CBSE Sample Papers for Class 12 Applied Mathematics Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Mathematics Set 5 with Solutions

Maximum Marks: 80 Marks
Time Allowed : 3 Hours

General Instructions:

1. This question paper contains five sections A, B, C, D and E. Each section is compulsory.
2. 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.
3. Section A: It comprises of 20 MCQs of 1 mark each.
4. Section B: It comprises of 5 VSA type questions of 2 marks each.
5. Section C: It comprises of 6 SA type of questions of 3 marks each.
6. Section D: It comprises of 4 LA type of questions of 5 marks each.
7. 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.
8. 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.
If the mean and variance of a binomial distribution are 4 and 2 respectively, then the probability of 1 success is: [1]
(a) \(\frac{1}{32}\)
(b) \(\frac{1}{16}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{4}\)
Solution:
(a) \(\frac{1}{32}\)

Explanation:
Here, np = 4 and npq = 2
So,q = \(\frac{1}{2}\) and hence p = \(\frac{1}{2}\) and n = 8
So. P(X = 1) = ⁸C₁ = (\(\frac{1}{2}\))¹ (\(\frac{1}{2}\))⁷ = 8(\(\frac{1}{2}\))⁸ = \(\frac{1}{132}\)

Question 2.
3% of the electric bulbs manufactured by a company are defective. The probability that a sample of 100 bulbs has no defective bulb is: [1]
(a) e^-3
(b) 1 – e^-3
(c) 3e^-3
(d) 1 – 3e^-3
Solution:
(a) e^-3

ExpLanation:
Here, p =0.03 and n = 100 and so, np = λ = 3
Thus, P(X = 0) = \(\frac{3^0}{0 !}\) . e^-3 = e^-3

Question 3.
A can run 45 m while B runs 50 m. In a 1 km race B beats A by how many metres? [1]
(a) 100 m
(b) 90 m
(c) 85 m
(d) 80 m
Solution:
(a) 100 m

Explanation:
B beats A in a race of 50 m = 5m
1000 × 1 m = \(\frac{5}{50}\) × 100
= 100 m

Question 4.
The probability distribution of a random variable X is given below: [1]

Xi	0	1	2	3
P(X = Xi)	k	\(\frac{k}{2}\)	\(\frac{k}{4}\)	\(\frac{k}{8}\)

Then, P( X > 2) is:
(a) \(\frac{14}{15}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{10}\)
Solution:
(b) \(\frac{1}{5}\)

Explanation:
Here,
k + \(\frac{k}{2}\) + \(\frac{k}{4}\) + \(\frac{k}{8}\) = 1 i.e., k = \(\frac{8}{15}\)
Now, P(X > 2) = P(X = 3) = 3 × \(\frac{k}{8}\) = \(\frac{1}{5}\)

Question 5.
The time taken by a carpenter to make a chair is \(\frac{3}{4}\) hour. What is the maximum number of chairs he can make if he has 7 hours to make them? [1]
(a) 5
(b) 7
(c) 8
(d) 9
Solution:
(d) 9

Explanation:
Let x be possible number of chairs he can make in 7 hours. Then, \(\frac{3}{4}\) x ≤ 7 This gives the maximum value of x as 9.

Question 6.
Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can fill it in 15 hours. If all the pipes are opened simultaneously, then in how much time the tank will be filled up? [1]
(a) 3′ hours
(b) 4′ hours
(c) 2′ hours
(d) 5′ hours
Solution:
(c) 2′ hours

Question 7.
If xy = f², then \(\frac{\left(d^2 y\right)}{d x^2}\) is : [1]
(a) \(\frac{-x}{(y)}\)
(b) \(\frac{-(y)}{x}\)
(c) \(\frac{2 y}{x^2}\)
(d) \(\frac{2 x}{y^2}\)
Solution:
(c) \(\frac{2 y}{x^2}\)

Explanation:
Given xy = k², we have

Question 8.
Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. Then the possible values of X are: [1]
(a) 0, 1, 3, 5
(b) 0, 2, 4, 6
(c) 0, 2, 5, 6
(d) 1, 3, 4, 5
Solution:
(b) 0, 2, 4, 6

Explanation:
We have in all 7 cases, namely
(i) 6T and OH. So, X = 6 – 0 = 6
(ii) 5T and 1H. So, X = 5 – 1 = 4
(iii) 4T and 2H. So, X = 4 – 2 = 2
(iv) 3T and 3H. So, X = 3 – 3 = 0
(v) 2T and 4H. So, X = 4 – 2 = 2
(vi) 1T and 5H. So, X = 5 – 1 = 4
(vi) 0T and 6H. So, X = 6 – 0 = 6
Thus, the possible values of X are 0, 2, 4 and 6.

Question 9.
In a game of 160 points, A can give 10 points to B and 30 points to C. How many points B can give to C in a game of 60 points? [1]
(a) 4 points
(b) 6 points
(c) 7 points
(d) 8 points
Solution:
(d) 8 points

Question 10.
The function f(x) = x³ + 3x is increasing on [1]
(a) (-∞, 0)
(b) (0, ∞)
(c) R
(d) (0, 1)
Solution:
(c) R

Explanation:
Here, f(x) = x³ + 3x
So, f'(x) = 3x² + 3, which is positive for real x
Hence, f(x) is increasing on R.

Question 11.
If the diameters of ball bearing are normally distributed with mean 0.6140 inches and standard deviation 0.0025 inches, then the percentage of ball bearings with diameter between 0.610 and 0.618 is; [1]
(a) 89.4%
(b) 11.51%
(c) 15.54%
(d) 18.02%
Solution:
(a) 89.4%

Explanation:
Here,
Z = \(\frac{X-0.6140}{0.0025}\)
P(0.610 < X < 0.618) = P( -1.6 < Z < 1.6)
= 2 × P(0 < Z < 1.6)
Hence, the percentage of ball bearings with diameter between 0.610 and 0.618 is 89.4%

Question 12.
When 10 coins are tossed simultaneously, the probability of getting at least 9 heads is: [1]
(a) \(\frac{11}{1024}\)
(b) \(\frac{3}{256}\)
(c) \(\frac{13}{1024}\)
(d) \(\frac{15}{1024}\)
Solution:
(a) \(\frac{11}{1024}\)

Explanation:
Here, P(X ≥ 9) = P(X = 9) + P(X = 10)
= ¹⁰C_p (\(\frac{1}{2}\))⁹ (\(\frac{1}{2}\))¹ + ¹⁰C10 (\(\frac{1}{2}\))¹⁰ (\(\frac{1}{2}\))⁰
= [10 + 1] (\(\frac{1}{2}\)⁸ = \(\frac{11}{256}\)

Question 13.
Pipe A can fill a tank in 10 hours, while another pipe B can empty it in 6 hours. When both the pipes are opened simultaneously, in how much time will the tank be empty: [1]
(a) 18 minutes
(b) 15 minutes
(c) 16 minutes
(d) 17 minutes
Solution:
(b) 15 minutes

Question 14.
\(\int\)(e^x + 2x – 3) dx is equal to: [1]
(a) x² – 3x + xe^x + C
(b) x² – 3x + e^x + C
(c) x² – 3x – e^x + C
(d) x² + 3x + xe^x + C
Solution:
(b) x² – 3x + e^x + C

Explanation:
\(\int\) (e^x + 2x – 3) dx = \(\int\)e^xdx + \(\int\)2x dx – \(\int\)3dx
= e^x + x² – 3x + C

Question 15.
Most preferred type of average for index number is: [1]
(a) arithmetic mean
(b) geometric mean
(c) harmonic mean
(d) none of the above
Solution:
(b) geometric mean

Question 16.
Index number is a: [1]
(a) measure of relative change
(b) a special type of an average
(c) a percentage relative
(d) all the above
Solution:
(d) all the above

Question 17.
The rise in prices before Deepawali is an example of [1]
(a) seasonal trend
(b) cyclical trend
(c) long term trend
(d) irregular trend
Solution:
(a) seasonal trend

Question 18.
\(\int(1-x) \sqrt{x}\) dx is equal to:
(a) \(\frac{2}{3}\) x^3/2 – \(\frac{2}{5}\) x^5/2 + C
(b) \(\frac{2}{5}\) x^5/2 – \(\frac{2}{3}\) x^3/2 + C
(c) x^3/2 – \(\frac{2}{5}\) x^5/2 + C
(d) \(\frac{2}{3}\) x^3/2 – x^5/2 + C
Solution:
(a) \(\frac{2}{3}\) x^3/2 – \(\frac{2}{5}\) x^5/2 + C

Explanation:
\((1-x) \sqrt{x} d x=\int \sqrt{x} d x-\int x \sqrt{x} d x\)
= \(\int\)x^1/2dx – \(\int\)x^3/2dx
= \(\frac{2}{3}\) x^3/2 – \(\frac{2}{5}\) x^5/2 + C

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 and (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): f(x) = is continuous at x = 0. [1]
Reason (R): Both h(x) = g x², g(x) = are continuous at x = 0.
Solution:
(c) (A) is true but (R) is false.

Explanation:
Assertion, f(0) = 0
\(\lim _{x \rightarrow 0}\) x² sin(\(\frac{1}{x}\)) = 0 x (finite value) = 0
∴ It is continuous at x = 0
Reason: h(x) = x² is continuous, but g(x) is not continuous.
\(\lim _{x \rightarrow 0}\) x² sin(\(\frac{1}{x}\)) = not defined value oscillates
\(\lim _{x \rightarrow 0}\) sin(\(\frac{1}{x}\)) ≠ 0
∴ It is not continuous.

Question 20.
Assertion (A): The intermediate solutions of constraint must be checked by substituting them back into objective function. [1]
Reason (R):

Here, (0, 2) (0, 0) and (3, 0) are all vertices of feasible region.
Solution:
(d) (A) is false but (R) is true.

Explanation:
The intermediate solutions of constraints must be checked substututing them back into constraint equations.

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]
6x + 2y ≥ 8; x + y ≤ 4;
x + 5y ≥ 4, x ≤ 3; y ≤ 3;
x ≥ 0, y ≥ 0
Solution:

Question 22.
By taking any negative real number x, verify that – |x| ≤ x ≤ |x| [2]
OR
Solve for x:
\(\frac{5(2-x)}{3}\) ≥\(\frac{3(x-2)}{5}\)
Solution:
Let x = -2.
Then, – |x| = – |-2|
= -2; |x| = |-2| = 2
So, -|x| ≤ x ≤ |x| is true for x = -2
OR
We have \(\frac{5(2-x)}{3}\) ≥\(\frac{3(x-2)}{5}\)
25(2 – x) ≥ 9(x – 2)
50 – 25x ≥ 9x – 18
34x ≤ 68, or x ≤ 2

Question 23.
Explain the difference between ‘parameter’ and ‘statistic’ with examples. [2]
Solution:
Parameter: A numerical value summarizing all the data of an entire population, e.g., Population Mean, population variance etc.

Statistic: A numerical value summarizing the sample data e.g., Sample Mean, sample variance etc.
Example: (i) Average income of all faculty members working at COMSATS is a parameter.
(ii) Average income of faculty members of Management Sciences Department at COMSATS is a statistic.

Question 24.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -4
\end{array}\right]\), then show that (A + B)^T = A^T + B^T [2]
OR
Show that AB ≠ BA, if A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-1 & 5 \\
4 & -2
\end{array}\right]\)
Solution:


From (1) and (2), we have (A + B)^T = A^T + B^T

OR

Here,
AB = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
-1 & 5 \\
4 & -2
\end{array}\right]\) = \(\left[\begin{array}{cc}
7 & 1 \\
13 & 7
\end{array}\right]\)
and
BA = \(\left[\begin{array}{cc}
-1 & 5 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
14 & 18 \\
-2 & 07
\end{array}\right]\)
Thus, AB ≠ BA

Question 25.
Which is a better investment: “6%, ₹ 100 share at ₹ 120” or “4%, ₹ 10 share at ₹ 8”? [2]
Solution:
Option 1: 6%, ₹ 100 share at ₹ 120
Here, income of an investment of ₹ 120 = ₹ 6
So, income on an investment of ₹120
= ₹ {\(\frac{6}{120}\) × 100} = ₹ 5
Option 2 :4% , ₹10 share at ₹ 8
Here, income of an investment of ₹ 8 = ₹ 0.4
So, income on an investment of ₹ 100
= ₹ {\(\frac{0.4}{8}\) × 100} = ₹ 5
Hence, both the investments yield the same return.

Section – C (18 Marks)

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

Question 26.
Given A = \(\left[\begin{array}{ll}
0.4 & 0.1 \\
0.7 & 0.6
\end{array}\right]\) and final demand D = \(\left[\begin{array}{l}
400 \\
200
\end{array}\right]\). Calculate the grass output X = \(\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]\) using input-output technique. [3]
OR
Using Cramer’s rule, solve the system of equations :
x + y + z + 1 = 0
x + 2y + 3z + 4 = 0
x + 3y + 4z + 6 = 0
Solution:
10.4 0.1
Here, the technology matrix A is \(\left[\begin{array}{ll}
0.4 & 0.1 \\
0.7 & 0.6
\end{array}\right]\)
and the demand matrix D is \(\left[\begin{array}{l}
400 \\
200
\end{array}\right]\)
We need to determine output motnx X = \(\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]\)
Following the steps of Ques. 24, we get
\(\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]\) = \(\frac{1}{0.17}\left[\begin{array}{ll}
0.4 & 0.1 \\
0.7 & 0.6
\end{array}\right]\left[\begin{array}{l}
400 \\
200
\end{array}\right]\)
= \(\left[\begin{array}{l}
1058.8 \\
2352.9
\end{array}\right]\)
Hence, x₁ = 1058.8
and x₂ = 2352.9
Hence, the gross output of two industries are 1058.8 units and 2352.9 units respectively.

OR

The given system of equations can be rewritten as:
x + y + z = -1; x + 2y + 3z = -4; x + 3y + 4z = -6

Since D = -1 ≠ 0, the system of equations has a unique solution given by
x = \(\frac{D_x}{D}=\frac{-1}{-1}\) = 1.
y = \(\frac{D_y}{D}=\frac{1}{-1}\) = -1
z = \(\frac{D_z}{D}=\frac{1}{-1}\) = -1
Hence, x = 1, y = -1, z = -1 is the solution of the given system of equations.

Question 27.
Prove that the area of a right -angled triangle of given hypotenuse is maximum when the triangle is isoscetes. [3]
Solution:
Let for the right angled triangle. hypotenuse be c and the other sides are x and y.

∴ A² and Hence A is maximum when x = \(\frac{c}{\sqrt{2}}\) and y = \(\sqrt{c^2-x^2}=\frac{c}{\sqrt{2}}\), i.e., when x and y are equal, or the given triangle is isosceles.

Question 28.
Find: \(\int \frac{3 x+1}{(x-1)^2(x+3)}\) dx [3]
OR
The demand and supply functions for a commodity are p = x² – 6x + 16 and p = \(\frac{1}{3}\) x² + \(\frac{4}{3}\) x + 4 respectively. Find each of the following assuming x ≤ 5:
(A) The equilibrium point
(B) CS. and P.S.at the equilibrium point.
Solution:
Let I = \(\int \frac{3 x+1}{(x-1)^2(x+3)}\) dx
Let \(\frac{3 x+1}{(x-1)^2 \cdot(x+3)}\)
= \(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}\)
⇒ 3x + 1A(x – 1) (x + 3) + B(x + 3) + C (x – 1)²
Solving these equations, we get A = \(\frac{1}{2}\), B = 1, C = \(\frac{1}{2}\)
Thus, I = \(\int \frac{3 x+1}{(x-1)^2(x+3)}\) dx
= \(\int\left[\frac{1}{2(x-1)}+\frac{1}{(x-1)^2}-\frac{1}{2(x+3)}\right]\) dx
= \(\frac{1}{2} \log |x-1|-\frac{1}{x-1}+\frac{1}{2} \log\) |x + 3| + C

OR

(A) We know that the equilibrium point (x₀, p₀) is the point where the demand curve and supply curve intersect Therefore, the equilibrium is obtained by setting D(x) = S(x)
Now, D(x) = S(x) gives x² – 6x + 16 = \(\frac{1}{3}\) x² + \(\frac{4}{3}\) x + 4
x² – 11x + 18 = 0
(x – 2)(x – 9) = 0
x = 2 or 9
Since x < 5, we take
x₀ = 2
For x₀ = 2; p₀ = (2)² x₀ = 2, p₀ = (2)² – 6(2) + 16 = 8
The equilibrium point is (2, 8)

Question 29.
The mean and variance of a random sample of 64 observations were computed as 160 and 100 respectively. Compute the 95% confidence interval for population mean. [3]
Solution:
Here, we have
n = 64, \(\bar{X}\) = 160 and S² = 100 \(\overline{P Q}\) S = 10
Therefore, Confidence limits (95%) are \(\bar{X}\) ± 1.96 \(\left(\frac{S}{\sqrt{n}}\right)\)
i.e., 160 ± 1.96 (\(\frac{10}{8}\)) = 160 ± 2.45
i.e., 157.55, 162.45

Question 30.
The price index of cosmetics was 110 in 2006 with base year 2000 and 120 in 2007 with base year 2006. It further increased by 30% in 2008 in relation to the price index of 2007 and decreased by 10% in 2009 as compared to its level in 2008. Find the index for 2009 with 2000 as the base year. [3]
Solution:
Here,
\(\frac{p_{06}}{p_{00}}\) × 100 = 100
⇒ \(\frac{p_{06}}{p_{00}}\) = 1.1
Similarly, \(\frac{p_{07}}{p_{06}}\) = 1.2
\(\frac{p_{08}}{p_{07}}\) = 1.3
and \(\frac{p_{09}}{p_{08}}\) = 0.9
Now, price index for 2009 with 2000 as base
= \(\frac{p_{09}}{p_{00}}\) × 100 = (\(\frac{p_{09}}{p_{09}}\) × \(\frac{p_{08}}{p_{07}}\) × \(\frac{p_{07}}{p_{06}}\) × \(\frac{p_{06}}{p_{00}}\)) × 100
= (0.9 × 1.3 × 1.2 × 1.1) × 100
= 1.5444 × 100 or 154.44

Question 31.
Mr. Ramesh wants to know the amount he should pay for a goldmine expected to yield an annual return of 4 lakh for the next 10 years, after which it will worthless. Find the amount he should pay for the mine if he wants to yield 18% annual return on his investment and also set up a skinning fund to replace the purchase price. Assume that the sinking fund earns 10% annually. (Use (1.1)¹⁰ = 2.5937) [3]
Solution:
Let X be the purchase price of the goldmine. Rate of return on investment = 18% = 0.18
So, rate of return on investment of ₹ X = 0.18 X
Net annual return = ₹ 4,00,000
Sinking fund is created to replace the purchase price.
Therefore, amount to be placed in sinking fund each year is ₹ (4,00,000 – 0.18X)
These payments must accumulate to ₹ X at the end of 10 years.
Now, A = \(\frac{S r}{(1+r)^n-1}\) gives
₹ (4,00,000 – 0.18X) = \(\frac{0.1 X}{(1.1)^{10}-1}\)
⇒ 4,00,000 = X[\(\frac{0.1}{1.5937}\) + 0.18]
⇒ X = \(\frac{4,00,000}{0.2427}\) = 1648125.25
Thus, the purchase price of the machine is ₹ 16,48,125.25

Section – D (20 Marks)

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

Question 32.
A manufacturer has three machines I, II III installed in his factory. Machines I and II and are capable of being operated for at most 12 hours whereas machine IN must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table:

Items	Number of hours required on machines
	I	II	III
M	1	2	1
N	2	1	1.25

She makes a profit of ₹ 600 and ₹ 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit. [5]
OR
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minute each for cutting and 10 minutes each for assembling. Souvening of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is ₹ 5 each for type A and each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?
Solution:
Let x and y be the number of items M and N respectively.
Then, the mathematical formulation of the given LPP is
Maximise Z = 600x+ 400y
subject to constraints
x + 2y ≤ 12, 2x + y ≤ 12, x + \(\frac{5}{4}\) y ≥ 5
x, y > 0
Let us draw the graph for the system inequalities representing constraints.

The feasible region is ABCDE shown (shaded) in Fig.5, which is bounded.
The coordinates of the corner points of the feasible region ABCDE are A(5, 0), B(6, 0), C (4, 4), D(0, 6) and E(0, 4)
Let us evaluate the objective function Z = 600 x + 400y at the comer points.

Corner point	Corresponding value of Z
A( 5, 0)	3000
B(6, 0)	3600
C (4, 4)	4000 (maximum)
D(0, 6)	2400
E(0, 4)	1600

Z is maximum at C(4, 4) and the maximum profit is ₹ 4000
Thus, the manufacturer should produce 4 items of each item to realise maximum profit of ₹ 4,000.

OR

Let the company manufacture x souvenirs of Type A, y souvenirs of Type B in a day and Z be the total profit of the company in a day.

	Cutting	Assembling	Profit
souvenirs of Type A	5 minutes	3 hours	₹ 5
souvenirs of Type B	8 minutes	2 hours	₹ 6
Time available	3 hours 20 minutes	20 hours

Then, the mathematical formulation of the given LPP is
Maximise Z = 5x+ 6y
subject to constraints
5x + 8y ≤ 200, 10x + 8y ≤ 240, x, y > 0
Let us draw the graph for the system of inequalities representing constraints.

The feasible region is OABC shown (shaded) in Fig. 8, which is bounded.
The coordinates of the comer points of the feasible region OABC are 0(0, 0), A(24, 0), B(8,20) and C (0, 25).
Let us evaluate the objective function Z = 5x + 6y at the corner points.

Corner point	Corresponding value of Z
0(0, 0)	0
A (24, 0)	120
B (8, 20)	160 (maximum)
C (0, 25)	150

Z is maximum at B(8, 20) and the maximum profit is ₹ 160.
Thus, the company should manufacture 8 souvenirs of Type A and 20 souvenirs of Type B to realise maximum profit of ₹ 160.

Question 33.
Given two matrices A and B, A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]\), find AB. Using this product AB, solve the following system of equations:
2x- y + z = -1, – x + 2y – z = 4, x – y + 2z = -3 [5]
OR
The daily cost C of operating a hospital is a linear function of the number of indoor patients (i) and outdoor patients (O) plus fixed cost (z), i.e., C = xl + yO + z. Find the values of costs x, y and z through a system of linear equations, using inverse of co¬efficient matrix method:

Days	C(in ₹)	Patients
		I	O
1	5200	30	20
2	5360	36	22
3	5650	15	45

Solution:

The given system of equations in the matrix form can be written as
AX = D ………………. (2)
where

⇒ x = 1 , y = 2 and z = – 1 is the required solution of the given system of equations.

OR

According to the problem, we have the following system of equations:
30x + 20y + z = 5200,
35x + 22y + z = 5360,
15x + 45y + z= 5650
So, D = \(\left|\begin{array}{lll}
30 & 20 & 1 \\
35 & 22 & 1 \\
15 & 45 & 1
\end{array}\right|\), D_x = \(\left|\begin{array}{lll}
5200 & 20 & 1 \\
5360 & 22 & 1 \\
5650 & 45 & 1
\end{array}\right|\)
D_y = \(\left|\begin{array}{lll}
30 & 5200 & 1 \\
35 & 5360 & 1 \\
15 & 5650 & 1
\end{array}\right|\), D_z = \(\left|\begin{array}{lll}
30 & 20 & 5200 \\
35 & 22 & 5360 \\
15 & 45 & 5650
\end{array}\right|\)
Here, D = 30(22 – 45) -20 (35 – 15) + 1 (1575 – 330) = 155 ≠ 0;
D_x = 5200 (22 – 45) – 20 (5360 – 5650) + 1 (241200 – 124300) = 3100;
D_y = 30(5360 – 5650) – 5200 (35 – 15) + 1 (197750 – 80400) = 4650;
D_z = 30(124300 – 241200) – 20 (197750 – 80400) + 5200 (1575 – 330) = 620000
Using Cramer’s rule, we get
x = \(\frac{D_x}{D}=\frac{3100}{155}\) = 20;
y = \(\frac{D_y}{D}=\frac{4650}{155}\) = 30;
z = \(\frac{D_z}{D}=\frac{620000}{155}\) = 4000
Thus, x = 20 ; y = 30 ; z = 4000

Question 34.
A bond of ₹ 1000 face value bond , bearing a coupon rate of 12% will mature after 7 years. What is the value of the bond if the required rate of return on the bond are 14% and 12%? [Given (1.14)^-7 = 0.3997] [5]
Solution:
Here,
F = Face value of the bond = ₹ 1000
n = number of periodic divident payments = 7
i = Annual yield rate = 0.14 and 0.12
R = F × i_d = ₹ (1000 × 0.12) = ₹ 120
Since the bond is to be redeemed at par,
C = Redemption price or Maturity value = Face value = ₹ 1000
Let V be the purchase value of the bond. Then,
V = \(R\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)^-n

(a) When i = 0.14
V = ₹ {120 \(\left[\frac{1-(1.14)^{-7}}{0.14}\right]\) + 1000(1.14)^-7}
= ₹ {857[1 – (1.14)^-7)] + 1000 (1.14)^-7)}
= ₹ {857[1 – 0.3997] + 1000 (0.3997)}
= ₹ {514.45 + 399.70}, or ₹ 914.15
Thus, the value of the bond is ₹ 914.15.

(b) When i = 0.12
V = {120 \(\left[\frac{1-(1.12)^{-7}}{0.12}\right]\) + 1000(1.12)^-7}
= ₹ {1000 [1 – (1.12)^-7)] + 1000 (1.12)^-7)}
= ₹ 1000
Thus, the value of the bond is ₹ 1000.

Question 35.
An investor is considering purchasing a new issue of 5-year bonds of ₹ 1,00,000 par value and an annual fixed coupon rate of 12% while coupon payments are made semi-annually. The minimum yield that the investor would accept is 6.75%. Find the fair value of the bond [Given (1.0675)^-10 = 0.52039) [5]
Solution:
Here
F = Face value of the bond = ₹ 1,00,000
n = number of periodic divident payments
i = Annual yield rate = 0.0675
R = F × i_d = ₹ (1,00,000 × 0.06) = ₹ 6000
Since the bond is to be redeemed at per,
C = Redemption price or maturity value = Face value = ₹ 1,00,000
Let V be the purchase value of the bond. Then,
V = \(R\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)^-n
⇒ V = ₹ {6000 \(\left[\frac{1-(1.0675)^{-10}}{0.0675}\right]\) + 1,00,000(1.0675)^-1}
= ₹ {88889 [1 – (1.0675)^-10)] + 1,00,000 (1.0675)^-10)}
= ₹ {88889 [1 – 0.52039] + 1,00,000(0.52039)}
= ₹ {42632 + 52039}, or ₹ 94671
Thus, the fair value of the bond is ₹ 94671.

Section – E (12 Marks)

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

Question 36.
The Aranmula Boat race the oldest river race festival in Kerala, the south western State of India is hel during Onam festival.

Let the speed of a boat in still water be u km/h and the speed of the current (stream) in a river be v km/h. When a boat is rowed in the direction of the current, then the flowing water in the river assists the motion of the boat and its speed increases, the increased speed is called the speed downstream, and is given by (u + v)km/h When a boat is rowed against the current, then the flowing water in the river resists the motion of the boat and its speed decreases, the decreased speed is called the speed upstream, and is given by (u – v)km/h

A man takes 8 hours to row a boat to a place and come back with a speed of 4 km/h in still water and the speed of the current is 2 km/h. Based on the above , answer the following questions:
(A) What is the speed of the boat downstream? [1]
(B) What is the speed of the boat up stream? [1]
(C) If one – side distance is ‘d’ km, how much time is taken to row the boat downstream and how much time is taken to row the boat upstream? [2]
OR
What is the value of ‘d’, where’d’ is the distance travelled?
Solution:
(A) Speed of the boat downstream = (4 + 2) km/h, i.e., 6 km/h
(B) Speed of the boat up stream= (4 – 2) km/h; i.e., 2 km/h.
(C) Time taken in traveling a distance of ‘d’ km
downstream = \(\frac{d}{62}\) hours.
Time taken in traveling a distance of ‘d’ km d ,
upstream = \(\frac{d}{2}\) hours.

OR

Since, time taken in downstream = \(\frac{d}{6}\)
Since, time taken in uperstream = \(\frac{d}{2}\)
A.T.Q. Here., \(\frac{d}{6}\) + \(\frac{d}{2}\) = 8
d = 12

Question 37.
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 the principal.
Based on this information, answer the following questions:
(A) If the principal increases continuously at the rate of r% per year, then find the value of r if ₹ 100 doubles itself in 10 years. (Use loge 2 = 0.6931) [1]
(B) If the principal increases continuously at the rate of 5% per year , in how many years ₹ 1000 double itself? [1] (Use loge 2 = 0.6931)
(C) In a bank, principal increases continuously at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it worth after 10 years? (e^0.5 = 1.648) [2]
OR
In a bank, principal increases continuously at the rate of 6% per year. An amount of ₹ 1000 is deposited with this bank, how much will it worth after 10 years? (e^0.6 = 1.822)
Solution:
Let P be the principal at any time t and the rate of interest be r% per annum compounded continuously. Then,
\(\frac{d \mathrm{P}}{d t}=\frac{\mathrm{Pr}}{100}\)
or, \(\frac{d \mathrm{P}}{\mathrm{P}}=\frac{r}{100}\) dt
⇒ \(\int \frac{d \mathrm{P}}{\mathrm{P}}=\int \frac{r}{100}\) dt
⇒ log P = \(\frac{\mathrm {r t}}{100}\) + C …………….. (1)
Let P₀ be the initiaL principal Then, from (1).
C = Log P₀ ………….. (2)
Thus log \(\frac{p}{p_0}\) = \(\frac{r t}{100}\) ……………… (3)
(A) Here, t = 10, P₀ = 100, P = 2P₀ = ₹ 200 and r% = ?
From (3). we have log 2 = \(\frac{10 r}{100}\)
⇒ r = 10 log 2 = 693.1

(B) Here, P₀ = ₹ 1000, P = 2P₀ = ₹ 2000, r% = 5 and t =?
From (3). we have log 2 = \(\frac{5 t}{100}\) .
⇒ t = 20 log 2 = 13.862 years

C)Here, P₀ = ₹ 1000, r% = 5, t = 10 and P=?
From (3). we have log \(\frac{P}{P_0}\) = \(\frac{50}{100}\)
log \(\frac{P}{P_0}\) = 0.5
⇒ P = P₀ e^0.5 = 1648.

OR

Here,P₀ = ₹ 1000, r% = 6, t = 10 and P = ?
From (3), we have
log \(\frac{P}{P_0}\) = \(\frac{60}{100}\)
⇒ log \(\frac{P}{P_0}\) = 0.6
⇒ P = P₀ e^0.6 = 1822.

Question 38.
A certain factory manufactures ball point pens. He sells pens in packets of 100 pieces. It is known that 6% of his product are defective.
{Use: e^-6 = 0.00279}

Based on this information, answer the following questions:
(A) In a consignment of 10,000 packets, find the number of packets containing no defective pen? [1]
(B) In a consignment of 10,000 packets, find the number of packets containing at most 4 defective pens. [1]
(C) In a consignment of 10,000 packets, find the number of packets containing at least 2 defective pens. [2]
OR
In a consignment of 10,000 packets, find the number of packets containing exactly 3 defective pens.
Solution:
Here, p (defective pen) = 0.06, n = 100 So, λ = np = 6
We know that
P(r) = e^-λ\(\frac{\lambda^r}{r !}\) = e^-6 \(\frac{(6)^r}{r !}\)
= 0.00279 × \(\frac{(6)^r}{r !}\)

(A) Probability of a packet containing no defective pen = P(0) = 0.00279
Thus, approximate number of packets have zero defective pens in the consignment = 10,000 × 0.00279 = 28

(B) Probability of a packet containing at most 4 defective pen
= P(0) + P(1) + P(2) +P(3) + P(4)
= 0.00279[ 1 + 6+18 + 36 + 54]
= 0.32085
Thus, approximate number of packets having at most 4 defective pens in the consignment = 10,000 × 0.32085 = 3209

(C) Probability of a packet containing at least 2 defective pen
= 1 – {P(0) + P(1)}
= 1 – {0.00279 [1 + 6]}
= 1-0.01953 = 0.98047
Thus, approximate number of packets having, at least 2defective pens in the consignment = 10,000 × 0.98047 = 9805

OR

Probability of a packet containing exactly 3 defective pen = P(3) = 0.00279 × \(\frac{(6)^r}{3 !}\)
= 0.00279 × 36
= 0.10044
Thus, approximate number of packets have zero defective pens in the consignment = 10,000 × 0.10044 = 1004.