CBSE Class 10 Sample Paper 2025 Solution for Maths Basic

CENTRAL BOARD OF SECONDARY EDUCATION

SAMPLE PAPER
CLASS 10
2025

Solutions

Marking Scheme
Class X Session 2024-25
MATHEMATICS BASIC (Code No.241)
TIME: 3 hours MAX.MARKS: 80

Q. No. Section A Marks

1. B) 90 1

2. A) consistent with unique solution 1

3. D) 7 1

4. C) 2 √𝑎2 + 𝑏 2 1

5. D) 145∘ 1

6. B) 15 cm 1

7. 5 1
A) 4

8. B) 𝛥𝐸𝐴𝐷 1

9. C) 3780 1

10. B) 40 1

11. D) 52∘ 1

12. B) 5 cm 1

13. A) cos 60∘ 1

14. (C) 3𝜋𝑟2 1

15. D) 4 1

16. B) real and equal 1

17. C) 30 - 40 1

18. D) 25𝑥 2 − 5𝑥 − 2 1

19. A) Both assertion (A) and reason (R) are true and reason (R) is the correct 1
explanation of assertion (A)

20. C) Assertion (A) is true but reason (R) is false. 1

Section B

1

21 (A). 𝑃𝐴2 = 𝑃𝐵 2
⟹ (𝑥 − 4)2 + (𝑦 − 3)2 = (𝑥 − 3)2 + (𝑦 − 4)2 1
⟹ 𝑥 = 𝑦 𝑜𝑟 𝑥 − 𝑦 = 0 1

OR

21 (B). AB = 6 cm = AC ½

OC = √36 − 9 = 3√3 cm 1
½
Point C is (3√3, 0)

22. Correct figure ½

O

A M B
½
AM = 4 cm

OM = √𝑂𝐴2 − 𝐴𝑀2
= √52 − 42
= 3 cm 1

23 (A). 12 ½
[2 × 20 + 11𝑑]=900
2
⇒ 𝑑 = 10 1
Also 𝑎12 = 20 + 11 × 10 = 130 ½

OR

23 (B). Putting 𝑛 = 1, 𝑆1 = 𝑎 = 6 − 12 = 5 . . . . . . . . . . . . . (𝑖) ½
Putting 𝑛 = 2, 𝑆2 = 2𝑎 + 𝑑 = 6 × 2 − 22 = 8 . . . . . . . . . . . . . (𝑖𝑖) 1
Solving (i) & (ii) 𝑑 = −2 ½

24. 1 ½
𝑠𝑖𝑛(𝐴 − 𝐵) = 2 ⇒ 𝐴 − 𝐵 = 30∘ ………..(i)
1
𝑐𝑜𝑠(𝐴 + 𝐵) = 2 ⇒ 𝐴 + 𝐵 = 60∘ ………..(ii) ½
Solving (i) & (ii) to get 𝐴 = 45∘ , 𝐵 = 15∘ ½+½

25. Class 5-10 10-15 15-20 20-25 25-30 30-35

Frequency 5 6 15 10 5 4

Modal class is 15-20. ½
15−6
𝑀𝑜𝑑𝑒 = 15 + 5 × (2×15−6−10) 1
= 18.21(approx.) ½

Section-C

2

26. Let √5 be a rational number.
p ½
∴ √5 = q , where q≠0 and p & q are coprime.

5q2 = p2 ⟹ p2 is divisible by 5
1
⟹ p is divisible by 5----- (i)
⟹ p = 3a, where ‘a’ is a postive integer
25a2 = 5q2 ⟹ q2 = 5a2 ⟹q2 is divisible by 5
⟹ q is divisible by 5 ----- (ii) 1

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

∴ √5 is an irrational number.

27(A). Let the required point on the y axis be P(0,y). ½

Let AP : PB be k : 1
−𝑘+4
Therefore, =0 1
𝑘+1
⇒k=4
Therefore, required ratio is 4:1 ½
8−5 3
&𝑦= 5 =5 ½
3
Hence point of intersection is (0,5). ½

OR

27 (B). Let the line 4𝑥 + 𝑦 = 4 intersects AB at 𝑃(𝑥1 , 𝑦1 ) such that AP: PB=𝑘:1

3𝑘−2 5𝑘−1 1
𝑥1 = 𝑘+1 and 𝑦1 = 𝑘+1
(𝑥1 , 𝑦1 ) lies on 4𝑥 + 𝑦 = 4

3𝑘−2 5𝑘−1
Therefore, 4( 𝑘+1 )+( 𝑘+1 )=4 ½
⇒ 𝑘=1 1

Required ratio is 1:1 ½

3

28. 1 1 ½
LHS= (𝑠𝑖𝑛𝐴 − 𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴 − 𝑐𝑜𝑠𝐴)
1−𝑠𝑖𝑛2 𝐴 1−𝑐𝑜𝑠2 𝐴
= 𝑠𝑖𝑛𝐴 × 1
𝑐𝑜𝑠𝐴
𝑐𝑜𝑠2 𝐴 𝑠𝑖𝑛2 𝐴
= 𝑠𝑖𝑛𝐴 × 𝑐𝑜𝑠𝐴
=𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐴 ½
𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐴
RHS = 𝑠𝑖𝑛2 𝐴+𝑐𝑜𝑠2𝐴
1
=𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐴 = LHS

29.
Class x frequency(f) 𝑥 − 25 𝑓𝑢
𝑢=
10
0-10 5 6 −2 −12

10-20 15 10 −1 −10

20-30 25 15 0 0
Correct
30-40 35 9 1 9 table
𝟏
40-50 45 10 2 20 𝟏
𝟐
∑ 𝑓 =50 ∑ 𝑓𝑢 = 7

7
𝑀𝑒𝑎𝑛 = 25 + 10 × (50) 1
= 26.4 ½

30 (A).

(i) 𝛥𝑂𝐴𝑃 ⩭ 𝛥𝑂𝐵𝑃
∠𝐴𝑃𝑂 = ∠𝐵𝑃𝑂
Or OP bisects ∠𝑃 1
(ii)𝛥𝐴𝑄𝑃 ⩭ 𝛥𝐵𝑄𝑃
⇒AQ=QB and ∠𝐴𝑄𝑃 = ∠𝐵𝑄𝑃 1
AB is a straight line
therefore ∠𝐴𝑄𝑃 = ∠𝐵𝑄𝑃 = 90∘
Hence OP is right bisector of AB 1

OR

30 (B). Correct Given, to prove, figure and construction 1
Correct proof 2

4

31. Let the two-digit number be 10𝑥 + 𝑦 ½
Therefore (10𝑥 + 𝑦) + (10𝑦 + 𝑥) = 99 ½
⇒ 𝑥 + 𝑦 = 9 ……….(i) ½
Also, 𝑥 = 3 + 𝑦……..(ii) ½
Solving (i) & (ii) to get 𝑦 = 3 , 𝑥 = 6 ½
Therefore, required number is 63 ½

Section D

32 (A). Let the number of books purchased be 𝑥
1920 1
Therefore, cost price of 1 book = 𝑥
1920 1920
Therefore − 𝑥+4 = 24 1
𝑥
⇒ 1920 × 4 = 24𝑥(𝑥 + 4)
or 𝑥 2 + 4𝑥 − 320 = 0
⇒ (𝑥 + 20)(𝑥 − 16) = 0 1
⇒ 𝑥 = 16, 𝑥 ≠ −20
Number of books bought=16 1
1920
Price of each book = 16 = ₹120 1

OR

32 (B). Let the initial average speed of the train be 𝑥 km/hr.
132 140 1
Therefore 𝑥 + 𝑥+4 = 4
⇒ 4𝑥 2 − 256𝑥 − 528 = 0 1
or 𝑥 2 − 64𝑥 − 132 = 0 1
⇒ (𝑥 − 66)(𝑥 + 2) = 0
⇒ 𝑥 = 66, 𝑥 ≠ −2 1
Initial average speed of train= 66 km/hr

132 140
Time taken to cover the distances separately= 66 & 70 i.e. 2 hours each 1

33. 𝟏
Correct Given, to prove, Construction and figure × 𝟒=2
𝟐
Correct Proof 3

34. 2𝜋𝑟𝜃
(i) Perimeter of sector = 2𝑟 + 360 = 73.12
2×3.14×24×𝜃 1
⇒ 2(24) + = 73.12
360
⇒ 𝜃 = 60∘ 1
3.14×24×24×60 1.73 2
(ii)Area of minor segment = ( − × 24 × 24) 𝑐𝑚 2
360 4
2
= (301.44 − 249.12) 𝑐𝑚
= 52.32 𝑐𝑚2 1

5

35 (A).

1 mark
for
correct
figure

Let AB be the building and CD be the tower.
ℎ
Here 𝑡𝑎𝑛60∘ = √3 = 𝑥 1
⇒ ℎ = 𝑥√3………..(i) ½
9
𝑡𝑎𝑛45∘ = 𝑥 = 1 1
⇒ 𝑥 = 9 m………….(ii) ( Distance between tower and building) ½

Solving (i) & (ii) to get ℎ = 9 × 1.732 = 15.588𝑚 ½

Therefore, the height of the tower = ℎ + 9 = 24.588 𝑚. ½

OR

35 (B).

1 mark
for
correct
figure

Let AB be the light house and C & D be positions of ships.
1 75
𝑡𝑎𝑛30∘ = = 1
√3 𝑥+𝑦
⇒ 𝑥 + 𝑦 = 75√3…………..(i) ½

75
𝑡𝑎𝑛45∘ =1= 𝑦 1
⇒ 𝑦 = 75………(ii) ½
Solving (i) & (ii) to get 𝑥 = 75(√3 − 1)
⇒ 𝑥 = 75 × 0.732
= 54.9 𝑚 1
Distance between the ships is 54.9 𝑚

Section E

36. (i) Number of students who do not prefer to walk = 200 − 120 = 80 ½
80 2
P (selected student doesn’t prefer to walk) = or
200 5 ½

6

(ii) Total number of students who prefer to walk or use bicycle = 120 + 50
= 170 ½
170 17
P (selected student prefers to walk or use bicycle) = or
200 20 ½

(iii) (A) 50% of walking students who used bicycle = 60 ½
Number of students who already use bicycle = 50 ½
110 11
P (selected student uses bicycle) = or 1
200 20
OR

(B) Number of students who preferred to be dropped by car
= 200 − (120 + 50 + 20)
= 10 students 1

10 1
P (selected student is dropped by car) = 200 or 20 1

37. (i) 1 and 4 1

(ii) 𝑥 = 5/2 1

(iii) (A) At 𝑥 = 5/2, 𝑝(𝑥) = 2.25 1
Therefore, ℎ = 0.10 + 2.25 = 2.35𝑚 1
OR
2
(B) −𝑥 + 5𝑥 − 4 = 2
𝑥 2 − 5𝑥 + 6 = 0 ½
(𝑥 − 2)(𝑥 − 3) = 0 ½
⇒ 𝑥 = 2 𝑎𝑛𝑑 𝑥 = 3 ½
Therefore, required points are (2,0) and (3,0) ½

38. (i) 𝑙 2 = (1.2)2 + (0.5)2 ½
= 1.44 + 0.25
⇒ 𝑙 = √1.69 = 1.3𝑐𝑚 ½

(ii) Curved surface area of sharpened part
= 𝜋 × 0.5 × 1.3 ½
= (0.65 𝜋) 𝑐𝑚2 ½

(iii) (A) Total surface area of pencil
= CSA of cylinder + CSA of cone + area of base circle
= 𝜋 × 0.5 × 0.5 × 21 + 0.65 𝜋 + 𝜋 × (0.5)2 ½
= (5.25 + 0.65 + 0.25)𝜋 1
= (6.15 𝜋) 𝑐𝑚2 ½
OR
(B) Length of cylindrical part of shortened pencil
= (21 − 8.2) 𝑐𝑚 = 12.8 𝑐𝑚 ½
So, volume of cylindrical part of shortened pencil
= 𝜋 × 0.5 × 0.5 × 12.8 1
= (3.2 𝜋 ) 𝑐𝑚3 ½

7

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