CBSE Class 12 Question Paper 2025 Solution Applied Mathematics

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior Secondary School Examination, 2025
APPLIED MATHEMATICS (241) PAPER CODE – 465
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Question Paper) has to be used. Please do not hesitate to award full marks if the answer
deserves it.
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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
APPLIED MATHEMATICS (Subject Code–241)
(PAPER CODE: 465)

Section A
Q. No. EXPECTED OUTCOMES/VALUE POINTS Marks
SECTION A
Questions no. 1 to 18 are multiple choice questions (MCQs) and questions number 19
and 20 are Assertion-Reason based questions of 1 mark each.
1.

Sol. (B) 4 1
2.

Sol. (D) a − c > b + c 1
3.

Sol. (D) skew symmetric matrix 1
4.

Sol. 1/6 1/6 1
(C) [ ]
−1/3 2/3
5.

Sol. (D) ± 3 1

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6.

Sol. (A) 4 1
7.

Sol. dP 1
(B) dt = (α − β)P

8.

Sol. 𝟏 1
(C) 𝟑

9.

Sol. (A) 0 1

10.

Sol. (C) seasonal variations 1
11.

Sol. (B) 140 1

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12.

Sol. (A) ₹ 2,000 1
13.

Sol. (D) ₹ 14 1
14.

Sol. (B) 100 1
15.

Sol. 𝟏 1
(B)
𝟓

16.

Sol. 𝑥̅ −𝑦̅ 1
(B) 𝑡 = 1 1
𝑠√ +
𝑛1 𝑛2

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17.

Sol. (B) 4.04 % 1
18.

Sol. (A) 49.53 % 1

19.

Sol. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct 1
explanation of the Assertion (A).
20.

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

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SECTION B
This section comprises very short answer (VSA) type questions of 2 marks each.
21(a).

Sol. Let the price per kg of mixture be ₹ 𝐱

𝟑𝟓−𝐱 𝟑 1
∴ 𝐱−𝟐𝟓 = 𝟐

⟹ 70 − 2x = 3x − 75
1
Solving, we get x = 29
∴ Price per kg of the mixture is ₹ 29
OR
21(b).

Sol. Let the required time taken be 𝑛 minutes
1 1 1 1
∴ 𝑛 = 60 + 90
1 3+2 5 1
⟹ 𝑛 = 180 = 180 = 36
3
∴ 𝑛 = 36 minutes or 5 hours 1

22.

Sol. Let the speed of the boat be 𝑥 km/h
∴ Speed of the boat upstream = (𝑥 − 1) km/h ½
and speed of the boat downstream = (𝑥 + 1) km/h
3.5 3.5 12 6
∴ 𝑥−1 + 𝑥+1 = 1 + 60 = 5 ½
⟹ 3.5 (2𝑥)5 = 6 (𝑥 2 − 1)
⟹ 6𝑥 2 − 35𝑥 − 6 = 0 ½
Solving, we get 𝑥 = 6 km/h ½
(rejecting the -ve value)

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23.

Sol.

Let the winning post be x metres away from the starting point.
x x−40
∴ = 1
3/2 1

x 3
⟹ 2 = 2 × 40 = 60 ⟹ x = 120 metres 1

24.

Sol. 1 1 −5 2 0 1
BA = [−5 1 −5] [3 4 5]
1 −2 4 0 2 3
Obtaining at least 4 correct entries 1
5 −6 −9
= [−7 −6 −15] 1
−4 0 3
25 (a).

Sol. 1 1 ½
Here, 𝑛 = 6, 𝑝 = 2 , 𝑞 = 2

P (at least 4 heads in 6 throws) = 𝑃(𝑋 ≥ 4)
1 4 1 2 1 5 1 1 1 6 1 0
= 6𝐶4 ( ) ( ) + 6𝐶5 ( ) ( ) + 6𝐶6 ( ) ( )
2 2 2 2 2 2 1
1 6 22 11
= 22 (2) = 64 𝑜𝑟 32 ½
OR
25 (b).

Sol. 𝑋−𝜇 15−9 1
(i) 𝑍 = = =2
𝜎 3
𝑋−9
(ii) 4 = ⟹ 𝑋 = 21
3
1

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SECTION C
This section comprises short answer (SA) type questions of 3 marks each.
26.

Sol. 72 = 49 ≡ −1(𝑚𝑜𝑑 10)
7295 = (72 )147 × 7 1
Now, (72 )147 ≡ (−1)147 (𝑚𝑜𝑑 10) ≡ −1(𝑚𝑜𝑑 10) 1

⟹ 7295 = (72 )147 × 7 ≡ −7(𝑚𝑜𝑑 10) ≡ 3(𝑚𝑜𝑑 10) ½

∴ Units digit is 3. ½

27.

Sol. Numbers on the dice are 1, 2, 3, 4, 5, 6
∴ X can take the values 1, 2, 3, 4, 5 ½
X 1 2 3 4 5
P(X) 5 4 3 2 1 1
15 15 15 15 15
X P(X) 5 8 9 8 5
1
15 15 15 15 15

35 7 ½
𝐸 (𝑋) = ∑ 𝑋 𝑃(𝑋) = 15 = 3

28 (a).

4 8
Sol. Mean = 𝑛𝑝 = , variance = 𝑛𝑝𝑞 = ½
3 9
8 3 2
⟹𝑞 = 9×4= 3 ½
2 1
∴ 𝑝 =1−3 =3 ½
1 4
𝑛×3 = 3 ⟹ 𝑛 = 4 ½
1 1 2 3 1 8 ½
𝑃(𝑥 = 1) = 4𝐶1 (3) (3) = 4 × 3 × 27
32
= 81 ½

OR

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28 (b).

Sol. Given, 𝑝 = 0.007, 𝑛 = 400 ½
∴ 𝜆 = 𝑛𝑝 = 400 × 0.007 = 2.8 1
(2.8)2 𝑒 −2.8 1
Now, 𝑃(X = 2) = 2!
7.84
= × 0.0608 = 0.2383 ½
2

29 (a).

Sol. Let 𝑥 kg of nitrogen and 𝑦 kg of phosphoric acid is used for minimum cost.
∴ the objective function is
Minimize 𝑍 = 6𝑥 + 5𝑦 ½
Subject to the constraints 10% × 𝑥 + 5% × 𝑦 ≥ 14 or 2𝑥 + 𝑦 ≥ 280 1
and 6% × 𝑥 + 10% × 𝑦 ≥ 14 or 3𝑥 + 5𝑦 ≥ 700 1

𝑥, 𝑦 ≥ 0 ½

Note: * Marks should be awarded for the formation of equations 2𝑥 + 𝑦 = 280 and
3𝑥 + 5𝑦 = 700 instead of inequations in Hindi medium only.
OR
29 (b).

10 | P a g e

Sol.

1½ for
correct
graph

Corner Points Value of Z 1 for
O (0,0) 0 correct
A (0, 17) 510 table
B (2,14) 520
C (9, 0) 450

∴ Maximum Z = 520 at B (2, 14) ½

30.

Sol. 𝑥̅ = 0.742, 𝜇 = 0.7
𝑛 = 10, 𝑠 = 0.04 ½
𝐻0 : Null hypothesis : If there is no significant difference between 𝑥̅ and 𝜇
𝐻1 : Alternate hypothesis : If there is a significant difference between 𝑥̅ and 𝜇
𝑥̅ −𝜇 0.742−0.7 1½
𝑡= 𝑠 = 0.04 = 3.15
√𝑛−1 √9

Df = 9 and 𝑡9 (0.05) = 2.262
Since |𝑡| = 3.15 > 2.262 ½
∴ Null hypothesis is rejected

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Hence the work is inferior ½
𝑥̅ −𝜇
Note:* 𝑡 = 𝑠 should also be accepted.
√𝑛

31.

Sol. 𝑃 = ₹ 5, 00, 000
𝑃𝑅𝑇
Interest =
100
500000×7.5×5
= = ₹ 1,87, 500 1
100

𝑛 = 5 years = 60 months ½
𝑃+𝐼
∴ EMI =
𝑛
500000+187500
= 1
60

= ₹ 11,458.33 ½
SECTION D
This section comprises of Long Answer (LA) type questions of 5 marks each.
32 (a).

Sol. |𝐴| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = −1 ≠ 0 ½
∴ 𝐴−1 exists
2 −3 5 x 11
Now, [3 2 −4] [y] = [−5]
1
1 1 −2 z −3
or AX = B ⟹ X = A−1 B
0 −1 2
adj(A) = [2 −9 23] 1½
1 −5 13
0 −1 2 0 1 −2
1 1
A−1 = |A| adj(A) = −1 [2 −9 23] = [− 2 9 −23] ½
1 −5 13 −1 5 −13
x 0 1 −2 11 1
∴ [y] = [− 2 9 −23] [−5] = [2] 1
z −1 5 −13 −3 3
So, x = 1, y = 2, z = 3 ½

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OR
32 (b).

Sol. (b + c)2 a2 a2
LHS = ∆= | b2 (c + a)2 b2 |
c2 c2 (a + b)2

Applying, C1 → C1 − C3
C2 → C2 − C3

(b + c)2 − a2 0 a2
∆= | 0 (c + a)2 − b2 b2 |
c − (a + b)2
2
c 2 − (a + b)2 (a + b)2 1+1

b+c−a 0 a2
= (a + b + c) 2|
0 c+a−b b2 | 1
c − a − b c − a − b (a + b)2

Applying, R1 → R 3 − R1 − R 2

b+c−a 0 a2
2|
= 2(a + b + c) 0 c+a−b b2 |
−b −a ab

Expanding, we get ∆ = 2abc(a + b + c)3 2

33.

Sol. p = 4 − 5𝑥 + 𝑥 2
p0 = 18, we have 18 = 4 − 5𝑥 + 𝑥 2
or 𝑥 2 − 5𝑥 − 14 = 0
⟹ (𝑥 − 7)(𝑥 + 2) = 0
∴ 𝑥 = 7, 𝑥 = −2 is rejected 1½

p0 𝑥0 = 18 × 7 = 126 ½
7
PS = p0 𝑥0 − ∫0 (𝑥 2 − 5𝑥 + 4) 𝑑𝑥 1
7
𝑥3 5𝑥 2
= 126 − [ 3 − 2 + 4𝑥] 1
0
119
= 126 −
6
637
= 6 or 106.17 approx. 1
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34(a).

Sol.
Year Index 4 yearly 4 yearly Centered Centered
Number moving moving Total moving 4 yearly
total Average average moving
2001 2450 total – 1

2002 1470 mark

7870 1967.5
2003 2150 3625 1812.5 4 yearly
moving
6630 1657.5
average
2004 1800 3435 1717.5
– 1½
7110 1777.5
marks
2005 1210 3592.5 1796.25
7260 1815
centered
2006 1950 3805 1902.5
total - 1
7960 1990
mark
2007 2300 4297.5 2148.75
9230 2307.5
centered
2008 2500 4797.5 2398.75 moving
9960 2490 average
2009 2480 - 1½
2010 2680 marks

OR
34(b).

Sol.

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Year Index 𝑥𝑖 − 𝐴 𝑋2 𝑋𝑌 𝑌𝑡 = 𝑎 + 𝑏𝑥
𝑋=
0.5
(𝑥𝑖 ) Number
𝑥𝑖 − 2013.5
(𝑌) =
0.5
2011 210 −5 25 −1050 234.17
+ (−5)1.64
= 225.97
2012 225 −3 9 −675 229.25
2013 275 −1 1 −275 232.53
2014 220 1 1 220 235.81
2015 240 3 9 720 239.09 2½ for

2016 235 5 25 1175 242.37 the
correct
𝑛=6 1405 ∑𝑋 = 0 ∑ 𝑋 2 = 70 ∑ 𝑋𝑌 = 115
table

∑𝑌 1405
𝑎= 𝑛 = = 234.17 (approx.) 1
6
∑ 𝑋𝑌 115
𝑏 = ∑ 𝑋2 = 70 = 1.64 (approx.) 1

Required line is 𝑌 = 𝑎 + 𝑏𝑥 = 234.17 + 1.64 𝑥 ½
35.

Sol. Amount needed after 12 years = ₹ 1,00,000 − ₹ 5,000 = ₹ 95,000 1
The payments into sinking fund consist of 12 annual payments at the rate of 5% per
year.
(1+𝑖)𝑛 −1
𝐴 = 𝑅[ ]
𝑖

(1.05)12 −1 2
95000 = 𝑅 [ ]
0.05
95000×0.05 4750 2
⟹𝑅= = ₹ 5968.84 (or = ₹ 5937.50 using approximations)
0.7958 0.8

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SECTION E
This section comprises of 3 case-study based questions of 4 marks each.
36.

Sol. (i) 𝑉 = 𝑥 (36 − 2𝑥)2 1

𝑑𝑉
(ii) = (36 − 2𝑥)2 + 2𝑥(36 − 2𝑥)(−2)
𝑑𝑥

= (36 − 2𝑥 )(36 − 2𝑥 − 4𝑥)
= (36 − 2𝑥 )(36 − 6𝑥 )
= 12 (18 − 𝑥)(6 − 𝑥) 1

𝑑𝑉
(iii) = 0 ⟹ 𝑥 = 18 or 𝑥 = 6 1
𝑑𝑥
Rejecting 𝑥 = 18, we have 𝑥 = 6
𝑑2 𝑉
and 𝑑𝑥 2 = 12(18 − 𝑥 )(−1) + 12 (−1)(6 − 𝑥 )
𝑑2 𝑉 ½
⟹ 𝑑𝑥 2 < 0 at 𝑥 = 6
∴ volume is maximum for 𝑥 = 6 ½
OR
𝑑𝑉
(iii) = 0 ⟹ 𝑥 = 18 or 𝑥 = 6 1
𝑑𝑥
Rejecting 𝑥 = 18, we have 𝑥 = 6
𝑑2 𝑉
and 𝑑𝑥 2 = 12(18 − 𝑥 )(−1) + 12 (−1)(6 − 𝑥 )
𝑑2 𝑉
⟹ 𝑑𝑥 2 < 0 at 𝑥 = 6
∴ Max V = 6 (36 − 12)2 = 6 (24)2 = 3456 𝑐𝑚3 1

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37.

Sol. (i) P to C = 9 − (𝑥 + 𝑦) 1

(ii) Q to A = (4 − 𝑥)
½
Q to B = (4 − 𝑦)
Q to C = 6 − [9 − 𝑥 − 𝑦] = (𝑥 + 𝑦 − 3) ½

(iii) 𝑍 = 160 𝑥 + 100 𝑦 + 150 (9 − 𝑥 − 𝑦) + 100 (4 − 𝑥 ) + 120(4 − 𝑦) +
100(𝑥 + 𝑦 − 3) 1
= 10𝑥 − 70𝑦 + 1930 1
OR

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(iii) 𝑥+𝑦≤9 ½
𝑥+𝑦≥3 ½
𝑥 ≤ 4, 𝑦 ≤ 4 1
𝑥, 𝑦 ≥ 0
38.

Sol. 𝑃𝑖 𝑃 𝑖 (1+𝑖)𝑛 1
(i) 𝐸 = 1−(1+𝑖)−𝑛 or (1+𝑖)𝑛−1

12
(ii) 𝑃 = ₹ 7, 00, 0000, 𝑖 = = 0.01, 𝑛 = 12 × 30 = 360 months 1
1200

7,00,000×0.01
(iii) 𝐸 = ₹ 1−(1.01)−360 1
7000 7000
= 0.97218332 = ₹ 7200.29 (or = ₹ 7216.49 using approximations) 1
0.97

OR

7,00,000×0.01
(iii) 𝐸 = ₹ 1−(1.01)−240 1
7000 7000
= 0.90819416 = ₹ 7707.60 (or = ₹ 7692.31 using approximations) 1
0.91

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