CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 2 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)

Question 1.
Evaluate: ∫\(\frac{e^{x}(x-3)}{(x-1)^{3}}\)dx
OR
The marginal cost of producing x units of a product is given by the marginal cost function as MC = \(\sqrt{x+1}\). The cost of producing 3 units is ₹ 7800. Find the cost function.
Answer:

Commonly Made Error:
Some candidates use product rule in ∫e^x {f(x) + f'(x)} in both parts which is wrong. Students should apply the rule that ∫e^x {f(x) + f'(x)} = e^xf(x) + C only.

Answering Tip: ∫e^x{f(x) + f'(x)} is specific situation in integration by parts. This needs to be sufficiently practiced by students.

OR

We know that,
Cost Function, C(x) = ∫MC(x)dx
where MC(x) is Marginal Cost, Function
C(x) = ∫\(\sqrt{x+1}\) dx ……… (1)
Let x + 1 = t dx = dt
So, C(t) = ∫\(\sqrt{t}\) dt

which is the required cost function

Question 2.
What is the present value of a sequence of payment of ₹ 1000 made at the end of every 6 months and continuing forever, if money is worth 8% per annum compounded semi-annually ?
Answer:
The given annuity is a perpetuity.
present value of perpetuity = \(\frac{\text { Cash flow }}{\text { Interest rate }}\)
Here, cash flow = ₹ 1000
interest rate = \(\frac{8 / 2}{100}\)
= \(\frac{4}{100}\) = 0.04
So, present value = \(\frac{1000}{0.04}\)
= ₹ 25,000

Question 3.
A Machine purchased two years ago. If its value depreciates at the annual rate of 10 %.The present value of the machine is ₹ 97200,then find the value of the machine after 3 years?
OR
If the present value of a perpetuity of ₹ 600 payable at end of every 6 months is ₹ 10,000, then find the rate of interest.
Answer:
Given, P = ₹ 97,200, i = 10% p.a.
⇒ i = \(\frac{10}{100}\) = 0.1
So, value after 3 years
= 97,200 × (1 – 0.1)³
= 97,200 × 0.729
= ₹ 70,858.80
= ₹ 70,859 (approx.)

OR

Let rate of interest be r% per annum,
then i = \(\frac{r}{200}\) …… (i)
Given, R = ₹ 600 and P = ₹ 10,000
Using P = \(\frac{R}{i}\)
⇒ i = \(\frac{R}{P}=\frac{600}{10,000}\) …. (ii)
From equations (i) & (ii)
⇒ \(\frac{r}{200}=\frac{600}{10,000}\) [∴ i = \(\frac{r}{200}\)]
⇒ r = \(\frac{120,000}{10,000}\)
= 12%
∴ Rate of interest is 12%

Question 4.
The standard deviation of a sample of size 50 is 6.3. Determine the standard error whose population’s standard deviation is 6.
Answer:
Sample size n = 50
Sample S.D. s = 6.3
Population S.D. σ = 6
The standard error for sample S.D. is given by

Thus standard error for samples S.D. = 0.6

Question 5.
From the following time series obtain trend value by 3 yearly moving averages.

Answer:
Calculation of trend values by three yearly moving average method.

Question 6.
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300. Formulate a LPP for finding how many of each should be produced daily to maximize the profit ? It is being given that at least one of each must be produced.
Answer:
Let x be the number of necklaces manufactured and y be the number of bracelets manufactured. The total number of necklaces and bracelets it can handle is at most 24.
x + y ≤ 24 …(i)
The x items takes x hours to manufacture and y items take \(\frac{y}{2}\) hours to manufacture and the y 2 maximum time available is 16 hours. So,
x + \(\frac{y}{2}\) ≤16
2x + y ≤ 32 …(ii)
The profit on one necklace is given as ₹ 100 and the profit on one bracelet is given as ₹ 300. Let the profit be Z. To maximize the profit,
Z = 100x + 300y
Therefore, the required LPP is
Max. Z = 100x + 300y
Subject to constraints,
x + y ≤ 24
2x + y ≤ 32
x, y ≥ 1
x > 0, y > 0

Section – B [3 marks each]

Question 7.
If the demand curve is given by D(x) = 50 – 0.06x². Find the surplus or profit of the consumers if the level of sale amounts to 20 units.
Answer:
As the number of units is 20, its price rises up = D (20)
= 50 – 0.06 × 20² = 26.
The profit of consumer’s:

= 50(20 – 0) – 0.06\(\left[\frac{20^{3}}{3}-0\right]\) – 520
= 1000 – 160 – 520
= 1000 – 680
= 320
The consumer’s gain is ₹ 320, if the level of sales is twenty units.

Question 8.
Fit a straight line trend by the method of least square to the following data on sales (₹ in lakhs) for the period 2011-2018.

OR
Calculate trend values from the following data assuming 5- yearly moving averages.

Answer:
x_i = t_i – A, Here A = 2014 + 2015/2 = 2014.5

Now,
a = \(\frac{\Sigma y_{i}}{n}\) = \(\frac{1052}{8}\) = 131.5
b = \(\frac{\Sigma x_{i} y_{i}}{\Sigma x_{i}^{2}}\) = \(\frac{1232}{168}\) = 7.33
Sp, tend equation is
y = 131.5 + 7.33x

OR

Calculation of Trend values by moving average method:

Commonly Made Error: Few students find the calculation tedious in such type of questions.

Answering Tip: Do not leave the question in the middle. Do calculation with little patience.

Question 9.
Find the student’s -t for the following variable values in a sample of eight:
– 4, – 2, – 2, 0, 2, 2, 3, 3 taking the mean of the universe to be zero.
Answer:

x	X – x̄	(x – x̄)2
-4	-4.25	18.0625
-2	-2.25	5.0625
-2	-2.25	5.0625
0	-0.25	0.0625
2	1.75	3.0625
2	1.75	3.0625
3	2.75	7.5625
3	2.75	7.5625
Σ  x =  2		Σ (x – x̄)2 =  49.5000

X̄ = mean
= \(\frac{\sum x}{n}\)
= \(\frac{2}{8}\)
= 0.25
Now, compute the standard deviation using formula as,

H₀ = The mean of universe, μ = 0, we get

Question 10.
A company borrowed ₹ 60,000 for renovation. The company plans to set up a sinking fund that will pay back the loan at the end of 5 years. Assuming a rate of 10% compounded semi-annually, Find the amount of sinking fund of the ordinary annuity. Calculate the amount of the sinking fund. [Given that (1.05)¹⁰ = 1.06288.]
Answer:
Given, P = ₹ 60,000, r = 10% or 0.10, No. of years, n = 5 years and No. of payments per year, m = 2 (Semi annually)
Sinking Fund is as:

which is the required amount

Section – C (4 marks each)

Question 11.
A manufacturing company makes two types of teaching aids A and B of Mathematics for class X. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of ₹ 80 on each piece of type A and ₹ 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.
Answer:
Let the number of two types of teaching aids A and B be x and y, respectively.
Writing the given data in tabular as

The profit on type A is ₹ 80 and type B is ₹ 120.
Thus, the required L.EE is
Max. Z = 80x + 120y …….. (i)
Subject to constraints
9x + 12y ≤ 180
or 3x + 4y ≤ 60
x + 3y ≤ 30
x ≥ 0, y ≥ 0
Consider the equation,
3x + 4y = 60

x	0	20
y	15	0

and
x + 3y = 30

x	0	30
y	10	0

On solving eq, (ii) and (iii), we get
x = 12 and y = 6
The graphical representation of the system of in equations is as shown below

From graph, feasible region is
OABCO, whose corner points are (0, 0), A(20, 0), B(12,6) and C(0,10).
The values of Z at corner points are as follows:

Corner Points	Value of Z = 80x + 120y
O(0,0).	Z = 80 × 0 + 120 × 0 = 0
A(20, 0)	Z = 80 × 20 + 120 × 0 = 1600
B(12, 6)	Z = 80 × 12 + 120 × 6 = 1680
C(0,10)	Z = 80 × 0 + 120 × 10 = 1200

From the table, the maximum value of Z is ₹ 1680.
Hence, 12 pieces of type A and 6 pieces of type B should be manufactured per week to get maximum profit of ₹ 1680 per week.

Commonly Made Error: A number of candidates could not form the correct inequalities subject to given constraints. Some noted incorrect corner points to find optimum value. A few candidates did not draw rough sketch of the feasible region.

Answering Tip: Do adequate practice to frame inequalities using given constraints and finding corner points by solving inequalities.

Question 12.
A bond has issued with the face (Par) value of ₹ 1,000 at 10% coupon for three years The required rate
of return is 8%. What is the value of the bond if the coupon amount is payable on half-yearly basis ? [Given (1.04)^-6 = 0.79031]
Answer:
Given, P = ₹ 1,000 Annual Coupon Payment
= ₹ 1,000 × 10% = ₹ 100
Semi-annual Coupon Payment,
C = ₹100 ÷ 2 = ₹ 50
r = 8% ÷ 2 = 4%
= 0.04
N = 3 years × 2 = 6 periods for semi-annual coupon payments

Bond value is ₹ 1052.42

Application for completing products and balance the equation calculator of chemical reactions.

Question 13.
Rohan purchased a laptop of worth ₹ 80,000. He paid ₹ 20,000 as cash down payment and remaining balance in equal monthly installments in 2 years. If bank charges 9.% p.a. compounded monthly. Calculate the EMI [Given (1.0075)²⁴ = 1.1964]
OR
A firm anticipates an expenditure of ₹ 5,00,000 for plant modernization at end of 10 years from now. How much should the company deposit at the end of year into a sinking fund earning interest 5% per annum. [Given log 1.05 = 0.0212, antilog (0.2120) = 1.629]
Answer:
Given, Cost of laptop = ₹ 80,000
Down payment = ₹ 20,000
∴ Balance = ₹ 60,000
So, p = ₹ 60,000,
i = \(\frac{9}{12 \times 100}\)
= 0.0075
and n = 2 × 12 = 24

The required value of EMI is ₹ 2741.24

OR

Given, A = ₹ 5,00,000, r = 5% and n = 10

Now, let x = (1.05)¹⁰ …(iii)
Taking log both sides, we get
log x = 10 log (1.05)
= 10 × 0.0212
= 0.2120
⇒ x = antilog (0,2120)
x = 1.629 ……… (iv)
Thus, (iii) (1.05)¹⁰ = 1.629
Now, From (ii) P = \(\frac{500000 \times 0.05}{1.629-1}\)
= \(\frac{25000}{0.629}\)
P = 39745.63
Hence, the company should deposit ₹ 39745.63 every year into the sinking fund.

CASE STUDY

Question 14.
A veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 p.m. which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation, it was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton’s law of cooling which is governed by the differential equation: \(\frac{d T}{d t}\) ∝ (T – 70), where 70°F is the room temperature and T is the temperature of the object at time t.
Substituting the two different observations of T and t made, in the solution of the differential equation \(\frac{d T}{d t}\) = k(T – 70) where k is a constant of proportion, time of death is calculated.

(i) What is the degree of the differential equation? Also, find the solution of the differential equation. (2)
Answer:
The given differential equation is
\(\frac{d T}{d t}\) = k(T – 70)
Since there is only one constant available in the given equation, therefore, degree of this differential equation is one.
The given equation can be written as:
\(\frac{d T}{(T-70)}\) = kdt
Integrating both sides we get
log|T – 70| = kt + C

(ii) Find the value of integration constant if it is given that t = 0, when T = 72. If the temperature was measured 2 hours after 11.30 p.m., will the time of death change ? (2)
Answer:
We have,
log |T – 70| = kt + C
Putting the value of t = 0, T = 72, we get
C = log 2
So the time of death is independent of all other factors.