CBSE Class 12 Marking Scheme 2022 for Applied Maths Term 2

Marking Scheme
Applied Mathematics
Term - II
Code-241
Q.N. Hints/Solutions Marks
Section – A
2
1 Given, 𝑀𝑅 = 9 + 2𝑥 − 6𝑥
𝑇𝑅 = ∫(9 + 2𝑥 − 6𝑥 2 )𝑑𝑥
𝑇𝑅 = 9𝑥 + 𝑥 2 − 2𝑥 3 + 𝑘
When 𝑥 = 0, 𝑇𝑅 = 0, so 𝑘 = 0
𝑇𝑅 = 9𝑥 + 𝑥 2 − 2𝑥 3 1
⇒ 𝑝𝑥 = 9𝑥 + 𝑥 2 − 2𝑥 3
⇒ 𝑝 = 9 + 𝑥 − 2𝑥 2 which is the demand function 1

OR
300
𝑇𝐶 = ∫(50 + )𝑑𝑥
𝑥+1
𝑇𝐶 = 50𝑥 + 300 log|𝑥 + 1| + 𝑘 1
If 𝑥 = 0, 𝑇𝐶 = ₹2000
So 2000 = 300(log 1) + 𝑘 ⇒ 𝑘 = 2000
So 𝑇𝐶 = 50𝑥 + 300 log(𝑥 + 1) + 2000
1
2 𝑅 = ₹600
0.08 1
𝑖 = 4 = 0.02
𝑅
Present value of perpetuity = 𝑃 =
𝑖
600
⟹ 𝑃 = 0.02 = ₹30,000
1
3 𝑟 𝑚 1
𝑟𝑒𝑓𝑓 = (1 + ) −1
𝑚
0.08 4
= (1 + ) −1
4

= (1.02)4 − 1 = 0.0824 𝑜𝑟 8.24% 1
So effective rate is 8.24% compounded annually.
OR
Present value of ordinary annuity 1
1−(1+𝑟)−𝑛
= 𝑅( )
𝑟
1−(1.06)−5
= 1000 ( )
0.06
1−0.7473 1
= 1000 ( ) = ₹4211.67
0.06

4 𝑬 (𝑋̅) = 60𝑘𝑔 1

𝜎 9
Standard deviation of 𝑋̅ = 𝑆𝐸 (𝑋̅) = = 6 = 1.5 𝑘𝑔
√ 𝑛
1
5 Year Y 3 yearly 3 yearly moving 1M for
moving total average(Trend) 3-yearly
2016 25 --- --- moving
2017 30 87 29 totals
2018 32 102 34
2019 40 117 39 1M for
2020 45 135 45 3-yearly
2021 50 --- --- moving
average

6

Corner Point Z=3x+2y
P (2, 2) 10 1
Q (3, 0) 9

The smallest value of Z is 9. Since the feasible region is
unbounded, we draw the graph of 3𝑥 + 2𝑦 < 9. The resulting
open half plane has points common with feasible region,
therefore Z = 9 is not the minimum value of Z. Hence the optimal 1
solution does not exist.

Section –B
7 Substituting, 𝑝0 = ₹48 in 𝑝 = 𝑥 2 + 4𝑥 + 3
We get 𝑥0 = 5 1
𝑥
𝑃𝑆 = 𝑝0 𝑥0 − ∫0 0 𝑔(𝑥)𝑑𝑥
5 1
= 48 × 5 − ∫0 (𝑥 2 + 4𝑥 + 3)𝑑𝑥
𝑥3 5 1
= 240 − [ 3 + 2𝑥 2 + 3𝑥] = ₹133.33
0

8

Year Quarters Y 4-Quarterly 4 Quarterly 1
1 2 for
Moving Total Moving average
(Centered) 4 quarterly
moving
𝑄1 12 --
totals
𝑄2 14 --
2018 𝑄3 18 64 16.75
𝑄4 20 70 17.75
72
𝑄1 18 18.25
74
2019 𝑄2 16 18.75
76 1
𝑄3 20 85 20.125 1 2 for 4
𝑄4 22 93 22.25
Quarterly
𝑄1 27 103 24.50 moving
2020 𝑄2 24 117 27.5 average
𝑄3 30 -- (Centered)
𝑄4 36 --

The trend value are given by 4 quarterly centered moving
average.
𝑶𝑹
𝑌𝑒𝑎𝑟 𝑌 𝑋 𝑋2 𝑋𝑌
= 𝑌𝑒𝑎𝑟 − 2017
2014 26 -3 9 -78
2015 26 -2 4 -52
2016 44 -1 1 -44
2017 42 0 0 0
2018 108 1 1 108
2019 120 2 4 240
2020 166 3 9 498 1
∑ 𝑌 = 532 ∑ 𝑋 2 = 28 ∑ 𝑋𝑌
= 672

∑𝑌 532 ∑ 𝑋𝑌 672
𝑎 = 𝑛 = 7 = 76, 𝑏 = ∑ 𝑋 2 = 28 = 24
1
𝑌𝑐 = 𝑎 + 𝑏𝑋, 𝑌𝑐 = 76 + 24𝑋
1
Estimated sales = 𝑌𝑐 for 2023 = 76 + 24 × 6 = ₹220 lacs

9 Define Null hypothesis 𝐻0 and alternate hypothesis 𝐻1 as follows:
𝐻0 : 𝜇 = 0.50 𝑚𝑚
𝐻1 : 𝜇 ≠ 0.50 𝑚𝑚 1

Thus a two-tailed test is applied under hypothesis 𝐻0 , we
have 1
𝑋̅−𝜇 0.53−0.50
𝑡 = 𝑆 √𝑛 − 1 = 0.03 × 3 = 3
Since the calculated value of 𝑡 = 3 does not lie in the internal
−𝑡0.025 to 𝑡0.025 i.e., -2.262 to 2.262 for 10-1= 9 degree of freedom
So we Reject 𝐻0 at 0.05 level. Hence we conclude that machine 1
is not working properly.
10
We know
1
𝐹𝑉 𝑛
CAGR=[( 𝐼𝑉 ) − 1] × 100, where, IV= Initial value of investment
FV=Final value of investment 1

1 1
25000 𝑛 5 𝑛
⇒ 8.88 = [( ) − 1] × 100 ⇒ 0.0888 = ( ) − 1
15000 3
1
1
⇒ 1.089 = (1.667)𝑛
1
⇒ 𝑛 log(1.667) = log(1.089) ⇒ 𝑛(0.037) = 0.2219
1
⇒ 𝑛 = 5.99 ≈ 6 𝑦𝑒𝑎𝑟𝑠

Section –C
11 Let the company produces 𝑥 and 𝑦 gallons of alkaline solution
and base oil respectively, also let 𝐶 be the production cost.
Min 𝐶 = 200𝑥 + 300𝑦
subject to constraints:
𝑥 + 𝑦 ≥ 3500 … . (1)
𝑥 ≥ 1250 … . (2) 1
1
2𝑥 + 𝑦 ≤ 6000. . . (3) 2
𝑥, 𝑦 ≥ 0

1
1
2

Corner Points 𝑪 = 𝟐𝟎𝟎𝒙 + 𝟑𝟎𝟎𝒚
P(1250, 2250) ₹9,25,000

Q(1250, 3500) ₹13,00,000
R(2500, 1000) ₹8,00,000
Minimum cost is 8,00,000 when 2500 gallons of alkaline
solutions & 1000 gallons of base oil are manufactured. 1

12 The amount of sinking fund S at any time is given by
(1+𝑖)𝑛 −1 1
𝑆 = 𝑅[ ]
𝑖
Where 𝑅 = 𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑎𝑦𝑚𝑒𝑛𝑡, 𝑖 = 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟 𝑝𝑒𝑟𝑖𝑜𝑑,
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑦𝑚𝑒𝑛𝑡𝑠
𝑆= Cost of machine – Salvage value
= 50,000-5000 = ₹45,000
8%
𝑖 = 4 = 0.02
(1+.02)40 −1 1
⟹ 45000 = 𝑅 [ ] 1
0.02 2
2.208−1
⟹ 45000 = 𝑅 [ 0.02 ]
900
⟹ 𝑅 = 1.208 ⟹ 𝑅 = ₹745.03 1
1
2
13. Amortized Amount i.e., P= Cost of house-Cash down payment
P= 15,00,000 – 4,00,000 = ₹11,00,000
0.09
𝑖 = 12 = 0.0075
1
𝑛 = 10 × 12 = 120
𝑃
EMI = 𝑅 = 𝑎 ,
𝑛¬𝑖
𝑃×𝑖 1
𝑅 = 1−(1+𝑖)−𝑛
11,00,000 ×0.0075 8250
= 1−(1.0075)−120 = 1−0.4079 1
8250
= 0.5921 = ₹13933.5
Total interest paid = 𝑛𝑅 − 𝑅 = 13933.5 × 120 − 11,00,000 1
=₹5,72,020
OR
Face value of bond, F = ₹2000
Redemption value C = 1.05 × 2000 = ₹2100
Nominal rate =8% 1
𝑅 = 𝐶 × 𝑖𝑑 = 2000 × 0.08 = ₹160 1
2
Number of periods before redemption i.e., n = 10
Annual yield rate, 𝑖 =10% or 0.1
1−(1+𝑖)−𝑛 1
Purchase price 𝑉 = 𝑅 [ ] + 𝐶(1 + 𝑖)−𝑛 1
𝑖
1−(1+0.1)−10 2
= 160 [ ] + 2100(1 + 0.1)−10
0.1
= 160 × 6.14 + 2100 × 0.3855
= 982.4 + 809.6 = 1792

Thus present value of the bond is ₹1792. 1

14 Case Study
a) 𝑑𝑥 𝑑𝑥
∵ ∝ 𝑥, ∴ = −𝑘𝑥
𝑑𝑡 𝑑𝑡
𝑑𝑥 1
⇒ ∫ 𝑥 = ∫ −𝑘 𝑑𝑡 ⇒ log 𝑥 = −𝑘𝑡 + 𝑐
⇒ 𝑥 = 𝑒 −𝑘𝑡+𝐶 ⇒ 𝑥 = 𝜆𝑒 −𝑘𝑡
Let 𝑥 = 𝑥0 𝑎𝑡 𝑡 = 0
∵ 𝑥0 = 𝜆 ⇒ 𝑥 = 𝑥0 𝑒 −𝑘𝑡 where 𝑥0 = original quantity 1

b) 𝑥 = 𝑥0 𝑒 −𝑘𝑡 … . . (1)
𝑥
Now, 20 = 𝑥0 𝑒 −5𝑘 (∵ half life = 5 hours)
1
1
⇒ 𝑒 −5𝑘 = 2 ⇒ 𝑒 𝑘 = 2 5
1
The quantity of propofol needed in a 50 years adult after
2 hours = 50 × 3 = 150 mg ⇒ 150 = 𝑥0 𝑒 −2𝑘 [ using… (1)]
⇒ 𝑥0 = 150 𝑒 2𝑘 ⇒ 𝑥0 = 150 (𝑒 𝑘 )2
1
⇒ 𝑥0 = 150(25 )2 = 150 × 1.3195 = 197.93 mg 1