CBSE Class 12 Sample Paper 2023 Solution Applied Maths

Sample Question Paper
CLASS: XII
Session: 2022-23
Applied Mathematics (Code-241)
Marking Scheme
Section – A
Each question carries 1-mark weightage
𝑥 ≡ 27(mod 4)
⇒ 𝑥 − 27 = 4𝑘, for some integer 𝑘
1. ⇒ 𝑥 = 31 as 27< 𝑥 ≤ 36
(C) option

2. (D) option
n = 26⇒ |𝑡| = 3.07 > 𝑡25 (0.05 ) = 2.06
3. (B) option

n = 34 ⇒ 𝑣 = 34 − 1 = 33
4. (B) option

Speed of boat downstream = u = 10 km/h
And, speed of boat upstream = v = 6 km/h
1
5. ⇒ Speed of stream = ( u – v) = 2 km/h
2
(B) option

6. (C) option
20×1500
Truck A carries water = 100 – ( 1000 ) = 70 𝑙
20×1000
7. Truck B carries water = 80 – ( 1000 ) = 60 𝑙
(C) option

Let the face value of the bond = 𝑥
10
Then, 200 𝑥 = 1800 ⇒ 𝑥 = 36000
8.
(D) option

9. (C) option
10. (D) option
𝐶−𝑆 480000−25000
D= 𝑛 = = 45500
10
11. (B) option

12. (A) option
𝑑𝑦 𝑑𝑥
∫ 𝑦 𝑙𝑜𝑔𝑦 = ∫ 𝑥
13. ⇒ log(𝑙𝑜𝑔𝑦) = log|𝑥| + log|𝐶|
⇒ log(𝑙𝑜𝑔𝑦) = log|𝐶𝑥|
⇒ 𝑦 = 𝑒 |𝐶𝑥|

(B) option

1
60000 4 4
[( ) − 1] × 100 = [ √ 6 − 1] × 100
14. 10000
(C) option

15.

⇒ 180 : 300 = 3 : 5
(C) option

16. (D) option
17. (C) option
18. (B) option
P(Win in one game) = P(Lose in one game) = ½
1 4 1
⇒ P (Beena to win in 3 out of 4 games) = 4 𝐶3 . (2) = 4 = 25%
19. Assertion is correct and Reason is the correct explanation for it
(A)option

Effective rate of interest = Nominal rate – inflation rate = 12.5 – 2 = 10.5%
Assertion is correct
20. Reason is true but not supportive of assertion
(B) option

Section – B
Each question carries 2-mark weightage
21. P = 250000, R = 7500, 𝑖 = 𝑟/400
7500 × 400 1
⇒ 250000 = ⇒ 𝑟 = 12
𝑟
⇒ 𝑟 = 12 1
22. a−8=1⇒𝑎 = 9
2
3b = −2 ⇒ 𝑏 = − 3 1
−c + 2 = −28 ⇒ 𝑐 = 30
⇒ 2a + 3b − c = − 14 1
OR
2 1
Expanding C1, we get ∆ = 1(2𝑥 + 4) − 2(−4𝑥 − 20) = 86

⇒ 𝑥 2 + 4𝑥 − 21 = 0
∴ 𝑥 = 3, −7 1

23. Let the number of hardcopy and paperback copies be x and y respectively
1
⇒ Maximum profit Z = (72x + 40y) −(9600 + 56x + 28y) = 16x + 12y− 9600

Subject to constraints: 1
x + y ≤ 960
5x + y ≤ 2400
x, y ≥ 0
24. Speed of boat in still waters = 𝑥 km/h 1
Speed of stream = 𝑦 km/h
Distance travelled = 𝑑 km
𝑑
Time taken to travel downstream = 𝑥+𝑦
𝑑
Time taken to travel upstream = 𝑥−𝑦
2𝑑 𝑑 1
Then, = 𝑥−𝑦 ⇒ 𝑥 ∶ 𝑦 = 3: 1
𝑥+𝑦
OR 1
Param runs 5 m in 3 seconds
3
⇒ time taken to run 200 m = × 200 = 120 seconds
5
1
Anuj ‘s time = 120 – 3 = 117 seconds

25. 𝑉𝑓 = 437500, 𝑉𝑖 = 350000 1
𝑉𝑓 −𝑉𝑖
Nominal rate = × 100
𝑉𝑖

437500 − 350000 1
= × 100 = 25%
350000
Section – C
Each question carries 3-mark weightage
′ (𝑥)
26. 𝑓 = 𝑥 − 6𝑥 2 + 11𝑥 − 6 = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3)
3
1

⇒ 𝑥 = 1,2,3
Strictly increasing in (1,2)∪(3,∞) 1
Strictly decreasing in (−∞,1)∪(2,3) 1
27.
2500 65
12700
Daily diet of team A = [2 3 ]
1 [1900 50] = [ ]
334 1.5
2000 54

Team A consumes 12700 calories and 334 g vitamin
2500 65
10300
[ ]
Daily diet of team B = 1 2 2 [1900 50] = [ ]
273
2000 54
1.5
Team B consumes 10300 calories and 273 g vitamin

28. 𝑑𝑥
∫
(1 + 𝑒 𝑥 )(1 + 𝑒 −𝑥 )
3
𝑒 𝑥 𝑑𝑥
= ∫ (1+𝑒 𝑥 )2

𝑑𝑡
= ∫ 𝑡 2 , where t = 𝑒 𝑥 + 1 and dt = 𝑒 𝑥 𝑑𝑥

−1
= +𝐶
𝑡
−1
= 1+𝑒 𝑥 + 𝐶
OR
𝑥 𝑙𝑜𝑔(1 + 𝑥 2 )𝑑𝑥
∫ 𝐼𝐼 , Integration by parts
𝐼
𝑑
= log (1 + 𝑥 2 ). ∫ 𝑥𝑑𝑥 − ∫[𝑑𝑥 log(1 + 𝑥 2 ) . ∫ 𝑥 𝑑𝑥] 𝑑𝑥

𝑥2 2𝑥 𝑥2
= 2 log (1 + 𝑥 2 ) − ∫[1+𝑥 2 . 2 ] 𝑑𝑥

𝑥2 𝑥3
= 2 log (1 + 𝑥 2 ) − ∫ 1+𝑥 2 𝑑𝑥

𝑥2 𝑥
= 2 log (1 + 𝑥 2 ) − ∫[𝑥 − 1 + 𝑥 2 ] 𝑑𝑥

𝑥2 𝑥2 1
= 2 log (1 + 𝑥 2 ) − 2 + 2 log (1 + 𝑥 2 ) + 𝐶

1
= 2 [(1 + 𝑥 2 )log (1 + 𝑥 2 ) − 𝑥 2 ] + 𝐶

29. Under pure competition, 𝑝𝑑 = 𝑝𝑠
8 𝑥+3
⇒ 𝑥+1 − 2 = 2
⇒ 𝑥 2 + 8𝑥 − 9 = 0 1.5
⇒ 𝑥 = −9, 1
∴𝑥=1

When 𝑥0 = 1 ⇒ 𝑝0 = 2
1
1 𝑥+3 𝑥 2 3𝑥 1
∴ Produce surplus = 2 − ∫0 2 𝑑𝑥 = 2 − [ 4 + 2 ] = 4 1.5
0

OR
2
𝑝 = 274 −𝑥
⇒ 𝑅 = 𝑝𝑥 = 274𝑥−𝑥 3
𝑑𝑅
= 274 − 3𝑥 2
𝑑𝑥 1.5
Given MR = 4 + 3𝑥
In profit monopolist market,
𝑑𝑅
MR = 𝑑𝑥 ⇒ 4 + 3𝑥 = 274 − 3𝑥 2
⇒ 𝑥 2 + 𝑥 − 90 = 0

⇒ 𝑥 = −10, 9
∴𝑥=9
When 𝑥0 = 9 ⇒ 𝑝0 = 193
9
∴ Consumer surplus = ∫0 (274 − 𝑥 2 )𝑑𝑥 − 193 × 9
9 1.5
𝑥3
= [274𝑥 − 3 ]
0
= 486
30. Purchase = ₹ 40,00,000
Down payment = 𝑥
Balance = 40,00,000 – 𝑥
9
𝑖= = 0.0075, n = 25 x 12 = 300 1
1200

E = ₹ 30,000

(4000000 − 𝑥) × 0.0075
⇒ 30000 =
1 − (1.0075)−300
(4000000 − 𝑥) × 0.0075
⇒ 30000 = 2
1 − 0.1062
⇒ 𝑥 = 424800
Down payment = ₹ 4,24,800

5
31. n = 10 x 2 = 20, S = 10,21,760, 𝑖 = = 0.025, R = ?
200
(1+𝑖)𝑛 −1 1.5
S =R [ ]
𝑖
(1+0.025)20 −1
⇒1021760 = R [ ]
0.025
1.6386−1
⇒1021760 = R [ ]
0.025
1021760×0.025
⇒R = [ ]
0.6386
⇒ R = ₹ 40,000 1.5
Mr Mehra set aside an amount of ₹ 40,000 at the end of every six months

Section – D
Each question carries 5-mark weightage
32. Probability of defective bucket = 0.03
n = 100
m = np = 100 x 0.03 = 3
Let X = number of defective buckets in a sample of 100 1
𝑚𝑟 𝑒 −𝑚
P (X = r) = , 𝑟 = 0,1,2,3, ….
𝑟!

30 𝑒 −3
(i) P (no defective bucket) = P(r = 0 ) = = 0.049 2
0!
(ii) P (at most one defective bucket) = P(r = 0, 1)
30 𝑒 −3 31 𝑒 −3 2
= +
0! 1!

= 0.049 + 0.147
= 0.196
OR
X = scores of students, 𝜇 = 45, 𝜎 = 5
𝑋 − 𝜇 𝑋 − 45 1
∴𝑍= =
𝜎 5

(i) When X = 45, 𝑍 = 0
P (X > 45) = P (Z > 0) = 0.5
2
⇒ 50% students scored more than the mean score

(ii) When X = 30, 𝑍 = −3 and when X = 50, 𝑍 = 1
P (30< X< 50) = P ( -3< Z < 1) = P (-3< Z ≤ 1)
= P ( − 3 < 𝑍 ≤ 0) + 𝑃 (0 ≤Z <1)
= P (0 ≤ 𝑍 < 3) + 𝑃 (0 ≤Z <1) 2
= 0.4987 + 0.3413 = 0.84
⇒ 84% students scored between 30 and 50 marks

33. Let 𝑥 be the number of guests for the booking
Clearly, 𝑥 > 100 to avail discount
200 2
∴ Profit, P = [4800 − 10 (𝑥 – 100)] 𝑥 = 6800𝑥 – 20𝑥 2

𝑑𝑃
⇒ 𝑑𝑥 = 6800 – 40 𝑥 ⇒ 𝑥 = 170
1
𝑑2 𝑃
As 𝑑𝑥 2 = −40 < 0, ∀ 𝑥
1

A booking for 170 guests will maximise the profit of the company
And, Profit = ₹ 5,78,000 1

OR
P(x) = R(x) – C(x) 2
= 5x – (100 + 0.025x2)
⇒P’(x) = 5 – 0.05 x ⇒ 𝑥 = 100 1
As P’’(x) = −0.05 < 0, ∀ 𝑥 1
∴ Manufacturing 100 dolls will maximise the profit of the company
And, Profit = ₹ 1,50,000 1

34. Let the number of tables and chairs be 𝑥 and 𝑦 respectively
(Max profit) Z = 22𝑥 + 18𝑦
Subject to constraints:
𝑥 + 𝑦 ≤ 20 1.5
3𝑥 + 2𝑦 ≤ 48
𝑥, 𝑦 ≥ 0

2

The feasible region OABCA is closed (bounded)
Corner points Z = 22 x + 18 y
O (0,0) 0
A (0,20) 360
B (8,12) 392 1.5
C (16,0) 352
Buying 8 tables and 12 chairs will maximise the profit
35.
1 2 3
A = [3 2 2]
2 3 2
2
⇒ |A| = 9 ⇒ A-1 exists
−2 5 −2
1
And A-1 = 9 [−2 −4 7 ]
5 1 −4

AX = B ⇒ 𝑋 = 𝐴−1 𝐵
−2 5 −2 85 15
1
⇒X = 9 [−2 −4 7 ] [105] = [20]
5 1 −4 110 10
3
⇒ 𝑝1 = 15, 𝑝2 = 20, 𝑝3 = 10

Section – E
Each Case study carries 4-mark weightage
36. CASE STUDY - I
2
Pipe C empties 1 tank in 20 h ⇒ 2/5 th tank in 5 × 20 = 8 hours
a) 1
1 1 1 1
Part of tank filled in 1 hour = 15 + 12 − 20 = 10 th
b) ⇒ time taken to fill tank completely = 10 hours 1

c) At 5 am, 2

Let the tank be completely filled in ‘t’ hours
⇒pipe A is opened for ‘t’ hours
pipe B is opened for ‘t−3’ hours
And, pipe C is opened for ‘t−4’ hours

⇒ In one hour,
𝑡
part of tank filled by pipe A = 15 th
𝑡−3
part of tank filled by pipe B = 15 th
𝑡−4
and, part of tank emptied by pipe C = 15 th
𝑡 𝑡−3 𝑡−4
Therefore 15 + 12 − 20 = 1
⇒ 𝑡 = 10.5
Total time to fill the tank = 10 hours 30 minutes

OR
6 am, pipe C is opened to empty ½ filled tank
Time to empty = 10 hours
Time for cleaning = 1 hour
1 1 3
Part of tank filled by pipes A and B in 1 hour= 15 + 12 = 20th tank
20
⇒ time taken to fill the tank completely = 3 hours
20
Total time taken in the process = 10 + 1 + 3 = 17 hour 40 minutes

37. CASE STUDY - II
a)
Year Y X X2 XY
2015 35 -2 4 -70
2016 42 -1 1 -42
2017 46 0 0 0
2018 41 1 1 41
2019 48 2 4 96
212 10 25

∑𝑌 212 ∑ 𝑋𝑌 25
a = 𝑛 = 5 = 42.4 and b = ∑ 𝑋 2 = 10 = 2.5 2

𝑌𝐶 = 42.4 + 2.5𝑋
OR
Year Y 3-year moving average
2015 35 -
2016 42 41
2017 46 43
2018 41 45
2019 48 -

b) For year 2022,
𝑌2022 = 42.4 + 2.5(2022 − 2017) = 54.9
1
⇒ the estimated sales for year 2022 = ₹ 54,900

c) 𝑌𝐶 = 42.4 + 2.5𝑋
⇒ 67.4 = 42.4 + 2.5𝑋
1
⇒ 𝑋 = 10
Sales will be ₹ 67,400 in year (2017+ 10) = year 2027
38. CASE STUDY - III
a)
𝑘 2𝑘 3(1−𝑘) 4𝑘 1
+ 6 + + 2 =1⇒𝑘 =4 1
6 6

b) P (getting admission on applying at least 2 weeks ahead of application
deadline)
= P (X = 2,3,4)
1 3 1 23 1
= 12 + 8 + 2 = 24
1 23
[alternate method: 1 – P (X = 1) = 1 - 24 = 24 ]

c) X = week applied ahead of application deadline
X 1 2 3 4
P(X) 1 1 3 1
24 12 8 2

XP(X) 1 1 9 2
24 6 8
80 1 2
∴ E(X) = 24 = 3 3 weeks
OR

X = Scholarship money awarded for the week applied in, before the
deadline
Week 1 2 3 4
applied in
X 9600 12000 20000 50000

P(X) 1 1 3 1
24 12 8 2
XP(X) 9600 12000 60000 50000
24 12 8 2

∴ E(X) = ₹ 33,900

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