CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 4 with Solutions

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Informatics Practices Term 2 Set 2 with Solutions

Time: 2 Hours
Maximum Marks: 40

General Instructions:

• The question paper is divided into 3 sections -A, B and C
• Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
• Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
• Section C comprises of 4 questions of 4 marks each. It contains one case study based question. Internal choice has been provided in one question.

Section – A (2 Marks each)

Question 1.
Evaluate:

OR
The marginal revenue function for a item is given by MR = 10 + 4x – 3x² – 4x³. Find the demand function.
Answer:
Put, 1 + x² = t
or 2x dx = dt
or x dx = \(\frac{d t}{2}\)

OR

Given, MR = 10 + 4x – 3x² – 4x³
TR = ∫(10 + 4x – 3x² – 4x³) dx
TR = 10x + 2x² – x³ – x⁴ + C
When x = 0, TR = 0, so C = 0
TR = 10x + 2x² – x³ – x⁴
⇒ px = 10x + 2x² – x³ – x⁴
⇒ p = 10 + 2x – x² – x³,
which is the demand function.

Question 2.
Find the present value of perpetuity of ₹ 900 at end of each quarter if money is worth 6 % compounded quarterly.
Answer:
R = ₹ 900
i = \(\frac{0.06}{4}\)
= 0.015
Present value of perpetuity, P = \(\frac{R}{i}\)
⇒ P = \(\frac{900}{0.015}\)
⇒ ₹ 60,000

Question 3.
What effective rate is equivalent to a nominal rate of 7% per annum compounded semi annually? 0 [Use (1.035)² = 1.071225]
OR
Find the present value of an annuity of ₹ 5000 payable at the end of each year for 7 years if money is worth 7% compounded annually. [Given (1.07)^-7 = 0.6227]
Answer:
Since, r_eff = \(\left(1+\frac{r}{m}\right)^{m}\) – 1
∴ r_eff = \(\left(1+\frac{0.07}{2}\right)^{2}\) – 1
= (1.035)² – 1
= 1.071225 – 1
= 0.071225
or 7.12 %
So, effective rate is 7.12 % compounded annually

OR

Present value of ordinary annuity

Question 4.
Find the sample size for the given standard deviation 10 and the standard error with respect of sample mean is 3.
Answer:

= 11.11
≅ 11,
The required sample size is 11.

Question 5.
Find the trend values using 3 yearly moving average for the education loans sanctioned to students for higher studies in abroad.

Answer:

Question 6.
In which quadrant, the bounded region for inequations x + y ≤ 1 and x – y ≤ 1 is situated ?
Answer:

As shown in graph drawn for x + y = 1 and x – y = 1, the origin included in the area. Hence, the bounded region situated in all four quadrants.

Section – B (3 marks each)

Question 7.
If the supply function for a commodity is p = 4 + x, and 12 units of goods are sold, then find the producer’s surplus.
Answer:
Given, the supply function is
p = 4 + x ……. (i)
and the market demand
x₀ = 12.
At equilibrium,
P₀ = 4 + x₀
Substituting this value of x₀ in (i), we get
P₀= 4 + 12
⇒ P₀ = 16
So, producer’s surplus

= 192 – (48 + 72) + 0
= 192 – 120
= 72 units

Question 8.
Calculate four-yearly moving averages of number of students studying in a higher secondary school in a particular city from the following data.

Year	Number of Students
2011	124
2012	120
2013	135
2014	140
2015	145
2016	158
2017	162
2018	170
2019	175

OR
Below are given the figures of production (in m. tonnes) of a rice factory:

(i) Fit the straight line trend to these figures and calculate trend values.
(ii) Estimate the likely sales of the company during 2019.
Answer:
Computation of four-yearly moving averages:

OR

(i) Filling straight line trend

Now, a = \(\frac{\Sigma y}{n}=\frac{630}{7}\) = 90
b = \(\frac{\Sigma x y}{\Sigma x^{2}}=\frac{56}{28}\) = 2
Thus, trend equation is: y_t = 90 + 2x
Trend values are
y₂₀₁₂ = 90 + 2 (- 3) = 84
y₂₀₁₃ = 90 + 2 (- 2) = 86
y₂₀₁₄ = 90 + 2 (- 1) = 88
y₂₀₁₅ = 90 + 2 (0) = 90
y₂₀₁₆ = 90 + 2 (1) = 92
y₂₀₁₇ = 90 + 2 (2) = 94
y₂₀₁₈ = 90 + 2 (2) = 96

(ii) Now,
For 2019, x would be 4.
Putting x = 4 in trend equation, we get
y₂₀₁₉ = 90 + 2(4) = 98 %
∴ Production in 2019 is 98m. tonnes

Question 9.
Country A has an average farm size of 191 acres, while Country B has an average farm size of 199 acres. Assume the data were attained from two samples with standard deviations of 38 and 12 acres and sample sizes of 8 and 10, respectively. Is it possible to infer that the average size of the farms in the two countries is different at α = 0.05 ? Assume that the populations are normally distributed.
Answer:
Hypothesis H₀: µ₁ = µ₂ and H₀: µ₁ ≠ µ₂ (claim) The test is two-tailed and a = 0.05, also variances are unequal, the degrees of freedom are the smaller of iq – 1 or n2 – 1. In this case, the degrees of freedom are 8 – 1 = 7.

Hence, from t-table F, the critical values are -2.365 and – 2.365.
t = \(\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}}{n_{2}}+\frac{s_{2}}{n_{2}}}}\)
= \(\frac{191-199}{\sqrt{\frac{38^{2}}{8}+\frac{12^{2}}{10}}}\)
= – 0.57
Do not reject the null hypothesis, since – 0.57 > – 2.365.
There is not enough evidence to support the claim that the average size of the farms is different.

Question 10.
Riya invested ₹ 20,000 in a mutual fund in year 2016. The value of mutual fund increased to ₹ 32,000 in year 2021. Calculate the compound annual growth rate of her investment. [Given, log(1.6) = 0.2041, antilog (0.04082) = 1.098]
Answer:
Given beginning value of investment = ₹ 20,000
Final value of the investment = ₹ 32,000
No. of years = 5

Taking log both sides, we get
log x = \(\frac{1}{5}\)log (1.6)
⇒ log x = \(\frac{1}{5}\) × 0.2041
⇒ log x = 0.04082
⇒ x = antilog (0.04082)
= 1.098
C.A.G.R. = 1.098 – 1 = 0.098 = 9.8%

Section – C (4 marks each)

Question 11.
A company manufacturers two types of products A and B. Each unit of A requires 3 gram of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 gram of chromium. The firm can produce 9 gram of nickel and 8 grams of chromium. The profit is ₹ 40 on each unit of product of type A and ₹ 50 on each unit of type B. How many units of each type should the company manufactures so as to earn maximum profit? Use linear programming to find the solution.
Answer:
Let x = Number of units of type A
y = Number of units of type B
Maximize Z = 40x + 50y
Subject to the constraints,
3 x + y < 9
x + 2y ≤ 8
and x ≥, 0, y ≥ 0
Consider the equation,
3x + y = 9

Shaded region in the diagram represents the feasible solution.
Now, we can determine the maximum value of Z by evaluating the value of Z at the four joints (vertices) as shown below.

Vertices	Z = 40x + 50y
(0,0)	Z = 40 × 0 + 50 × 0 = 0
(3,0)	Z = 40 × 3 + 50 × 0 = ₹ 120
(0,4)	Z = 40 × 0 + 50 × 4 = ₹ 200
(2,3)	Z = 40 × 2 + 50 × 3 = ₹ 230 Max

From graph,
Maximum profit, Z = ₹ 230
∴ Number of units of type A is 2 and number of units of type B is 3.

Commonly Made Error: In some cases, all the constraints were not used and hence coordinates of all feasible points were not obtained.

Answering Tip: Students should solve the constraints equations in pairs to obtain all feasible points leading to maximum or minimum value of desired function.

Question 12.
A machine costing ₹ 200000 has effective life 7 years and its scrap value is ₹ 30000. What amount should the company put into a sinking fund earning 5% per annum, so that it can replace the machine after its useful life ? Assume that a new machine will cost ₹ 300000 after 7 years. [Given log (1.05) = 0.0212 and antilog (0.1484) = 1.407]
Answer:
Cost of new machine = ₹ 3,00,000
Scrap value of old machine = ₹ 30,000
Hence,the money required for new machine after 7 years = ₹ 300000 – ₹ 30,000 = ₹ 2,70,000
So, we have A = ₹ 2,70,000

Now,let (1.05)⁷ = x
Taking log both sides, we get
7 log(1.05) = log x
⇒ log x = 7 × 0.0212 = 0.1484
⇒ x = antilog (0.1484) = 1.407
So, (1.05)⁷ = 1.407
Thus, P = \(\frac{270000 \times 0.05}{1.407-1}\)
⇒ P = \(\frac{13500}{0.407}\)
⇒ P = ₹ 33169.53
Hence, the company should deposit ₹ 33170 (Approx) at the end of each year for 7 years.

Question 13.
Calculate the monthly payment on ₹ 2,00,000 at a 6% annual interest rate that is amortized over 10 years. [Given (1.005)¹²⁰ = 1.819]
OR
Seema purchased bonds for 3 years of ₹ 1000 with period coupon rate 10%. The minimum per-annum yield that she would accept is 14%. Find the fair-value of bond. Find the bond value.
Answer:

OR

= 100 × 0.87719 + 100 × 0.769467 + 100 × 0.674972 + 1000 × 0.674972
= 87.719 + 76.947 + 67.479 + 674.97
= 907.125

Case Study

Question 14.
Riya a biologist runs an experiment in the laboratory involving a culture of bacteria. She is doing this experiment first time. She notices that the mass of the bacteria in culture increases exponentially. The bacteria count in the culture is 1,00,000. The rate of growth of bacteria is proportional to the number of bacteria present in the culture. Also, the number of bacteria is increased by 10% in 2 hours.

(i) If y be the number of bacteria at time t, then find the differential equation that models this scenario. Also, find the general solution of the differential equation. (2)
Answer:
If y be the number of bacteria at time t, then
\(\frac{d y}{d t}\) ∝ y
⇒ \(\frac{d y}{d t}\) = ky
\(\frac{d y}{d t}\) = ky
⇒ \(\frac{d y}{y}\) = k dt
Integrating both sides, we get
log y = kt + c

(ii) Find the particular solution of differential equation and in how many hours will the Count reach and in how many hours will the count reach 2,00,000. (2)
Answer:
Put t = 0 and y = 1,00,000 in equation
log y = kt + c, we get
log 1,00,000 = k × 0 + c
⇒ c = log 1,00,000
∴ Particular solution is log y = kt + log 1,00,000
Now, bacteria count after 2 hours i.e.,
When, t = 2 hours
y = 1,00,000 + 10% × 1,00,000
= 1,10,000
Now Putting t = 2, y = 1,10,000
log 1,10,000 = 2k + log 1,00,000
k = \(\frac{1}{2}\) log \(\frac{1,10,000}{1,00,000}\)
= \(\frac{1}{2}\) log \(\frac{11}{10}\)
∴ log y = \(\frac{1}{2}\) log \(\left(\frac{11}{10}\right)\)t + log 1, 00, 000
log 2,00,000 = \(\frac{1}{2}\) log \(\left(\frac{11}{10}\right)\)t + log 1, 00, 000
\(\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\) = t
= \(\frac{0.69315 \times 2}{0.09531}\)
= 14.55 years (Approx)