CBSE Class 12 Question Paper 2023 Solution Applied Mathematics

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior School Certificate Examination, 2023
SUBJECT NAME: APPLIED MATHEMATICS (SUBJECT CODE S46547A)
(PAPER CODE 465)
General Instructions: -

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>
Set 4
M.S. Code S46547A
MARKING SCHEME

APPLIED MATHEMATICS

Section A
Q. EXPECTED OUTCOMES/VALUE POINTS Marks
No.
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. (c) 6 (1)
2.

Sol. (c) 6 (1)
3.

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Sol. −5 10 ] (1)
(d) [
0 −25
4.

Sol. (a) 𝒙 = 𝟑, 𝒚 = 𝟓 (1)
5.

Sol. (b) 8 : 7 (1)
6.

Sol. (b) 𝒙 ∈ (−𝟏, ∞) (1)
7.

Sol. (d) −𝟖𝟏 (1)

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8.

Sol. (a) −𝟏 (1)
9.

Sol. 𝒅(𝑨𝑪) 𝟏 (1)
(d) = 𝒙 (𝑴𝑪 − 𝑨𝑪)
𝒅𝒙

10.

Sol. (c) −𝐱𝐞−𝐱 + 𝐂 (1)
11.

Sol. (b) 𝒙𝒚 = 𝑪 (1)

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12.

Sol. 𝟏 (1)
(b) 𝒆

13.

Sol. (b) accepted (1)
14.

Sol. (c) 𝒏𝟏 + 𝒏𝟐 − 𝟐 (1)
15.

Sol. (a) 𝒚𝒄 = 𝒂 + 𝒃𝒙 (1)

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16.

Sol. 𝑹 (1)
(c) 𝒊

17.

Sol. (b) 10.38% (1)
18.

Sol. (a) I quadrant (1)

19.

Sol. (d) Assertion (A) is false and Reason (R) is true. (1)

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20.

Sol. (d) Assertion (A) is false and Reason (R) is true. (1)
SECTION B
This section comprises very short answer (VSA) type questions of 2 marks
each.
21(a).

Sol. Let B be closed after n minutes. Then, pipe A runs for 18 minutes and B runs
for n minutes to fill the tank.
𝟏𝟖 𝐧
 𝟐𝟒 + 𝟑𝟐 = 1 (1)

3 n
 4 + 32 = 1  n = 8. (1)

Hence, pipe B must be closed after 8 min

OR
21(b).

Sol.
Suppose A takes ‘t’ seconds to run 1 km race. Then, B takes (t + 30)
seconds and C takes (t + 30 + 15) seconds, i.e. (t + 45) seconds.

We find A beats C by (30 + 15) seconds = 45 seconds and it is given that
A beats C by 180 metres.
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 C runs 180 m in 45 seconds 𝟏
( 𝟐)

45
 C runs 1000 m in (180 × 1000) seconds = 250 seconds.
(1)
 t + 45 = 250  t = 205 𝟏
( 𝟐)
Hence, A takes 205 seconds to run 1 km
22.

𝐱+𝟑 –𝐱+𝟕 x–7
Sol. − 𝟐 ≤ 𝟎  𝐱 – 𝟐  0 or x – 2  0 (1)
𝐱–𝟐
Thus, the solution set is (– , 2) ∪ [7, ∞) (1)
23(a).

Sol. 2 – 1 𝟏
( 𝟐)
Here, D = | | = 13
3 5
17 – 1 𝟏
D1 = | | = 91 ( 𝟐)
6 5
2 17 𝟏
D2 = | | = – 39 ( 𝟐)
3 6
D D
Thus, x = D1 = 7; y = D2 = – 3 𝟏
( 𝟐)

OR
23(b).

Sol.
A is singular gives

𝑥 + 1 – 3 4 𝟏
| – 5 𝑥 + 2 2 |=0 ( 𝟐)
4 1 𝑥– 6

i.e. (x + 1) [(x + 2) (x – 6) – 2] + 3[– 5x + 30 – 8] + 4[– 5 – 4x – 8] = 0

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i.e. (x + 1) (x2 – 4x – 14) – 15x + 66 – 52 – 16x = 0

i.e. x3 – 3x2 – 49x = 0 (1)

3±√205
x = 0, 2 𝟏
Hence, x = 0 is the only integral value. ( 𝟐)

24.

Sol. Here, 6y = x3 + 2
𝟏
dy dx ( 𝟐)
 6 dt = 3x2 dt
𝟏
dy dx
As dt = 8 dt , we have ( 𝟐)

𝟏
dx dx ( 𝟐)
48 = 3x2  x = 4, – 4
dt dt
𝟏
– 31 ( 𝟐)
when x = 4, y = 11; when x = – 4, y = .
3
– 31
 Points on the curve are (4, 11), (– 4, )
3

25.

Sol. 2 𝟏
Let p be the probability that an item is defective so, p = = 0·02. ( )
100 𝟐
𝟏
Here n = 100  m = np = 2 ( 𝟐)

mr 2r e–2
P(X = r) = r! e–m =
𝟏
r!
( 𝟐)
𝟏
𝟐𝟑 𝐞–𝟐 𝟒 ( 𝟐)
 P(X = 3) = =  0·135 = 0·18
𝟑! 𝟑

SECTION C

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

Sol. Let the original quantity of dettol be x litres and the quantity of Dettol
replaced by water be y litres.
x y 3 (1)
So, y = 3. After 3 operations the quantity of dettol left = x(1 – x) .

x 3
After 3 operations the quantity of water in the bottle = x – x(1 – 3x) (1)

x 3 x 3
Hence, the required ratio is x (1 – 3x) ∶ [x − x (1 – 3x) ]

1 3 1 3
= (1 – 3) ∶ [1 − (1 – 3) ]

8 19
= 27 ∶ 27
= 8 : 19 (1)
26(b).

Sol.
Here, 𝑛𝐴 = 3, 𝑛𝐵 = 7 and 𝑛𝐶 = 10.
1 1 1 1
= − −
𝑛 𝑛𝐴 𝑛𝐵 𝑛𝐶
(2)
1 1 1 1
 = – –
𝑛 3 7 10
1 19 1 (1)
 =  n = 11
𝑛 210 19
1
Hence, the tank is filled in 11 hours.
19

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27.

Sol. y = x3 – 6x2 + 9x – 8

 dx = 3x2 – 12x + 9
dy
(1)

dy
 dx = 3(x − 1)(x − 3)

Critical points are 1, 3 (1)
𝟏
Showing, x=1 is a point of local maxima. ( 𝟐)
𝟏
Showing, x=3 is a point of local minima. ( 𝟐)

28.

Sol. Let A be the event of obtaining two sixes in the first five throws of a die. Let
B be the event of obtaining a six in the sixth throw of a die.
Then required probability = P(AB) = P(A) P(B)

1 1 2 5 3 625
Here, P(B) = 6 and P(A) = 5C2 (6) (6) = 3888 (2)

625 1 625
Thus, Required probability = 3888  6 = 23328 (1)

29.

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Sol. We are given

–
 = 50, x = 55, SD = 10, 𝑛 = 20
(1)
H0:  = 50

H1:  > 50

𝑥̅ −𝜇 55 – 50
t = 𝑆𝐷 = 10 = 2·236 (2)
√𝑛 √20

t cal value > t tab value

Hence H0 is rejected.
So, Advertising Campaign was successful.
30(a).

Sol. Here C = ₹ 4,50,000

S = ₹ 1,00,000

and n = 5 years.

C–S (2)
Annual depreciation D = = ₹ 70,000
n

Thus, yearly depreciation schedule is as follows:
Book value at the Book value at the
Depreciation
Years beginning of the year end of the year
(in ₹)
(in ₹) (in ₹)

1 4,50,000 70,000 3,80,000

2 3,80,000 70,000 3,10,000

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(1 for
3 3,10,000 70,000 2,40,000
correct
4 2,40,000 70,000 1,70,000 table)

5 1,70,000 70,000 1,00,000

30(b).

15 𝟏
Sol. Here P = ₹ 9,50,000, i = 1200 = 0·0125 ( 𝟐)
n = 48 months 𝟏
( 𝟐)

Using the reducing balancing method,
𝑃𝑖 9,5,0000 × 0·0125 (1)
E =1 – (1 + 𝑖)–𝑛 = 1 – (1 + 0·0125)–48

11875 11875 𝟏
=1 – (1.0125)–48 = 1 – 0.5508565 ( 𝟐)
𝟏
( )
𝟐
= ₹ 26,439·21

31.

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Sol.

(2)

Corner Points Value of Z
O (0,0) 0
(1)
A (16,0) 4800
B (8,16) 5440 → Max Value
C (0,24) 4560

So Z is maximum at B (8,16)

Max Value of Z = 5440

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

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32(a).

Sol. Here, |A| = – (– 4 – 3) – (12 + 1) + 2(9 – 1)
(1)
= 7 – 13 + 16 = 10  0

– 7 – 13 8 𝑇 – 7 2 3 𝟏
 adj(A) = [ 2 (2𝟐)
– 2 2 ] = [– 13 – 2 7]
3 7 – 2 8 2 – 2

1 – 7 2 3
Hence A–1 = [– 13 – 2 7] 𝟏
( 𝟐)
10 8 2 – 2

−1 1 2 −7 2 3 1 0 0
1 (1)
A𝐴−1 = 10 [ 3 −1 1] [−13 −2 7 ] = [ 0 1 0]
−1 3 4 8 2 −2 0 0 1

OR
32(b).

Sol.
The matrix equation AX = B is

1 −1 1 𝑥 4 𝟏
[2 1 −3] [𝑦]= [0] ( )
𝟐
1 1 1 𝑧 2

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|A| = 10
(1)

4 −5 1 ′ 4 2 2
adj A = [2 0 −2] = [– 5 0 5]
2 5 3 1 – 2 3
(2)
4 2 2
–1 1
Here A = [– 5 0 5]
10 1 – 2 3 𝟏
( 𝟐)

𝑥 4 2 2 4 2
1
So, [𝑦] = 10 [−5 0 5 ] [0 ] = [– 1 ]
𝑧 1 −2 3 2 1
(1)
Thus, x = 2, y = – 1, z = 1

33(a).

Sol. Let the two parts be x and 15 – x. Then, let y = x2(15 – x)3 (1)

𝑑𝑦
 𝑑𝑥 = x(15 – x)2 (– 5x + 30) (1)

𝑑𝑦 𝟏
= 0 gives x = 0, 15, 6 (1𝟐)
𝑑𝑥

Rejecting x = 0, 15. Hence x = 6

Showing, 𝑥 = 6 is a point of maxima (1)

So, y is maximum when x = 6.

Hence two parts are 6 and 9 𝟏
( 𝟐)

OR
33(b).

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Sol.
Let P (x, y) be the required point which is nearest to Q (1, 4). Then
𝟏
distance PQ should be minimum and hence (PQ)2 should be minimum. ( 𝟐)

2
𝑦2
Now, (PQ)2 = (x – 1)2 + (y – 4)2 = ( 2 – 1) + (y – 4)2 (1)

𝑦 4 – 32𝑦 +68
= (1)
4

𝑦 4 – 32𝑦 +68
Let D = 4
𝟏
𝑑𝐷 ( 𝟐)
= y3 – 8
𝑑𝑦

𝑑𝐷
=0⇒y=2 (1)
𝑑𝑦

𝟏
Showing, y = 2 is a point of minima ( 𝟐)
𝟏
Thus, the point is (2, 2) ( 𝟐)

34.

Sol. Consider year 2014 as the year of origin. Calculation of trend values by
method of least squares.
Squares of
Sales Deviations Deviations Sales
Year
(in lakh ₹) y from 2014 (x) deviation (xy)
(x2)
2010 65 –4 16 – 260

2012 68 –2 4 – 136

2013 70 –1 1 – 70

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2014 72 0 0 0

2015 75 1 1 75

2016 67 2 4 134 (2 for

2019 73 5 25 365 correct
table)
n=7 y = 490 x = 1 x2 = 51 xy = 108

The equation of the straight-line trend is

yc = a + bx

Two normal equations are

y = na + bx
xy = ax + bx2

 490 = 7a + b and 108 = a + 51b
(1)
 a = 69·9 and b = 0·75

yc = 69.9 + 0.75x (1)

Thus, trend values are

𝑦2010 = 69·9 + 0·75(– 4) = 66·90

𝑦2012 = 69·9 + 0·75(– 2) = 68·40

𝑦2013 = 69·9 + 0·75(– 1) = 69·15
(1 for
𝑦2014 = 69·9 + 0·75(0) = 69·90
correct
𝑦2015 = 69·9 + 0·75(1) = 70·65 trend
𝑦2016 = 69·9 + 0·75(2) = 71·40 values)

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𝑦2019 = 69·9 + 0·75 (5) = 73·65

35.

Sol. CAGR is the mean annual growth rate of an investment over a specified (1)
period of time longer than one year.
1
Ending investment amount no.of years (1)
CAGR = [ ] −1
Start amount

P.V. = ₹ 10,000

F.V. = ₹ 14,000 (1)

n = 6 years

14000 1/6
So, CAGR = (10000 ) – 1 = (1·4)1/6 – 1 𝟏
( 𝟐)
𝟏
= 1·058 – 1 ( 𝟐)
𝟏
= 0·058 ( 𝟐)

Hence, CAGR = 5·8% 𝟏
( 𝟐)

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

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36.

Sol. 6
n= 100, p = 100 , m= np
6
Here m = 100  100 = 6.

mr
P(r) = 𝑒 −𝑚 r!

𝑚0
(i) P (0) = 𝑒 −𝑚 0! = e–6 = 0·0024 (1)

𝑚2
(ii) P (2) = 𝑒 −𝑚 2! = e–6  2 = 0.0432
36
(1)
1
(iii)(a) P(0) + P(1) = e–6 + e–6 1! = e–6 + 6e–6 = 7e–6 = 0·0168
𝑚 (1+1)

OR

(iii)(b) Mean = Variance = m = np = 6
(1+1)

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37.

Sol. (i) Z = 10x + 20y (1)
(1)
(ii) x + 3y  24
(iii) (a) other constraints are
2x +y ≤ 28
𝑥≥0
𝑦≥0

(1)

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Corner Points Value of Z
O (0,0) 0
A (14,0) 140
B (12,4) 200 → Max value
C (0,8) 160 (1)

 P is maximum at B (12,4); which is ₹ 200
OR

(iii) (b)

(1)

Corner Points Value of Z
O (0,0) 0
A (14,0) 140
B (12,4) 200 → Max value
C (0,8) 160 (1)

12 bats and 4 rackets
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38.

Sol. 7·5
Given P = ₹ 30,00,000, i = 1200 = 0·00625

and n = 12  20 = 240 months
𝑃𝑖
(i) EMI = 1− (1+𝑖)−𝑛
30,00,000 × 0·00625 𝟏
= 1− (1·00625)−240 – 1 ( 𝟐)

30,00,000 × 0·00625 × 4·4608
= 𝟏
3·4608 ( 𝟐)
= ₹ 24167.82

(ii) Interest paid on 150th instalment

EMI × [(1 + i)240– 150 + 1 – 1]
= (1 + i)240– 150 + 1

24167 × [1·7629 – 1] 𝟏
= ( 𝟐)
1·7629
= ₹ 10458.70

 Principal paid in 150th instalment = EMI – interest

=₹ (24167.82 – 10458.70) 𝟏
( 𝟐)

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= ₹ 13709.12

(iii) (a) Total Interest paid = n  EMI – P

= ₹ (240  24167.82 – 30,00,000) (1)
= ₹ 28,00,276.80 (1)
OR

(iii) (b) Total amount paid = n x EMI
(1)
= 240 x 2416.81

= ₹ 5800276.8 (1)

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