CBSE Class 10 Question Paper 2025 Solution Mathematics Basic

Marking Scheme Strictly Confidential
(For Internal and Restricted use only) Secondary School Examination, 2025
SUBJECT NAME MATHEMATICS (BASIC) (Q.P. CODE 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. 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 totaled up and written in the left-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 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.
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.
430/1/1 1 P.T.O.

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 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 totaled 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.

430/1/1 2

430/1/1
MARKING SCHEME
MATHEMATICS (BASIC)

Answer : (D) ab 1

Answer : (B) an irrational number 1

Answer : (B) 17 1

Answer : (A) 5 1

Answer : (D) 8 1

Answer : (C) 1 : 1 1

430/1/1 3 P.T.O.

Answer : None of the given options is correct. 1
Note – One mark to be given to all students who have attempted this question.

Answer : (C)  P = 40 1

Answer : (D) 5√3 cm 1

430/1/1 4

Answer : (C) sin 30 = cos 30 1

Answer : (D) – 1 1

Answer : (C) z 1

430/1/1 5 P.T.O.

Answer : (C) l + 2r 1

Answer : (C) 1 : 4 1

Answer : (D) Sphere 1

Answer : (D) 62·5 1

430/1/1 6

Answer : (B) 4000 1

Answer : (D) Getting a natural number < 7 1

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

430/1/1 7 P.T.O.

Answer : (D) Assertion (A) is false, but Reason (R) is true. 1

Solution : Solving the two equations to get x = 2, y = – 3 1+1

Solution : (a) As PQ ∥ RS
P=S
Alternate interior angles
Q=R 1
 POQ  SOR (by AA similarity criterion) 1
OR
(b)  OSR = 125 – 70 = 55 [ by exterior angle property] 1
As  OSR   OQP
 OSR =  OQP (Corresponding angles of similar triangles) ½
½
 OQP = 55

430/1/1 8

Solution:

Correct
figure
½

BC2 = OB2 – OC2
⇒ BC2 = 102 – 62 = 64
⇒ BC = 8 cm 1
AB = 8  2 = 16 cm ½

Solution: (a) A + B = 45 ½
A – B = 30 ½
Solving and getting A = 37·5 and B = 7·5 1
OR
(b) Consider an isosceles right  ABC ½
Using angle sum property A =  C = 45 ½
𝐴𝐵 𝑥
Clearly, tan 45 = = =1 1
𝐵𝐶 𝑥

Solution: Radius of circle = 10 cm = r
 OPQ is an equilateral triangle. ½
1 2 – √3  (10)2
Area of segment =  (3.14)  (10) 1
6 4
109
= sq. cm or 9.08 sq. cm
12 ½

430/1/1 9 P.T.O.

p
Solution: (a) Let √3 be a rational number such that √3 = (p and q are co-prime numbers, q  0) ½
q
√3q = p  3q2 = p2
3 divides p2  3 divides p as well
1
Let, p = 3m (for some integer m)
3q2 = 9m2  q2 = 3m2
3 divides q2  3 divides q as well 1
p and q have a common factor 3, which is a contradiction as p and q are co-prime.
 our assumption is wrong ½
Hence, √3 is an irrational number
OR
(b) b = 7 ½
a=3 ½
y = 420 ½
x = 840 ½
1
x = 840 = 23  3  5  7

Solution: Polynomial is x2 – 0 (x) + (– 9) = x2 – 9 1
For zeroes :
x2 – 9 = (x + 3) (x – 3) 1
Zeroes are – 3, 3 1

430/1/1 10

Solution: (a) Correct graph of each equation 1+1

Solution is x = 6, y = 0 or (6, 0) 1
OR
(b) x + y = 90 1
2x = y 1
Solving to get x = 30, y = 60 ½+½

Solution:

Correct
figure ½

As the length of tangents from an external point to a circle are equal
Thus,
AP = AS
BP = BQ
DR = DS
CR = CQ 1
Adding the above equations,
AB+CD=BC+AD
As AB =CD & BC = AD (opp. sides of rectangle)
 AB = AD 1
 ABCD is a square ½

𝑐𝑜𝑠2 𝐴 𝑠𝑖𝑛2 𝐴+𝑐𝑜𝑠2 𝐴
1+ 𝑐𝑜𝑠 2 𝐴
𝑠𝑖𝑛2 𝐴 𝑠𝑖𝑛2 𝐴 1
Solution: LHS = 𝑠𝑖𝑛2 𝐴
= 𝑐𝑜𝑠2 𝐴+𝑠𝑖𝑛2 𝐴
=
1+ 𝑠𝑖𝑛2 𝐴
𝑐𝑜𝑠2 𝐴 𝑐𝑜𝑠2 𝐴
𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴 2
𝑐𝑜𝑠 2 𝐴 𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴 2 𝑠𝑖𝑛𝐴 1½
= (𝑐𝑜𝑠𝐴−𝑠𝑖𝑛𝐴) = ( 𝑐𝑜𝑠𝐴−𝑠𝑖𝑛𝐴 )
𝑠𝑖𝑛2 𝐴
𝑐𝑜𝑠𝐴
1−𝑐𝑜𝑡𝐴 2
=( ) = RHS ½
1−𝑡𝑎𝑛𝐴
430/1/1 11 P.T.O.

20 1 1
Solution: P (customer will not buy the pen) = 200 = 10
After mixing the two lots
Total pens = 200 + 100 = 300
Number of good pens = 180 + 80 = 260 1
260 13
P (customer will buy the pen) = 300 or 15 1

Solution: (a) Let the smaller number be y and greater number be x.
A.T.Q.
x2 – y2 = 180 1
y2 = 8x 1
 x2 – 8x = 180
x2 – 8x – 180 = 0 1
(x – 18) (x + 10) = 0
x = 18, x = – 10 (rejected) 1
∴ The numbers are 18 and 12 1

OR
(b) For equal roots; b2 – 4ac = 0
k2 – 24 = 0 1
 k =  2√6 1
Equations are
2x2 + 2√6x + 3 = 0; 2x2 – 2√6x + 3 = 0
3 3 3 3
½+½
Roots are x = −√2, −√2; x = √2 , √2
1+1
430/1/1 12

Solution: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two
sides in distinct points, the other two sides are divided in the same ratio. 1
𝐴𝑂 𝐶𝑂 Correct
Given : ABCD is a quadrilateral in which 𝐵𝑂 = 𝐷𝑂 figure,
given, to
To Prove : AB ∥ CD prove and
construction
Construction : Draw OE ∥ AB
1
Proof : In  DAB, OE ∥ AB
𝐷𝐸 𝐷𝑂
∴ 𝐴𝐸 = 𝐵𝑂 (by BPT) 1
𝐴𝑂 𝐶𝑂
Also 𝐵𝑂 = 𝐷𝑂 (given)
𝐷𝑂 𝐶𝑂
 𝐵𝑂 = 𝐴𝑂

𝐷𝐸 𝐶𝑂
 𝐴𝐸 = 𝐴𝑂 ½
In  ADC,
𝐷𝐸 𝐶𝑂
=
𝐴𝐸 𝐴𝑂
 OE ∥ CD (by converse of BPT) 1
As OE ∥ AB and OE ∥ CD
 AB ∥ CD ½
Hence, ABCD is a trapezium

Solution: (a) Radius = r = 5 cm
Height of cone = h = 10 cm
Volume of toy = volume of hemisphere + volume of cone
2
= r3+ 3r2h
1
3
2 22 1 22
= 3  7  5  5 5 + 3  7  5  5  10 2+2
5500 5500
= 21 + 21
430/1/1 13 P.T.O.

11000 1
= cu. cm or 523.81 cu. cm
21
OR
(b) Edge of cube = a = 3·5  2 = 7 cm 1
Total surface area of solid
= 6 a2 + 2r2 – r2
= 6 a2 + r2
22 1½ + 1½
= 6  7  7 + 7  3·5  3·5
665 1
= 2 sq. cm or 332·5 sq. cm

Solution:
𝑥 –𝑎
CI xi fi ui = 𝑖20 fiui
0 – 20 10 10 –2 – 20 Correct
20 – 40 30 35 –1 – 35 table
40 – 60 50 = a 50 0 0 2
60 – 80 70 60 1 60
80 – 100 90 30 2 60
100 – 120 110 15 3 45
Total 200 110
110
Mean = 50 + 200  20 2
Mean = 61 1
Thus the mean lifetime of the electrical components is 61 hours.

430/1/1 14

Solution: (i) Let the length of the ladder be ‘a’
15
= sin 60 ½
𝑎
30
a= or 10√3 ½
√3
30
Thus the length of the ladder is m or 10√3 m
√3

(ii) Let the distance of the point on the ground be ‘x’
15
= tan 60 ½
𝑥
15
x= or 5√3 ½
√3
15
Thus, the distance of the point on the ground is m or 5√3 m
√3

(iii) (a) Let the width of the road be y.

Correct
figure
1

430/1/1 15 P.T.O.

15
= tan 30
𝑦 ½
y = 15√3
Thus, the width of the road is 15√3 m. ½
OR
(b) Let the length of the ladder be l.

15
= sin 30
𝑙
l = 30 1
Thus, the length of the ladder is 30 m. 1

430/1/1 16

Solution: (i) a13 = 650 cm 1
(ii) an = 500
50 + (n – 1)50 = 500
n = 10 1
(iii) (a) a = 10, d = 10
11 1½
S11 = 2 [20 + 10  10]
= 660 ½
OR
(b) a = 10, d = 10
𝑛 1
450 = 2 [20 + (n – 1) 10]
½
n2 + n – 90 = 0
n=9 ½

10 + 50 20 + 50
Solution: (i) Co-ordinates of C are ( , ) = C(30, 35) 1
2 2
(ii) Radius = √(30 – 10)2 + (35 – 20)2 = 25 1
(iii) (a) P divides AB in the ratio 1 : 2, ½

430/1/1 17 P.T.O.

1 × 50 + 2 × 10 1 × 50 + 2 × 20 ½+½
co-ordinates of P are ( , )
3 3

70 ½
i.e. ( 3 , 30)
OR
(b) Distance AB = 2  25 = 50 ½
2 2
AQ = 3AB = 3 50 1
100
AQ = 3 ½

430/1/1 18