CBSE Class 10 Sample Paper 2025 Solution for Maths Standard

CENTRAL BOARD OF SECONDARY EDUCATION

SAMPLE PAPER
CLASS 10
2025

Solutions

Marking Scheme
Class X Session 2024-25
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80

Q.No. Section A Marks

1. D) -6,6 1

2. B) -5 1

3. D) From a point inside a circle only two tangents can be drawn. 1

4. A) 7 1

5. B) 20 cm 1

6. 11 1
A)
9
7. C) 140𝑂 1

8. B) 8𝑥 2 - 20 1

9. C) 30 1

10. B) isosceles and similar 1

11. A) Irrational and distinct 1

12. 3 1
C)
√3
13. 594 1
B)
7
14. 3 1
B)
8
15. B) (-4, 0) 1

16. A) median 1

17. C) (3,0) 1

18. 3 1
D)
26
19. B) 1

20. D) 1

1

Section B

21. (A) 480 = 25 x 3 x 5 ½
720 = 24 x 32 x 5 ½

LCM (480,720) = 25 x 32 x 5 = 1440 ½

HCF (480, 720) = 24 x 3x 5 = 240 ½

OR

85 = 5x17, 238 = 2x7x17
(B)
HCF( 85, 238) = 17 1
17 = 85xm -238
m=3 1

22.(A) Total number of possible outcomes = 6x6=36 ½
For a product to be odd, both the numbers should be odd.
Favourable outcomes are (7,7) (7,9) (7,11) (9,7) (9,9) (9, 11) (11,7) (11,9)
(11,11)
no. of favourable outcomes = 9
9 1 1
P (product is odd) = or ½
36 4

OR
(B) ½
Total number of three-digit numbers = 900.
Numbers with hundredth digit 8 & and unit’s digit 5 are 805,815,
1
825,....,895
Number of favourable outcomes = 10 ½
10 1
P(selecting one such number) = or
900 90

23. √3 2 1 1½
2 ( ) − ( )2
2 √3
2
(√2)
7 ½
=
12
24 Let the required point be (x,0) ½

√(8 − 𝑥)2 + 25 = √41 ½
=> (8 − 𝑥)2 = 16
=> 8 - x =土4
=> x = 4 , 12
Two points on the x-axis are (4,0) & (12,0). 1

2

25. AB = √(3 + 5)2 + (0 − 6)2 = 10 ½
BC = √(9 − 3)2 + (8 − 0)2 = 10
½
AC = √(9 + 5)2 + (8 − 6)2 = 10√2
½

Since AB = BC, therefore 𝛥 ABC is isosceles ½

Section C

26.(A)

3 1
2 4 5 6

Since D, E, F are the mid points of BC, CA, AB respectively
Therefore, EF||BC, DF||AC, DE||AB 1
BDEF is a parallelogram
∠ 1= ∠ 2 & ∠ 3 = ∠ 4
1
𝛥 FBD ～ 𝛥 DEF
Also, DCEF is a parallelogram
∠ 3= ∠ 6 & ∠ 1 = ∠ 2 ( proved above)
1
𝛥 DEF ～ 𝛥 ABC

OR

A

P R Q

(B)
B D C

Since PQ//BC therefore 𝛥 APR ～ 𝛥 ABD
𝐴𝑃 𝑃𝑅
=> = …….. (i) 1
𝐴𝐵 𝐵𝐷

3

𝛥 AQR ～ 𝛥 ACD
𝐴𝑄 𝑅𝑄
=> = …….. (ii)
𝐴𝐶 𝐷𝐶
1
𝐴𝑃 𝐴𝑄
Now, = ……….(iii)
𝐴𝐵 𝐴𝐶
𝑃𝑅 𝑅𝑄
Using (i), (ii) & (iii), = 1
𝐵𝐷 𝐷𝐶
But, BD = DC
=> PR = RQ or AD bisects PQ

27. Let the numbers be x and 18-x. ½
1 1 9 1
+ =
𝑥 18−𝑥 40
=> 18×40 = 9x(18-𝑥)
=> 𝑥 2 -18 𝑥 + 80=0
=> （𝑥-10)(𝑥-8）=0
1
=> 𝑥=10, 8.
=> 18-𝑥 =8, 10 ½
Hence two numbers are 8 and 10.

28. 5 1 1
From given polynomial 𝛼 + 𝛽 = , 𝛼𝛽 =
6 6
5 2 1 13 1
𝛼2 + 𝛽2 = ( ) - 2 x =
6 6 36

1 2 1 ½
2 2
And 𝛼 𝛽 =( ) =
6 36

13 1
𝑥2 - 𝑥+
36 36 ½
 Required polynomial is 36𝑥 2 -13 𝑥+1

29. (𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛 𝜃) 2 + (𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛 𝜃) 2 = 2 ( 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 ) = 2
=> (1)2 + (𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛 𝜃) 2 = 2 1½
=> (𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛 𝜃) 2 = 1 1
=> 𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛 𝜃 = 土 1 ½

30.(A) Angle described by minute hand in 5 min = 30°.
length of minute hand =18 cm = r.
Area swept by minute hand in 35 minutes
22 30
=( 7 x18x18×360) x 7 2
1
= 594 𝑐𝑚2 .
OR

(B) Area of minor segment = Ar. Sector OAB- Ar. 𝛥 OAB
90 22 √3 2
= x × 14x14 - x 14x14
360 7 4 1
= 69.23 cm2

4

31. Let √3 be a rational number.
𝑝 ½
∴ √3 = , where q≠0 and let p & q be co-prime.
𝑞
3q2 = p2 ⟹ p2 is divisible by 3 ⟹ p is divisible by 3 ----- (i) 1
⟹ p = 3a, where ‘a’ is some integer
9a2 = 3q2 ⟹ q2 = 3a2 ⟹q2 is divisible by 3 ⟹ q is divisible by 3----- (ii) 1

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

Section D

32.(A) x+2y=3, 2x-3y+8=0
Correct graph of each equation 2+2 = 4
Solution x=-1 and y=2 1

OR

(B) Let car I starts from A with speed x km/hr and car Il Starts from B with
speed y km/hr (x>y)

Case I- when cars are moving in the same direction.
Distance covered by car I in 9 hours = 9x.
Distance covered by car II in 9 hours = 9y
Therefore 9 (x-y) = 180
=> x-y= 20 ……………. (i) 2
case II- when cars are moving in opposite directions.

Distance covered by Car I in 1 hour = x
Distance covered by Car II in 1 hour = y
2
Therefore x + y=180 ………….. (ii)
Solving (i) and (ii) we get, x=100 km/hr, y=80 km/hr. 1

33. Correct given, to prove, construction, figure 1

Correct proof 2

AR = AQ = 7cm ½
BP = BR = AB-AR = 3cm ½
CP =CQ = 5cm ½
BC = BP+PC = 3+5 = 8 cm ½

5

34. Correct
B figure
C 1mark

h
h
60˚
30˚
A F
x G
1.35 m
D E
Let A be the eye level & B, C are positions of balloon
Distance covered by balloon in 12 sec = 3x12 = 36 m
1
BC = GF = 36 m

ℎ
tan 600 = √3 = 𝑥 1
=> h = 𝑥 √3 …….. (i)

1 ℎ 1
tan 300 = = 𝑥+ 36
√3
𝑥+36
=> h = …….. (ii)
√3
Solving (i) and (ii) h= 18√3 = 31.14 m 1
Height of balloon from ground = 1.35 + 31.14 = 32.49 m

35. Correct
𝑥−102.5 table
Class x f u= fu cf
5 2marks

85-90 87.5 15 -3 -45 15

90-95 92.5 22 -2 -44 37

95-100 97.5 20 -1 -20 57

100-105 102.5 18 0 0 75

105-110 107.5 20 1 20 95

110-115 112.5 25 2 50 120

𝛴f = 120 𝛴fu = -39

39
Mean = 𝑥 = 102.5 - 5 x
120 1
= 100.875 ½
Median class is 100-105 ½
5
Median = 100 + 18 (60-57) = 100.83 1

OR

6

Monthly Expenditure fi xi fixi Correct
table
1000-1500 24 1250 30,000 2marks
1500-2000 40 1750 70,000
2000-2500 33 2250 74,250
2500-3000 X=28 2750 77,000
3000-3500 30 3250 97,500
3500-4000 22 3750 82,500
4000-4500 16 4250 68,000
4500-5000 7 4750 33,250
1
172+x=200
X=28
532500 1
Mean= 200

= 2662.5 1

Section E

36.(i) First term a = 3, A.P is 3, 6, 9, 12…..,24 ½
common difference d = 6-3 = 3 ½

(ii) 34 = 3+ (n-1)3
1 ½
=> n = 34/3 = 11 which is not a positive integer.
3
Therefore, it is not possible to have 34 jars in a layer if the given pattern is
½
continued.
(iii)(A) 𝑛 ½
𝑆𝑛 = 2 [ 2x3 + (n-1) 3]
𝑛
= 2 [ 6 + 3n-3] 1
𝑛
= [ 3+3n]
2
𝑛
=3 [1+n]
2
8 ½
𝑠8 = 3 x 2 (1+8)
= 108
OR

A.P will be 6, 9, 12, …..
(iii) (B) ½
a= 6, d=3

𝑡5 = 6 + (5-1)3
1
= 6 + 12
= 18
½

37. (i) ∠DPQ = ∠DEF
∠PDQ =∠EDF
1
(ii) Therefore 𝛥 DPQ ∼ 𝛥 DEF
DE = 50 + 70 = 120 cm ½
𝐷𝑃 𝑃𝑄
=
𝐷𝐸 𝐸𝐹

7

𝑃𝑄 50 5 ½
Therefore 𝐸𝐹 = 120 or 12

(iii) (A)
𝐴𝐵 5 𝐵𝐶 𝐴𝐶
𝐷𝐸
= 2 = 𝐸𝐹 = 𝐷𝐹
5 1
⇒ AB = 2 DE
5
𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 △𝐴𝐵𝐶 (𝐷𝐸 + 𝐸𝐹 + 𝐹𝐷) 5
= 2 𝐷𝐸 + 𝐸𝐹 + 𝐹𝐷 = 2 ( Constant) 1
𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 △𝐷𝐸𝐹

OR

A

D

(iii)(B) Correct
fig. ½
P Q mark

B M C
E N F
𝐴𝐵 𝐵𝐶 𝐵𝐶/2 𝐵𝑀
= = =
𝐷𝐸 𝐸𝐹 𝐸𝐹/2 𝐸𝑁
1
Also ∠B =∠E

½
Therefore △ ABM ∼ △ DEN.

38. (i) l= √𝑟 2 + ℎ2
½
= √(1.5)2 + (2)2
= √2.25 + 4
= √6.25
= 2.5 m ½

(ii) CSA of cone = 𝛱rl
22 ½
= 7 x 1.5 x 2.5
= 11.78 𝑚2 ½

(iii) (A) CSA of cylinder = 2𝛱 rh
22 1
= 2 x 7 x 1.5 x 7
= 66 𝑚2
Cost of metal sheet used = 66 x 2000 1
= ₹1,32,000

OR
Volume of cylinder = 𝛱 𝑟2 h
(iii) (B) 22
= 7 x (1.5)2 x 7
= 49.5 𝑚3 ½

8

1
Volume of cone = 3 𝛱 𝑟 2 h
1 22
=3 x 7 x (1.5)2 x 2
= 4.71 𝑚3 1

Total capacity = 49.5 + 4.71 = 54.21 𝑚3 ½

9

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