5 Marks Questions

Question 15 Marks

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.
a. If the interest is compounded continuously at 5% p.a., in how many years will ₹ 100 double?
b. At what interest rate will ₹ 100 double itself in 10 years?
[Given: loge 2 = 0.6931]

Answer

a. We know the formula, We know the formula, $\frac{d P}{d t}=\frac{P}{100}$.
$
\begin{array}{l}
\Rightarrow \int \frac{d P}{P}=\int \frac{r}{100} d t \\
\Rightarrow \log P=\frac{r t}{100}+c \\
\text { Let, } t=0, P=P_0 \\
\Rightarrow \log P_0=0+c \Rightarrow c=\log P_0
\end{array}
$
Therefore the equation becomes: $\log P =\frac{r t}{100}+\log P _0$
$
\begin{array}{l}
\Rightarrow \log P-\log P_0=\frac{r t}{100} \\
\Rightarrow \log \frac{P}{P_0}=\frac{r t}{100}
\end{array}
$
According to given, $P _0=100, P =2 P _0=200 \& r =5$
$
\begin{array}{l}
\Rightarrow \log \frac{200}{100}=\frac{5 t}{100} \\
\Rightarrow 20 \times \log 2=t
\end{array}
$
Now, $\log 2=0.6931$
$
t=20 \times 0.6931=13.8 \text { years }
$
b. Let principal $=p$
Given $p$ increases at the rate $r \%$ per year
$
\therefore \frac{d p}{d e}=r \% \times p \Rightarrow \frac{d p}{p}=\frac{r}{100} dt
$
integrating both side
$
\begin{array}{l}
\int \frac{d p}{p}=\frac{r}{100} \int dt \Rightarrow \log p=\frac{r t}{100}+\log c \\
\Rightarrow \log p=\log c=\frac{r t}{100} \\
\Rightarrow \log \frac{p}{c}=\frac{r t}{100} \Rightarrow \frac{p}{c}=e \frac{r t}{100} \ldots(i) \\
\text { put } t=0, p=100 \text { in (i) } \\
\frac{100}{c}=e \frac{r \times o}{c}
\end{array}
$
$
\therefore c=100
$
Now keep value of c in eq (i)
$
\frac{p}{100}=e \frac{r t}{100}
$
Now put $t =10, p =200$ in eq(given ₹ 100 double in)
$
\begin{array}{l}
\frac{200}{100}=e \frac{10 r}{100} \\
z=e^{\frac{r}{10}}
\end{array}
$
$
\log 2=\frac{r}{10} \Rightarrow 0.6931=\frac{r}{10}
$
$\therefore$ Rate of interest $r =6.931 \%$

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Question 25 Marks

A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹ 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Answer

i. When no head and 4 tails appear. Let A be the event money lost = = ₹ (4 X 1.50) = ₹ 6.00.
There is only one way of getting no head and 4 tails i.e., (TTTT) $\Rightarrow n(A)=1$ $n(S)=16$, since there are 16 possible outcomes
$
\therefore P(A)=\frac{n(A)}{n(S)}=\frac{1}{16}
$
ii. Let B be the event when 1 head and 3 tails appear.
$
\begin{array}{l}
\therefore B=\{\text { HTTT, THTT, TTHT, TTTH }\} \\
\Rightarrow n(B)=4
\end{array}
$
Money lost = ₹ (3 X 1.50 - 1 X1) = ₹ 3.50
$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{4}{16}=\frac{1}{4}$
iii. Let C be the event that 2 head and 2 tail appear.
Money lost = ₹ (2X 1.50 - 2X 1)
= ₹ 1
C $=\{$ HHTT, HTHT, HTTH, THHT, THTH, TTHH $\}$
$
\begin{array}{l}
\Rightarrow n(C)=6 \\
\therefore P(C)=\frac{n(C)}{n(S)}=\frac{6}{16}=\frac{3}{8}
\end{array}
$
iv. Let D be the event that 3 head and 1 tail appear.
$
\begin{array}{l}
\therefore D=\{H H H T, \text { HHHT, THHH, HHTH }\} \\
\Rightarrow n(D)=4
\end{array}
$
Money gained = ₹ (3 X1 - 1X 1.5) = ₹ 1.50
$P ( E )=\frac{n(D)}{n(S)}=\frac{4}{16}=\frac{1}{4}$
v. Let E be the event that all heads appear.
$
\therefore E=\{H H H H\} \Rightarrow n(E)=1
$
Money gained = ₹ (4X 1) = ₹ 4
Also, $P ( E )=\frac{n(E)}{n(S)}=\frac{1}{16}$.

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Question 35 Marks

A die is tossed twice. A success is getting an odd number on a random toss. Find the variance of the number of successes.

Answer

Let X be a random variable denoting the number of successes in two tosses of a die. Then, X can take values $0,1,2$.
Let $S_i$ and $F_i$ denote the success and failure respectively in $i ^{\text {th }}$ toss. Then, we have,
$P \left( S _{ i }\right)=$ Probability of getting an odd number in $i ^{ th }$ toss $=\frac{3}{6}=\frac{1}{2}$
and $P \left( F _{ i }\right)=$ Probability of not getting an odd number in $i ^{\text {th }}$ toss $=\left(1-\frac{1}{2}\right)=\frac{1}{2}$
Now, $P ( X =0)=$ Probability of getting no success in two tosses of a die
$
\begin{array}{l}
\Rightarrow P(X=0)=P\left(F_1 \cap F_2\right) \\
\Rightarrow P(X=0)=P\left(F_1\right) P\left(F_2\right) \text { [by Multiplication Theorem] } \\
\Rightarrow P(X=0)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\left[\because P\left(F_1\right)=P\left(F_2\right)=\frac{1}{2}\right]
\end{array}
$
$P ( X =1)=$ Probability of getting one success in two tosses of a die
$
\begin{array}{l}
\Rightarrow P(X=1)=P\left(\left(S_1 \cap F_2\right) \cup\left(F_1 \cap S_2\right)\right) \\
\Rightarrow P(X=1)=P\left(S_1 \cap F_2\right)+P\left(F_1 \cap S_2\right)=P\left(S_1\right) P\left(F_2\right)+P\left(F_1\right) P\left(S_2\right)=\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}=\frac{1}{2}
\end{array}
$
and, $P ( X =2)=$ Probability of getting two successes in two tosses of a die
$
\Rightarrow P(X=2)=P\left(S_1 \cap S_2\right)=P\left(S_1\right) P\left(S_2\right)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}
$
Therefore, the probability distribution of X i.e. the number of successes in two tosses of a die is as follows:

X	0	1	2
P(X)	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

Computation of variance:

xi	pi = P(X = x_i)	p_ix_i	pix²_i
0	$\frac{1}{4}$	0	0
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{1}{4}$	$\frac{1}{2}$	1
		Sigmap_(i)x_(i)=1	$\Sigma p_i x_i^2=\frac{3}{2}$

Therefore, we have $\Sigma p_i x_i=1$ and $\Sigma p_i x_i^2=\frac{3}{2}$
$
\therefore \operatorname{Var}(X)=\Sigma p_i x_i^2-\left(\Sigma p_i x_i\right)^2=\frac{3}{2}-1=\frac{1}{2}
$

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Question 45 Marks

A company manufactures cassettes and its cost and revenue functions for a week are $C =300+\frac{3}{2} x$ and $R =2 x$ respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold for the company to realize a profit?

Answer

We have been a week's data
Cost of cassette, $C =300+\frac{3}{2} x$
Revenue, $R=2 x$
Where $x =$ number of cassettes produced and sold in a week.
We know that profit is given by, Profit = Revenue - Cost . . (i)
Revenue is the income that a business has from its normal business activities, usually from the sale of goods and services to customers.
A cost is the value of money that has been used up to produce something or deliver a service and hence is not available for use anymore.
And Profit is the gain in the business.
So, it is justified that profit in any business would be measured by the difference in the capital generated by the business and the capital used up in the business.
Profit generated by the company manufacturing cassettes is given by,
Profit $=$ R - C (from (i))
Where, $R =$ Revenue
C = Cost of cassette
Here,
If $R < C$, then
Profit $<0$
$\Rightarrow$ There is a loss.
If $R = C$, then
Profit $=0$
$\Rightarrow$ There is no profit no loss.
If $R>C$, then
Profit $>0$
$\Rightarrow$ There is a profit.
We need to find the number of cassettes sold to make a profit. That is, we need to find $x$.
So, $R > C$ (to realize a profit)
Substituting values of R and C . We get
$
\begin{array}{l}
2 x>300+\frac{3}{2} x \\
\Rightarrow 2 x-\frac{3}{2} x>300 \\
\Rightarrow \frac{4 x-3 x}{2}>300 \\
\Rightarrow \frac{x}{2}>300 \\
\Rightarrow x>300 \times 2 \\
\Rightarrow x>600
\end{array}
$
This means that $x$ must be greater than 600 .
Thus, the company must sell more than 600 cassettes to realize a profit.

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Question 55 Marks

A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of carboard is in stock, and if the profits on the large and small boxes are ₹3 and ₹2 per box, how many of each should be made in order to maximize the total profit?

Answer

Let x large boxes and y small boxes be manufactured.
The number of boxes cannot be negative. Therefore, $x \geq 0, y \geq 0$
The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box and if 60 sq. metre of cardboard is stock.
$
4 x+3 y \leq 60
$
The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes.
$
\begin{array}{l}
x \geq 3 \\
y \geq 2 x
\end{array}
$
If the profits on the large and small boxes are ₹ 3 and ₹ 2 per box. Therefore, profit gained by him on $x$ large boxes and $y$ small boxes is $₹ 3 x$ and $₹ 2 y$ respectively.
$
\text { Total profit }=Z=3 x+2 y
$
The mathematical formulation of the given problem is
$
\begin{array}{l}
\text { Max } Z=3 x+2 y \\
\text { subject to } \\
4 x+3 y \leq 60 \\
x \geq 3 \\
y \geq 2 x \\
x \geq 0, y \geq 0
\end{array}
$
First we will convert inequations into equations as follows:
$
4 x+3 y=60, x=3, y=2 x, x=0 \text { and } y=0
$
The region represented by $4 x+3 y \leq 60$ :
The line $4 x+3 y=60$ meets the coordinate axes at $A(15,0)$ and $B(0,20)$ respectively. By joining these points we obtain the line $4 x+3 y=60$. Clearly $(0,0)$ satisfies the $4 x+3 y=60$. So, the region which contains the origin represents the solution set of the inequation $4 x+3 y \leq 60$.
Region represented by $x \geq 3$ :
The line $x=3$ is the line passes through $(3,0)$ and is parallel to Y-axis. The region to the right of the line $x=3$ will satisfy the inequation
$
x \geq 3
$
Region represented by $y \geq 2 x$ :
The line $y=2 x$ is the line that passes through $(0,0)$. The region above the line $y=2 x$ will satisfy the inequation $y \geq 2 x$. Like if we take an example taking a point $(5,1)$ below the line $y=2 x$. Here, $1<10$ which does not satisfies the inequation $y \geq 2 x$.
Hence, the region above the line $y=2 x$ will satisfy the inequality $y \geq 2 x$.
Region represented by $x \geq 0$ and $y \geq 0$ :
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations $x \geq 0$, and $y \geq 0$.
The feasible region determined by the system of constraints $4 x+3 y \leq 60, x \geq 3, y \geq 2 x, x \geq 0$ and $y \geq 0$ are as follows

The corner points are E (3, 16), D(6, 12) and C(3, 6). The values of Z at the corner points are

Corner points	Z=3x+2y
E	41
D	42
C	21

The maximum value of Z is 42 which is at D(6, 12).
Thus, for a maximum profit is ₹42, 6units of large boxes and 12 units of smaller boxes should be manufactured.

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Question 65 Marks

Maximise Z = 8x + 9y subject to the constraints given below:
$2 x+3 y \leq 6 ; 3 x-2 y \leq 6 ; y \leq 1 ; x, y \geq 0$

Answer

Given constraints are
$
\begin{array}{l}
2 x+3 y \leq 6 \\
3 x-2 y \leq 6 \\
y \leq 1 \\
x, y \geq 0
\end{array}
$
For graph of $2 x +3 y \leq 6$
We draw the graph of $2 x+3 y=6$

X	0	3
Y	2	0

$2 \times 0+3 \times 0<6 \Rightarrow(0,0)$ satisfy the constraints.
Hence, feasible region lie towards origin side of line.
For graph of $3 x-2 y \leq 6$
We draw the graph of $3 x-2 y=6$

X	0	2
Y	-3	0

$
\begin{array}{l}
3 \times 0-2 \times 0 \leq 6 \\
\Rightarrow \text { Origin }(0,0) \text { satisfy } 3 x-2 y=6
\end{array}
$
Hence, feasible region lie towards origin side of line.

For graph of $y<1$
We draw the graph of line $y =1$, which is parallel to x -axis and meet y -axis at 1 .
$0 \leq 1 \Rightarrow$ feasible region lie towards origin side of $y=1$.
Also, $x \geq 0, y \geq 0$ say feasible region is in 1 st quadrant.
Therefore, OABCDO is the required feasible region, having corner point $0(0,0), A (0,1) B \left(\frac{3}{2}, 1\right), C \left(\frac{30}{13}, \frac{6}{13}\right) O (2,0)$.
Here, feasible region is bounded. Now the value of objective function $Z=8 x+9 y$ is obtained as.

Comer Points	Z=8x+9y
O(0,0)	0
A(0,1)	9
$B \left(\frac{3}{2}, 1\right)$	21
$C \left(\frac{30}{13}, \frac{6}{13}\right)$	22.6 larr Maximum
D(2,0)	16

Z is maximum when $x =\frac{30}{13}$ and $y =\frac{6}{13}$

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Applied Maths 5 Marks Questions

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