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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 = 4 𝑘, for some integer 𝑘
⇒ 𝑥 = 31 as 27 < 𝑥 ≤ 36
(C) option
- (D) option
n = 26⇒ |𝑡| = 3. 07 > 𝑡
25
(B) option
n = 34 ⇒ 𝑣 = 34 − 1 = 33
(B) option
Speed of boat downstream = u = 10 km/h
And, speed of boat upstream = v = 6 km/h
⇒ Speed of stream =
1
2
( u – v) = 2 km/h
(B) option
- (C) option
Truck A carries water = 100 – (
20 × 1500
1000
Truck B carries water = 80 – (
20 × 1000
1000
(C) option
Let the face value of the bond = 𝑥
Then,
10
200
(D) option
- (C) option
- (D) option
D =
𝐶−𝑆
𝑛
480000 − 25000
10
(B) option
- (A) option
𝑑𝑦
𝑦 𝑙𝑜𝑔𝑦
𝑑𝑥
𝑥
⇒ log
= log
⇒ log
= log
|𝐶𝑥|
(B) option
[(
60000
10000
)
1
4
− 1 ] × 100 = [ √
6
4
− 1 ] × 100
(C) option
(C) option
- (D) option
- (C) option
- (B) option
P(Win in one game) = P(Lose in one game) = ½
⇒ P (Beena to win in 3 out of 4 games) =
4
3
1
2
4
1
4
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
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/
7500 × 400
- a − 8 = 1 ⇒ 𝑎 = 9
3b = − 2 ⇒ 𝑏 = −
2
3
−c + 2 = − 28 ⇒ 𝑐 = 30
⇒ 2a + 3b − c = − 14 1
OR
Expanding C 1
, we get ∆ = 1
2
2
- Let the number of hardcopy and paperback copies be x and y respectively
⇒ Maximum profit Z = (72x + 40y) −(9600 + 56x + 28y) = 16x + 12y− 9600
𝑑𝑡
𝑡
2
, where t = 𝑒
𝑥
𝑥
− 1
1 +𝑒
𝑥
OR
2
, Integration by parts
= log (1 + 𝑥
2
[
𝑑
𝑑𝑥
log( 1 + 𝑥
2
𝑥 𝑑𝑥] 𝑑𝑥
𝑥
2
2
log (1 + 𝑥
2
[
2 𝑥
1 +𝑥
2
𝑥
2
2
] 𝑑𝑥
𝑥
2
2
log (1 + 𝑥
2
𝑥
3
1 +𝑥
2
𝑥
2
2
log (1 + 𝑥
2
) − ∫[𝑥 −
𝑥
1 + 𝑥
2
] 𝑑𝑥
𝑥
2
2
log (1 + 𝑥
2
𝑥
2
2
1
2
log (1 + 𝑥
2
1
2
[( 1 + 𝑥
2
)log (1 + 𝑥
2
2
] + 𝐶
- Under pure competition, 𝑝
𝑑
𝑠
8
𝑥+ 1
𝑥+ 3
2
2
When 𝑥
0
0
∴ Produce surplus = 2 − ∫
𝑥+ 3
2
1
0
= 2 − [
𝑥
2
4
3 𝑥
2
]
1
4
OR
2
3
𝑑𝑅
𝑑𝑥
2
Given MR = 4 + 3 𝑥
In profit monopolist market,
MR =
𝑑𝑅
𝑑𝑥
2
2
When 𝑥
0
0
∴ Consumer surplus = ∫
2
9
0
− 193 × 9
= [ 274 𝑥 −
𝑥
3
3
]
- Purchase = ₹ 40,00,
Down payment = 𝑥
Balance = 40,00,000 – 𝑥
9
1200
= 0. 0075 , n = 25 x 12 = 300
E = ₹ 30,
( 4000000 − 𝑥) × 0. 0075
− 300
× 0. 0075
Down payment = ₹ 4 , 24 , 800
n = 10 x 2 = 2 0, S = 10,21,760, 𝑖 =
5
200
= 0.0 25 , R =?
S =R [
( 1 +𝑖
)
𝑛
− 1
𝑖
]
⇒ 1021760 = R [
( 1 + 0. 025
)
20
− 1
- 025
]
⇒ 1021760 = R [
6386 − 1
025
]
⇒R = [
1021760 × 0. 025
- 6386
]
⇒ R = ₹ 40 , 000
Mr Mehra set aside an amount of ₹ 40,000 at the end of every six months
Section – D
Each question carries 5 - mark weightage
- Probability of defective bucket = 0.
n = 100
m = np = 100 x 0.03 = 3
Let X = number of defective buckets in a sample of 100
P (X = r) =
𝑚
𝑟
𝑒
−𝑚
𝑟!
(i) P (no defective bucket) = P(r = 0 ) =
3
0
𝑒
− 3
0!
(ii) P (at most one defective bucket) = P(r = 0, 1)
3
0
𝑒
− 3
0!
3
1
𝑒
− 3
1!
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
C (16,0) 352
Buying 8 tables and 12 chairs will maximise the profit
A = [
]
⇒ |A| = 9 ⇒ A
exists
And A
1
9
[
]
AX = B ⇒ 𝑋 = 𝐴
− 1
⇒X =
1
9
[
] [
] = [
]
1
2
3
Section – E
Each Case study carries 4-mark weightage
36. CASE STUDY - I
a)
Pipe C empties 1 tank in 20 h ⇒ 2 / 5 th tank in
2
5
× 20 = 8 hours
b)
Part of tank filled in 1 hour =
1
15
1
12
1
20
1
10
th
⇒ time taken to fill tank completely = 10 hours
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,
part of tank filled by pipe A =
𝑡
15
th
part of tank filled by pipe B =
𝑡− 3
15
th
and, part of tank emptied by pipe C =
𝑡− 4
15
th
Therefore
𝑡
15
𝑡− 3
12
𝑡− 4
20
Total time to fill the tank = 1 0 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
Part of tank filled by pipes A and B in 1 hour=
1
15
1
12
3
20
th
tank
⇒ time taken to fill the tank completely =
20
3
hours
Total time taken in the process = 10 + 1 +
20
3
= 17 hour 40 minutes
37. CASE STUDY - II
a)
Year Y X X
2
XY
a =
∑ 𝑌
𝑛
212
5
= 42. 4 and b =
∑ 𝑋𝑌
∑ 𝑋
2
25
10
𝐶
OR
Year Y 3 - year moving average
P(X)
XP(X)
∴ E(X) = ₹ 33 , 900
