Class 10 Question Paper 2025 Maths Standard

ANNUAL EXAMINATION

Question Paper
2025
NCERT BASED SYLLABUS

FOR CBSE AND STATE BOARD
FOLLOWING NCERT

KVS QUESTION PAPERS

Series : GE1FH SET~1

. - >
Roll No. Q.P. Code 30/1/1
- -
-
Candidates must write the Q.P. Code
on the title page of the answer-book.
> NOTE
(I) - (I) Please check that this question paper
27 contains 27 printed pages.
(II) - - (II) Q.P. Code given on the right hand side
- - of the question paper should be written
on the title page of the answer-book by
the candidate.
(III) - 38 (III) Please check that this question paper
contains 38 questions.
(IV) (IV) Please write down the Serial Number of
, - the question in the answer-book at the
given place before attempting it.

(V) - 15 (V) 15 minute time has been allotted to read
- this question paper. The question paper
will be distributed at 10.15 a.m. From
10.15 10.15 10.15 a.m. to 10.30 a.m., the candidates
10.30 - will read the question paper only and
- will not write any answer on the
answer-book during this period.

#

J{UV (_mZH$)
MATHEMATICS (STANDARD)
3 80
Time allowed : 3 hours Maximum Marks : 80

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(i) - 38

(ii) - , , ,
(iii) 1 18 (MCQ) 19 20
1

(iv) 21 25 - (VSA) 2

(v) 26 31 - (SA) 3

(vi) 32 35 - (LA) 5

(vii) 36 38 4
2

(viii) - , 2 , 2 ,
2 3

(ix) = ,

(x)

20 (MCQ) , 1 20 1=20

1. , 3x2 + 6x + k + + = , k

(A) 8

(B) 8

(C) 4

(D) 4

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General Instructions :
Read the following instructions very carefully and strictly follow them :
(i) This question paper contains 38 questions. All questions are compulsory.
(ii) This question paper is divided into five Sections A, B, C, D and E.
(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs)
and questions number 19 and 20 are Assertion-Reason based questions of
1 mark each.
(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type
questions, carrying 2 marks each.
(v) In Section C, Questions no. 26 to 31 are short answer (SA) type
questions, carrying 3 marks each.
(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions
carrying 5 marks each.
(vii) In Section E, Questions no. 36 to 38 are case study based questions
carrying 4 marks each. Internal choice is provided in 2 marks questions in
each case study.
(viii) There is no overall choice. However, an internal choice has been provided
in 2 questions in Section B, 2 questions in Section C, 2 questions in
Section D and 3 questions in Section E.
(ix) Draw neat diagrams wherever required. Take = wherever required,
if not stated.
(x) Use of calculator is not allowed.

SECTION A
This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each. 20 1=20

1. If and are the zeroes of polynomial 3x 2 + 6x + k such that
+ + = , then the value of k is :

(A) 8
(B) 8
(C) 4
(D) 4

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2. x=1 y = 2, 2x 3y + a = 0 2x + 3y b=0
,
(A) a = 2b
(B) 2a = b
(C) a + 2b = 0
(D) 2a + b = 0

3. P( 4, 5) Q(4, 6) -
(A) x-
(B) y-
(C) -
(D) x- y-

4. 7 + 4 sin =9 , :
(A) 90
(B) 30
(C) 45
(D) 60

5. tan2 :

(A) 1
(B) 0
(C) 1
(D) 2

6. HCF(98, 28) = m LCM(98, 28) = n , n 7m :
(A) 0
(B) 28
(C) 98
(D) 198

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2. If x = 1 and y = 2 is a solution of the pair of linear equations 2x 3y + a = 0
and 2x + 3y b = 0, then :
(A) a = 2b
(B) 2a = b
(C) a + 2b = 0
(D) 2a + b = 0

3. The mid-point of the line segment joining the points P( 4, 5) and Q(4, 6)
lies on :
(A) x-axis
(B) y-axis
(C) origin
(D) neither x-axis nor y-axis

4. If is an acute angle and 7 + 4 sin = 9, then the value of is :
(A) 90
(B) 30
(C) 45
(D) 60

5. The value of tan2 is :

(A) 1
(B) 0
(C) 1
(D) 2

6. If HCF(98, 28) = m and LCM(98, 28) = n, then the value of n 7 m is :
(A) 0
(B) 28
(C) 98
(D) 198

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7. -
(A)
(B)
(C)
(D)

8. ABC DEF , B= E, F= C AB = 3 DE ,
(A)
(B)
(C)
(D)

9. ( 1)n + ( 1)8 = 0 , n:
(A)
(B)
(C)
(D)

10. ,

(A) 3 (B) 5
(C) 2 (D) 4

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7. The tangents drawn at the extremities of the diameter of a circle are
always :
(A) parallel
(B) perpendicular
(C) equal
(D) intersecting

8. In triangles ABC and DEF, B= E, F= C and AB = 3 DE. Then,
the two triangles are :
(A) congruent but not similar
(B) congruent as well as similar
(C) neither congruent nor similar
(D) similar but not congruent

9. If ( 1)n + ( 1)8 = 0, then n is :
(A) any positive integer
(B) any negative integer
(C) any odd number
(D) any even number

10. Two polynomials are shown in the graph below. The number of distinct
zeroes of both the polynomials is :

(A) 3 (B) 5
(C) 2 (D) 4

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11. m 2m2 + 3m

(A) 10

(B) 9

(C) 12

(D) 4

12. 15x 18x

(A) x

(B) 11x

(C) 17x

(D) 34x

13. 52

(A)

(B)

(C)

(D)

14. - ?
(A) 1·4142387954012 . . . .

(B)

(C)

(D) 1·857142

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11. If the sum of first m terms of an AP is 2m 2 + 3m, then its second term is :

(A) 10

(B) 9

(C) 12

(D) 4

12. Mode and Mean of a data are 15x and 18x, respectively. Then the median
of the data is :

(A) x

(B) 11x

(C) 17x

(D) 34x

13. A card is selected at random from a deck of 52 playing cards. The
probability of it being a red face card is :

(A)

(B)

(C)

(D)

14. Which of the following is a rational number between 3 and ?

(A) 1·4142387954012 . . . .

(B)

(C)

(D) 1·857142

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15. 40 72 ,

(A) 200

(B) 100

(C) 20

(D) 10

16. , O P - PA
POB = 115 , APO :

(A) 25
(B) 65
(C) 90
(D) 35

17. 150 m
30 :
(A) 100 m
(B) 300 m
(C) 150 m
(D) 150 m

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15. If a sector of a circle has an area of 40 sq. units and a central angle of
72 , the radius of the circle is :

(A) 200 units
(B) 100 units
(C) 20 units
(D) 10 2 units

16. In the given figure, PA is a tangent from an external point P to a circle
with centre O. If POB = 115 , then APO is equal to :

(A) 25

(B) 65
(C) 90

(D) 35

17. A kite is flying at a height of 150 m from the ground. It is attached to a
string inclined at an angle of 30 to the horizontal. The length of the
string is :
(A) 100 m

(B) 300 m

(C) 150 m
(D) 150 m

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60
18. 20 cm cm

:

(A) 30

(B) 60

(C) 90

(D) 50

19 20
(A) (R) (A), (B),
(C) (D)

(A) (A) (R) (R), (A)

(B) (A) (R) , (R), (A)

(C) (A) , (R)

(D) (A) , (R)

19. (A) 1 20 1

(R) E , P(E) = 1 , E

20. (A) ,

(R) r 3 r2

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18. A piece of wire 20 cm long is bent into the form of an arc of a circle of
radius cm. The angle subtended by the arc at the centre of the circle

is :

(A) 30

(B) 60

(C) 90

(D) 50

Questions number 19 and 20 are Assertion and Reason based questions. Two
statements are given, one labelled as Assertion (A) and the other is labelled as
Reason (R). Select the correct answer to these questions from the codes (A), (B),
(C) and (D) as given below.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the
correct explanation of the Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not
the correct explanation of the Assertion (A).

(C) Assertion (A) is true, but Reason (R) is false.

(D) Assertion (A) is false, but Reason (R) is true.

19. Assertion (A) : The probability of selecting a number at random from the
numbers 1 to 20 is 1.
Reason (R): For any event E, if P(E) = 1, then E is called a sure event.

20. Assertion (A) : If we join two hemispheres of same radius along their
bases, then we get a sphere.
Reason (R): Total Surface Area of a sphere of radius r is 3 r2.

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5 - (VSA) , 2 5 2=10

21. x cos 60 + y cos 0 + sin 30 cot 45 = 5 , x + 2y

22. p(x) = x2 + x

23. (2a, a 7) (11, 9)
10 ,

24. ABC PQR AB = 6 cm, BC = 4 cm, AC = 8 cm
PR = 6 cm , (PQ + QR)

, = 1= 2 , PQS TQR.

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SECTION B

This section has 5 Very Short Answer (VSA) type questions carrying 2 marks
each. 5 2=10

21. (a) If x cos 60 + y cos 0 + sin 30 cot 45 = 5, then find the value of
x + 2y.

OR

(b) Evaluate :

22. Find the zeroes of the polynomial p(x) = x2 + x .

23. The coordinates of the centre of a circle are (2a, a 7). Find the value(s)
9) and has diameter
10 units.

24. (a) If ABC PQR in which AB = 6 cm, BC = 4 cm, AC = 8 cm and
PR = 6 cm, then find the length of (PQ + QR).
OR

(b) In the given figure, = and 1= 2, show that

PQS TQR.

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25. 26 m P

A B 10 m PA PB
-

6 - (SA) , 3 6 3=18

26. ,O BCD C -
BAC + ACD = 90 .

-

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25. A person is standing at P outside a circular ground at a distance of 26 m
from the centre of the ground. He found that his distances from the
points A and B on the ground are 10 m (PA and PB are tangents to the
circle). Find the radius of the circular ground.

SECTION C

This section has 6 Short Answer (SA) type questions carrying 3 marks each. 6 3=18

26. (a) In the given figure, O is the centre of the circle and BCD is tangent

to it at C. Prove that BAC + ACD = 90 .

OR

(b) Prove that opposite sides of a quadrilateral circumscribing a circle

subtend supplementary angles at the centre of the circle.

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tan
27. : + = 1 + sec cosec

: + =

28. y- (5, 6) ( 1, 4)

29.

30.

, m3 ,
( = )

31. 2

4 - (LA) , 5 4 5=20

32. A B , 8% 9% ,
- 1,860 ,
, 20

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27. (a) Prove that : + = 1 + sec cosec

OR

(b) Prove that : + =

28. Find the ratio in which the y-axis divides the line segment joining the
points (5, 6) and ( 1, 4). Also find the point of intersection.

29. Prove that is an irrational number.

30. A room is in the form of a cylinder surmounted by a hemispherical dome.
The base radius of the hemisphere is half of the height of the cylindrical
part. If the room contains m3 of air, find the height of the

cylindrical part. (Use = ).

31. Two dice are thrown at the same time. Determine the probability that the
difference of the numbers on the two dice is 2.

SECTION D

This section has 4 Long Answer (LA) type questions carrying 5 marks each. 4 5=20

32. Vijay invested certain amounts of money in two schemes A and B, which
offer interest at the rate of 8% per annum and 9% per annum,
respectively. He received 1,860 as the total annual interest. However,
had he interchanged the amounts of investments in the two schemes, he
would have received 20 more as annual interest. How much money did
he invest in each scheme ?

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33. ABCD BD, AE F ,
E BC DF EF = FB FA.

ABC , AD BC AD2 = BD DC ,
BAC = 90 .

34. 60 cm 25 cm

480 km 8 km/h
, 3

35. , 18

11 13 7

13 15 6

15 17 9

17 19 13

19 21 f

21 23 5

23 25 4

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33. (a) The diagonal BD of a parallelogram ABCD intersects the line
segment AE at the point F, where E is any point on the side BC.
Prove that DF EF = FB FA.

OR

(b) In ABC, if AD BC and AD2 = BD DC, then prove that
BAC = 90 .

34. (a) The perimeter of a right triangle is 60 cm and its hypotenuse is
25 cm. Find the lengths of other two sides of the triangle.

OR

(b) A train travels 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 speed of the train.

35. in the following table, if the mean of the
given data is 18. Hence find the mode.

Daily Allowance Number of Children

11 13 7

13 15 6

15 17 9

17 19 13

19 21 f

21 23 5

23 25 4

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3 4 3 4=12

1
36.
,
300 ,
50 ,
350 , 400
10

(i) , 1

(ii) 8 1

(iii) 10 2

(iii) 6 ,
2

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SECTION E

This section has 3 case study based questions carrying 4 marks each. 3 4=12

Case Study 1
36. A school is organizing a charity run to raise funds for a local hospital.
The run is planned as a series of rounds around a track, with each round
being 300 metres. To make the event more challenging and engaging, the
organizers decide to increase the distance of each subsequent round by
50 metres. For example, the second round will be 350 metres, the third
round will be 400 metres and so on. The total number of rounds planned
is 10.

Based on the information given above, answer the following questions :
(i) Write the fourth, fifth and sixth term of the Arithmetic Progression
so formed. 1
(ii) Determine the distance of the 8th round. 1
(iii) (a) Find the total distance run after completing all 10 rounds. 2
OR
(iii) (b) If a runner completes only the first 6 rounds, what is the
total distance run by the runner ? 2

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2
37. ,
,

, 35 mm
5 , 10 ,

(i) 1

(ii) ACB 1

(iii) 2

(iii) 2

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Case Study 2

37. A brooch is a decorative piece often worn on clothing like jackets, blouses
or dresses to add elegance. Made from precious metals and decorated
with gemstones, brooches come in many shapes and designs.

One such brooch is made with silver wire in the form of a circle with
diameter 35 mm. The wire is also used in making 5 diameters which
divide the circle into 10 equal sectors as shown in the figure.

Based on the above given information, answer the following questions :

(i) Find the central angle of each sector. 1

(ii) Find the length of the arc ACB. 1
(iii) (a) Find the area of each sector of the brooch. 2
OR
(iii) (b) Find the total length of the silver wire used. 2

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3
38.
60 , 40
45

(i) CD h , BD 1
(ii) BC 1
(iii) CE [ = 1·73 ] 2

(iii) AC = 100 m , AE 2

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Case Study 3
38. Amrita stood near the base of a lighthouse, gazing up at its towering
height. She measured the angle of elevation to the top and found it to be
60 . Then, she climbed a nearby observation deck, 40 metres higher than
her original position and noticed the angle of elevation to the top of
lighthouse to be 45 .

Based on the above given information, answer the following questions :
(i) 1
(ii) 1
(iii) (a) Find the height CE of the lighthouse [Use = 1·73] 2
OR
(iii) (b) Find distance AE, if AC = 100 m. 2

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Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2025
MATHEMATICS (Standard) (Q.P. CODE 30/1/1)
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.
2. “Evaluation policy is a confidential policy as it is related to the confidentiality of the
examinations conducted, Evaluation done and several other aspects. It’s leakage to public in
any manner could lead to derailment of the examination system and affect the life and future
of millions of candidates. Sharing this policy/document to anyone, publishing in any magazine
and printing in News Paper/Website etc. may invite action under various rules of the Board
and IPC.”
3. 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 due 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, due
marks should be awarded.
4. The Marking scheme carries only suggested value points for the answers.
These are in the nature of Guidelines only and do not constitute the complete answer. The students
can have their own expression and if the expression is correct, the due marks should be awarded
accordingly.
5. 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. If there is any variation, the same should be zero after deliberation and
discussion. 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.
6. Evaluators will mark (✓) wherever answer is correct. For wrong answer CROSS ‘X” be marked.
Evaluators will not put right (✓) while evaluating which gives an impression that answer is correct
and no marks are awarded. This is most common mistake which evaluators are committing.
7. 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 totalled up and written on the left-hand margin and
encircled. This may be followed strictly.
8. If a question does not have any parts, marks must be awarded on the left-hand margin and encircled.
This may also be followed strictly.

MS_X_Mathematics_041_30/1/1_2024-25
1

9. If a student has attempted an extra question, answer of the question deserving more marks should
be retained and the other answer scored out with a note “Extra Question”.
10. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
80
11. A full scale of marks __________ (example 0 to 80/70/60/50/40/30 marks as given in Question
Paper) has to be used. Please do not hesitate to award full marks if the answer deserves it.
12. 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). This is in view of the reduced syllabus and number
of questions in question paper.
13. 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 totalling of marks awarded to an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totalling on the title page.
● Wrong totalling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● 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.
14. 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.
15. Any un assessed 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.
16. The Examiners should acquaint themselves with the guidelines given in the “Guidelines for spot
Evaluation” before starting the actual evaluation.
17. Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title
page, correctly totalled and written in figures and words.
18. The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once
again reminded that they must ensure that evaluation is carried out strictly as per value points for
each answer as given in the Marking Scheme.

MS_X_Mathematics_041_30/1/1_2024-25
2

MARKING SCHEME
MATHEMATICS (Subject Code–041)
(PAPER CODE: 30/1/1)

Q. No. EXPECTED OUTCOMES/VALUE POINTS Marks
SECTION A
This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
1.

Sol. (D) 4 1
2.

Sol. (B) 2a = b 1
3.

Sol. (B) y – axis 1
4.

Sol. (B) 300 1

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3

5.

Sol. (C) −1 1
6.

Sol. (C) 98 1
7.

Sol. (A) parallel 1
8.

Sol. (D) similar but not congruent 1
9.

Sol. (C) any odd number 1

MS_X_Mathematics_041_30/1/1_2024-25
4

10.

Sol. (C) 2 1
11.

Sol. (B) 9 1
12.

Sol. (C) 17x 1
13.

3
Sol. (D) 1
26

MS_X_Mathematics_041_30/1/1_2024-25
5

14.

Sol. (D) 1.857142 1
15.

Sol. (D) 10√2 units 1
16.

Sol. (A) 25° 1
17.

Sol. (B) 300 m 1

MS_X_Mathematics_041_30/1/1_2024-25
6

18.

Sol. (B) 60° 1

19.

Sol. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct 1
explanation of the Assertion (A).
20

Sol. (C) Assertion (A) is true, but Reason (R) is false. 1
SECTION B
This section has 5 Very Short Answer (VSA) type questions carrying 2 marks each.
21.(a)

Sol. 1 1
x ( ) + y (1) + – 1 = 5 1½
2 2
⟹ x + 2y = 11 ½
OR

MS_X_Mathematics_041_30/1/1_2024-25
7

21. (b)

Sol. 2
(√3)
2 2
1½
√3 √3
( 2 ) +( 2 )

= 2 ½
22.

Sol. 1
(3x2 + 4x – 4) ½
3
1
= (3x2 + 6x – 2x – 4) ½
3
1
= (3x – 2)(x + 2) ½
3
2
Zeroes are ,–2 ½
3
23.

Sol. radius = 5√2 units ½
(2a − 11)2 + (a − 7 + 9)2 = 50 ½
⟹ a2 − 8a + 15 = 0 ½
⟹ (a − 5)(a − 3) = 0
⟹ a = 5, 3 ½
24.(a)

Sol. 6 4 8
= = ½
PQ QR 6
9
⟹ PQ = cm or 4.5 cm ½
2
and QR = 3 cm ½
∴ PQ + QR = 7.5cm ½
OR

MS_X_Mathematics_041_30/1/1_2024-25
8

24.(b)

Sol. In  PQR,  1 =  2  PR = PQ ½
QR QT QR QT ½
 = ⇒ =
QS PR QS PQ
Also,  1 =  1 ½
  PQS   TQR ½
25.

Sol.  OAP = 90 ½
In right  OAP,
(26)2 = OA2 + (10)2 1
⟹ OA = √576= 24 ½
 radius = 24 m

MS_X_Mathematics_041_30/1/1_2024-25
9

SECTION C
This section has 6 Short Answer (SA) type questions carrying 3 marks each.
26. (a)

Sol. In  OAC,
OA = OC
  OCA =  OAC 1
Now,  OCD = 90 1
⟹ OCA +  ACD = 90 ½
⟹ OAC +  ACD = 90 ½
or  BAC +  ACD = 90
OR
26.(b)

Sol.

Correct Figure ½
 OAP   OAS 1
1=2
Similarly,  3 =  4 ,  5 =  6 ,  7 =  8 ½
Also,  1 +  2 +  3 +  4 +  5 +  6 +  7 +  8 = 360
 2 ( 1 +  4 +  5 +  8) = 360 ½
  AOB +  COD = 180 ½
Similarly,  BOC +  AOD = 180

MS_X_Mathematics_041_30/1/1_2024-25
10

27. (a)

Sol. tan θ cot θ
LHS = +
1−cot θ 1−tan θ
sin θ cos θ
cos θ sin θ ½
= cos θ + sin θ
1− 1−
sin θ cos θ
sin2 θ cos2 θ 1
= −
cos θ(sin θ−cos θ) sin θ (sin θ−cos θ)
1 sin3 θ − cos3 θ
= [ ] ½
(sin θ−cos θ) sin θ cos θ
(sin θ−cos θ)(sin2 θ+ sin θ cos θ+cos2 θ)
= ½
(sin θ−cos θ) sin θ cos θ
(1+ sin θ cos θ)
=
sin θ cos θ
= 1+ sec  cosec  = RHS ½
OR
27.(b)

Sol. sin A + cos A sin A – cos A
LHS = +
sin A – cos A sin A + cos A
(sin A + cos A) + (sin A − cos A)2
2
= 1
(sin A – cos A)(sin A + cos A)
sin2 A+cos2 A + 2 sin A cos A + sin2 A+cos2 A− 2 sin A cos A 1
=
sin2 A− cos2 A
1+ 1
= ½
sin2 A− (1− sin2 A)
2
= = RHS ½
2sin2 A− 1
28.

Sol.

Let the ratio be k:1 and point on y- axis be P(0, y) ½
−k+5
0=
k+1
k=5 1
Hence, ratio is 5:1 ½
−4(5)−6 −26 −13
y= = = ½
5+1 6 3
13
Coordinates of point of intersection are P(0, − 3 ) ½

MS_X_Mathematics_041_30/1/1_2024-25
11

29.

Sol. 1
Let be a rational number.
√5
1 p
∴ = , where q ≠ 0 and let p & q be the co-primes. ½
√5 q
5p = q2 ⟹ q2 is divisible by 5.
2

⟹ q is divisible by 5. ----- (i) 1
let q = 5a, where ‘a’ is some integer.
25a2 = 5p2 ⟹ p2 = 5a2 ⟹ p2 is divisible by 5.
⟹ p is divisible by 5 ----- (ii) 1
(i) and (ii) leads to contradiction as p and q are coprimes. ½
1
∴ is an irrational number
√5
30.

Sol. Let r is the radius of hemisphere and cylinder and h is the height of cylinder
h = 2r ½
2
Volume of air in room = πr 3 + πr 2 h
3
1408 2
= πr 3 + πr 2 (2r) ½
21 3
1408 8 22
= × × r3
21 3 7
r3 = 8 1
∴ r=2m ½
and h = 4 m ½
31.

Sol. Total outcomes = 36 1
Number of Outcomes with difference of the numbers on the two dice is 2 = 8 1
(1,3) (3,1) (4,2) (2,4) (5,3) (3,5) (4,6) (6,4)
8 2
P (difference of the numbers on the two dice is 2) = or 1
36 9

MS_X_Mathematics_041_30/1/1_2024-25
12

SECTION D
This section has 4 Long Answer (LA) type questions carrying 5 marks each.
32.

Sol. Let Vijay invested ₹ x at 8% rate of interest & ₹ y at 9% rate of interest
ATQ,
8x 9y
+ = 1860 1½
100 100
or 8x + 9y = 186000 …………..(i)
9x 8y
+ = 1880 1½
100 100
or 9x + 8y = 188000 …………..(ii)
On solving (i) and (ii), we get
x = 12000 1
y = 10000 1
Hence, money invested in scheme A is ₹ 12000 and scheme B is ₹ 10000.
33.(a)

Sol.

Correct figure 1
In  ADF and  EBF,
 DFA =  EFB
 ADF =  FBE
  ADF   EBF 2
DF FA
 = 1
FB EF
 DF ×EF = FB ×FA 1
OR

MS_X_Mathematics_041_30/1/1_2024-25
13

33.(b)

Sol.

Correct figure 1
AD = BD  DC
2

AD BD 1
=
DC AD
Also,  ADB =  ADC
  DBA   DAC 1
 DBA =  DAC
1
 BAD =  DCA
Adding both
 DBA +  DCA =  DAC +  BAD
∴  BAC = 90° 1
34.(a)

Sol. Let the other two sides be x cm and y cm
ATQ
x + y + 25 = 60 1
y = 35 – x ½
Now,
x2 + y2 = (25)2
x2 + (35 – x)2 = 625 1
x2 – 35x + 300 = 0 1
(x – 20) (x – 15) = 0
 x = 20, 15 1
x = 20  y = 15
½
x = 15  y = 20
Hence sides are 15 cm and 20 cm.
OR
34.(b)

Sol. Let the speed of train be x km/h
Reduced speed of train = (x – 8) km/h ½
ATQ
480 480
− =3 1½
x−8 x

MS_X_Mathematics_041_30/1/1_2024-25
14

x2 – 8x – 1280 = 0 1½
(x – 40) (x + 32) = 0 1
 x = 40 ½
 Speed of train = 40 km/h
35.

Sol.
Daily Allowance 𝑥𝑖 𝑓𝑖 𝑓𝑖 𝑥𝑖
11 – 13 12 7 84
13 – 15 14 6 84
15 – 17 16 9 144
17 – 19 18 13 234
19 – 21 20 f 20f
21 – 23 22 5 110
23 – 25 24 4 96
Total 44 + f 752 + 20f
Correct table 1½

752+20f
Mean = 1
44+f
752+20f
⟹ 18 =
44+f
∴ f = 20 ½
modal class is 19 – 21 ½
20−13
mode = 19 + ×3 1
40−13−5
= 19.95 approx. ½

MS_X_Mathematics_041_30/1/1_2024-25
15

SECTION E
This section has 3 case study based carrying 4 marks each.
36.

Sol. A.P formed is 300, 350, 400……
(i) a4 = 450
a5 = 500 1
a6 = 550
(ii) a8 = 300 +7 × 50 ½
= 650 m ½
10
(iii) (a) S10 = 2 × (2 × 300 + 9 × 50) 1
= 5250 m 1
OR
6
(iii) (b) S6 = 2 × (2 × 300 + 5 × 50) 1
= 2250 m 1

MS_X_Mathematics_041_30/1/1_2024-25
16

37.

Sol. 3600
(i) central angle = = 36° 1
10
1 22 35
(ii) length of arc ACB = ×2× × = 11mm 1
10 7 2
1 22 35 35
(iii)(a) Area of each sector of the brooch = × × × 1
10 7 2 2
385
= mm2 or 96.25 mm2 1
4
OR
22 35
(iii) (b) length of silver wire used = 2 × × + 5 × 35 1
7 2
= 285 mm 1

MS_X_Mathematics_041_30/1/1_2024-25
17

38.

Sol. h
(i) = tan 45° = 1 ½
BD
⟹ BD = h m ½
h 1
(ii) = sin 45° = ½
BC √2
⟹ BC = √2h m ½
EC
(iii)(a) tan 60° =
AE
h+40 1
⟹ √3 =
h
⟹ h = 20 (√3 + 1) = 20 × 2.73 = 54.6 m ½
∴ CE = 54.6 + 40 = 94.6 m ½
OR

MS_X_Mathematics_041_30/1/1_2024-25
18

AE
(iii)(b) cos 60° =
AC
1 AE 1
⟹ =
2 100
∴ AE = 50 m 1

MS_X_Mathematics_041_30/1/1_2024-25
19

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