CBSE Class 10 Question Paper 2026 Solution Mathematics Standard

FOR CBSE CLASS 10 EXAM PREPARATION

CBSE Class 10 2026
Question Paper
Solution · Mathematics
Standard
EXAM YEAR TYPE SUBJECT

CBSE Class 10 2026 Question Paper Solution Mathematics Standard

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Strictly Confidential ag
Marking Scheme

(For Internal and Restricted use only)
Secondary School Examination, 2026
MATHEMATICS (STANDARD) (041) (PAPER CODE 30/1/1)

General Instructions: -

1. You are aware that evaluation is the most important process in the actual and correct assessment of
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the candidates. A small mistake in evaluation may lead to serious problems which may affect the
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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
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Guidelines carefully.
2.
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“Evaluation policy is a confidential policy as it is related to the confidentiality of the
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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 BNS.”
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

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strictly adhered to and religiously followed. However, while evaluating, answers which are

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based on latest information or knowledge and/or are innovative, they may be assessed for

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their correctness otherwise and due marks be awarded to them. In Class-X, while evaluating

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the Competency-based questions, please try to understand given answer and even if reply is

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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
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discussion. The remaining answer books meant for evaluation shall be given only after ensuring

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that there is no significant variation in the marking of individual evaluators.
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6.. Evaluators will mark (✓) wherever answer is correct. For wrong answer e
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s em Evaluators will not put right (✓) while evaluating which gives an impression
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la and no marks are awarded. This is most common mistake whichaevaluators are committing.
ag 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.
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”.

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10. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

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11. A full scale of marks 0 to 80 (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 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.
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.

2

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MARKING SCHEME
se MATHEMATICS (Subject Code–041)
(PAPER CODE: 30/1/1) ag l
Q. No. EXPECTED OUTCOMES/VALUE POINTS Step Marks

SECTION – A

Question Numbers 1 to 20 are multiple choice questions of 1 mark each.

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1.
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Sol. (a) 48 las ag 1
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2.

Sol. (a) a prime number 1

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3.
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Sol. (b) 6
a 1

4.

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c. o (a) 0 m.co
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Sol. 1

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a

Sol. (d) an infinite number of solutions 1

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6.

Sol. (a) √2 1

7.

Sol. (d) 16 cm 1

8.

Sol. (a) 𝑥-axis 1

9.

Sol. √𝑏2 −𝑎2 1
(c)
𝑏

10.

Sol. 5 1
(b)
4

4

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a

Sol. (d) 60° 1

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12.
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Sol.
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(a) 60 l° a ag 1
a
13.

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Sol. (d) 35°
s em 1

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14. a

Sol. (a) 22 cm 1

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15.
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a Sol. (c) 35° 1

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16.

Sol. (a) 3 𝜋 𝑑 2 1

17.

Sol. (c) 15 1

18.

Sol. 5 1
(c)
6

19.

Sol. (c) Assertion (A) is true, but Reason (R) is false. 1

20.

Sol. (b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is 1
not the correct explanation of the Assertion (A).
6

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se SECTION – B
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Question Numbers 21 to 25 are Very Short Answer (VSA) type
questions, carrying 2 marks each.

21.

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Sol.
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Here, α + β = 3, αβ = −1 I 1
e m
1 1
s emα+β 3 as
II gl ½
α
+ =
β
l a =
αβ −1
a
ag = −3 III ½

22 (A).

Sol.

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s e
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a

AD AE
Since DE ∥ BC ⟹ = I 1
DB EC

𝑥 𝑥+2
⟹ =
𝑥−2 𝑥−1
o mII ½

mSolving, we get 𝑥 = 4 . c
c. o s e m III ½

s em g l a
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a

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22 (B).

Sol. AB BC AC
∆ ABC~ ∆XYZ ⟹ = =
XY YZ XZ

4 6 𝑦
⟹ = = I 1
𝑥 7.2 6

Solving, we get 𝑥 = 4.8 cm, 𝑦 = 5 cm II ½+½

23.

Sol. √(𝑥 − 7 + 9)2 + (2𝑥 − 11)2 = 5√2 I ½

⟹ (𝑥 + 2)2 + (2𝑥 − 11)2 = 50

⟹ 5𝑥 2 − 40𝑥 + 75 = 0 or 𝑥 2 − 8𝑥 + 15 = 0
II ½
⟹ (𝑥 − 5)(𝑥 − 3) = 0

∴ 𝑥 = 3,5 III ½+½

24 (A).

Sol. 24 P
tan θ = =
7 B

24 7
Getting, sin θ = and cos θ = I 1½
25 25

24 7
∴ sin θ + cos θ = +
25 25

31
=
25 II ½

8

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24 (B)

Sol. (1+sin θ)(1−sin θ) 1−sin2 θ
=
(1+cos θ)(1−cos θ) 1−cos2 θ I 1
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cos2 θ
=m = cot 2 θ
.co m
sin2 θ II ½

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s e 7 2 49
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= ( ) =
g l a 8 64 III ½

25.
a

Sol.

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c. oCorrect Figure
em
I ½

OM ⊥ AB
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AM = √52 − 42 = 3 cm ag II 1

AB = 2 × 3 = 6 cm III ½

SECTION – C

Question numbers 26 to 31 are Short Answer (SA) type questions,
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carrying 3 marks each.
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26.

as g l
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𝐩
∴ √3 = , where 𝑞 ≠ 0 and 𝑝 & 𝑞 are coprime. I ½
𝐪

3𝑞2 = 𝑝2 ⟹ 𝑝2 is divisible by 3 ⟹ 𝑝 is divisible by 3 ----- (i) II 1

Let 𝑝 = 3𝑎, where ‘𝑎’ is some integer
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9𝑎2 = 3 𝑞2 ⟹ 𝑞2 = 3 𝑎2 ⟹ 𝑞2 is divisible by 3

⟹ 𝑞 is divisible by 3 ----- (ii) III 1

(i) and (ii) leads to a contradiction as ‘𝑝’ and ‘𝑞’ are coprime. IV ½

∴ √3 is an irrational number.

27.

Sol.

Let coordinates of P be (𝑥, 0) and P divides the line segement AB in the
ratio k ∶ 1 I ½

−4k−6 −k+5
( , ) = (𝑥, 0) II 1
k+1 k+1

−k+5
⟹ =0⟹k=5 III ½
k+1

Hence the required ratio is 5: 1 IV ½
−4×5−6 13
∴ Coordinates of P are ( , 0) = (− , 0) V ½
5+1 3

28 (A).

Sol. 𝑥−ℎ
𝑥 = ℎ + 𝑎 cos θ ⟹ = cos θ I 1
𝑎

𝑦−𝑘
𝑦 = 𝑘 + 𝑏 cos θ ⟹ = sin θ II 1
𝑏

𝑥−ℎ 2 𝑦−𝑘 2
∴ LHS = (
𝑎
) + ( 𝑏 ) = cos 2 θ + sin2 θ = 1 = RHS III 1

OR

10

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Sol. sin A sin A
tan A tan A
LHS = − = cos A
1 − cos A
1 I 1
1+sec A 1−sec A 1+ 1−
cos A cos A

sin A sin A
= − II ½
cos A+1 cos A−1
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= sin A (
−2
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III 1
−sin2 A

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2
sin A s
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= = 2 cosec A = RHS
g l a IV ½

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29 (A).

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Sol.
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a

AC = √32 + 42 = 5 cm I ½

Let BE = BD = 𝑥 cm II ½
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AD = 4 − 𝑥 = AF, CE = 3 − 𝑥 = CF
om c. o III ½

. c e m
e m AF + CF = AC ⟹ 4 − 𝑥 + 3 − 𝑥 = 5
l as
las ∴𝑥=1
ag IV 1
ag BD = BE = 1 and ∠B = 90°

Hence radius of circle = 𝑥 = 1 cm V ½

ALTERNATE SOLUTION:

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AC = √32 + 42 = 5 cm I ½

Let 𝑟 be the radius of the circle
1
ar(∆𝐴𝐵𝐶) = × 4 × 3 = 6 𝑐𝑚2 II ½
2

1 1 1
Also, ar(∆𝐴𝐵𝐶) = ( × 𝑟 × 4) + ( × 𝑟 × 3) + ( × 𝑟 × 5) III 1
2 2 2

⟹ 6𝑟 = 6 ⟹ 𝑟 = 1 IV 1

Hence the radius of the circle is 1 cm.

OR

29 (B)

Sol.

PM = PN

QS = QM I 1½

RS = RN

PM + PN = PQ + QM + PR + RN II ½

2 PM = PQ + QS + PR + RS

= PQ + QS + RS + PR III ½

12

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= PQ + QR + PR
se 1
∴ PM = (PQ + QR + PR) ag l IV ½
2

30.

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Sol. Radius of cylinder = radius of hemisphere = cm
m
I ½
2
s e
s em l a
ag
Total height of solid = 20 cm

g l a
a of cylinder = (20 − 72 − 72) cm = 13 cm
Height II ½

22 7 2 2 22 7 3
Total volume of solid = [ × ( ) × 13 + 2 × × ×( ) ] III 1
7 2 3 7 2

22 7 2 14
= × ( ) × (13 + )
7 2 3

4081
m
c. o
= = 680.1 cm3 (approx.) IV 1
6

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31.
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ag

Sol. Possible outcomes are

(1,1)(1,2), (1,3), (1,4), (1,5), (1,6)
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m (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) .co
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s e (3,1)(3,2), (3,3), (3,4), (3,5), (3,6)
g l a
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I 1
(4,1)(4,2), (4,3), (4,4), (4,5), (4,6) a
ag
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
6 1
(i) P(same number appear on both the dice) = or II 1
36 6

1 5

om .
(ii) P(different number appear on both the dice) = 1 − = III 1
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6 6

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SECTION – D

Question numbers 32 to 35 are Long Answer (LA) type questions,
carrying 5 marks each.

32.

Sol.

Correct graph of the line 2𝑥 − 3𝑦 + 6 = 0 I 1

Correct graph of the line 2𝑥 + 3𝑦 − 18 = 0 II 1

Correct graph of the line 𝑥 = 0 III ½

Coordinates of vertices of the triangle are A(0,2), B(3,4) and C(0,6) IV 1½
1
Area of ∆ ABC = × 4 × 3 = 6 sq. units V 1
2

33 (A).

Sol. Let the speed of faster train be 𝑥 km/h

∴ speed of slower train = (𝑥 − 10) km/h

According to the question,
200 200
− =1 I 2
𝑥−10 𝑥

14

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⟹ 𝑥 2 − 10𝑥 − 2000 = 0
a
II 1
s g l
⟹ (𝑥 − 50)(𝑥 + 40) = 0 a III 1

∴ 𝑥 = 50 IV ½
𝑥 = −40 (Rejected)

Hence, speed of faster train = 50 km/h

and speed of slower train = 40 km/h V ½
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33 (B).

Sol. Let the sides of the two squares be 𝑥 m and 𝑦 m (𝑥 > 𝑦)

𝑥 2 + 𝑦 2 = 640 I 1

and 4𝑥 − 4𝑦 = 64 ⟹ 𝑦 = 𝑥 − 16 II 1
m
∴ 𝑥 2 + (𝑥 − 16)2 = 640
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⟹ 𝑥 2 − 16𝑥 − 192 = 0 s e
la
III 1

⟹ (𝑥 − 24)(𝑥 + 8) = 0 ag IV 1
∴ 𝑥 = 24
V ½
𝑥 = −8 (Rejected)

⟹ 𝑦 = 24 − 16 = 8 VI ½

Hence the sides of the two squares are 24 m and 8 m
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34 (A).

s g l a
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a to intersect the other two sides in distinct points, the other two sides are
I 1

divided in the same ratio.

For Correct given , to prove, construction and figure II 2

For correct proof III 2

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OR

34 (B).

Sol. (i) ∆ ABC~∆PQR ⟹ ∠A = ∠P I ½

AB AC 2 AM AC
= ⟹ = (as CM and RN are the medians) II 1½
PQ PR 2 PN PR

∴ ∆ AMC~∆PNR III ½

(ii) ∆ ABC~∆PQR ⟹ ∠B = ∠Q IV ½

AB BC 2 MB BC
= ⟹ = (as CM and RN are the medians) V 1½
PQ QR 2 NQ QR

∴ ∆ CMB~∆RNQ VI ½

35.

Sol.

16

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Class Interval Frequency (𝑓𝑖 ) 𝑥𝑖
a 𝑓𝑥 𝑖 𝑖

0 − 10 1 5 5

10 − 20 𝑥 15 15 𝑥

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20 − 30 m 125 .co
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5 25
e m
em l as
la s
30 − 40 7 35 245
ag
g
a 40 − 50 𝑦 45 45 𝑦

50 − 60 3 55 165

60 − 70 1 65 65

m
Total 25
m .co 605 + 15𝑥 + 45𝑦

s e
g la
a
Correct Table I 2

∑ 𝑓𝑖 = 25 ⟹ 𝑥 + 𝑦 = 8 … (i) II ½
605+15𝑥+45𝑦
Mean (𝑥̅ ) = = 35 III ½
25

⟹ 15𝑥 + 45𝑦 = 270 or 𝑥 + 3𝑦 = 18 … (ii) IV 1

Solving (i) and (ii), we get 𝑥 = 3, 𝑦 = 5
o mV ½+½

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c. o s e m
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a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 17 of 21

SECTION – E

Question numbers 36 to 38 are Case Study Based questions,
carrying 4 marks each

36.

Sol. (i) Required distance for the first potato = 5 + 5 = 10 m I 1

(ii) Required distance for the second potato = 8 + 8 = 16 m I 1

(iii) (a) Distances covered form an A.P. with 𝑎 = 10 and 𝑑 = 6 I 1
10
∴ 𝑆10 = [2 × 10 + 9 × 6]
2

= 5 × 74 = 370 m II 1
OR
(iii) (b) Distances covered form an A.P. with 𝑎 = 10 and 𝑑 = 6 I ½
10
∴ 𝑆10 = [2 × 10 + 9 × 6]
2

= 5 × 74 = 370 m II 1
370
Time = = 74 seconds III ½
5

18

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se 37.
g l a
a

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m .co s e m
s e l a
g l a ag
a

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s e
g la
Sol.
a

m
m (i) c. o I
.co
√3 6

em
cos 30° = = ½
2 OB

em l as
las ⟹ OB =
12
or 4√3 m
ag II ½
g
√3
a (ii) cos 60° = =
1 6
2 OA I ½

⟹ OA = 12 m II ½
1 BP
(iii) (a) tan 30° = =
√3 6

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⟹ BP = 2√3 m I 1

AP
tan 60° = √3 =
6

⟹ AP = 6√3 m II ½

AB = AP − BP = 6√3 − 2√3 = 4√3 m III ½

OR
1 BP
(iii) (b) tan 30° = =
√3 6

⟹ BP = 2√3 m I 1
1
ar(∆OPB) = × BP × OP
2

1
= × 2 √3 × 6 = 6 √ 3 𝑚 2 II 1
2

38.

Sol. 35
(i) 𝑟= cm = 17.5 cm I 1
2

22 35
(ii) Circumference = 2 × × = 110 cm I 1
7 2

(iii) (a) Total length of wire required = (5 × 35 + 110) cm I 1

= 285 cm II 1

OR

20

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360
a
(b) Central angle of each sector = = 36° I 1
l
(iii)
s 36 22
10

35 a
35
g
Area of each sector = × × ×
360 7 2 2

385
= or 96.25 𝑐𝑚2 II 1
4

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a

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a

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a

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