CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 3 with Solutions

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Informatics Practices Term 2 Set 3 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)

Question 1.
Evaluate: ∫¹₀ e^x2 xdx
OR
Find the area of region bounded by the curve x = 2y + 3, Y-axis and the lines y = 1 and y = -1.
Answer:
Let, I = ∫₀¹ e^x2 xdx
Put, x² = t
or
xdx = \(\frac{1}{2}\)dt

Also when x = 0 or t = 0 and when x = 1 ⇒ t = 1
∴ I = \(\frac{1}{2}\) ∫₀¹ e^tdt
or
I = \(\frac{1}{2}\)[e^t]₀¹
= \(\frac{1}{2}\)(e – 1)

OR

From the figure, area of the shaded region,
A = ∫¹_-1 (2y + 3)dy
= [y² + 3y]¹_-1
= [1 + 3 – 1 + 3]
= 6 sq. units

Question 2.
A machine costing ₹ 50,000 has a useful life of 4 years. If the estimated scrap value is ₹ 10,000, then find the annual depreciation.
Answer:
Annual depreciation = \(\frac{\text { Original cost }-\text { Scrap value }}{\text { Useful life }}\)
= \(\frac{50,000-10,000}{4}\)
= \(\frac{40,000}{4}\)
= ₹ 10,000

Question 3.
What is the sum of money needed now, so as to get ₹ 6000 at the beginning of every month forever, when the money is worth 6% per annum compounded monthly?
OR
Assume that Shyam holds a perpetual bond that generates an annual payment of ₹ 500 each year. He believes that the borrower is creditworthy and that an 8% interest rate will be suitable for this bond. Find the present value of this perpetuity.
Answer:
Given R = ₹ 6000
r = \(\frac{6}{12}\)% = 0.5 per month

So, i = \(\frac{0.5}{100}\) = 0.005
So, P = R + \(\frac{R}{i}\)
= 6000 + \(\frac{6000}{0.005}\)
= 6000 + 1200000
= ₹ 1206000

OR

P.V. of perpetuity = \(\frac{\text { Annual Payment/Cash flow }}{\text { Interest rate/yield }}\)

= ₹ 6250

Question 4.
A wholesaler in apples claims that only 4% of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning good apples.
Answer:
Sample Size = 600
No. of defective apples = 36
Sample proportion p = \(\frac{36}{600}\) = 0.06
Population proportion P = probability of defective apples = 4% = 0.04
Q = 1 – P
= 1 – 0.04 = 0.96
The S.E. for sample proportion is given by S.E.

Question 5.
Calculate three-yearly moving averages of number of students studying in a higher secondary school in a particular village from the following data.

Year	Number of Students
2011	332
2012	317
2013	357
2014	392
2015	402
2016	405
2017	410
2018	427
2019	435
2020	438

Answer:
Computation of three-yearly moving averages.

Question 6.
Draw the feasible region for given inequation system y ≤ 6, x + y ≤ 3, x ≥ 0, y ≥ 0.
Answer:

Thus, feasible region is OAB.

Section – B (3 marks each)

Question 7.
Find the producer’s surplus defined by the supply curve S(x) = 4x + 8 for the supply of 5 units.
Answer:
Here, S(x) = 4x + 8 and Q_e = 5
P_e = S(5) = 4(5) + 8 = 28

Question 8.
Given below are the data relating to the sales of a product in a district.
Fit a straight line trend by the method of least squares and tabulate the trend values.

OR
From the following data calculate the 4- yearly moving averages and determine the trend values.

Answer:
Computation of trend values by the method of least squares.
In case of EVEN number of years, let us consider
X = \(\frac{(x \text {-Arithimetic mean of two middle years })}{0.5}\)

a = \(\frac{\Sigma Y}{n}=\frac{47.8}{8}\) = 5.975
b = \(\frac{\Sigma X Y}{\Sigma X^{2}}=\frac{8.6}{168}\) = 0.05119
Therefore, the required equation of the straight line trend is given by
Y = a + bX; Y = 5.975 + 0.05119X.

When X = 2015, X= 1995, Y_t = 5.975 + 0.05119\(\left(\frac{2015-2018.5}{0.5}\right)\) = 5.6167
When X = 2016, X = 1996, Y_t = 5.975 + 0.05119\(\left(\frac{2016-2018.5}{0.5}\right)\) = 5.7190
Similarly, other values can be obtained.
OR
Calculation of Trend values by it four yearly Moving Averages:

Question 9.
Ten individuals are chosen at random from the population and their heights are found to be in inches 63, 63, 64, 65, 66, 69, 69, 70, 70, 71. Discuss the freedom value of student’s height and 5 % level of significance is 2.62.
Answer:

x̄ = mean
= \(\frac{\sum x}{n}\)
= \(\frac{670}{10}\) = 67

Now, compute the standard deviation using formula as,

The number of degree of freedom = n – 1 = 9
Given that the tabulated value for 9 d.f. at level of significance is 2.62.

Since calculated value of f is less than the tabulated value i.e., 2.02 < 2.62, the error has arisen due to fluctuations and we may conclude that the data are consistent with the assumption of mean of height in the universe of 65 inches.

Question 10.
A company ABC Ltd. has raised funds in the form of 1,000 zero-coupon bonds worth ₹ 1,000 each. The company wants to set up a sinking fund for repayment of the bonds, which will be after 10 years. Determine the amount of the periodic contribution if the annualized rate of interest is 5%, and the contribution will be done half-yearly. Given that (1.025)²⁰ = 1.6386.
Answer:
Sinking Fund, A = ₹ 1,000 x 1000 = ₹ 1,000,000, r = 5% or 0.05, No. of years, n = 10 years and No. of payments per year, m = 2 (Half Yearly)

Periodic Contribution,

Therefore, the company will be required to contribute a sum of ₹ 39,148 half-yearly in order to build the sinking fund to retire the zero-coupon bonds after 10 years.

Section – C (4 marks each)

Question 11.
A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After soil test, it is found that atleast 7 kg of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹ 5.00 per kg and the type B costs ₹ 8.00 per kg. Using Linear Programming, find how many kilograms of each type of the fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.
Answer:
Let the fertilizer of type A be x kg and type 3 be ykg.

According to Question:

∴ Min.Z = 5x + 8y
Subject to constraints
2x + y ≥ 140
3x + 5y ≥ 350
x ≥ 0, y ≥ 0
2x + y = 140 ……..(i)

3x + 5y = 350 ……………(ii)

On solving eq (i) and (ii), we get x = 50 and y – 40

We draw the lines 2x + y = 140 and 3x + 5y = 350 and obtain the feasible region (unbounded and convex) as shown in figure. Thus, corner points are A(0, 140), B(50, 40),

The values of Z at these points are given in following table:

x	114	0
y	0	71.25

Since, there is no common point between the feasible region and 5x + 8y < 570
Hence, the cost will be minimum, if Fertilizer of type A used = 50 kg
Fertilizer of type B used = 40 kg
Minimum cost = ₹ 570

Commonly Made Error:
Many candidates made errors in calculating the conditions using constraints. Some candidates were unable to represent feasible region on the graph as such they may not have clarity of the concept of solving linear equations graphically. Some candidates solved it in terms of decimals and mostly went incorrect in simplifications.

Answering Tip:
The optimum function and all possible constraints in the form of inequations must be put down from what is stated in the problem.

Question 12.
Find the value of a ₹ 1,000 corporate bond with an annual interest rate of 5%, making semi-annual interest payments for 2 years, after which the bond matures and the principal must be repaid. Assume a Y.T.M. (yield to maturity) of 3%.
Answer:
Given, P = ₹ 1000 for corporate bond Annual Coupon Payment
= ₹ 1,000 × 5% = ₹ 50

Semi-annual Coupon Payment,
C = ₹ 50 – 2 = ₹ 25
N = 2 years × 2
= 4 periods for semi-annual coupon payments
r = Y.M.T.
= 3% = 0.03

Commonly Made Error
Sometimes there were calculation mistakes while simplification.

Answering Tip
When computing the bond value, you may have to round off the bond value up to nearest hundred to ensure that enough money in the given amount of time. Always recheck your solution.

Question 13.
The cost of a T.V depreciates by ₹ 800 during the second year and by ₹ 700 during the third year. Calculate:
(i) the rate of depreciation per annum.
(ii) the original cost of the T.V.
(ii) the value of the T.V at the end of third year.
OR
A machine costs ₹ 97,000 and its effective life is estimated to be 12 years. If scrap realises ₹ 2,000 only, what amount should be retained out of profits at the end of each year to accumulate at compound interest of 5% per annum in order to buy a new machine after 12 years ? (use 1.0512 = 1.769)
Answer:
(i) Let the original cost of the T.V. be ₹ P and the rate of depreciation be r% p.a. Then the value of T.V (in ₹) after 1 year, 2 years and 3 years are P(1 – i), P(1 – i)² and P(1 – i)³, respectively,
Where i = \(\frac{r}{100}\)

According to Question,
P(1 – i) – P(1 – i)² = 800
and P(1 – i)² – P(1 – i)3 = 700
⇒ P(1 – i)²[1 – (1 – i)] = 800
and P(1 – i)² [1 – (1 – i)] = 700
⇒ P(1 – i)i = 800 ……….(i)
and P(1 – i)²i = 700 …(ii)

On dividing eq. (ii) by eq (i), we get

Hence, the rate of depreciation = 12.5% p.a.

(ii) Putting i = \(\frac{1}{8}\) in eq. (i), we get

Hence, the original cost of TV is ₹ 7314.

(iii) The value of TV at the end of 3rd = P(1 – i)³

Thus, the value of TV at the end of the 3rd year is ₹ 4900.
OR
Cost of machine = ₹ 97,000
Value of scrap = ₹ 2,000
Required money = ₹ 95,000

Now, M = \(\frac{A}{r}\)95,000, n = 12, r = 5% = 0.05

Case Study

Question 14.
Mr. Lai was 75 year old businessman. He had two sons and three daughters. On the silver jubilee of his company, he gifted an equal amount of money among his children. One of his daughters buys jewellery from the gifted money whereas the other daughter invested his money in bank.
It is known that if the interest is compounded continuously, the principal changes at the rate equal to the product of the rate of bank interest per annum and principal.

(i) If P denotes the principal at time t and the rate of interest be r% per annum compounded continuously, then find the differential equation that models this scenario. Also, find the particular solution of the differential equation. (2)
Answer:
If P denotes the principal at time t and the rate of interest be r% per annum compounded continuously, then according to the law given in the problem, we get

Let P0 be the initial principal i.e. at t = 0, P = P₀.
Putting P = P₀ in eq. (i),
we get log P₀ = C
Putting C = log P₀ in eq. (i), we get
log P = \(\frac{r t}{100}\) + log P₀
⇒ log\(\left(\frac{P}{P_{0}}\right)\) = \(\frac{r t}{100}\)

(ii) At what rate of interest will X 100 double itself in 10 years ? (2)
(Use log_e2 = 0.6931)
Answer:
We have,
log\(\left(\frac{P}{P_{0}}\right)\) = \(\frac{r t}{100}\)
Put t = 10, P₀ = ₹ 100 and P = ₹ 200 = 2P₀ in the above equation, we get
log 2 = \(\frac{10r}{100}\)
⇒ r = 10 log_e2
= 10 × 0.6931
= 6.931% or 6.93% per annum (Approx)