal to the sum of the squares of
the other two sides, then prove that the angle opposite to the first side
is a right angle.
1
Ans: For correct given, To prove, construction and figure. 1
2
1
For correct proof. 1
2
33. If sin θ + cos θ = 3 , then prove that tan θ + cot θ = 1.

( 3)
2
Ans: sin θ + cos θ = 3 ⇒ (sin θ + cos θ)2 = 1
sin2θ + cos 2θ + 2sin θ cos θ = 3 ⇒ sin θ cos θ = 1 1
sin θ cos θ 1
L.H.S = tan θ + cot θ = cos θ + sin θ = cos θ sin θ = 1 = R.H.S 1

34. A cone of base radius 4 cm is divided into two parts by drawing a plane
through the mid-point of its height and parallel to its base. Compare the
volume of the two parts.
h BC
Ans: ΔABC ~ ΔADE, = cor. fig 1/2
A 2h 4
∴ BC = 2 cm 1
Ratio of volumes of two parts
h
B C 1
π× 22 × h
3
= 1 1
h π× ( 22 + 4 2 + 2× 4)× h
D E 3
4 cm
4 1
= = or 1 : 7 (accept 7 : 1 also) 1/2
28 7

.30/1/1. 9

SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. Show that the square of any positive integer cannot be of form
(5q + 2) or (5q +3) for any integer q.
Ans: Let a be any positive integer. Take b = 5 as the divisor.
∴ a = 5m + r , r = 0,1,2,3,4 1
⎫
Case-1 : a = 5m ⇒ a2 = 25m2 = 5(5m2) = 5q ⎪
⎪
1/2
⎪
Case-2 : a = 5m+1 ⇒ a2 = 5(5m2 + 2m) + 1 = 5q + 1 ⎪ for
⎪
⎪
Case-3 : a = 5m+2 ⇒ a2 = 5(5m2 + 4m) + 4 = 5q + 4 ⎬ each
⎪
⎪
Case-4 : a = 5m+3 ⇒ a2 = 5(5m2 + 6m + 1) + 4 = 5q + 4 ⎪ case
⎪ 1
⎪
Case-5 : a = 5m+4 ⇒ a2 = 5(5m2 + 8m + 3) + 1 = 5q + 1 ⎪
=2
⎭ 2
Hence square of any positive integer cannot be of the form
(5q + 2) or (5q + 3) for any integer q. 1/2
OR
Prove that one of every three consecutive positive integers is divisible by 3.
Ans: Let n be any positive integer. Divide it by 3.
∴ n = 3q + r, r = 0, 1, 2 1
Case-1 : n = 3q (divisible by 3)
⎫
n + 1 = 3q + 1, n + 2 = 3q + 2 ⎪ 1 for
⎪
⎪
Case-2 : n = 3q + 1 ⇒ n + 1 = 3q + 2, n + 2 = 3q + 3 (divisible by 3) ⎬ each
⎪
⎪
Case-3 : n = 3q + 2 ⇒ n + 1 = 3q + 3 (divisible by 3), n + 2 = 3q + 4 ⎪⎭ case = 3
36. The sum of four consecutive numbers in AP is 32 and the ratio of product
of the first and last terms to the product of two middle terms is 7:15. Find
the numbers.
Ans: Let four consecutive number be a – 3d, a – d, a + d, a + 3d 1/2
Sum = 32 ∴ 4a = 32 ⇒ a = 8 1/2
(a − 3d)(a + 3d) 7
= ⇒ 15(64 − 9d 2 ) = 7(64 − d 2 ) 1
(a − d)(a + d) 15
∴ d2 = 4 ⇒ d = ±2 1
Four numbers are 2, 6, 10, 14. 1
OR
Solve: 1+ 4 + 7 + 10 +…+ x = 287
Ans: x = a n = 1 + 3n − 3 = 3n − 2 1

n
Sn = 287 ⇒ [1 + 3n − 2] = 287 1
2
∴ 3n 2 − n − 574 = 0 1/2
(n − 14)(3n + 41) = 0 ⇒ n = 14 1
∴ x = 3n – 2 = 40 1/2

.30/1/1. 10

37. Draw a line segment AB of length 7 cm. Taking A as centre, draw
a circle of radius 3 cm and taking B as centre, draw another circle of
radius 2 cm. Construct tangents to each circle from the centre of the
other circle.
Ans: Constructing the circles of radii 3 cm and 2 cm. 1
Constructing the tangents. 3
38. A vertical tower stands on a horizontal plane and is surmounted by a
vertical flag-staff of height 6 m. At a point on the plane, the angle of
elevation of the bottom and top of the flag-staff are 30° and 45° re-
spectively. Find the height of the tower. (Take 3 = 1.73)
Ans: cor. fig 1
A h
= tan 30° 1
x
6m ⇒ x=h 3
6+h
= tan 45° ⇒ 6 + h = x 1
B x

h 6
∴h = = 3( 3 + 1) = 3 × 2.73 m
3 −1
45° 30°
C D = 8.19 m 1
x

39. A bucket in the form of a frustum of a cone of height 30 cm with radii of its
lower and upper ends as 10 cm and 20 cm respectively. Find the capacity of
the bucket. Also find the total cost of milk that can completely fill the
⎛ 22 ⎞
bucket at the rate of `40 per litre. ⎜ Use π = ⎟
⎝ 7 ⎠
1
( 2 2
Ans: Capacity of bucket = πh r1 + r2 + r1r2
3
)
1 22
( 2 2
= × × 30 10 + 20 +10 × 20 cm
3 7
) 3
1

1
= 22000 cm3 1
2
= 22l 1/2
Cost of milk = `40 × 22 = `880 1
40. The following table gives production yield per hectare (in quintals)
of wheat of 100 farms of a village:
Production yield/hect. 40-45 45-50 50-55 55-60 60-65 65-70
No. of farms 4 6 16 20 30 24
Change the distribution to 'a more than' type distribution and draw its ogive.

.30/1/1. 11

Ans: Production yield/hectare No. of forms

More than or equal to 40 100
More than or equal to 45 96
More than or equal to 50 90 2
More than or equal to 55 74
More than or equal to 60 54
More than or equal to 65 24
Total

(40, 100)
(45, 96)
100
(50, 90)
90
80 (55, 74)
Number of farms

70
60 (60, 54)
50
40 2
30 (65, 24)
20
10
0
40 45 50 55 60 65
Production yield/hectare

.30/1/1. 12

OR
The median of the following data is 525. Find the values of x and y,
if total frequency is 100:

900-1000
100-200

200-300

300-400

400-500

500-600

600-700

700-800

800-900
Class :

0-100
Frequency: 2 5 x 12 17 20 y 9 7 4

Ans:

Classes Frequency Cumulative
frequency
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19 + x
400-500 17 36 + x
500-600 20 56 + x Median class 2
600-700 y 56 + x + y
700-800 9 65 + x+ y
800-900 7 72 + x+ y
900-1000 4 76 + x+ y
Total 100

76 + x + y = 100 ⇒ x + y = 24 ... (i) 1/2

500 – 600 is the median class

n
− cf
Median = l + 2 ×h
f
50 − 36 − x
⇒ 525 = 500 + × 100
20
Solving we get, x = 9 1
From (i), y = 15 1/2

.30/1/1. 13

QUESTION PAPER CODE 30/1/2
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Q. NO. 1 to 10 are multiple choice type question of 1 mark each.
Select the correct option.
Q.No. Marks
1. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3,140 (b) 12,420 (c) 3,420 (d) 420,3
Ans: (c) 3,420 1
2. The value of x for which 2x,(x + 10) and ( 3x + 2 ) are the three
consecutive terms of an AP, is
(a) 6 (b) –6 (c) 18 (d) –18
Ans: (a) 6 1
3. The value of k for which the system of equations x + y – 4 = 0 and
2x + ky = 3 has no solution, is
(a) –2 (b) ≠2 (c) 3 (d) 2
Ans: (d) 2 1
4. The first term of an AP is p and the common difference is q, then its 10 term is
th

(a) q + 9p (b) p – 9q (c) p + 9q (d) 2p + 9q
Ans: (c) p + 9q 1
5. The quadratic polynomial, the sum of whose zeroes is –5 and their product is
6, is
(a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2 – 5x – 6 (d) –x2 + 5x + 6
Ans: (a) x2 + 5x + 6 1
6. The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ – b cos θ), is
(a) a2 + b2 (b) a2 – b2 (c) a 2 +b 2 (d) a 2 − b2
Ans: (c) a 2 + b2 1
7. The total number of factors of a prime number is
(a) 1 (b) 0 (c) 2 (d) 3
Ans: (c) 2 1
8. If the point P(k, 0) divides the line segment joining the points A(2, –2) and
B(–7, 4) in the ratio 1 : 2, then the value of k is,
(a) 1 (b) 2 (c) –2 (d) –1
Ans: (d) –1 1
9. The value of p, for which the points A(3, 1), B(5, p) and C(7, –5) are collinear, is
(a) –2 (b) 2 (c) –1 (d) 1
Ans: (a) –2 1
10. If one of the zeroes of the quadratic polynomial x2 + 3x + k is 2, then the value
of k is
(a) 10 (b) –10 (c) –7 (d) –2
Ans: (b) – 10 1

.30/1/2. 14

In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.
11. ABC is an equilateral triangle of side 2a, then length of one of its
altitude is _______.
Ans: 3 a 1
12. In Fig. 1, Δ ABC is circumscribing a circle, the length of BC is _______cm.

A

4 cm
11 cm

P R
3 cm

B Q C
Fig. 1

Ans: 10 1
⎛ 1 ⎞
13. The value of ⎜ sin 2 θ + ⎟ = _________.
⎝ 1 + tan θ ⎠
2

Ans: 1 1
OR
The value of (1 + tan2θ) (1 – sin θ) (1 + sin θ) =_______.
Ans: 1 1
2 2
⎛ sin 35° ⎞ ⎛ cos 43° ⎞
14. ⎜ ⎟ +⎜ ⎟ − 2 cos 60° = _____________.
⎝ cos 55° ⎠ ⎝ sin 47° ⎠

Ans: 1 1
15. ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. Ratio of the areas of triangles ABC and BDE is _____________.

Ans: 4 : 1 1
Q. Nos. 16 to 20 are short answer type questions of 1 mark each.
16. A die is thrown once. What is the probability of getting a number less than 3?
2 1
Ans: P (number less than 3) = or 1
6 3
OR
If the probability of winning a game is 0.07, what is the probability of losing it?
Ans: P (losing) = 1 – 0.07 1/2
= 0.93 1/2
17. If the mean of first n natural number is 15, then find n.
n(n + 1)
Ans: 2 = 15 1/2
n
∴ n = 29 1/2

.30/1/2. 15

18. Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1.
What is the ratio of their volumes?
r1 3 h1 1
Ans: r = 1 , h = 3 1/2
2 2
1 2
πr1 h1
3 = 3 :1
∴ Ratio of volumes = 1 2 1/2
πr2 h 2
3
19. The ratio of the length of a vertical rod and the length of its shadow
is 1: 3 . Find the angle of elevation of the sun at that moment?
1
Ans: tan θ = ⇒ θ = 30° 1/2+1/2
3
20. A die is thrown once. What is the probabiliy of getting an even prime number?

Ans: Number of even prime numbers on a die is 1 (i.e. 2) 1/2
1
∴ P (even prime number) = 1/2
6
SECTION – B
Q. Nos. 21 to 26 carry 2 marks each.
BE BC
21. In Fig. 2 DE ||AC and DC ||AP. Prove that = .
EC CP
A

D

B E C P
Fig 2

BD BE
Ans: In ΔABC, DE || AC, ∴ = ... (i) 1
DA EC
BD BC
In ΔABP, DC || AP, ∴ = ... (ii) 1/2
DA CP
BE BC
From (i) & (ii), = 1/2
EC CP
OR
P
In Fig. 3, two tangents TP and TQ are drawn
to a circle with centre O from an external
point T. Prove that ∠PTQ = 2∠OPQ. T O

Fig 3 Q

.30/1/2. 16

Ans: Let ∠OPQ = θ
∴ ∠TPQ = ∠TQP = 90° – θ 1/2
In ΔTPQ, 2(90° – θ) + ∠PTQ = 180° 1
∴ ∠PTQ = 2θ 1/2
= 2∠OPQ
22. The rod AC of a TV disc antenna is fixed at right
angle to the wall AB and a rod CD is supporting A
C
the disc as shown in Fig. 4. If AC = 1.5m long
and CD = 3m, find (i) tanθ (ii) secθ + cosecθ.
θ
AC 1
Ans: = sin θ ⇒ sin θ = ⇒ θ = 30° D 1/2
CD 2 B
1 Fig 4
(i) tan θ = tan 30° = Wall
1/2
3
(ii) sec θ + cosecθ = sec30° + cos ec30° 1
2 2(3 + 3)
=
+ 2 or
3 3
23. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3.
What is the probability that x2 ≤ 4?
Ans: Total number of outcomes = 7
Favourable outcomes are –2, –1, 0, 1, 2, i.e., 5 1

5
∴ P(x 2 ≤ 4) = 1
7
24. Find the mean of the following distribution:
Class: 3-5 5-7 7-9 9-11 11-13
Frequency: 5 10 10 7 8

Ans: Classes xi fi fxxi 1½
3–5 4 5 20
5–7 6 10 60
7–9 8 10 80
9 – 11 10 7 70
11 – 13 12 8 96
Total 40 326

x=
∑ fi xi = 326 = 8.15
∑ fi 40 1/2

.30/1/2. 17

OR
Find the mode of the following data:
Class: 0-20 20-40 40-60 60-80 80-100 110-120 120-140
Frequency: 6 8 10 12 6 5 3
Ans: Modal class : 60 – 80 1/2

f1 − f 0 12 − 10
Mode = l + 2f − f − f × h = 60 + 24 − 10 − 6 × 20 1
1 0 2

= 60 + 5 = 65 1/2
25. Find the sum of first 20 terms of the following AP:
1, 4, 7, 10, …
20 1
Ans: S20 = [ 2 ×1 + 19 × 3] 1
2 2
= 10 × 59 = 590 1/2
26. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the
area of the sector.
2π(5.2)θ 360 × 6
Ans: 2 × 5.2 + = 16.4 ⇒ θ = 1
360° 2π× 5.2
π× (5.2) 2 360 × 6
Area of sector = × = 15.6 cm 2 1
360° 2π× 5.2
SECTION – C
Question numbers 27 to 34 carry 3 marks each.
27. A cone of base radius 4 cm is divided into two parts by drawing a plane
through the mid-point of its height and parallel to its base. Compare the
volume of the two parts.
Ans: cor. fig 1/2

A h BC
ΔABC ~ ΔADE, =
2h 4
∴ BC = 2 cm 1
h
C Ratio of volumes of two parts
B
1
π× 22 × h
h 3
= 1 1
D 4 cm E 2 2
π× ( 2 + 4 + 2× 4)× h
3
4 1
=
= or 1 : 7 (accept 7 : 1 also) 1/2
28 7
28. In a triangle, if square of one side is equal to the sum of the squares of
the other two sides, then prove that the angle opposite to the first side
is a right angle.

.30/1/2. 18

1
Ans: For correct given, To prove, construction and figure. 1
2
1
For correct proof. 1
2
29. Find the area of triangle PQR formed by the points
P(–5, 7), Q(– 4, –5) and R(4, 5).
1
Ans: ar(PQR) = ⎡⎣ −5 ( −5 − 5 ) − 4 ( 5 − 7 ) + 4 ( 7 + 5 ) ⎤⎦ sq. units 2
2
1
= [50 + 8 + 48] sq. units
2

= 53 sq. units 1
OR
If the point C(–1, 2) divides internally the line segment joining A(2, 5)
and B(x, y) in the ratio 3 : 4, find the coordinates of B.
⎛ 3x + 8 3y + 20 ⎞
⎟ = ( −1, 2 )
1
Ans: Coordinates of C arc ⎜ , 1
⎝ 7 7 ⎠
3:4 2
⇒ x = –5, y = –2 A B 1
(2, 5) (-1, 2) (x, y)
∴ Coordinates of B are (–5, –2) 1/2
30. Find the quadratic polynomial whose zeroes are reciprocal of the zeroes
of the polynomial f (x) = ax2 + bx + c, a ≠ 0, c ≠ 0.
Ans: f (x) = ax 2 + bx + c
b c
α + β = − , αβ = 1/2
a a
1 1 b
New sum of zeroes = + =− 1
α β c
1 1 a
New product of zeroes = × = 1
α β c

∴ Required quardratic polynomial = x +
2 b
c
a
(
x + or cx 2 + bx + a
c
) 1/2

OR
Divide the polynomial f(x) = 3x2 – x3 – 3x + 5 by the polynomial
g(x) = x – 1 – x2 and verify the division algorithm.
Ans: 2
3 2
− x 2 + x − 1 − x + 3x − 3x + 5 x−2
−x 3 + x 2 − x
+ – +
2x 2 − 2x + 5
2x 2 − 2x + 2
– + –
3

.30/1/2. 19

Divisor × Quotient + Remainder
= ( − x 2 + x − 1)(x − 2) + 3
3 2
= − x + 3x − 3x + 5 = Dividend 1
31. Determine graphically the coordinates of the vertices of a triangle,
the equations of whose sides are given by 2y – x = 8, 5y – x = 14
and y – 2x = 1.

Ans:
y
2y–x=8
6
5 C(2,5) 5y–x=14
(0,4) 4
A(–4,2) 3 (6,4) 2y – x = 8
B(1,3)
2
1 (0,1) x 0 2 –4
x′ x y 4 5 2
–6 –5 –4 –3 –2 –1
–1 O1 2 3 4 5 6
5y – x = 14
–2
x 1 6 –4
y 3 4 2
y–2x=1
y – 2x = 1
y′ x 1 2 0
y 3 5 1

1
Drawing 3 lines 1
2

Coordinates of the vertices of the triangle are A(– 4, 2),
1
B(1, 3) and C(2, 5) 1
2
OR
If 4 is the zero of the cubic polynomial x3 – 3x2 – 10x + 24,
find its other two zeroes.
Ans: x – 4 is a factor of given polynomial.

x−4 x 3 − 3x 2 − 10x + 24 x 2 + x − 6
x 3 – 4x 2
– +
x 2 − 10x + 24
x 2 – 4x 2
– +
−6x + 24
−6x + 24
+ –
0

x 2 + x − 6 = (x + 3)(x − 2)
∴ Other than zeroes are –3 and 2. 1

.30/1/2. 20

32. A train covers a distance of 480 km at a uniform speed. If the speed had
been 8 km/h less, then it would have taken 3 hours more to cover the same
distance. Find the original speed of the train.
Ans: Let the speed of train be x km/hr
480 480
∴ − =3 1
x −8 x
⇒ x2 – 8x – 1280 = 0 1
(x – 40) (x + 32) = 0
x = 40, – 32 (Rejected) 1
∴ Speed of train = 40 km/hr
33. Prove that the parallelogram circumscribing a circle is a rhombus.
Ans: AP = AS ⎫ A P B
BP = BQ ⎪
⎪ 1
DR = DS ⎬ S Q
⎪
CR = CQ ⎪⎭ D R C
Adding, we get (AP + BP) + (DR + CR) + (AS + DS) + (BQ + CQ)
⇒ AB + CD = BC + AD 1
Since ABCD is a ||gm ∴ 2AB = 2 BC
⇒ AB = BC 1
34. Prove that : 2(sin6θ + cos6θ) – 3(sin4 θ + cos4 θ) + 1 = 0
Ans: 2(sin6θ + cos6θ) – 3(sin4 θ + cos4 θ) + 1

= 2 ⎡⎢( sin θ ) + ( cos θ ) ⎤⎥ − 3 ( sin θ + cos θ ) + 1
2 3 2 3 4 4
1
⎣ ⎦

( 2 2
)(
4 4 2 2
) (
4 4
)
= 2 ⎡⎣ sin θ + cos θ sin θ + cos θ − sin θ cos θ ⎤⎦ − 3 sin θ + cos θ + 1 1

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

= − ⎡⎢( sin θ + cos θ ) − 2 sin θ cos θ ⎤⎥ − 2 sin θ cos θ + 1 ⎫
2 2 2 2 2 2 2
⎣ ⎦ ⎬ 1
= –1 + 1 = 0 ⎭
SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. The following table gives production yield per hectare (in quintals)
of wheat of 100 farms of a village:
Production yield/hect. 40-45 45-50 50-55 55-60 60-65 65-70
No. of farms 4 6 16 20 30 24
Change the distribution to 'a more than' type distribution and draw its ogive.

.30/1/2. 21

Ans: Production yield/hectare No. of forms 2
More than or equal to 40 100
More than or equal to 45 96
More than or equal to 50 90
More than or equal to 55 74
More than or equal to 60 54
More than or equal to 65 24
Total
(40, 100)
100
(45, 96) 2
(50, 90)
90
80 (55, 74)
Number of farms

70
60 (60, 54)
50
40
30 (65, 24)
20
10
0
40 45 50 55 60 65
Production yield/hectare
OR
The median of the following data is 525. Find the values of x and y,
if total frequency is 100:
900-1000
100-200

200-300

300-400

400-500

500-600

600-700

700-800

800-900

Class :
0-100

Frequency: 2 5 x 12 17 20 y 9 7 4

Ans:
Classes Frequency Cumulative
frequency
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19 + x
400-500 17 36 + x
500-600 20 56 + x Median class 2
600-700 y 56 + x + y
700-800 9 65 + x+ y
800-900 7 72 + x+ y
900-1000 4 76 + x+ y
Total 100

.30/1/2. 22

76 + x + y = 100 ⇒ x + y = 24 ... (i) 1/2
500 – 600 is the median class
n
− cf
Median = l + 2 ×h
f
50 − 36 − x
⇒ 525 = 500 + × 100
20
Solving we get, x = 9 1
From (i), y = 15 1/2
36. A vertical tower stands on a horizontal plane and is surmounted by a
vertical flag-staff of height 6 m. At a point on the plane, the angle of
elevation of the bottom and top of the flag-staff are 30° and 45° re-
spectively. Find the height of the tower. (Take 3 = 1.73)
Ans: cor. fig 1
A h
= tan 30° 1
x
6m
⇒ x=h 3
6+h
B = tan 45° ⇒ 6 + h = x 1
x
h 6
∴h = = 3( 3 + 1) = 3 × 2.73 m
45° 30° 3 −1
C x D = 8.19 m 1

37. Show that the square of any positive integer cannot be of form
(5q + 2) or (5q +3) for any integer q.
Ans: Let a be any positive integer. Take b = 5 as the divisor.
∴ a = 5m + r , r = 0, 1, 2, 3, 4 1
⎫
Case-1 : a = 5m ⇒ a2 = 25m2 = 5(5m2) = 5q ⎪ 1/2
⎪
⎪
Case-2 : a = 5m+1 ⇒ a = 5(5m + 2m) + 1 = 5q + 1
2 2 ⎪ for
⎪
⎪
Case-3 : a = 5m+2 ⇒ a = 5(5m + 4m) + 4 = 5q + 4
2 2 ⎬ each
⎪
⎪
Case-4 : a = 5m+3 ⇒ a = 5(5m + 6m + 1) + 4 = 5q + 4 ⎪
2 2 case
⎪
Case-5 : a = 5m+4 ⇒ a2 = 5(5m2 + 8m + 3) + 1 = 5q + 1 ⎪⎪
1
=2
⎭ 2
Hence square of any positive integer cannot be of the form
(5q + 2) or (5q + 3) for any integer q. 1/2

.30/1/2. 23

OR
Prove that one of every three consecutive positive integers is divisible by 3.
Ans: Let n be any positive integer. Divide it by 3.
∴ n = 3q + r, r = 0, 1, 2 1
Case-1 : n = 3q (divisible by 3) ⎫
⎪
n +1 = 3q + 1, n + 2 = 3q + 2 ⎪ 1 for
⎪
Case-2 : n = 3q+1 ⇒ n+1 = 3q + 2, n+2 = 3q+3 (divisible by 3) ⎬ each
⎪
⎪
Case-3 : n = 3q+2 ⇒ n+1 = 3q + 3 (divisible by 3), n + 2 = 3q + 4 ⎪⎭ case = 3
38. The sum of four consecutive numbers in AP is 32 and the ratio of product
of the first and last terms to the product of two middle terms is 7:15. Find
the numbers.
Ans: Let four consecutive number be a – 3d, a – d, a + d, a + 3d 1/2
Sum = 32 ∴ 4a = 32 ⇒ a = 8 1/2
(a − 3d)(a + 3d) 7
= ⇒ 15(64 − 9d 2 ) = 7(64 − d 2 ) 1
(a − d)(a + d) 15
∴ d2 = 4 ⇒ d = ±2 1
Four numbers are 2, 6, 10, 14. 1
OR
Solve: 1+ 4 + 7 + 10 +…+ x = 287
Ans: x = a n = 1 + 3n − 3 = 3n − 2 1

n
Sn = 287 ⇒ [1 + 3n − 2] = 287 1
2
∴ 3n 2 − n − 574 = 0 1/2
(n − 14)(3n + 41) = 0 ⇒ n = 14 1
∴ x = 3n – 2 = 40 1/2

39. A bucket in the form of a frustum of a cone of height 16 cm with radii of its
lower and upper circular ends as 8 cm and 20 cm respectively. Find cost of
milk which can completely fill the bucket, at the rate of `40 per litre.
(Use π = 3.14)
1
3
(
Ans: Capacity of bucket = πh r1 + r2 + r1r2
2 2
) 1

1
3
(
= × 3.14 × 16 8 + 20 + 8× 20 cm
2 2
) 3
1
1
2
= 10449.92 cm3
= 10.45 l (approx.) 1/2
Cost of milk = `40 × 10.45 = `418 1

.30/1/2. 24

40. Construct a triangle with sides 4 cm, 5 cm and 6 cm. Then construct
2
another triangle whose sides are times the corresponding sides of
3
the first triangle.
Ans: Construction of Δ ABC with given dimensions 1
Construction of similar triangle 3

.30/1/2. 25

QUESTION PAPER CODE 30/1/3
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Q. NO. 1 to 10 are multiple choice type question of 1 mark each.
Select the correct option.
Q.No. Marks

1. The value of k for which the system of equations x + y – 4 = 0 and
2x + ky =3 has no solution, is
(a) –2 (b) ≠2 (c) 3 (d) 2
Ans: (d) 2 1
2. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3,140 (b) 12,420 (c) 3,420 (d) 420,3
Ans: (c) 3,420 1
3. The value of x for which 2x,(x + 10) and ( 3x + 2 ) are the three
consecutive terms of an AP, is
(a) 6 (b) –6 (c) 18 (d) –18
Ans: (a) 6 1
4. The first term of an AP is p and the common difference is q, then its 10th term is
(a) q + 9p (b) p – 9q (c) p + 9q (d) 2p + 9q
Ans: (c) p + 9q 1
5. If one of the zeroes of the quadratic polynomial x2 + 3x + k is 2, then the value
of k is
(a) 10 (b) –10 (c) –7 (d) –2
Ans: (b) – 10 1
6. The total number of factors of a prime number is
(a) 1 (b) 0 (c) 2 (d) 3
Ans: (c) 2 1
7. The quadratic polynomial, the sum of whose zeroes is –5 and their product is
6, is
(a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2 – 5x – 6 (d) –x2 + 5x + 6
Ans: (a) x2 + 5x + 6 1
8. The value of p, for which the points A(3, 1), B(5, p) and C(7, –5) are collinear, is
(a) –2 (b) 2 (c) –1 (d) 1
Ans: (a) –2 1
9. The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ – b cos θ), is
(a) a2 + b2 (b) a2 – b2 (c) a 2 +b 2 (d) a 2 − b2
Ans: (c) a 2 + b2 1
10. If the point P(k, 0) divides the line segment joining the points A(2, –2) and
B(–7, 4) in the ratio 1 : 2, then the value of k is,
(a) 1 (b) 2 (c) –2 (d) –1
Ans: (d) –1 1

.30/1/3. 26

In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.

AB 1 ar ( ΔABC )
11. Given Δ ABC ~ Δ PQR, if = , then = __________.
PQ 3 ar ( ΔPQR )
1
Ans: 1
9
12. In Fig. 1, Δ ABC is circumscribing a circle, the length of BC is _______cm.

A

4 cm
11 cm

P R
3 cm

B Q C
Fig. 1

Ans: 10 1

⎛ 2 1 ⎞
13. The value of ⎜ sin θ + ⎟ = _________.
⎝ 1 + tan θ ⎠
2

Ans: 1 1
OR
The value of (1 + tan2θ) (1 – sin θ) (1 + sin θ) =_______.
Ans: 1 1

14. A ladder 10 m long reaches a window 8 m above the ground. The distance
of the foot of the ladder from the base of the wall is _________ m.

Ans: 6 1

2 cos 67° tan 40°
15. − − cos 0° = _____________.
sin 23° cot 50°

Ans: 0 1

Q. Nos. 16 to 20 are short answer type questions of 1 mark each.
16. If the mean of first n natural number is 15, then find n.
n(n + 1)
Ans: 2 = 15 1/2
n
∴ n = 29 1/2
17. A die is thrown once. What is the probability of getting a number less than 3?
2 1
Ans: P (number less than 3) = or 1
6 3

.30/1/3. 27

OR
If the probability of winning a game is 0.07, what is the probability of losing it?
Ans: P (losing) = 1 – 0.07 = 0.93 1
18. The ratio of the length of a vertical rod and the length of its shadow
is 1: 3 . Find the angle of elevation of the sun at that moment?
1
Ans: tan θ = ⇒ θ = 30° 1/2+1/2
3
19. Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1.
What is the ratio of their volumes?
r1 3 h1 1
Ans: r = 1 , h = 3 1/2
2 2

1 2
πr1 h1
3 = 3 :1
∴ Ratio of volumes = 1 2 1/2
πr2 h 2
3
20. A pair of dice is thrown once. What is the probabiliy of getting a doublet?
Ans: Number of favourable outcomes are 6

i.e. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) 1/2

6 1
∴ P (doublet) = or 1/2
36 6
SECTION – B
Q. Nos. 21 to 26 carry 2 marks each.
BE BC
21. In Fig. 2 DE ||AC and DC ||AP. Prove that = .
EC CP
A

D

B E C P
Fig 2

BD BE
Ans: In ΔABC, DE || AC, ∴ = ... (i) 1
DA EC
BD BC
In ΔABP, DC || AP, ∴ = ... (ii) 1/2
DA CP

BE BC
From (i) & (ii), = 1/2
EC CP

.30/1/3. 28

OR
In Fig. 3, two tangents TP and TQ are drawn to a circle with centre O from
an external point T. Prove that ∠PTQ = 2∠OPQ.

P

T O

Fig 3 Q

Ans: Let ∠OPQ = θ
∴ ∠TPQ = ∠TQP = 90° – θ 1/2
In ΔTPQ, 2(90° – θ) + ∠PTQ = 180° 1
∴ ∠PTQ = 2θ 1/2
= 2∠OPQ
22. The rod AC of a TV disc antenna is fixed at right C
A
angle to the wall AB and a rod CD is supporting
the disc as shown in Fig. 4. If AC = 1.5m long
and CD = 3m, find (i) tanθ (ii) secθ + cosecθ. θ
D
AC 1
Ans: = sin θ ⇒ sin θ = ⇒ θ = 30° B 1/2
CD 2
Wall Fig 4
1
(i) tan θ = tan 30° = 1/2
3
(ii) sec θ + cosecθ = sec30° + cos ec30° 1
2 2(3 + 3)
= + 2 or
3 3
23. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3.
What is the probability that x2 ≤ 4?
Ans: Total number of outcomes = 7
Favourable outcomes are –2, –1, 0, 1, 2, i.e., 5 1

5
∴ P(x 2 ≤ 4) = 1
7
24. Find the mean of the following distribution:
Class: 3-5 5-7 7-9 9-11 11-13
Frequency: 5 10 10 7 8

.30/1/3. 29

Ans: Classes xi fi fxxi 1½
3–5 4 5 20
5–7 6 10 60
7–9 8 10 80
9 – 11 10 7 70
11 – 13 12 8 96
Total 40 326

x=
∑ fi xi = 326 = 8.15
∑ fi 40 1/2
OR
Find the mode of the following data:
Class: 0-20 20-40 40-60 60-80 80-100 110-120 120-140
Frequency: 6 8 10 12 6 5 3
Ans: Modal class : 60 – 80 1/2

f1 − f 0 12 − 10
Mode = l + 2f − f − f × h = 60 + 24 − 10 − 6 × 20 1
1 0 2

= 60 + 5 = 65 1/2
25. The minute hand of a clock is 12 cm long. Find the area of the face of
the clock described by the minute hand in 35 minutes.
360
Ans: Angle swept in 35 min = × 35 = 210° 1
60
22 210
Area of face of clock = × 12 × 12 × = 264 cm 2 1
7 360

22 35
Accept: Area = × (12) 2 × = 264 cm 2 1+1
7 60
26. The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is
161. Find the AP.
7
Ans: S7 = 63 ⇒ (2a + 6d) = 63
2
∴ a + 3d = 9 …(i) 1/2
14
S14 – S7 = (2a + 13d) − 63 = 161
2
⇒ 2a + 13d = 32 …(ii) 1/2
Solving (i) and (ii), a = 3, d = 2 1/2
∴ AP is 3, 5, 7 … 1/2

.30/1/3. 30

SECTION – C
Question numbers 27 to 34 carry 3 marks each.
27. Determine graphically the coordinates of the vertices of a triangle,
the equations of whose sides are given by 2y – x = 8, 5y – x = 14
and y – 2x = 1.

Ans:
y
2y–x=8
6
5 C(2,5) 5y–x=14
(0,4) 4 2y – x = 8
A(–4,2) 3 (6,4)
B(1,3) x 0 2 –4
2
1 (0,1) y 4 5 2
x′ x
–6 –5 –4 –3 –2 –1 O1 2 3 4 5 6 5y – x = 14
–1
–2 x 1 6 –4
y 3 4 2
y–2x=1 y – 2x = 1
x 1 2 0
y′ y 3 5 1

1
Drawing 3 lines 1
2

Coordinates of the vertices of the triangle are A(– 4, 2),
1
B(1, 3) and C(2, 5) 1
2
OR
If 4 is the zero of the cubic polynomial x3 – 3x2 – 10x + 24,
find its other two zeroes.
Ans: x – 4 is a factor of given polynomial.

x−4 x 3 − 3x 2 − 10x + 24 x 2 + x − 6
x 3 – 4x 2
– +
x 2 − 10x + 24
x 2 – 4x 2
– +
−6x + 24
−6x + 24
+ –
0
x 2 + x − 6 = (x + 3)(x − 2)

∴ Other than zeroes are –3 and 2. 1

.30/1/3. 31

28. Find the area of triangle PQR formed by the points
P(–5, 7), Q(– 4, –5) and R(4, 5).
1
Ans: ar(PQR) = ⎡⎣ −5 ( −5 − 5 ) − 4 ( 5 − 7 ) + 4 ( 7 + 5 ) ⎤⎦ sq. units 2
2
1
= [50 + 8 + 48] sq. units
2

= 53 sq. units 1
OR
If the point C(–1, 2) divides internally the line segment joining A(2, 5)
and B(x, y) in the ratio 3 : 4, find the coordinates of B.
⎛ 3x + 8 3y + 20 ⎞
⎟ = ( −1, 2 )
1
Ans: Coordinates of C arc ⎜ , 1
⎝ 7 7 ⎠ 2
3:4
⇒ x = –5, y = –2 A B 1
(2, 5) (-1, 2) (x, y)
∴ Coordinates of B are (–5, –2) 1/2
29. Find the quadratic polynomial whose zeroes are reciprocal of the zeroes
of the polynomial f (x) = ax2 + bx + c, a ≠ 0, c ≠ 0.
Ans: f (x) = ax 2 + bx + c
b c
α + β = − , αβ = 1/2
a a
1 1 b
New sum of zeroes = + =− 1
α β c
1 1 a
New product of zeroes = × = 1
α β c

∴ Required quardratic polynomial = x +
2 b
c
a
(
x + or cx 2 + bx + a
c
) 1/2

OR
2 3
Divide the polynomial f(x) = 3x – x – 3x + 5 by the polynomial
g(x) = x – 1 – x2 and verify the division algorithm.
Ans: 3 2 2
− x 2 + x − 1 − x + 3x − 3x + 5 x−2
−x 3 + x 2 − x
+ – +
2x 2 − 2x + 5
2x 2 − 2x + 2
– + –
3
Divisor × Quotient + Remainder
= ( − x 2 + x − 1)(x − 2) + 3
3 2
= − x + 3x − 3x + 5 = Dividend 1

.30/1/3. 32

30. In a triangle, if square of one side is equal to the sum of the squares of
the other two sides, then prove that the angle opposite to the first side
is a right angle.
1
Ans: For correct given, To prove, construction and figure. 1
2
1
For correct proof. 1
2
31. A cone of base radius 4 cm is divided into two parts by drawing a plane
through the mid-point of its height and parallel to its base. Compare the
volume of the two parts.
Ans: cor. fig 1/2
h BC
A ΔABC ~ ΔADE, =
2h 4
∴ BC = 2 cm 1
h Ratio of volumes of two parts
B C
1
π× 22 × h
3
h = 1 1
2 2
D E π× ( 2 + 4 + 2× 4)× h
4 cm 3
4 1
= = or 1 : 7 (accept 7 : 1 also) 1/2
28 7
32. A man can row a boat downstream 20 km in 2 hours and upstream 4 km in
2 hours. Find his speed of rowing in still water. Also find the speed of the
stream.
Ans: Let the speed of rowing in still water x km/hr
and speed of stream be y km/hr.
20
= 2 ⇒ x + y = 10 (i) 1
x+y
4
=2 ⇒ x−y=2 (ii) 1/2
x−y
Solving (i) & (ii), x = 6, y = 4 1
∴ Speed of rowing in still water = 6 km/hr ⎫
⎬ 1/2
and speed of stream = 4 km/hr ⎭
33. In given Fig. 5, two circles touch each other at the point C. Prove that the
common tangent to the circles at C, bisects the common tangent at P and Q.

P T Q
A C B

.30/1/3. 33

Ans: PT = TC 1
and TQ = TC 1
∴ PT = TQ
Hence TC bisects PQ 1
cot θ + cosec θ − 1 1 + cos θ
34. Prove that : =
cot θ − cosec θ + 1 sin θ

cot θ + cosec θ − 1 (cot θ + cosec θ) − (cosec 2 θ − cot 2 θ)
Ans: cot θ − cosec θ + 1 = cot θ − cosec θ + 1
1

(cosec θ + cot θ ) (1 − cosecθ + cot θ)
= (cot θ − cosecθ + 1) 1

1 + cos θ
= cosecθ + cot θ = 1
sin θ

SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. The following table gives production yield per hectare (in quintals)
of wheat of 100 farms of a village:
Production yield/hect. 40-45 45-50 50-55 55-60 60-65 65-70
No. of farms 4 6 16 20 30 24
Change the distribution to 'a more than' type distribution and draw its ogive.

Ans: Production yield/hectare No. of forms

More than or equal to 40 100
More than or equal to 45 96
More than or equal to 50 90
More than or equal to 55 74 2
More than or equal to 60 54
More than or equal to 65 24
Total
(40, 100)
(45, 96)
100
(50, 90)
90
80 (55, 74)
Number of farms

70
60 (60, 54)
50
2
40
30 (65, 24)
20
10
0
40 45 50 55 60 65
Production yield/hectare

.30/1/3. 34

OR
The median of the following data is 525. Find the values of x and y,
if total frequency is 100:

900-1000
Class :

100-200

200-300

300-400

400-500

500-600

600-700

700-800

800-900
0-100
Frequency: 2 5 x 12 17 20 y 9 7 4

Ans:

Classes Frequency Cumulative
frequency
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19 + x
400-500 17 36 + x
500-600 20 56 + x Median class 2
600-700 y 56 + x + y
700-800 9 65 + x+ y
800-900 7 72 + x+ y
900-1000 4 76 + x+ y
Total 100

76 + x + y = 100 ⇒ x + y = 24 ... (i) 1/2

500 – 600 is the median class

n
− cf
Median = l + 2 ×h
f
50 − 36 − x
⇒ 525 = 500 + × 100
20
Solving we get, x = 9 1
From (i), y = 15 1/2
36. A bucket in the form of a frustum of a cone of height 30 cm with radii of its
lower and upper ends as 10 cm and 20 cm respectively. Find the capacity of
the bucket. Also find the total cost of milk that can completely fill the bucket
⎛ 22 ⎞
at the rate of ` 40 per litre. ⎜ Use π = ⎟
⎝ 7 ⎠
1 2
Ans: Capacity of bucket = πh r1 + r2 + r1r2
3
2
( )

.30/1/3. 35

=
1 22
( )
× × 30 102 + 20 2 +10 × 20 cm3
3 7
1

1
= 22000 cm3 1
2
= 22l 1/2
Cost of milk = `40 × 22 = `880 1
37. Show that the square of any positive integer cannot be of form
(5q + 2) or (5q +3) for any integer q.
Ans: Let a be any positive integer. Take b = 5 as the divisor.
∴ a = 5m + r , r = 0,1,2,3,4 1
⎫
Case-1 : a = 5m ⇒ a2 = 25m2 = 5(5m2) = 5q ⎪ 1/2
⎪
⎪
Case-2 : a = 5m+1 ⇒ a2 = 5(5m2 + 2m) + 1 = 5q + 1 ⎪ for
⎪
⎪
Case-3 : a = 5m+2 ⇒ a2 = 5(5m2 + 4m) + 4 = 5q + 4 ⎬ each
⎪
Case-4 : a = 5m+3 ⇒ a2 = 5(5m2 + 6m + 1) + 4 = 5q + 4 ⎪⎪ case
⎪ 1
Case-5 : a = 5m+4 ⇒ a2 = 5(5m2 + 8m + 3) + 1 = 5q + 1 ⎪ =2
⎪ 2
⎭
Hence square of any positive integer cannot be of the form
(5q + 2) or (5q + 3) for any integer q. 1/2
OR
Prove that one of every three consecutive positive integers is divisible by 3.
Ans: Let n be any positive integer. Divide it by 3.
∴ n = 3q + r, r = 0,1,2 1
Case-1 : n = 3q (divisible by 3) ⎫
⎪
n +1 = 3q + 1, n + 2 = 3q + 2 ⎪ 1 for
⎪
Case-2 : n = 3q+1 ⇒ n+1 = 3q + 2, n+2 = 3q+3 (divisible by 3) ⎬ each
⎪
Case-3 : n = 3q+2 ⇒ n+1 = 3q + 3 (divisible by 3), n + 2 = 3q + 4 ⎪⎪ case = 3
⎭
38. The sum of four consecutive numbers in AP is 32 and the ratio of product
of the first and last terms to the product of two middle terms is 7:15. Find
the numbers.
Ans: Let four consecutive number be a – 3d, a – d, a + d, a + 3d 1/2
Sum = 32 ∴ 4a = 32 ⇒ a = 8 1/2
(a − 3d)(a + 3d) 7
= ⇒ 15(64 − 9d 2 ) = 7(64 − d 2 ) 1
(a − d)(a + d) 15
∴ d2 = 4 ⇒ d = ±2 1
Four numbers are 2,6,10,14. 1

.30/1/3. 36

OR
Solve: 1+ 4 + 7 + 10 +…+ x = 287
Ans: x = a n = 1 + 3n − 3 = 3n − 2 1

n
Sn = 287 ⇒ [1 + 3n − 2] = 287 1
2
∴ 3n 2 − n − 574 = 0 1/2
(n − 14)(3n + 41) = 0 ⇒ n = 14 1
∴ x = 3n – 2 = 40 1/2
39. Draw a Δ ABC with BC = 7 cm, ∠ B = 45° and ∠ A = 105°. Then construct
3
another triangle whose sides are times the corresponding sides of Δ ABC.
4
Ans: Construction Δ ABC with given measurement. 1
Construction of similar triangle 3
40. From the top of a 7 m high building the angle of elevation of the top of a
tower is 60° and the angle of depression of its foot is 45°. Determine the
height of the tower.
A

h cor. fig 1
Ans: From the figure, =tan 60° h
x
⇒ h = x 3 …(i) 60°
45°
x
B 1
7 7m 7m
and = tan 45°
x D C
x
⇒ x=7 1
From (i), h = 7 3

(
∴ Height of tower = 7 3 + 7 m or 7 ) ( 3 + 1) m 1

.30/1/3. 37