CBSE Class 12 Question Paper 2026 Solution Applied Mathematics

FOR CBSE CLASS 12 EXAM PREPARATION

CBSE Class 12 2026
Question Paper
Solution · Applied
Mathematics
EXAM YEAR TYPE SUBJECT

CBSE Class 12 2026 Question Paper Solution Applied Mathematics

Notes · Sample Papers · Previous Year Papers · Mock Tests

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m .co Marking Scheme
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Strictly Confidential
(For Internal and Restricted use only)
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Senior Secondary School Examination, 2026 (XII) a
APPLIED MATHEMATICS (241) QUESTION PAPER CODE – 465
General Instructions: -

1 The CBSE has decided to introduce On Screen Marking (OSM) for the evaluation of Class
XII answer Book with the 2026 Examination.
2 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
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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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“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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public in any manner could lead to derailment of the examination system and affect
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4 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,

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answers which are based on latest information or knowledge and/or are innovative,

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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
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understand given answer and even if reply is not from marking scheme but correct

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competency is enumerated by the candidate, due marks should be awarded.
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
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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

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after deliberation and discussion. The remaining answer books meant for evaluation shall

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be given only after ensuring that there is no significant variation in the marking of individual

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o evaluators.
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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 an impression that
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ag answer is correct and no marks are awarded. This is most common mistake which
evaluators are committing.
8 If a question has parts, please award marks on the right-hand side for each part in the OSM
Portal. Marks awarded for different parts of the question will be totaled up by the OSM
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OSM Portal. This may also be followed strictly.
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once.
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11 A full scale of marks ___80____ (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.
13 Ensure that you do not make the following common types of errors committed by the
Examiner in the past :-
● 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.
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for Spot Evaluation” before starting the actual evaluation.
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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.
17 If a candidate attempts both alternatives/options in a question where only one option/
alternative is required to be attempted, the Evaluator shall award marks in both the
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then the evaluator shall mark “NA” (Not attempted) against the option that has not
been attempted by the candidate.

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MARKING SCHEME
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(PAPER CODE: 465)
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APPLIED MATHEMATICS (Subject Code–241)

Section A
Q. No. EXPECTED OUTCOMES/VALUE POINTS Step Marks

SECTION A
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This section comprises 20 Multiple Choice Questions (MCQs) of 1 mark
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each.
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1.
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Sol. (C) x > 4 I 1

2.

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Sol. (D) 64
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3.
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Sol. (D) 2B I 1

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4.
c. o s e m
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a

Sol. (D) 0 I 1

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5.

Sol. (B) y = e|cx| I 1

6.

Sol. (B) 9 I 1

7.

Sol. 3 I 1
(B)
4𝑡

8.

Sol. (B) 12 𝜋 I 1

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Sol. (B) E(X 2 ) − [E(X)]2 I 1

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Sol. (C) m I 1

11.

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Sol. (A) 1
a I 1

12.

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Sol. (B) 33 I 1

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Sol. (D) (−∞, ∞) I 1

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14.

Sol. R I 1
(C)
i

15.

Sol. (B) ₹ 800 I 1

16.

4
Sol. (C) ( √6 − 1) × 100 I 1

17.

Sol. (B) ₹ 7,000 I 1

18.

Sol. (C) 31 I 1

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19.

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

Note: Full 1 mark may be awarded to the students who attempted the question. I 1

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20.

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Sol. (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct I 1
explanation of the Assertion (A).

SECTION B
This section comprises 5 very Short Answer (VSA) type questions of 2
marks each.

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21.
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s em g l a
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a Sol. Let 𝑥 km/h be the speed of the boat in still water.

∴ Speed of the boat downstream = (𝑥 + 3) km/h
I ½
Speed of the boat upstream = (𝑥 − 3) km/h
12 12
According to question, + =3 II ½
m .
𝑥+3 𝑥−3

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⟹ 3𝑥 2 − 24𝑥 − 27 = 0 or 𝑥 2 − 8𝑥 − 9 = 0 III ½

⟹ (𝑥 − 9)(𝑥 + 1) = 0

Rejecting the −ve value, we get

𝑥=9 IV ½

So, the speed of the boat in still water = 9 km/h

22 (a).

Sol. dy dx I ½
∫ 2−y = ∫ x+1

⟹ − log|2 − y| = log|x + 1| − log c II 1

⟹ log|(2 − y)(x + 1)| = log c III ½

or |(2 − y)(x + 1)| = c

OR

22 (b).

b b
Sol. x4 x3 2
Here, | | = 0 and | | =
4 a 3 a 3 I ½
1 1 2
⟹ (b4 − a4 ) = 0 and (b3 − a3 ) = II ½
4 3 3

Solving the two, we get a = −1, b = 1 III 1

23 (a).

Sol. e−m (m)r
P (X = r ) =
r!

e−1 (1)1
P (r = 1) =
1! I 1
1
= 𝑒 −1 or II 1
e

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OR
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23 (b).
a

1 4
Sol. Mean = 𝑛𝑝 = 4 × = I 1
3 3

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1 2 √8 2√2
S.D = √𝑛𝑝𝑞 = √4 × × =
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or II 1
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3 3 3 3

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24.
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Sol.

Year (t) Variable (x) 3-yearly 3-yearly
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moving total moving

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2013 3
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2014 5 15 5

2015 7 22 7.33

2016 10 29 9.67

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2017 12 36 12

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2018 14 41 13.67
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2019 15 45
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a 2020 16 - -

Correct Table I 2

(Remark: 1 mark for any 3 correct entries)

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25.

Sol. 𝑃𝑖
EMI =
1−(1+𝑖)−𝑛

15
Here, P = ₹ 2,00,000, 𝑖 = = 0.0125, 𝑛 = 4 × 12 = 48
1200

200000 ×0.0125
EMI = I 1½
1−(1.0125)−48

2500
= = ₹ 5555.56 (approx.) II ½
1−0.55

SECTION C
This section comprises 6 short answer (SA) type questions of 3 marks each.

26 (a).

Sol. (i) (17 + 13) mod 30 = 30 mod 30 = 0 I 1½

(ii) (11 − 3) mod 8 = 8 mod 8 = 0 II 1½

OR

26 (b).

1
Sol. Part of the tank filled by pipes A, B and C in 1 hour = I ½
8

2 1
Part of the tank filled by pipes A, B and C in 2 hours = = II ½
8 4

1 3
Remaining part to be filled = 1 − = III ½
4 4

3 1 1
Part of unfilled tank filled by pipes A and C in 1 hour = × = IV ½
4 9 12

1 1 1
So, part of tank filled by pipe B alone in 1 hour = − = V ½
8 12 24

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Hence the time taken by pipe B alone = 24 hours
g l a VI ½
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27.

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Sol. 2 −3 5
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D = |3 2 −4| = −1
e
I ½
1
em
1 −2
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s
la −3 5
Since D ≠ 0 ∴ the system of equations is consistent.
ag
a g
11
𝐷 = |−5 2 −4| = −1
1 II ½
−3 1 −2
2 11 5
𝐷2 = |3 −5 −4| = −2 III ½
1 −3 −2
2 −3 11

om
𝐷3 = |3 2 −5| = −3 IV ½
1 1 −3
. c
𝐷1 𝐷2 𝐷3
e m
𝑥= = 1, 𝑦 = = 2, 𝑧 = =3
s
V 1

la
𝐷 𝐷 𝐷

28 (a).
ag

Sol. f ′ (x) = 4x 3 − 4x = 4x(x − 1)(x + 1) I 1

f ′ (x) = 0 ⟹ x = 0, 1, −1 II 1

Sign of f’(x)
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s em f(x) is decreasing for x ϵ (−∞, −1) ∪ (0,1) and increasing
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g la x ϵ (−1,0) ∪ (0, ∞). a
a
OR f(x) is decreasing for x ϵ (−∞, −1] ∪ [0,1] and increasing for III 1
x ϵ [−1,0] ∪ [0, ∞).

OR

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28 (b).

2x+1
Sol. I = −∫ 2 dx I 1
x +4x−18

2x+4 3
= −∫ 2 dx + ∫ 2 dx II 1
x +4x−18 x +4x−18

3
== − log|x 2 + 4x − 18| + ∫ dx III 1
(x+2+2√22)(x+2−2√22)

3 x+2−√22
= − log|x 2 + 4x − 18| + log | |+𝑐
2√22 x+2+√22

29.

Sol. 𝑥̅ = 147, 𝜇 = 140

𝑛 = 26, 𝑠 = 16

𝐻0 : Null hypothesis : If there is no significant difference between 𝑥̅ and 𝜇

𝐻1 : Alternate hypothesis : If there is a significant difference between 𝑥̅ I ½
and 𝜇
𝑥̅ −𝜇 147−140
𝑡= 𝑠 = 16 = 2.187 II 1½
√𝑛−1 √25

Df = 25 and 𝑡25 (0.05) = 2.06

Since |𝑡 | = 2.187 > 2.06 III ½

∴ Null hypothesis is rejected

Hence the advertisement campaign is effective. IV ½

̅−𝝁
𝒙
Note:* 𝒕 = 𝒔 = 2.23 should also be accepted.
√𝒏

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Sol. Here, F = ₹ 10,000, n = 10, i = 0.08

R = 8.5% of ₹ 10000 = ₹ 850, C = F = ₹ 10,000 I 1
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Let V be the purchase price of the bond, then
c o m
] .+ C(1 + i) s e
1−(1+i)−n
V = R[
e m −n

l a
l asi

ag
ag[
1−(1+0.08)−10
= 850 ] + 10000(1 + 0.08)−10 II 1
0.08

= ₹(5703.57 + 4631.93) = ₹ 10,335.50 (approx.) III 1

31.

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s e
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a
Sol.

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s e g l a
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Correct graph I 1½
Corner Points Value of Z

O (0,0) 0

A (0,600) 3000

B (800, 600) 5400

C (1000,500) 5500 Maximum Value II 1½

D (1500, 0) 4500

∴ Maximum Z = ₹ 5500 at B (1000, 500)

SECTION D
This section comprises 4 Long Answer (LA) type questions of 5 marks
each.

32.

Sol. Given equation can be written as

(2x − 1) − 4(x − 11) < 3(3x + 1) I 2

⟹ −2x + 43 < 9x + 3 II 2

⟹ 11 x > 40
40
⟹𝑥> III 1
11

40 40
Hence the solution is 𝑥 > or ( , ∞)
11 11

33 (a).

Sol. For equilibrium, p = p0 and x = x0

25 − x0 − x0 2 = 19 I ½

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⟹ (x0 + 3)(x0 − 2) = 0
g l a II 1

Rejecting the −ve value
a
x0 = 2 III ½

p0 x0 = 38 IV ½
2
CS = ∫0 (25 − x − x 2 )dx − p0 x0
V 1
m
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2
x2 x3
= 25x − − ⌉ − 38
e m
s
VI 1
em
2 3 0

s l
g ½a
= 50 − 2la
VIIa
8 22
− − 38 =
a g 3 3

OR

33 (b).

m
Sol.
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The given equation can be written as
s e
𝑑𝑦
=
2𝑥
𝑑𝑥 g la I 1
𝑦 𝑥−1
a
𝑑𝑦 2𝑥
∫ 𝑦 = ∫ 𝑥−1 𝑑𝑥

𝑑𝑦 1
⟹∫ = 2 ∫ (1 + ) II 1½
𝑦 𝑥−1

⟹ log|𝑦| = 2(𝑥 + log|𝑥 − 1|) + 𝑐 III 1

m
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⟹ log|𝑦| − log(𝑥 − 1)2 = 2𝑥 + 𝑐
m
c. oWhen 𝑥 = 2, 𝑦 = 1 then 𝑐 = −4 s e m
em
IV 1
s g l a
l a So, the solution is
a
ag log|𝑦| − log(𝑥 − 1)2 = 2𝑥 − 4 V ½

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34 (a).

Sol.
Let a random variable X denote the number of aces.

Then X can take the values 0, 1 and 2

Let 𝑝 be the probability of getting an ace.

4 1 12
p= = ⟹q= I 1
52 13 13

X 0 1 2

II 2
12 12 144 12 1 24 1 1 1
P(X) × = 2× × = × =
13 13 169 13 13 169 13 13 169

1 2
E(X) = np = 2 × = III 1
13 13

1 12 24
Var (X) = npq = 2 × × = IV 1
13 13 169

OR

34(b).

Sol. Let X be the random variable and 𝒑 be the probability of defective screws
2
Then 𝑝 = = 0.02, 𝑛 = 100 I 1
100

𝑚 = 𝑛𝑝 = 2 II 1

(i) Probability that the packet has no defective screw

𝑚0
= 𝑃(𝑋 = 0) = 𝑒 −𝑚 = 𝑒 −2 III 1½
0!

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m
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m .co (ii) Probability that the packet has one defective screw as
e m
se 1 a g l
−𝑚 𝑚 −2
= 𝑃 (𝑋 = 1) = 𝑒 = 2𝑒 IV 1½
1!

35.

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Sol.
l a
Let the purchase price of the goldmine be ₹ 𝑥
ag
Returnagon investment = 18% of 𝑥 = 0.18 𝑥
Sinking fund is created to replace the purchase price

∴ Amount to be placed in sinking fund = ₹ (400000 − 0.18𝑥)
𝑆𝑟
Now, 𝐴 = (1+𝑟)𝑛
−1
m
⟹ (400000 − 0.18 𝑥) =
0.1 𝑥
m .co
(1.1)10 −1
s e
0.1
+ 0.18) gl
a
⟹ 400000 = 𝑥 (
1.5937 a
Solving, we get 𝑥 = 1648125.25 (approx.)

Thus, the purchase price of the goldmine is ₹ 1648125.25

Note: Full 5 marks may be awarded to the students who attempted the I 5
question.
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SECTION E
This section comprises 3 case-study based questions of 4 marks each.

36.

Sol.
(i) P = [25 50 10] I 1
50 30 35
(ii) S = [60 35 40] I 1
40 50 25

(iii) (a) Funds collected by School B
30
= [25 50 10] [35] I 1
50

= [750 + 1750 + 500]

= [3000] II 1
OR
(iii) (b) Funds collected by School A

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m
m .co

m .co s e m
se = [25 50
50
10] [60] g l a
40 a
= [1250 + 3000 + 400]

= ₹ 4650 I 1

Funds collected by School C
35
m
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= [25 50 10] [40]

o m 25
m
c
=. [875 + 2000 + 250] s e
e m l a
l as = ₹ 3125 II a
g ½
agTotal amount of funds collected by School A and C
= 4650 + 3125 = ₹ 7775 III ½

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s e
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a

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37.

Sol.

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m
m .co

m .co s e m
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Year Sales (in ₹ A = 2021 X2
XY
l a
Trend value
(xi) thousands) X = xi – A ag Y = a + bX t

𝑌
2018 80 −3 9 −240 90 + 2(−3) =
84
2019 90 −2 4 −180 86
−1 −92
2020 92 1 88
m
2021 83
o m 0 0 0 90
m .co
94.c e
2022
em99
1 1 94 92
l as
2023
l as 2 4 198 94
ag
2024 g
a 92 3 9 276 96
𝑛=7 ∑𝑌 = 630 ∑𝑋 = 0 ∑𝑋 2 ∑𝑋𝑌
= 28 = 56

∑𝑌 630 ∑𝑋𝑌 56
(i) 𝑎= = = 90; 𝑏 = = =2 I ½
𝑛 7 ∑𝑋 2 28

m
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Trend line = Yt = 𝑎 + 𝑏 𝑋 = 90 + 2 𝑥 II ½

s em
la
(ii) Correct Table I 1
g
a = b = 2 × 1000 = ₹ 2000
Average change in sales

(iii) (a) ∑ (Y − Yt ) = −4 + 4 + 4 − 7 + 2 + 5 − 4 = 0 I 2

OR

(iii) (b) Trend value for the year 2025 = 90 + 2 × 4 = 98
o m I 1
m . c
c. o ∴ Expected sales for 2025 = ₹ 98,000
s e m
m
II 1

s e g l a
g la a
a

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38.

Sol.
(i) Objective function 𝑍 = 100𝑥 + 90𝑦 I 1

(ii) Constraints are

𝑥 + 𝑦 ≤ 10
I 1
1800 𝑥 + 1200 𝑦 ≤ 15000 𝑖. 𝑒 3𝑥 + 2𝑦 ≤ 25

𝑥 ≥ 0 ,𝑦 ≥ 0

(½ marks for writing two correct constraints)

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(iii) (a)
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a

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s e l a
g l a ag
a

m
c. o
Correct graph I 1

Corner Points
e
Z = 100 x + 90 y m
s
la0
O (0, 0)
g
a 900
A (0, 10)
B (5,5) 950 Maximum Value II 1
25
C ( , 0) 2500
3
3

To get max profit, a man should buy 5 bags of rice and 5 bags
m
.co
of wheat.
m
m .co OR
s e m
s e g l a
g la a
a

m .
.co s e m
s em l a
la ag
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g
a For more Question Papers, Sample Papers, Notes & Syllabus visit Page 23 of 24

(iv) (b)

Correct graph I 1

Corner Points Z = 100 x + 90 y
(0, 0) 0
(0, 10) 900
(5,5) 950 Maximum Value II 1
25 2500
( , 0)
3 3

Maximum profit = ₹ 950

Note: Alternatively, marks to be given for solving without graph in part
(iii)

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