5 Marks Questions

Question 15 Marks

The cost of a car purchased 2 years ago, depreciates at the rate of 20 % every year. If its present worth is ₹315600, find:
i. its purchase price
ii. its value after 3 years.

Answer

It is given that,
Present value of car $=315600$
Rate of depreciation $( r )=20 \%$
i. We know that
Value of car 2 years ago $= A \div\left(1-\frac{r}{100}\right)^n$
Substituting the values
$
=315600 \div\left(1-\frac{20}{100}\right)^2
$
By further calculation
$
\begin{array}{l}
=315600 \times \frac{5}{4} \times \frac{5}{4} \\
=493125
\end{array}
$
ii. We know that
Value of car after 3 years $=315600 \times\left(1-\frac{20}{100}\right)^3$
By further calculation
$
\begin{array}{l}
=315600 \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \\
=161587.20
\end{array}
$

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Question 25 Marks

Two positive numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and the variance of the distribution.

Answer

The number of ways of selecting two numbers from the first five positive integers $={ }^5 C_2=10$.
So, the sample space $S$ of the random experiment has 10 equally likely outcomes.
The outcomes are:
$
1,2 ; 1,3 ; 1,4 ; 1,5 ; 2,3 ; 2,4 ; 2,5 ; 3,4 ; 3,5 ; 4,5 .
$
As the random variable $X$ denote the larger of the two numbers. $X$ can take values $2,3,4,5$. Note that in a sample space $S$, we have

Larger than any number	Number of outcomes
2	1
3	2
4	3
5	4

$
P(X=2)=\frac{1}{10}, P(X=3)=\frac{2}{10}, P(X=4)=\frac{3}{10}, P(X=5)=\frac{4}{10}
$
$\therefore$ Probability distribution of X is

X	2	3	4	5
P(X)	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

$\begin{array}{l}\text { Mean }=\Sigma p_i x_i=\frac{1}{10} \times 2+\frac{2}{10} \times 3+\frac{3}{10} \times 4+\frac{4}{10} \times 5=4 \\ \text { Variance }=\Sigma p_i x_i^2-\left(\Sigma p_i x_i\right)^2 \\ =\frac{1}{10}\left(1 \times 2^2+2 \times 3^2+3 \times 4^2+4 \times 5^2\right)-(4)^2 \\ =\frac{170}{10}-16=17-16=1\end{array}$

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Question 35 Marks

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Compute the
variance of the number of aces.

Answer

Let $A _{ i }( i =1,2)$ denote the event of getting an ace in ith draw. Since the cards are drawn with replacement. Therefore, $P \left( A _{ i }\right)=$ Probability of getting an ace in $i ^{\text {th }}$ draw $=\frac{{ }^4 C_1}{{ }^{52} C_1}=\frac{4}{52}=\frac{1}{13}$ and $P\left(\overline{A_i}\right)=1- P \left( A _{ i }\right)=1-\frac{1}{13}=\frac{12}{13}, i =1,2$
Let X denote the number of aces in two draws. Then, X can take values $0,1,2$.
Now, $P ( X =0)=$ Probability of getting no ace in two draws
$
\begin{array}{l}
\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}\right)=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169} \\
\Rightarrow P(X=1)=\text { Probability of getting an ace in either 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}\right)+P\left(\overline{A_1}\right) P\left(A_2\right)=\frac{1}{13} \times \frac{1}{13}=\frac{1}{169}
\end{array}
$
Thus, the probability distribution of X is given by:

X	0	1	2
P(X)	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

$
\begin{array}{l}
\therefore \Sigma p_i x_i=0 \times \frac{144}{169}+1 \times \frac{24}{169}+2 \times \frac{1}{169}=\frac{26}{169} \\
\text { and, } \Sigma p_i x_i^2=0 \times \frac{144}{169}+1 \times \frac{24}{169}+4 \times \frac{1}{169}=\frac{28}{169} \\
\text { Hence, } \bar{X}=\operatorname{Mean}=\Sigma p_i x_i=\frac{26}{169}=\frac{2}{13} \\
\text { and, } \operatorname{Var}(X)=\sum p_i x_i^2-\left(\Sigma p_i x_i\right)^2=\frac{28}{169}-\left(\frac{2}{13}\right)^2=\frac{24}{169} \\
\therefore \text { S.D. }=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{24}{169}}=\frac{2 \sqrt{6}}{13}
\end{array}
$
Hence, Mean $=\frac{2}{13}$ and S.D. $=\frac{2 \sqrt{6}}{13}$

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Question 45 Marks

A person can row a boat at 5 km/h in still water. It takes him thrice as long to row upstream as to row
downstream. Find the rate at which the stream is flowing.

Answer

Let the distance covered be d km . and y be speed of stream speed of boat $=5 km / h$
speed of stream $=y km / h$
speed of boat in upstram(u): $x-y km / h$
$
=5-ykm / h
$
speed of boat in downstream (v) $=x+y km / h$
$
=5+y km / h
$
ATQ.
$
\begin{array}{l}
\frac{d}{5-y}=3\left(\frac{d}{5+y}\right)\left[\because T=\frac{D}{S}\right] \\
\frac{1}{5-y}=\frac{3}{5+y} \\
5+y=3(5-y) \\
5+y=15-3 y \\
y+3 y=15-5 \\
4 y=10 \\
y=\frac{10}{4} \\
y=\frac{5}{2} km / h \\
y=2 \frac{1}{2} km / h
\end{array}
$
speed of stream is $2.5 km / h$

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Question 55 Marks

If $A=\left[\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]$, find $A^{-1}$ Using $A^{-1}$ solve the system of linear equations $x-2 y=10,2 x-y-z=8,-2 y+z=7$.

Answer

$\begin{array}{l}\text { Here, }| A |=\left|\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right|=1(-1-2)-2(-2-0) \\ =-3+4=1 \neq 0 \\ \Rightarrow A^{-1} \text { exists and } A ^{-1}=\frac{1}{|A|} \text { adj } A \\ \text { Here, } A _{11}=\left|\begin{array}{rr}-1 & -2 \\ -1 & 1\end{array}\right|=-3, A_{12}=-\left|\begin{array}{rr}-2 & -2 \\ 0 & 1\end{array}\right|=2, A_{13}=\left|\begin{array}{rr}-2 & -1 \\ 0 & -1\end{array}\right|=2 ; \\ A _{21}=-\left|\begin{array}{rr}2 & 0 \\ -1 & 1\end{array}\right|=-2, A_{22}=\left|\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right|=1, A_{23}=-\left|\begin{array}{rr}1 & 2 \\ 0 & -1\end{array}\right|=1 ; \\ A _{31}=\left|\begin{array}{rr}2 & 0 \\ -1 & -2\end{array}\right|=-4, A_{32}=-\left|\begin{array}{rr}1 & 0 \\ -2 & -2\end{array}\right|=2, A_{33}=\left|\begin{array}{rr}1 & 2 \\ -2 & -1\end{array}\right|=3 \\ \left.\therefore \text { adj } A =\left\lvert\, \begin{array}{rrr}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right.\right]=\left[\begin{array}{rr}-3 & -2 \\ 2 & 1 \\ 2 & -4 \\ 2 & 1 \\ 3\end{array}\right] \\ \therefore A ^{-1}=\frac{1}{|A|} \text { adj } A =\frac{1}{1}\left[\begin{array}{rrr}2 & 1 & 2 \\ 2 & -2 & -4 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{rrr}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right]\end{array}$
The given system of equations can be written as
$
\left[\begin{array}{rrr}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]
$
or $A ^{\prime} X = B$, where $X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{r}10 \\ 8 \\ 7\end{array}\right]$
Now $\left|A^{\prime}\right|=|A|=1 \neq 0 \Rightarrow\left(A^{\prime}\right)^{-1}$ exists
$\Rightarrow$ the given system has a unique solution $X=\left(A^{\prime}\right)^{-1} B$
$
\begin{array}{l}
\text { or } X=\left(A^{-1}\right)^{\prime} B\left(\because\left(A^{\prime}\right)-1=\left(A^{-1}\right)^{\prime}\right) \\
\Rightarrow X=\left[\begin{array}{rrr}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]^{\prime}\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]\left[\begin{array}{r}
10 \\
8 \\
7
\end{array}\right] \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
-30+16+14 \\
-20+8+7 \\
-40+16+21
\end{array}\right]=\left[\begin{array}{r}
0 \\
-5 \\
-3
\end{array}\right] \Rightarrow x=0, y=-5, z=-3
\end{array}
$
Hence, the solution of the given system of equations is $x=0, y=-5, z=-3$.

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Question 65 Marks

A firm produces three products $P _1, P _2$ and $P _3$ requiring the mix-up of three materials $M _1, M _2$ and $M _3$. The perunit requirement of each product for each material is as follows:

	$M _1$	$M _2$	$M _3$
$P _1$	2	4	5
$P _2$	3	2	4
$P_3$	1	3	2

Using matrix algebra, find:
i. The total requirement of each material if the firm produces 100,200 and 300 units of products $P_1, P_2$ and $P_3$ respectively.
ii. The per-unit cost of production of each product if the per-unit costs of materials $M _1, M _2$ and $M _3$ are ₹ 10 , ₹ 15 and ₹ 12 respectively.

Answer

The matrix showing per unit requirement of materials $M_1, M_2$ and $M_3$ in producing three products $P_1, P_2$ and $P_3$ is
$
A=\begin{array}{l}
\\
M_1 \\
M_2 \\
M_3
\end{array}\left[\begin{array}{lll}
P_2 & P_3 & P_3 \\
4 & 3 & 1 \\
4 & 2 & 3 \\
5 & 4 & 2
\end{array}\right]
$
i. (i) The matrix representing the requirements of products $P _1, P _2$ and $P _3$ is
$
B=\begin{array}{r}
P_1 \\
P_2 \\
P_3
\end{array}\left[\begin{array}{l}
100 \\
200 \\
300
\end{array}\right]
$
So, the requirements of each material for producing the given quantities of three products is given by the product
$
AB=\begin{array}{l}
\\
\\
\\
M_1 \\
M_2
\end{array}\left[\begin{array}{lll}
2 & P_3 & P_3 P_2
\end{array}\right] \begin{array}{r}
P_1 \\
M_3
\end{array}\left[\begin{array}{lll}
100 \\
200 \\
M_3 & 2 & 3 \\
500
\end{array}\right]
$
$
\Rightarrow AB=\begin{array}{l}
M_1 \\
M_2 \\
M_3
\end{array}\left[\begin{array}{l}
200+600+300 \\
400+400+900 \\
500+800+600
\end{array}\right]=\begin{array}{c}
M_1 \\
M_2 \\
M_3
\end{array}\left[\begin{array}{l}
1100 \\
1700 \\
1900
\end{array}\right]
$
Thus, 1100 units of material $M _1, 1700$ emits of material $M _2$ and 1900 units of material $M _3$ are required to produce 100 units of $P_1, 200$ units of $P_2$ and 300 units of $P_3$.
ii. The matrix representing per-unit costs of materials $M_1, M_2$ and $M_3$ is as given below:
$
C=\begin{array}{l}
M_1 \\
M_2 \\
M_3
\end{array}\left[\begin{array}{l}
10 \\
15 \\
12
\end{array}\right]
$
The matrix exhibiting the materials $M _1, M _2$ and $M _3$ in three products $P _1, P _2$ and $P _3$ is
$
\begin{array}{l}
D= M_1 M_2 M_3 \\
P_1 \\
P_2 \\
P_3
\end{array}\left[\begin{array}{lll}
2 & 4 & 5 \\
3 & 2 & 4 \\
1 & 3 & 2
\end{array}\right], ~ l
$
So, the per emit cost of each product is given by the matrix product
$
\begin{aligned}
DC= & \begin{array}{lll}
M_1 & M_2 & M_3
\end{array} M_2
\end{aligned}\left[\begin{array}{l}
10 \\
15 \\
12
\end{array}\right]=\begin{array}{l}
P_1 \\
P_2 \\
P_3
\end{array}\left[\begin{array}{c}
20+60+60 \\
30+30+48 \\
10+45+24
\end{array}\right]=\begin{array}{l}
P_1 \\
P_2 \\
P_3
\end{array}\left[\begin{array}{c}
140 \\
108 \\
79
\end{array}\right]
$
Hence, per unit cost of production of products P₁, P₂ and P₃ are ₹ 140, ₹ 180 and ₹ 79 respectively.
iii. The total cost of product is given by the matrix product
$\left.\begin{array}{cccc}P_1 & P_2 & P_3 & P_1 \\ {[100} & 200 & 300\end{array}\right] \begin{array}{c}P_2 \\ P_3\end{array}\left[\begin{array}{c}140 \\ 108 \\ 79\end{array}\right]=(14,000+21,600+23,700)=$ ₹ 59,300
Hence, the total cost of product is ₹ 59,300

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Applied Maths 5 Marks Questions

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