5 Marks Questions

Question 15 Marks

A start-up company invested ₹ 3,00,000 in shares for 5 years. The value of this investment was ₹ 3,50,000 at the end of second year, ₹ 3,80,000 at the end of third year and on maturity, the final value stood at ₹ 4,50,000. Calculate the Compound Annual Growth Rate (CAGR) on the investment. [Given that $\left.(1.5)^{\frac{1}{5}}=1.084\right]$

Answer

Given that,
Beginning Value $=$ BV $=300000$
Ending Value $=$ EV $=450000$
Number of Years = $n =5$
$
\therefore C A G R=\left(\frac{E V}{B V}\right)^{\frac{1}{n}}-1
$
$
=\left(\frac{450000}{3000000}\right)^{\frac{1}{5}}-1
$
$
=\left(\frac{3}{2}\right)^{\frac{1}{5}}-1
$
$
=(1.5)^{\frac{1}{5}}-1
$
$
=1.084-1
$
$
=0.084
$
CAGR% $=0.084 \times 100=8.4 \%$
Hence the compound Annual Growth Rate (CAGR) on the investment is $8.4 \%$.

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Question 25 Marks

Two cards are drawn successively without replacement from a well-shuffled deck of 52 cards. Compute the variance of the number of aces.

Answer

Let $A _{ i }$ denote the event of getting an ace in $i ^{\text {th }}$ draw, where $i =1,2$.
Further, let X be a random variable denoting the number of aces in two draws. Then, X can take values $0,1,2$. Then, we have, $P ( X =0)=$ Probability of getting no ace in two successive draws
$
\Rightarrow P(X=0)=P\left(\overline{A_1} \cap \overline{A_2}\right)=P\left(\overline{A_1}\right) P\left(\overline{A_2} / \overline{A_1}\right)=\frac{48}{52} \times \frac{47}{51}=\frac{564}{663}
$
$P ( X =1)=$ Probability of getting an ace in one of the two draws
$\Rightarrow P ( X =1)=P\left(\left(A_1 \cap \overline{A_2}\right) \cup\left(\overline{A_1} \cap A_2\right)\right)$
$\Rightarrow P ( X =1)=P\left(A_1 \cap \overline{A_2}\right)+P\left(\overline{A_1} \cap A_2\right)$
$\Rightarrow P ( X =1)=P\left(A_1\right) P\left(\overline{A_2} / A_1\right)+P\left(\overline{A_1}\right) P\left(A_2 / \overline{A_1}\right)=\frac{4}{52} \times \frac{48}{51}+\frac{48}{52} \times \frac{4}{51}=\frac{96}{663}$
$P ( X =2)=$ Probability of getting an ace in each draw
$
\Rightarrow P(X=2)=P\left(A_1 \cap A_2\right)=P\left(A_1\right) P\left(A_2 / A_1\right)=\frac{4}{52} \times \frac{3}{51}=\frac{3}{663}
$
Therefore, the probability distribution of X is as follows:

X	0	1	2
P(X)	$\frac{564}{663}$	$\frac{96}{663}$	$\frac{3}{663}$

$\begin{array}{l}\therefore \Sigma p_i x_i=0 \times \frac{564}{663}+1 \times \frac{96}{663}+2 \times \frac{3}{663}=\frac{102}{663} \\ \text { and, } \Sigma p_i x_i^2=\frac{564}{663} \times 0+\frac{96}{663} \times 1+\frac{3}{663} \times 4=\frac{108}{663} \\ \text { Hence, } \operatorname{Var}( X )=\Sigma p_i x_i^2-\left(\Sigma p_i x_i\right)^2=\frac{108}{663}-\left(\frac{102}{663}\right)^2=\frac{108 \times 663-(102)^2}{(663)^2}=\frac{61200}{663 \times 663}=\frac{400}{2873}\end{array}$

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Question 35 Marks

A class XII has 20 students whose marks (out of 30) are 14, 17, 25, 14, 21, 17, 17, 19, 18, 26, 18, 17, 17, 26, 19, 21, 21, 25, 14 and 19. If random variable X denotes the marks of a selected student given that the probability of each student to be selected is equally likely.
a. Prepare the probability distribution of the random variable X.
b. Find mean, variance and standard deviation of X.

Answer

a. Let us prepare the following frequency table:

Marks obtained	14	17	18	19	21	25	26
No. of students	3	5	2	3	3	2	2

Total number of students $=20$
Given that $X =$ marks of a selected student.
So, $P ( X =14)=\frac{3}{20}$;
$P ( X =17)=\frac{5}{20}=\frac{1}{4}$;
$P(X=18)=\frac{2}{20}=\frac{1}{10}$;
$P(X=19)=\frac{3}{20}$;
$P ( X =21)=\frac{3}{20}$;
$P ( X =25)=\frac{2}{20}=\frac{1}{10}$;
$P(X=26)=\frac{2}{20}=\frac{1}{10}$
Hence, the required probability distribution is

X	14	17	18	19	21	25	26
P(X)	$\frac{3}{20}$	$\frac{1}{4}$	$\frac{1}{10}$	$\frac{3}{20}$	$\frac{3}{20}$	$\frac{1}{10}$	$\frac{1}{10}$

b. To calculate mean, variance and standard deviation, we construct the following table:

x1	pi	$p _{ i } x _{ i }$	$p_i x_i^2$
14	$\frac{3}{20}$	$\frac{42}{20}$	$\frac{147}{5}$
17	$\frac{1}{4}$	$\frac{17}{4}$	$\frac{289}{4}$
18	$\frac{1}{10}$	$\frac{18}{10}$	$\frac{165}{5}$
19	$\frac{3}{20}$	$\frac{57}{20}$	$\frac{1083}{20}$
21	$\frac{3}{20}$	$\frac{63}{20}$	$\frac{1323}{20}$
25	$\frac{1}{10}$	$\frac{25}{10}$	$\frac{125}{2}$
26	$\frac{1}{10}$	$\frac{26}{10}$	$\frac{338}{5}$
Total		$\frac{385}{20}$	$\frac{7689}{20}$

$\therefore$ Mean $\mu=\Sigma p_i x_i=\frac{385}{20}=\frac{77}{4}=19.25$
Variance $\sigma^2=\Sigma p_i x_i^2-\mu^2=\frac{7689}{20}-\left(\frac{77}{4}\right)^2=\frac{7689}{20}-\frac{5929}{16}$
$\begin{array}{l}=\frac{30756-29645}{80}=\frac{1111}{80}=13.89 \\
\text { Standard deviation } \sigma=\sqrt{\text { Variance }}=\sqrt{13.89}=3.73\end{array}$

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Question 45 Marks

Solve the following system of inequalities graphically:
3y - 2x < 4, x + 3y > 3 and x + y ≤ 5.

Answer

The given inequalities are
$
\begin{array}{l}
3 y-2 x<4 \ldots \text { (i) } \\
x+3 y>3 \ldots \text { (ii) } \\
\text { and } x+y \leq 5 \ldots \text { (iii) }
\end{array}
$
To draw the graph of $3 y-2 x<4$ :
We draw the straight line $3 y-2 x=4$ which passes through the points $(-2,0)$ and $\left(0, \frac{4}{3}\right)$. The line divides the plane into two parts. Further, as $O(0,0)$ satisfies the inequality $3 y-2 x<4$.
$(\because 3 \times 0-2 \times 0=0<4)$, therefore, the graph consists of that part of the plane divided by the line $3 y-2 x=4$ which contains the origin.
Similarly, draw the graphs of other two inequalities $x+3 y>3$ and $x+y \leq 5$.

Shade the common part of the graphs of all the three given inequalities (i), (ii) and (iii).
The solution set consists of all the points in the shaded part of the coordinate plane shown in fig. The points on the line segment BC are included in the solution.

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Question 55 Marks

Solve the following LPP graphically:
$
\text { Minimize } Z=3 x+5 y
$
Subject to
$
\begin{array}{l}
-2 x+y \leq 4 \\
x+y \geq 3 \\
x-2 y \leq 2 \\
x, y \geq 0
\end{array}
$

Answer

Converting the inequations into equations, we obtain the lines -2x + y = 4, x + y = 3, x - 2y = 2, x = 0 and y = 0. These lines are drawn on a suitable scale and the feasible region of the LPP is shaded in Figure.

Now, give a value, say 15 equal to (1 c.m. of 3 and 5) to $Z$ to obtain the line $3 x+5 y=15$. This line meets the coordinate axes at $P_1(5,0)$ and $Q_1(0,3)$. Join these points by a dotted line. Move this line parallel to itself in the decreasing direction towards the origin so that it passes through only one point of the feasible region. Clearly, $P _3 Q _3$ is such a line passing through the vertex P of the feasible region. The coordinates of $P$ are obtained by solving the lines $x-2 y=2$ and $x+y=3$. Solving these equations, we get $x=\frac{8}{3}$ and $y=\frac{1}{3}$. Putting $x=\frac{8}{3}$ and $y=\frac{1}{3}$ in $Z=3 x+5 y$, we get
$
Z=3 \times \frac{8}{3}+5 \times \frac{1}{3}=\frac{29}{3}
$
Hence, the minimum value of Z is $\frac{29}{3}$ at $x =\frac{8}{3}, y =\frac{1}{3}$

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Question 65 Marks

Solve the following LPP graphically:
Maximize $Z=5 x+3 y$
Subject to $3 x+5 y \leq 15$
$5 x+2 y \leq 10$
and, $x, y \geq 0$

Answer

Converting the given inequations into equations, we obtain the following equations:
$3 x+5 y=15,5 x+2 y=10, x=0$ and $y=0$
Region represented by $3 x+5 y \leq 15$: The line $3 x+5 y=15$ meets the coordinate axes at $A_1(5,0)$ and $B_1(0,3)$ respectively. Join these points to obtain the line $3 x+5 y=15$. Clearly, $(0,0)$ satisfies the inequation $3 x+5 y \leq 15$. So, the region containing the origin represents the solution set of the inequation $3 x+5 y<15$.
Region Represented by $5 x+2 y \leq 10$: The line $5 x+2 y=10$ meets the coordinate axes at $A_2(2,0)$ and $B_2(0,5)$ respectively. Join these points to obtain the graph of the line $5 x+2 y=10$. Clearly, $(0,0)$ satisfies the inequation $5 x+2 y \leq 10$. So, the region containing the origin represents the solution set of this inequation.
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 shaded region $OA _2 PB_1$ in figure represents the common region of the above inequations. This region is the feasible region of the given LPP.
The coordinates of the vertices (comer-points) of the shaded feasible region are $O (0,0), A _2(2,0), P \left(\frac{20}{19}, \frac{45}{19}\right)$ and $B _1(0,3)$.

These points have been obtained by solving the equations of the corresponding intersecting lines, simultaneously. The values of the objective function at these points are given in the following table:

Point (x,y)	Value of the objective function Z=5x+3y
O(0,0)	$Z=5 \times 0+3 \times 0=0$
$A _2(2,0)$	$Z=5 \times 2+3 \times 0=10$
$P\left(\frac{20}{19}, \frac{45}{19}\right)$	$Z=5 \times \frac{20}{19}+3 \times \frac{45}{19}=\frac{235}{19}$
$B _1(0,3)$	$Z=5 \times 0+3 \times 3=9$

Clearly, Z is maximum at $P \left(\frac{20}{19}, \frac{45}{19}\right)$. Hence, $x =\frac{20}{19}, y =\frac{45}{19}$ is the optimal solution of the given LPP and the optimal value of Z is $\frac{235}{19}$.

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Applied Maths 5 Marks Questions

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