CBSE Sample Papers for Class 12 Applied Mathematics Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Mathematics Set 8 with Solutions

Maximum Marks: 80 Marks
Time Allowed : 3 Hours

General Instructions:

This question paper contains five sections A, B, C, D and E. Each section is compulsory.
Section – A carries 20 marks weightage, Section – B carries 10 marks weightage, Section – C carries 18 marks weightage, Section – D carries 20 marks weightage and Section – E carries 3 case-based with total weightage of 12 marks.
Section A: It comprises of 20 MCQs of 1 mark each.
Section B: It comprises of 5 VSA type questions of 2 marks each.
Section C: It comprises of 6 SA type of questions of 3 marks each.
Section D: It comprises of 4 LA type of questions of 5 marks each.
Section E: It has 3 case studies. Each case study comprises of 3 case-based questions, where 2 VSA type questions are of 1 mark each and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks question in each case-study.
Internal choice is provided in 2 questions in Section – B, 2 questions in Section – C, 2 questions in Section – D. You have to attempt only one of the alternatives in all such questions.

Section – A (20 marks)

(All questions are compulsory. No internal choice is provided in this section.

Question 1.
The shaded region shown in the figure, is given by the inequalities: [1]

(a) 14x + 5y ≥ 70, y < 14, x – y ≤ 5 (b) 14x + 5y ≥ 70, y ≤ 14, x – y ≥ 5 (c) 14x + 5y ≤ 70, y ≤ 14, x – y ≥ 5 (d) 14x + 5y ≥ 70, y > 14, x – y ≥ 5
Solution:
(d) 14x + 5y ≥ 70, y > 14, x – y ≥ 5

Explanation:
The equations of the three sides of the triangle are:
x – y = 5, y = 14
and 14x + 5y = 70
So. the shaded area is formed with the inequalities:
x – y ≤ 5, y ≤ 14 and 14x + 5y ≥ 70
So. the correct option is (a).

Question 2.
The rate of change of the area of a circle with respect to its radius r when r = 6 cm, is: [1]
(a) 10π
(b) 12π
(c) 8π
(d) 11π
Solution:
(b) 12n

Explanation:
Let A denote the area of the circle having radius r. Then,
A = πr²
⇒ \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2πr
We need to determine \(\frac{\mathrm{dA}}{\mathrm{dr}}\) at r = 6
So, (\(\frac{\mathrm{dA}}{\mathrm{dr}}\))_r=6 = (2πr)_r=6 = 12π
Thus, (b) is the correct option.

Question 3.
If A is a 2 × 3 matrix and AB is a 2 × 5 matrix, then, the order of matrix B is: [1]
(a) 3 × 3
(b) 3 × 5
(c) 5 × 3
(d) 5 × 5
Solution:
(b) 3 × 5

Explanation:
Let the order of B be m × n.
Since the order of BA is 2 × 5, we have m = 3 and n = 5
Thus, the order of B is 3 × 5

Question 4.
An outlet pipe can empty a cistern in 30 minutes. What part of the cistern will be emptied in 12 minutes? [1]
(a) 0. 35
(b) 0.4
(c) 0.28
(d) 0.3
Solution:
(b) 0.4

Explanation:
We know that, when a pipe empties a cistern in ‘n’ min, then the part emptied by the pipe in 1 min = \(\frac{\mathrm{1}}{\mathrm{n}}\)
Here, n = 30
Therefore, Required part of the tank emptied in
12 min = \(\frac{12}{30}\) part = 0.4

Question 5.
The maximum value of [x (x – 1) + 1]^1/3, 0 ≤ x ≤ 1 is: [1]
(a) 0
(b) 1
(c) (\(\frac{1}{2}\))^1/3
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

Explanation:
Let
f(x) = [x(x – 1) + 1]^1/3
= [x² – x + 1]^1/3
Then.f'(x) = [x² – x + 1]^-2/3 [2x – 1]
Now, f'(x) = 0 implies \(\frac{1}{3}\) [x^-2 – x + 1]^-2/3[2x – 1] = 0
⇒ 2x – 1 =0
Thus. r = \(\frac{1}{2}\) ∈ [0, 1] is o criticaL point.
Now we shall find f(0), f(1) and f(\(\frac{1}{2}\))
f(0) = [0² – 0 + 1]^1/3 = 1:
f(\(\frac{1}{2}\)) = [(\(\frac{1}{2}\))² – \(\frac{1}{2}\) + 1]^1/3
= (\(\frac{3}{4}\))^1/3
So, the maximum value of [x(x – 1) + 1]^1/3 is 1.
Thus, (b) is the correct option.

Question 6.
A The probability distribution of a random variable X denoting the number of tails in simultaneous toss of 3 coins is given below: [1]

X_i	0	1	2	3
P(X = X_i)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	α	β

Then, (α, β) is:
(a) (\(\frac{1}{8}\), \(\frac{1}{8}\))
(b) (\(\frac{1}{4}\), \(\frac{1}{4}\))
(c) (\(\frac{3}{8}\), \(\frac{1}{8}\))
(d) (\(\frac{1}{8}\), \(\frac{1}{8}\))
Solution:
(c) (\(\frac{3}{8}\), \(\frac{1}{8}\))

Explanation:
We know that P(a tail) = \(\frac{1}{2}\)
P(a head) = \(\frac{1}{2}\)
So, P(2 tails) = P(HTT) + P(THT) + (TTH)
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{3}{8}\)
⇒ α = \(\frac{3}{8}\)
P(3 tails) = P(TTT)
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\)
⇒ β = \(\frac{1}{8}\)

Question 7.
If x is a real number and |x| < 7, then [1] (a) – x > – 7
(b) – 7 < x < 7
(c) x ≥ 7
(d) x ≤ – 7
Solution:
7. (b) -7 < x < 7

Explanation:
|x| < 7 gives, by definition,
-7 < x < 7.

Question 8.
If A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\) then A² is: [1]
(a) A
(b) 2A
(c) 3A
(d) unit matrix
Solution:
(c) 3A

Explanation:
Every element of A² is (1 + 1 + 1), i.e., 3.

Question 9.
In a binomial distribution with parameters n and p, if n = 5 and P(X = 2) = 9 P(X = 3), then its mean is: [1]
(a) \(\frac{1}{10}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{1}{2}\)
(d) 1
Solution:
(c) \(\frac{1}{2}\)

Explanation:
Since it is given that P(X = 2) = 9 P(X = 3), we have
⁵C₂ (p)²(q)³ = 9 × ⁵C₃ (p)³(q)²
10q = 90p
q = 9p
Hence, 1 – p = 9p
p = \(\frac{1}{10}\)
Thus, mean = np = 5 × \(\frac{1}{10}\) = \(\frac{1}{2}\).

Question 10.
Let F be the face value of a bond and i_d be the nominal rate of interest per period stating on the bond. Then, periodic dividend payment R periodic – interest) is given by: [1]
(a) R = F × i_d
(b) R = F + i_d
(c) R = \(\frac{\mathrm{F}}{i_d}\)
(d) R = F^id
Solution:
(a) R = F × i_d

Question 11.
If \(\left[\begin{array}{cc}
4 & x+2 \\
2 x-3 & x+1
\end{array}\right]\) is a symmetric matrix, then the value of x is : [1]
(a) 3
(b) 5
(c) 6
(d) 7
Solution:
(b) 5

Explanation:
Since the given matrix is a symmetric matrix,
a₁₂ = a₂₁
x + 2 = 2x – 3
x = 5

Question 12.
If Z is a standard random variable, then P( Z > 1.7) is equal to: [1]
(a) 0.4554
(b) 0.9554
(c) 0. 0446
(d) 0. 4460
Solution:
(c) 0. 0446

Explanation:
Here,
P(Z > 1.7) = 0.5 – P(0 < Z < 1.7)
= 0.5 – 0.4554
= 0.0446

Question 13.
If \(\int_0^{40} \frac{d x}{2 x+1}\) = log a, then a equals: [1]
(a) 6
(b) 0
(c) 3
(d) 4
Solution:
(a) 6

Explanation:
\(\int_0^{40} \frac{d x}{2 x+1}\) = \(\left[\frac{\log (2 x+1)}{2}\right]_0^{40}\)
= \(\frac{1}{2}\) log 81
= log 81^1/2
= log 9
log 9 = log a
a = 9

Question 14.
In the equation of trend line y = a + bx, b denotes. [1]
(a) mean of y
(b) slope of the trend line
(c) total of x
(d) total of xy
Solution:
(b) slope of the trend line

Question 15.
If \(\left[\begin{array}{cc}
x+y & -5 \\
0 & x-y
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & -5 \\
0 & 1
\end{array}\right]\), then (x, y) is: [1]
(a) (2, 3)
(b) (3, 2)
(c) (1, 2)
(d) (2, 1)
Solution:
(d) (2, 1)

Explanation:
Setting corresponding elements equal we get
x + y = 3
and x – y = 1
x = 2
and y = 1
So, (x, y) = (2, 1)

Question 16.
Which of the following index numbers satisfy the time reversal test? [1]
(a) Laspeyre’s index
(b) Passche’s index
(c) Fisher’s index
(d) Dorbish and Bowley’s index
Solution:
(c) Fisher’s index

Question 17.
At a telephone enquiry system, the number of phone calls regarding relevant enquiry follows Poisson distribution with an average of 5 phone calls during 10 minutes time intervals. The probability that there is at most one phone call during a 10 minute time period is: [1]
(a) \(\frac{5}{6}\) e^-5
(b) 5e^-5
(c) \(\frac{1}{6}\) e^-3
(d) 6e^-5
Solution:
(d) 6e^-5

Explanation:
Here,
mean = λ= 5
Thus, P(X ≤ 1) = P(X = 0) + P(X = 1)
= \(\frac{5^0}{0 !}\) . e^-5 + \(\frac{5^1}{1 !}\) . e^-5
= 6e^-5

Question 18.
The area of the region bounded by the curve y = |x – 1| and y = 1 is: [1]
(a) \(\frac{3}{2}\)
(b) \(\frac{1}{2}\)
(c) 1
(d) 2
Solution:
(c) 1

Explanation:
Required Area = \(\int_0^2\) 1 dx – \(\int_1^2\) (x – 1) dx
\(\int_0^1\) (1 – x) dx

= (2 – \(\frac{1}{2}\) – \(\frac{1}{2}\)) = 1 sq unit.

Assertion-Reason Questions

Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both (A) and (R) are true but (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 19.
Assertion (A): Order of a differential equation represents number of arbitrary constants in the general solution. [1]
Reason (R): Degree of differential equation represents number of family of curves.
Solution:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
Order and degree of a differential equation represent number of arbitrary constants and family of curves.

Question 20.
Assertion (A): For a binomial distribution B (n, P), Mean > Variance [1]
Reason (R): Probability is less than or equal to 1.
Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Since, Mean = np and variance = npq
= npq < np (∴ q < 1)
And probability is always less than or equal to 1.

Section – B (10 Marks)

(All questions are compulsory. In case of internal choice, attempt any one question only)

Question 21.
Graph the following system of constraints and shade the feasible region: [2]
x + 2y < 8 ; x – y < 1 ; 2x + y > 2
x > 0, y > 0
Solution:

Question 22.
A random variable has the following probability distribution: [2]

X	4	5	6	8
P(X)	0.1	0.3	0.4	0.2

Find the expectation and the standard deviation of the random variable.
OR
The probability that a bomb dropped from a plane strikes the target is \(\frac{1}{5}\). Find the probability that out of six bombes doped atleast 2 bombs strike the target.
Solution:
E(X) = Σx_ip_i
= (4 × 0.1) + (5 × 0.3) + (6 × 0.4) + (8 × 0.2)
= 5.9
E(X²) = Σx_i²p_i
= (16 × 0.1) + (25 × 0.3) + (36 × 0.4) + (64 × 0.2)
= 36.3
S.D.(X) = \(\sqrt{E\left(X^2\right)-[E(X)]^2}\)
= \(\sqrt{36.3-(5.9)^2}\)
= \(\sqrt{1.49}\) = 1.22

Here, p = \(\frac{1}{5}\)
q = \(\frac{4}{5}\)
and n = 6
We need to determine P(X ≥ 2)
Now, P(X ≥ 2) = 1 – {P(X = 0) + P(X = 1)}
= 1 – {⁶C₀ p⁰q⁶ + ⁶C₁ p¹q⁵}
= 1 – \(\left\{\left(\frac{4}{5}\right)^6+6\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^5\right\}\)
= 1 – \(\left\{\frac{4}{5}+\frac{6}{5}\right\}\left(\frac{4}{5}\right)^5\)
= 1 – \(2\left(\frac{4}{5}\right)^5\)

Question 23.
Rajni obtained 70 and 75 marks in first two unit tests. Find the number of minimum marks that she should obtain in the third unit test to have an average of at least 60 marks. [2]
Solution:
Let x be the number to be obtained by Rajni in the third unit test Then,
\(\frac{70+75+x}{3}\) ≥ 60
⇒ 145 + x ≥ 180
⇒ x ≥ 35
Thus the possible values of breadth are 35 cm or more.

Question 24.
How much money is needed to endure a series of lectures costing ₹ 2500 at the beginning of each year indefinitely, if money is worth 5% compounded annually? [2]
OR
A man is retiring at the end of next year. He would like to ensure that his heirs receive payments of ₹ 75,000 a year forever, starting when he retires. How much does he need to invest in the beginning of this year to produce the desired cash flow, if money is worth 6% compounded annually?
Solution:
We know that the present value of perpetuity of ₹ R payable in the beginning of every period forever at a rate of i per period is given by R + \(\frac{\mathrm{R}}{\mathrm{i}}\), i.e., P_∞ = R + \(\frac{\mathrm{R}}{\mathrm{i}}\).
Here, R = 2500.
i per period = \(\frac{5}{100}\) = 0.05
So, P_∞ = R + \(\frac{\mathrm{R}}{\mathrm{i}}\)
= 2500 + \(\frac{2500}{0.05}\)
= ₹ 52500
So, an amount of ₹ 52,500 is needed.

We know that the present value of perpetuity of ₹ R payabLe at the end of every period forever at a rate of i per period is given by \(\frac{\mathrm{R}}{\mathrm{i}}\), L. P_∞ = \(\frac{\mathrm{R}}{\mathrm{i}}\)
Here, R = 75.000,
i per period = \(\frac{6}{100}\) = 0.06
So, P_∞ = \(\frac{\mathrm{R}}{\mathrm{i}}\)
= \(\frac{75000}{0.06}\) = ₹ 1250,000
So, he need is to invest ₹ 12,50,000.

Question 25.
Identify the below statement a biased or unbiased statement. Justify your answer “For a survey about daily mobile uses by students, random selection of 20 students from a school” [2]
Solution:
It is unbiased statement, since it is random sampling.

Section – C (18 Marks)

(All questions are compulsory. In case of internal choice, attempt any one question only)

Question 26.
A company set aside a sum of ₹ 5000 at the end of each year for 10 years to pay off a debenture issue of ₹ 60,000. If the money earns interest at 5% p.a., find the surplus after the full redemption of the debenture issue. [3]
Solution:
If ₹ A be the sum of money set aside periodically. S is the amount accumulated after n periods and r is the rate of interest per period, then S is given by
S = \(\frac{\mathrm{A}}{\mathrm{r}}\) [(1 + r)n -1]
Here, A = 5000;
r = 0.05
and n = 10
So. S = \(\frac{\mathrm{A}}{\mathrm{r}}\) [(1 + r) n -1]
S = \(\frac{5000}{0.05}\) {(1.05)10 – 1}
= 100000 (0.629) = 62900
Thus, the amount in the fund is ₹ (62900 – 60,000), i.e., ₹ 2900.

Question 27.
In a sample of 1000 cases, the mean of a certain test is 14 and S.D. is 2.5. Assuming the distribution to be normal, find: [3]
(A) how many students score between 12 and 15?
(B) how many score above 18?
(C) how many score below 8?
OR
In a certain factory turning out razor blades, there is a small chance of 0.002 for any blade to be defective. The blades are supplied in packets of 10. Calculate the number of packets containing no defective, one defective and two defective blades in a consignment of 10,000 packets.
{Given e^-0.02 = 0.9802}
Solution:
According to the problem,
Z = \(\frac{X-14}{2.5}\)
(A) Here, P(12 ≤ X ≤ 15) = P(- 0.8 ≤ Z ≤ 0.4)
= P(0 ≤ Z ≤ 0.8) + P(0 ≤ Z ≤ 0.4)
= 0.2881 + 0.1554, i.e., 0.4435
So, the number of student who score between 12 and 15 is 1000 × 0.4435 = 444

(B) Here,
P(X > 18) =P(Z > 1.6)
= 0.5 – P(0 ≤ Z ≤ 1.6)
= 0.5 – 0.4452, i.e., 0.0548
So, the number of student who score below 8 is 1000 × 0.0548 = 55 (Approx.)

(C) Here,
P(X < 8) = P( Z ≤ -2.4) = P(Z ≥ 2.4)
= 0.5 – P(0 ≤ Z ≤ 2.4)
= 0.5 – 0.4918, i.e., 0.0082
So, the number of student Who score below 8 is 1000 × 0.0082 = 8 (Approx.)

Here, p = 0.002,
n = 10.
So, np = λ = 0.02
Probability of having no defective blade
P(X = 0) = \(\frac{(0.2)^0}{0 !}\).e^-(0.2)
= e^-(0.2) = 0.9802
⇒ Approximate number of packets having no defective blade in the consignment of 10,000 packets = 0.9802 × 10,000 = 9802
Probability of having one defective blade
P(X= 1) = \(\frac{0.02}{1}\) . P(x = 0)
= 0.02 × 0.9802
= 0.019604
⇒ Approximate number of packets having one defective blade in the consignment of 10,000 packets = 0.019604 × 10,000 = 196
Probability of having two defective blade
P(X = 2) = \(\frac{0.02}{2}\) . P(X = 1)
= 0.01 × 0.019604
= 0.000196
⇒ Approximate number of packets having two defective blades in the consignment of 10,000 packets = 0.000196 × 10,000 = 1.96 or 2.

Question 28.
The number N of bacteria in a culture grew at a rate proportional to N. The value of N was initially 100 and increased to 332 in one hour. Find the value of N after two hours. [3]
OR
Solve: \(\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}\) = 1, x ≠ 0
Solution:
Let N denote the number bacteria in a culture at any time t. Then, \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = k N, where k is constant of proportionality.
\(\frac{\mathrm{dN}}{\mathrm{N}}\) = k dt
Integrating both sides, we have
\(\int \frac{d N}{N}=\int k d t\)
log N = kt + C …………… (1)
Let N₀ = 100 be the number of bacteria in the culture at t = 0. Then , from (1), C = log N₀
∴From (1), we get log \(\left(\frac{N}{N_0}\right)\) = kt ………….. (2)
Now y = 332 when t = 1. So, by (2),
we have log \(\left(\frac{332}{100}\right)\) = k
k = log (3.32)
Thus, (2) gives log \(\left(\frac{N}{N_0}\right)\) = log (3.32) t
Let N₁ be the number of bacteria in the culture at t = 2. Then,
log \(\left(\frac{N_1}{100}\right)\) = 2 log (3.32)
⇒ log \(\left(\frac{N_1}{100}\right)\) = log (11.0224)
\(\frac{N_1}{100}\) = 11.0224
N₁ = 1102
Thus, the value of N after 2 hours is 1102.

Question 29.
Using Cramer’s rule, show the following system of equations: [3]
x – y + 3z = 6
x + 3y – 3z = -4
5x + 3y + 3z = 10
Solution:
You can try to get the following yourself.
D = 0, D_x = 0, D_y = 0, D_z = 0
Since D = D_x = D_y = D_z = 0, the system of equations may have infinite number of solutions or no solution.
For Solution
Consider first to equations, we get
x – y = 6 – 3z
x + 3y = 3z – 4
Considering ‘z’ as constant, we solve the above equations by Cramer’s rule:
Here, D = 4, D_x = 14 – 6z, D_y = 6z – 10
Since D ≠ 0, the system of equations has a unique solution,
x = \(\frac{D_x}{D}=\frac{14-6 z}{4}=\frac{7-3 z}{2}\)
y = \(\frac{D_y}{D}=\frac{6 z-10}{4}=\frac{3 z-5}{2}\)
Putting these values in third equation, we have
5 \(\left(\frac{7-3 z}{2}\right)\) + 3 \(\left(\frac{3 z-5}{2}\right)\) + 3z = 10
⇒ 35 – 15z + 9z -15 + 6z = 20
⇒ 20 = 20, which is true.
Hence, the given system of equations has in¬finite number of solutions, namely
x = \(\frac{7-3 z}{2}\)
y = \(\frac{3 z-5}{2}\)
z = k, where k is any arbitrary constant.

Question 30.
A null matrix of order 3 If A = \(\left[\begin{array}{ccc}
-1 & 0 & 5 \\
2 & 4 & -3 \\
7 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
3 & 1 \\
0 & -2 \\
5 & -4
\end{array}\right]\),then find: [3]
(A) a₁₁ + b₂₁
(B) a₃₁ + b₁₁
(C) a₂₂ + b₃₁
Solution:
Here,
(A) a₁₁ + b₂₁ = -1 + 0 = -1
(B) a₃₁ + b₁₁ = 7 + 3 = 10
(C) a₂₂ + b₃₁ = 4 + 5 = 9

Question 31.
The ratio of the speed of a motor boat to that of the speed of the stream in a river is 36 : 5. The boat along with the stream in 5 hours 10 minutes. How much will it take to come back? [3]
Solution:
Let the speed of the boat in still water be 36x km/h; and the speed of the stream be 5x km/h.
Then,
Downstream speed = (36x + 5x) km/h
= 41 × km/h
Upstream speed = (36x – 5x) km/h
= 31 × km/h
Suppose the boat goes ‘d’ km along the stream.
Then,\(\frac{d}{41 x}\) – 5\(\frac{1}{6}\)
d = \(\frac{31}{6}\) × 41 x
Suppose the motor boat takes‘t’ hours to come back with the speed of 31 x km/h. d
So, \(\frac{d}{31 x}\) = t
t = \(\frac{1}{31 x}\) \(\frac{31}{6}\) 41x
= \(\frac{41}{6}\) hours
= 6 hours 50 minutes

Section – D (20 Marks)

(All questions are compulsory. In case of internal choice, attempt any one question only)

Question 32.
The cost function of a firm is C = 5x² + 28x + 5, where C is the cost and x is the level of output. A tax at the rate of ₹ 2 per unit output is imposed and the producer ads it to his cost. Find minimum value of average cost. [5]
OR
Find the interval in which the function f(x) = (x + 2)e^-x is strictly increasing or decreasing.
Solution:
Here, Total cost function C(x) is given by
C(x) = 5x² + 28x + 5 + 2x, i.e., 5x² + 30x + 5
AC = \(\frac{5 x^2+30 x+5}{x}\) = 5x + 30 + \(\frac{5}{x}\)
⇒ \(\frac{d(\mathrm{AC})}{d x}\) = 5 \(\frac{5}{x^2}\) and \(\frac{d^2(\mathrm{AC})}{d x^2}=\frac{10}{x^3}\)
For maximum or minimum value of AC, we must have \(\frac{d^2(\mathrm{AC})}{d x}\) = 0, i.e., x = 1
At x = 1, \(\frac{d^2(\mathrm{AC})}{d x^2}=\frac{10}{(1)^3}\) = 10 > 0
Thus, AC is minimum when x = 1.
Hence, minimum average cost = 5(1)² + 30(1) + 5, i.e., ₹ 40

Here, f(x) = (x + 2)e^-x.
So, f ‘(x) = e^-x – (x + 2)e^-x = –\(\frac{x+1}{e^x}\)
The function f(x) will be strictly increasing if f ‘(x) > 0
i.e., if –\(\frac{x+1}{e^x}\) > 0, or \(\frac{x+1}{e^x}\) < 0
⇒ if x < -1 {Because e^x > 0 for all real x}
Thus, the function f(x) will be strictly increasing on (-∞, -1).
The function f(x) will be strictly decreasing if f'(x) < 0
i.e., if –\(\frac{x+1}{e^x}\) < 0, or \(\frac{x+1}{e^x}\) > 0
⇒ if x > -1 {Because e^x > 0 for all real x}
Thus, the function f(x) will be strictly increasing on (-1, ∞).

Question 33.
Calculate Fisher’s Ideal index number for the following data. Also, show that Fisher’s index is the geometric mean of Laspegre’s and Passche’s indices. [5]

Commodity	2019		2020
Commodity	P_bq_b		P_cq_c
A	6	50	1056
B	2	100	2120
C	4	60	660
D	10	30	12	24
E	8	40	12	36

OR
Coded monthly sales figures of a particular band of T.V. for 18 months commencing January 2020 are as follows:

Month	Sale of T.V.	Year	Sale of T.V.
Jan. 2020	18	Oct. 2020	27
Feb. 2020	16	Nov. 2020	28
Mar. 2020	23	Dec. 2020	24
Apr. 2020	27	Jan. 2021	24
May. 2020	28	Feb. 2021	28
June. 2020	19	Mar. 2021	29
July. 2020	31	Apr. 2021	30
Aug. 2020	29	May. 2021	29
Sept. 2020	35	June. 2021	22

Calculate 6-monthly moving averages and display these and the original figures on the same graph, using the same axes for both.
Solution:
We have the table

Laspeyre’s index number (L)
= \(\frac{\Sigma p_c q_b}{\Sigma p_b q_b}=\frac{1900}{1360}\) × 100
= 139.70
Passche’s index number (P)
= \(\frac{\Sigma p_c q_b}{\Sigma p_b q_b}\) × 100 = \(\frac{1880}{1344}\) × 100 = 139.88
Now, \(\sqrt{L \times P}\) = \(\sqrt{139.7 \times 139.88}\)
= \(\sqrt{19541.236}\)
= 139.79 = F
Thus, proved

Thus, the 6 – monthly centred moving averages are 22.92, 25.08, 27.17, 28.17, 28.17, 28.59, 28.42, 27.75, 27.17, 26.92, 27.25 and 27.17

Question 34.
An investor is considering the purchase of a 8% ₹ 1000 bond redeemable after 5 years at par. The investor’s required rate of return is 10%. What should he be willing to pay now to purchase the bond? (Given (1.1)^-5 = 0.62092) [5]
Solution:
Here,
F = Face value of the bond = ₹ 1000
n = number of periodic dividend payments = 5
i = AnnuaL yield rate = 0.1
R = F × i_d = ₹ (1000 × 0.08) = ₹ 80
Since the bond is to be redeemed at par,
C = Redemption price or Maturity value = Face value = ₹ 1000
Let V be the purchase price of the bond. Then,
V = R\(\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)^-n
= ₹ {80 \(\left[\frac{1-(1.1)^{-5}}{0.1}\right]\) + 1000(1.1)^-5}
= ₹ {800[1 – (1.1)^-5] + 1000 (1.1)^-5
= ₹ {800[1 – 0.62092] + 1000 (0.62092)}
= ₹ {303.26 + 620:92}, or ₹ 924.18
Hence, It is advisable to purchase the bond at ₹ 950.

Question 35.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ₹ 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ?200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag? [5]
Solution:
Let x be the number of bags of brand P, y be the number of bags of brand Q, and Z be the minimum cost of the mixture per bag.

	A	B	C	Cost
One bag of brand P	3	2.5	2	₹ 250
One bag of brand Q	1.5	11.25	3	₹ 200
Minimum requirement	18	45	24

Then, the mathematical formulation of the given LPP is
Minimise Z= 250 x + 200y
subject to constraints
3x + 1.5y
≥ 18 or 2x + y ≥ 12
2.5x + 11.25y ≥ 45 or x + 9y ≥ 36
2x + 3y ≤ 24
x, y ≥ 0
Let us draw the graph for the system of inequalities representing constraints.

The feasible region is ABCD shown (shaded) in Fig. 9, which is unbounded.
The coordinates of the corner points of the feasible region ABCD are A(18, 0), B(9, 2), C (3, 6) and D (0, 12).
Let us evaluate the objective function Z = 250x + 200y at the corner points.

Corner point	Corresponding value of Z
A (18, 0)	4500
B (9, 2)	2650
C (3, 6)	1950 (minimum)
D (0,12)	2400

Z is minimum at C (3, 6) and minimum cost is ₹ 1950

Since the feasible region is unbounded, so we have to determine whether Z = 1950 is minimum cost or not.

Now we draw the graph of the inequality 250x + 200y < 1950, or 5x + 4y < 39
The open half plane of 5x + 4y < 39 and the feasible region has no common points.

So, the least cost of the mixture is ₹ 1950 when 3 bags of brand P and 6 bags of brand Q are mixed.

Section – E (12 marks)

(All questions are compuLsory. In case of internal choice, attempt any one question only)

Question 36.
For providing water to the families of a colony, a Large water tank with two inlet pipes A and B and an outlet pipe C, is installed.

Pipes A and B can fill the tank in 10 hours and 12 hours respectively; whereas the pipe C can empty the tank in 13 hours.
Based on the above, answer the following questions:
(A) If both the pipes A and B are opened together, then in how much time the tank will be filled? [1]
(B) If both the pipes A and C are opened together, then in how much time the tank will be filled? [1]
(C) If all the three pipes are opened together, then in how much time the tank will be filled? [2]
OR
If both the pipes A and B are opened together for some time and then pipe B is turned off. If the tank is filled in 6 hours, then after how many hours the pipe B is turned off?
Solution:
(A) Part of the tank filled by pipe A in one hour = \(\frac{1}{10}\)
Part of the tankfilled by pipe B in one hour = \(\frac{1}{12}\)
Part of the tank filled by pipes A and B in one hour = (\(\frac{1}{10}\) + \(\frac{1}{12}\)) = \(\frac{11}{60}\)
So, two pipes A and B together fill the tank in \(\frac{60}{11}\) hours, i.e. 5 \(\frac{5}{11}\) hours

(B) Part of the tank filled by pipe A in one hour = \(\frac{1}{10}\)
Part of the tank emptied by pipe C in one hour = \(\frac{1}{15}\)
Part of the tank filled by pipes A and C in one hour = (\(\frac{1}{10}\) – \(\frac{1}{15}\)) = \(\frac{1}{30}\)
So, two pipes A and C together fill the tank in 30 hours

(C) Part of the tank filled by pipe A in one hour = \(\frac{1}{10}\)
Part of the tank filled by pipe B in one hour = \(\frac{1}{12}\)
Part of the tank emptied by pipe C in one hour = \(\frac{1}{15}\)
Part of the tank filled by three pipes A, B and C in one hour = (\(\frac{1}{10}\) + \(\frac{1}{12}\) + \(\frac{1}{15}\)) = \(\frac{7}{60}\)
So, three pipes A, B and C together fill the tank in \(\frac{60}{7}\) hours, i.e., 8 \(\frac{4}{7}\) hours.

Let the pipe B is turned off after’t’ hours. Then, (A + B) are opened for ‘t‘ hours and A alone is open for (6 – t) hours.
⇒ (\(\frac{1}{10}\) + \(\frac{1}{12}\))t + \(\frac{1}{10}\) (6 – t) = 1
⇒ t = 4.8 hours

Question 37.
There is a farmer in a village he has a part of land to plough. He has two cows which he uses for this purpose. An engineer came to his field and observed that the paths travelled by the two cows is given by the equations x² + y² = 1 and (x – 1)² + y² = 1. The farmer wants to know the common area grazed by the cows.

Based on the information, answer the following questions:
(A) Find one of the points of intersection of the two curves? [1]
(B) What is the area of common region? [1]
(C) What is the area of 1st circle? [2]
OR
What is the area of 2nd circle?
Solution:
(A) From the figure, we have the point of intersection as (\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\))

(B) Area of common region

Question 38.
A machine costs a company ₹ 2, 30, 000 and its effective life is estimated to be 15 years. The scrap value of the machine at the end of its life is expected to realise ₹ 10,000 only. In order to provide money at that time for a new machine costing the same amount, a sinking fund is set up. Equal amount of ₹10,000 is contributed to the fund at the end of every year and the funds earns an interest at 6% per annum compounded annually.

Based on the above information, answer the
following:
(A) What amount of money is needed to buy the new machine at the end of 15 years? [1]
(B) What amount of money is contributed to the sinking fund at the end of every year? [1]
(C) What amount of interest will be earned by the sinking fund over a period of 15 years? [2]
OR
If a discount of 5% on the purchase of the new machine and VAT of 18% is added to the price, what is the net cost of the new machine?
Solution:
(A) Amount of money is needed to buy the new machine at the end of 15 years
₹ 2,30,000 – ₹ 10, 000 = ₹ 2, 20,000

(B) Amount of money that is contributed to the sinking fund at the end of every year
= ₹ 10,000.

(C) Here, amount (A) accumulated in the sinking fund over 15 years is given by
A = P\(\left[\frac{(1+i)^n-1}{i}\right]\)
where P = ₹ 10,000; i = 0.06 and n = 15 years
So, A = 10,000 \(\left[\frac{(1+0.06)^{15}-1}{0.06}\right]\)
= ₹ 2, 32,759
Thus, amount of interest in the sinking fund = ₹ 2, 32, 759 – (₹ 10,000 × 15) = ₹ 82, 759

Cost of new machine after discount
= ₹ 2,30,000 – 5 % of ₹ 2,30,000
= ₹ 2,18,5000
New cost of new machine after adding VAT
= ₹ 2,18,500 + 18% of ₹ 2,18,500
= ₹ 2,57,8300.