5 Marks Questions

Question 15 Marks

A person amortizes a loan of ₹ 1500000 for renovation of his house by 8 years mortgage at the rate of $12 \%$ p.a. compounded monthly. Find
i. the equated monthly installment
ii. the principal outstanding at the beginning of 40th month.
iii. the interest paid in $40^{\text {th }}$ payment.
$
\text { [Given } \left.(1.01)^{96}=2.5993,(1.01)^{57}=1.7633\right]
$

Answer

Given, $\mathrm{P}=$ ₹ $1500000, \mathrm{i}=\frac{12}{12 \times 100}=\frac{1}{100}=0.01$
and $\mathrm{n}=8 \times 12=96$
$\begin{array}{l}\text { i. } EMI =\frac{1500000 \times 0.01 \times(1.01)^{96}}{(1.01)^{96}-1} \\ =\frac{1500000 \times 0.01 \times 2.5993}{2.5993-1} \\ =\frac{1500000 \times 0.01 \times 2.5993}{1.5993}\end{array}$
= ₹ 24379.10
ii. Principal outstanding at the beginning of $40^{\text {th }}$ month
$\begin{array}{l}=\frac{E M I\left[(1+i)^{96-40+1}-1\right]}{i(1+i)^{96-40+1}} \\ =\frac{24379.10 \times\left[(1.01)^{57}-1\right]}{0.01(1.01)^{57}} \\ =\frac{24379.10 \times(1.7633-1)}{0.01 \times 1.7633} \\ =\frac{24379.10 \times 0.7633}{0.017633}\end{array}$
= ₹ 1,055,326.20
iii. Interest paid in $40^{\text {th }}$ payment
$
\begin{aligned}
& =\frac{E M I\left[(1+i)^{96-40+1}-1\right]}{(1+i)^{96-40+1}} \\
& =\frac{24379.10\left[(1.01)^{57}-1\right]}{(1.01)^{57}}
\end{aligned}
$
$
\begin{aligned}
& =\frac{24379.10 \times 0.7633}{1.7633} \\
& \end{aligned}$
= ₹ 10553.26

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Question 25 Marks

Let X denote the number of hours a person watches television during a randomly selected day. The probability that X can take the values $\mathrm{x}_{\mathrm{i}}$ has the following form, where k is some unknown constant.
$\mathrm{P}\left(\mathrm{X}=\mathrm{x}_{\mathrm{i}}\right)= \begin{cases}0.2, & \text { if } x_{i}=0 \\ k x_{i}, & \text { if } x_{i}=1 \text { or } 2 \\ k\left(5-x_{i}\right), & \text { if } x_{i}=3 \\ 0, & \text { otherwise }\end{cases}$
a. Find the value of $k$.
b. What is the probability that the person watches two hours of television on a selected day?
c. What is the probability that the person watches at least two hours of television on a selected day?
d. What is the probability that the person watches at most two hours of television on a selected day?
e. Calculate mathematical expectation.
f. Find the variance and standard deviation of random variable X.

Answer

From the given information, we find that the probability distribution of X is

X	0	1	2	3
P(X)	0.2	k	2k	2k

a.We know that $\sum p_{i}=1$
$
\begin{aligned}
& \Rightarrow 0.2+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}=1 \\
& \Rightarrow 5 \mathrm{k}=0.8 \Rightarrow \mathrm{k}=\frac{4}{25}
\end{aligned}
$
b. Probability that the person watches two hours of television
$=\mathrm{P}(\mathrm{X}=2)=2 \mathrm{k}=2 \times \frac{4}{25}=\frac{8}{25}$
c. P (the person watches at least two hours of television)
$=\mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)$
$=2 \mathrm{k}+2 \mathrm{k}=4 \mathrm{k}$
$=4 \times \frac{4}{25}=\frac{16}{25}$
d. P (the person watches at most two hours of television)
$=\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$
$=0.2+\mathrm{k}+2 \mathrm{k}$
$=0.2+3 \mathrm{k}=\frac{1}{5}+\frac{12}{25}=\frac{17}{25}$
e. We construct the following table:

$x _{ i }$	$p _{ i }$	$p _{ i } x _{ i }$	$P _{ i } x _{ i }{ }^2$
0	0.2	0	0
2	$\frac{4}{25}$	$\frac{4}{25}$	$\frac{4}{25}$
1	$\frac{8}{25}$	$\frac{16}{25}$	$\frac{32}{25}$
3	$\frac{8}{25}$	$\frac{24}{25}$	$\frac{72}{25}$
Total		$\frac{44}{25}$	$\frac{108}{25}$

$
\mathrm{E}(\mathrm{X})=\Sigma p_{i} x_{i}=\frac{44}{25}=1.76
$
f. Variance $\sigma^{2}=\Sigma p_{i} x_{i}^{2}-\left(\Sigma p_{i} x_{i}\right)^{2}$
$=\frac{108}{25}-\left(\frac{44}{25}\right)^{2}=\frac{108}{25}-\frac{1936}{625}=\frac{764}{625}=1.22$
and standard deviation $\sigma=\sqrt{\text { Variance }}=\sqrt{1.22}=1.1$

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Question 35 Marks

A die is tossed twice. Success is defined as getting an odd number on a random toss. Find the mean and variance of the number of successes.

Answer

Let $x$ be the random variable denoting the number of times an odd number (the number of successes) when a die is tossed twice. Then $x$ takes the values $0,1,2$
Let $\mathrm{P}(\mathrm{X}=0)$ be probability of getting no odd number (both times showing even).
$\therefore \mathrm{P}(\mathrm{X}=0)=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$
Let $\mathrm{P}(\mathrm{X}=1)$ be probability of getting odd number once.
$\therefore \mathrm{P}(\mathrm{X}=1)={ }^{2} \mathrm{C}_{1} \frac{3}{6} \times \frac{3}{6}=\frac{6}{6} \times \frac{3}{6}=\frac{1}{2}$
Let $\mathrm{P}(\mathrm{X}=2)$ be probability of getting odd number twice.
$\therefore \mathrm{P}(\mathrm{X}=2)=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$
Thus the probability distribution of X is given by
$X=x: x=0, x=1, x=2$
$\mathrm{P}(\mathrm{X}=\mathrm{x}) \frac{1}{4} \frac{1}{2} \frac{1}{4}$
We know that mean $\mathrm{E}(\mathrm{X})=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=0 \times \frac{1}{4}+1 \times \frac{1}{2}+2 \times \frac{1}{4}$
$\therefore \mathrm{E}(\mathrm{X})=0+\frac{1}{2}+\frac{1}{2}=1$
Thus mean $\mathrm{E}(\mathrm{X})=1$
We know that $\operatorname{var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}$
$\mathrm{E}\left(\mathrm{X}^{2}\right)=\sum x_{i}^{2} \mathrm{p}_{\mathrm{i}}=0 \times \frac{1}{4}+1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{4}$
$\therefore \mathrm{E}\left(\mathrm{X}^{2}\right)=0+\frac{1}{2}+4 \times \frac{1}{4}=\frac{3}{2}$
Thus $\operatorname{var}(\mathrm{X})=\frac{3}{2}-[1]^{2}=\frac{3}{2}-1=\frac{1}{2}$
Hence mean is 1 and variance is $\frac{1}{2}$

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Question 45 Marks

Solve the system of inequations graphically: $2 x+y \geq 8, x+2 y \geq 8, x+y \leq 6$

Answer

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality. You can choose any value but find the two mandatory values which are at $\mathrm{x}=0$ and $\mathrm{y}=0$, i.e., x and y -intercepts always.
$2 x+y \geq 8$

x	0	2	4
y	8	4	0

$x+2 y \geq 8$

x	0	4	8
y	4	2	0

$x+y \leq 6$

x	0	3	6
y	6	3	0

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Question 55 Marks

A small manufacturer has employed 5 skilled men and 10 semi-skilled men and makes an article in two qualities deluxe model and an ordinary model. The making of a deluxe model requires 2 hrs. work by a skilled man and 2 hrs. work by a semi-skilled man. The ordinary model requires 1 hr by a skilled man and 3 hrs. by a semi-skilled man. By union rules no man may work more than 8 hrs per day. The manufacturers clear profit on deluxe model is ₹15 and on an ordinary model is ₹10. How many of each type should be made in order to maximize his total daily profit.

Answer

Let x articles of deluxe model and y articles of an ordinary model be made
Number of articles cannot be negative
Therefore, $\mathrm{x}, \mathrm{y} \geq 0$
According to the question, the making of a deluxe model requires 2 hrs. work by a skilled man and the ordinary model requires 1 hr by a skilled man
$2 x+y \leq 40$
The making of a deluxe model requires 2 hrs. work by a semi-skilled man ordinary model requires 3 hrs. work by a semi-skilled man
$2 x+3 y \leq 80$
Total profit $=\mathrm{Z}=15 \mathrm{x}+10 \mathrm{y}$ which is to be maximised
Thus, the mathematical formulation of the given linear programming problem is
Max $Z=15 x+10 y$
subject to
$2 \mathrm{x}+\mathrm{y} \leq 40$
$2 x+3 y \leq 80$
$x \geq 0$
$\mathrm{y} \geq 0$
The feasible region determined by the system of constraints is

The corner points are $\mathrm{A}\left(0, \frac{80}{3} \mathrm{~B}(10,20), \mathrm{C}(20,0)\right.$
The values of Z at these corner points are as follows

Corner Point	$Z=15 x+10 y$
A	$\frac{800}{3}$
B	350
C	300

The maximum value of Z is 300 which is attained at $\mathrm{C}(20,0)$
Thus, the maximum profit is ₹300 obtained when 10units of deluxe model and 20 unit of ordinary model is produced.

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Question 65 Marks

A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food $\mathbf{I}$ contains 2 units$/ \mathrm{kg}$ of vitamin A and 1 unit/kg of vitamin C while food $\mathbf{l l}$ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ₹5.00 per kg to purchase food I and ₹7.00 per kg to produce food II. Determine the minimum cost to such a mixture. Formulate the above as an LPP and solve it.

Answer

Let the dietician mix x kg of food $\mathbf{I}$ with y kg of food II. Then, the mathematical model of the LPP is as follows:
Minimize $Z=5 x+7 y$
Subject to $2 x+y \geq 8$
$x+2 y \geq 10$
and, $\mathrm{x}, \mathrm{y} \geq 0$
To solve this LPP graphically, we first convert the inequations into equations to obtain the following lines.
$2 \mathrm{x}+\mathrm{y}=8, \mathrm{x}+2 \mathrm{y}=10, \mathrm{x}=0, \mathrm{y}=0$
The line $2 x+y=8$ meets the coordinate axes at $A_{1}(4,0)$ and $B_{1}(0,8)$. Join these points to obtain the line represented by $2 x+y=$ 8. The region not containing the origin is represented by $2 x+y \geq 8$.
The line $x+2 y=10$ meets the coordinate axes at $A_{2}(10,0)$ and $B_{2}(0,5)$. Join these points to obtain the line represented by $x+2 y$ $=10$. Clearly, $\mathrm{O}(0,0)$ does not satisfy the inequation $x+2 y \geq 10$. So, the region not containing the origin is represented by this inequation.
Clearly, $\mathrm{x} \geq 0$ and, $\mathrm{y} \geq 0$ represent the first quadrant.
Thus, the shaded region in figure is the feasible region of the LPP. The coordinates of the comer-points of this region are $\mathrm{A}_{2}(10$, $0), \mathrm{P}(2,4)$ and $\mathrm{B}_{1}(0,8)$.

The point $P(2,4)$ is obtained by solving $2 x+y=8$ and $x+2 y=10$ simultaneously.
The values of the objective function $Z=5 x+7 y$ at the corner points of the feasible region are given in the following table:

Point (x, y)	Value of the objective function $\mathbf{Z}=\mathbf{5 x}+\mathbf{7 y}$
$\mathrm{A}_{2}(10,0)$	$\mathrm{Z}=5 \times 10+7 \times 0=50$
$\mathrm{P}(2,4)$	$\mathrm{Z}=5 \times 2+7 \times 4=38$
$\mathrm{~B}_{1}(0,8)$	$\mathrm{Z}=5 \times 0+7 \times 8=56$

Clearly, Z is minimum at $\mathrm{x}=2$ and $\mathrm{y}=4$. The minimum value of Z is 38 . We observe that open half-plane represented by $5 \mathrm{x}+7 \mathrm{y}$ $<38$ does not have points in common with the feasible region. So, Z has a minimum value equal to 38 at $\mathrm{x}=2$ and $\mathrm{y}=4$. Hence, the optimal mixing strategy for the dietician will be to mix 2 kg of food $\mathbf{I}$ and 4 kg of food II. In this case, his cost will be minimum and the minimum cost will be ₹ 38.00

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Applied Maths 5 Marks Questions

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