CBSE Class 10 Question Paper 2026 Solution Mathematics Basic

FOR CBSE CLASS 10 EXAM PREPARATION

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

CBSE Class 10 2026 Question Paper Solution Mathematics Basic

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Marking Scheme
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Strictly Confidential
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(For Internal and Restricted use
Secondary School Examination, 2026 (X)
SUBJECT NAME- MATHEMATICS BASIC (Q.P. CODE /Set No. 241/ 430/1/1)
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.
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To avoid mistakes, it is requested that before starting evaluation, you must read and
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understand the spot evaluation guidelines carefully.

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2 “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. Its leakage to

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the public in any manner could lead to derailment of the examination system and
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affect the life and future of millions of candidates. Sharing this policy/document to
anyone, publishing in any magazine and printing in Newspaper/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. The 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
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them. In class-XII, while evaluating the competency-based questions, please try to
understand the given Ans.wer and even if reply is not from a marking scheme but

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correct competency is enumerated by the candidate, due marks should be awarded.
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la value points for the Answers.
These are Guidelines only and dognot constitute the complete Answer. The students can
4 The Marking Scheme carries only suggested

have their own expression and a 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.
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m Evaluators will mark (√) wherever Answer is correct. For wrong Answer CROSS ‘X’ be
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marked. Evaluators will not put right (✓) while evaluating which gives the impression that

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the Answer is correct, and no marks are awarded. This is the most common
a for each part. Marks
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las 7 If a question has parts, please award marks on the right-hand g
a up and written in the left-
ag awarded for different parts of the question should then be totaled
hand margin and encircled. This may be followed strictly.
8 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.
9 If a student has attempted an extra question, Answer to 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.
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 must 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 totaling of marks awarded on an Answer.
● 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/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 the 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 total error
detected by the candidate shall damage the prestige of all the personnel engaged in the
evaluation work as also of the Board. Hence, 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 totaled and written in figures and words.
18 The candidates are entitled to obtain a 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.

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MARKING SCHEME – 430/1/1
MATHEMATICS BASIC (Subject Code 241)
Q.No. EXPECTED ANSWERS / VALUE POINTS Marks
SECTION-A
This section comprises multiple choice questions (MCQs) of 1 mark each.

1.
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Ans.
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2.
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Ans. 1

3.

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Ans. 1

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Ans. 1

5.

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1
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Ans.

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a Ans.
1

7.

Ans. 1

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8.

Ans. 1

9.

Ans. 1

10.

1
Ans.

11.

Ans. 1

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12.

Ans. 1

13.

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Ans.
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14.
a
Ans. 1

15.

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Ans. a 1

16.

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Ans.. s e
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1
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a

1
Ans.

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18.

Ans. 1

19.

Ans. (a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct
1
explanation of the Assertion (A).

20.

Ans. (d) Assertion (A) is false, but Reason (R) is true. 1
SECTION-B
This section comprises 5 Very Short Answer (VSA) type questions of 2 marks each.

21.

Ans.
Let the coordinates of point be (x, y)
2×3 + 7×(−6) 2×(−8)+ 7×10 1+1
x= = −4 , y = =6
9 9

Coordinates of the point are (−4, 6)

22.

Ans. (A) Let the zeroes be a, 2a ½
a + 2a = −6 gives a = −2 1
Polynomial is 𝑥 2 + 6𝑥 + 8 ½
OR
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(B) As −1 is the zero, so 1 + 5 – c = 0 gives c = 6 a 1
Sum of zeroes = 5 ½
Thus, other zero is 6 ½

23.

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Ans. Let AC = 𝑥 cm .co s e m
s em l a
AP a 1 𝑥−3
= gl ⟹ =
AQ ag 1 ½
AB a AC 2 𝑥
𝑥 = 6 ∴ AC = 6 cm ½

24.

Ans. 1 2 1 2 1 2
1½
(A) Given expression = ( ) − ( ) + ( )
2 √2 √3
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1
=

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12 ½
OR
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(B) sin (A + 2B) = 1
A + 2B = 90o ag ½
½
Using A = 3B, we get
B = 18o
½
A = 54o ½
25.

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a (i) P(box will be rejected) = or
60 12
15 1 1
(ii) P(clock has minor defect) = or
60 4
SECTION-C
This section comprises 6 Short Answer (SA) type questions of 3 marks each.
26.

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Ans. Let 3 + 2 √5 = r be a rational number ½
r−3 1
So √5 =
2
RHS is a rational number 1
So LHS is a rational number which is a contradiction
Hence 3 + 2√5 is an irrational number. ½
27.

Ans. (A) x + 3y = 6 2x – 3y = 12
x 0 6 x 0 6
y 2 0 y −4 0
y
x

x
Correct
x Graph:
2

½
Solution is x = 6, y = 0 or (6, 0)
1
Required area = × 2 × 6 = 6 ½
2
OR
(B) Let the angles be x and y
x + y = 180o 1
x – y = 120o 1
solving we get
x = 150o ½
y = 30o ½

28.

Ans. √(x − 3)2 + (y − 6)2 = √(x + 3)2 + (y − 4)2 1

gives –12x – 4y = – 20 or 3x + y = 5 1

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As P lies on x axis so y = 0 gives x =
5 a ½
3
5
Coordinates of P are ( , 0) ½
3

29.

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m sinA s e m
s esinA l a
Ans.
g l a −
cosA ag 1
a sinA + cosA
(A) LHS = sinA

1
sinA( 1 − cosA)
= 1
1
sinA( 1 + )
cosA
1−sec A
= = RHS 1
1+sec A
OR

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cosx √1 − sin2 x
(B) (i) cotx = = 1
sinx sinx
e m
= s
√1−𝑝2
la2
½
2
1+tan x
𝑝
2 g
sec x a sin x
(ii) = = ½+½
1+cot2 x cosec2 x cos2 x
𝑝2
= ½
1−𝑝2

30.

Ans. Correct
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Let PA and PB are tangents drawn from
figure:
the external point P to the circle with centre O.
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½
To prove: PA = PB
e m ½
em
Construction: Join OA, OB and OP
Proof: In  AOP and  BOP
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ag
ag OP = OP (common)
 OAP =  OBP (each 90)
1½

  AOP   BOP (RHS congruency)

Hence, PA = PB (CPCT) ½

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31.

Ans. ∠DAB = 90o , ∠CBA = 90o (Angle between tangent and radius at point of 1
o
contact is 90 )
∠DAB + ∠CBA = 180o 1
Hence, p ∥ 𝑞 as co-interior angles are supplementary
∠ADC = 180o − 110o = 70o 1
SECTION-D
This section comprises 4 Long Answer (LA) type questions of 5 marks each.

32.

Ans. (A) Given equation can be written as
24(18 + x) – 24(18 – x) = 324 – x2 1
i.e., x2 + 48x – 324 = 0 1
D = 482 – 4(–324) = 3600 1
−48 ±60
Roots are
2 1
i.e., 6, –54 1
OR
(B) Let the numbers be x, x + 2 ½
x2 + (x + 2)2 = 100 1½
simplifying we get
2x2 + 4x – 96 = 0 or x2 + 2x – 48 = 0 1

which gives (x + 8) (x – 6) = 0 1
x = 6, – 8 ½
As x > 0 thus, numbers are 6, 8 ½

33.

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Ans. (i) As ∆ABE ≅ ∆ACD a
AE = AD, AB = AC (CPCT) ½
AD AE 1
Getting =
AB AC
½
and ∠A = ∠A (common)
∴ ∆ADE ~ ∆ABC (by SAS Similarity) ½

(ii) As ∆ABE ≅ ∆ACD m
∠ABE = ∠ACD (CPCT)
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and ∠BOD = ∠COE (Vertically opposite angles)
m s
1
e m
e
∴ ∆BOD ~ ∆COE (by AA Similarity)
s l a ½

34. g l a ag
a

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s e
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a

Ans. (A) ∠AOB = 60o ½
Thus, angle of sector corresponding to shaded region = 300o
m . com 1

c. o
300 1
So, 𝜋𝑟 2 = 750𝜋
e m
s
360

e m 𝑟 2 = 900
l a
l as r = 30 cm
ag 1

ag
300 1
Perimeter of shaded region = × 2 × 𝜋 × 30 + 2 × 30
360
1520 ½
= 50𝜋 + 60 = cm or 217.14 cm
7
OR
60 √3 2
(B) Area of unshaded segments in figure (i) = 2 ( πr2 − r ) 1
360 4

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Getting length of rectangle = √3𝑟 𝑎nd width = r 1
Area of rectangle = √3𝑟 2
60 √3 2 1
Area of shaded region in figure (i) = πr 2 − 2 ( πr 2 − r ) − √3r 2
360 4
2
= πr 2 √3
− r2 1
3 2
60 √3 1 √3
Area of shaded region in figure (ii) = 2 ( πr 2 − 4 r 2 ) = πr 2 − 2 r 2 ½
360 3
2 √3
πr2 − r2 4π−3√3
Required ratio = 3
1
2
√3
= 2π−3√3 or (4π − 3√3): (2π − 3√3)
3
πr2 − r2
2 ½
or (88 − 21√3): (44 − 21√3)

35.

Ans. Family Size fi xi fi.xi Cf Correct
1-3 7 2 14 7 Table:
3-5 8 4 32 15 2
5-7 2 6 12 17
7-9 2 8 16 19
9-11 1 10 10 20
84
Mean =
20
= 4.2 1+½
10−7
Median = 3 + ×2 1
8
= 3.75 ½

Aliter: 3median = 2(4.2) + 3.286 or 3(3.75) = 2mean + 3.286 1
median = 3.895 or mean = 3.982 ½
SECTION-E
This section comprises 3 case study-based questions of 4 marks each
36.

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a

Ans. (i) 101 + 20d = 1 ½
d=–5 ½
(ii) a15 = 101 + 14(–5) ½
= 31 ½
1½
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(iii) (a) S21 = [202 + 20(–5)]
2
½
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= 1071
OR
e m
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(iii) (b) 1 + (n–1) 5 = 101 + (n–1) (–5)
l as1½

l as n = 11
ag
½
37.
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s e
g la
a

Ans. (i) Radius, r = 0.7 m 1
m
m 2
.co
.co
3
× 𝜋 × (0.7)
m
(ii) Volume of water in hemispherical depression = ½
me
3
πs
343

em = 0.228 𝜋 m approx. or
1500 a
½
3 3

s g l
g la a
a (iii)(a) Required area = 2 × 𝜋 × (0.7) × (1.4) + 2 × 𝜋 × (0.7) × (0.7) 1½
231 ½
= 2.94 𝜋 m2 or 9.24 m2 or m2
25
OR
2
(iii)(b) Volume of log of wood = 𝜋 × (0.7) × (0.7) × (1.4) − × 𝜋 × (0.7)3 1½
3
539 3
= 0.457 𝜋 m or 1.437 m or
3 3
m ½
375

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38.

Ans. 10 ½
(i) = tan 30o
x ½
x = 17.32 m
17.32 ½
(ii) = tan𝜃
17.32 ½
θ = 45o
10
(iii) (a) = sin30o ½
𝑙1
gives l1 = 20 m ½
½
Similarly, l2 = 17.32 √2 = 24.248 m
½
Total length of rope needed = 44.248 m
OR
10
(iii) (b) = sin300 ½
𝑙1
Gives l1 = 20 m ½
Similarly, l2 = 17.32 √2 = 24.248 m ½
l2 is longer than l1 by 4.248 m ½

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