5 Marks Questions

Question 15 Marks

How much amount has to be deposited at the end of each year for 4 years at 6% p.a compounded annually so as to get a sum of ₹ 80,000. [Use $\left.(1.06)^4=1.262\right]$

Answer

Sum of annuity = ₹ 80,000, Each annuity = ₹ x
$r=6 \%$ p.a. $\Rightarrow i =0.06$
n = 4 years
$\begin{array}{l}\therefore 80000=\frac{x}{0.06}\left[(1+0.06)^4-1\right] \\ \Rightarrow 4800= x \left[(1.06)^4-1\right] \\ \Rightarrow 4800= x (1.262-1) \Rightarrow 4800=0.262 x \end{array}$
$\Rightarrow x=\frac{4800}{0.262}=18320.61$
Amount of each annuity = ₹ 18320.61 ~ ₹ 18321

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Question 25 Marks

Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces drawn. Also find the mean and standard deviation of the distribution.

Answer

Let E be the event 'drawing an ace from a pack of 52 cards'.
Then $p = P ( E )=\frac{4}{52}=\frac{1}{13}$, so $q=1-\frac{1}{13}=\frac{12}{13}$.
As two cards are drawn successively with replacement, events are independent, therefore, it is a problem of binomial distribution.
As 2 cards are drawn, n = 2.
Let X denote the number of aces drawn, then X can take values 0, 1, 2.
$\begin{array}{l} P (0)={ }^2 C_0 q^2=1 \cdot\left(\frac{12}{13}\right)^2=\frac{144}{169}, \\ P (1)={ }^2 C_1 p q=2 \cdot \frac{1}{13} \cdot \frac{12}{13}=\frac{24}{169}, \\ P (2)={ }^2 C_2 p^2=1 \cdot\left(\frac{1}{13}\right)^2=\frac{1}{169},\end{array}$
$\therefore$ Required probability distribution is $\left(\begin{array}{ccc}0 & 1 & 2 \\ \frac{144}{169} & \frac{24}{169} & \frac{1}{169}\end{array}\right)$.
$\begin{array}{l}\text { Mean }=\mu=\Sigma p_i x_i=\frac{1}{169}(144 \times 0+24 \times 1+1 \times 2)=\frac{26}{169}=\frac{2}{13}. \\ \text { Now, } \Sigma p_i x_i^2=\frac{1}{169}\left(144 \times 0^2+24 \times 1^2+1 \times 2^2\right)=\frac{28}{169}. \\ \text { Variance }=\Sigma p_i x_i^2-\mu^2=\frac{28}{169}-\left(\frac{2}{13}\right)^2=\frac{28}{169}-\frac{4}{169}=\frac{24}{169} . \\ \text { Standard deviation }=\sqrt{\text { Variance }}=\sqrt{\frac{24}{169}}=\frac{2}{13} \sqrt{6} .\end{array}$

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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 P(X = 0) be probability of getting no odd number (both times showing even).
$\therefore P(X=0)=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$
Let P(X = 1) be probability of getting odd number once.
$\therefore P ( X =1)={ }^2 C _1 \frac{3}{6} \times \frac{3}{6}=\frac{6}{6} \times \frac{3}{6}=\frac{1}{2}$
Let P(X = 2) be probability of getting odd number twice.
$\therefore P ( X =2)=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$
Thus the probability distribution of X is given by
X = x: x = 0, x = 1, x = 2
$P ( X = x ) \frac{1}{4} \frac{1}{2} \frac{1}{4}$
We know that mean $E ( X )=\sum x _{ i } p _{ i }=0 \times \frac{1}{4}+1 \times \frac{1}{2}+2 \times \frac{1}{4}$
$\therefore E ( X )=0+\frac{1}{2}+\frac{1}{2}=1$
Thus mean E(X) = 1
We know that var(X) $= E \left( X ^2\right)-[ E ( X )]^2$
$\begin{array}{l} E \left( X ^2\right)=\sum x_i^2 p _{ i }=0 \times \frac{1}{4}+1^2 \times \frac{1}{2}+2^2 \times \frac{1}{4} \\ \therefore E \left( X ^2\right)=0+\frac{1}{2}+4 \times \frac{1}{4}=\frac{3}{2}\end{array}$
Thus $\operatorname{var}( X )=\frac{3}{2}-[1]^2=\frac{3}{2}-1=\frac{1}{2}$
Hence mean is 1 and variance is $\frac{1}{2}$

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Question 45 Marks

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

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Question 55 Marks

Following equations are consistent? If consistent, solve the:
$x + y - 2z = 4$
$x - 2y + z = -2$
$5x - 5y + z = -2$

Answer

$\begin{array}{l}\text { Here, } D =\left|\begin{array}{rrr}2 & 1 & -2 \\ 1 & -2 & 1 \\ 5 & -5 & 1\end{array}\right|=2(-2+5)-1(1-5)-2(-5+10)=0 \\ D _1=\left|\begin{array}{rrr}4 & 1 & -2 \\ -2 & -2 & 1 \\ -2 & -5 & 1\end{array}\right|=4(-2+5)-1(-2+2)-2(+10-4)=0 \\ D _2=\left|\begin{array}{lrr}2 & 4 & -2 \\ 1 & -2 & 1 \\ 5 & -2 & 1\end{array}\right|=2(-2+2)-4(1-5)-2(-2+10)=0\end{array}$
and $D_3=\left|\begin{array}{rrr}2 & 1 & 4 \\ 1 & -2 & -2 \\ 5 & -5 & -2\end{array}\right|=2(4-10)-1(-2+10)+4(-5+10)=0$.
Thus, $D = D _1= D _2= D _3=0$, therefore, the given system may or may not be consistent. Let us solve first two equations for x and y in terms of z.
These equations can be written as
2x + y = 4 + 2z
x - 2y = -2 - z
To solve these equations we use Cramer's rule,
$\begin{array}{l} D =\left|\begin{array}{rr}2 & 1 \\ 1 & -2\end{array}\right|=-4-1=-5 \neq 0 \\ D _1=\left|\begin{array}{rr}4+2 z & 1 \\ -2-z & -2\end{array}\right|=-8-4 z +2+ z =-6-3 z \\ D _2=\left|\begin{array}{rr}2 & 4+2 z \\ 1 & -2-z\end{array}\right|=-4-2 z -4-2 z =-8-4 z \end{array}$
By Cramer's rule, $x =\frac{ D _1}{ D }=\frac{6+3 z}{5}, y =\frac{ D _2}{ D }=\frac{8+4 z}{5}$
Let z = k where k is any real number, then we get
$x =\frac{6+3 k}{5}, y =\frac{8+4 k}{5}, z = k$, where $k \in R$.
Note that these values satisfy the third equation i.e. 5x - 5y + z = -2 of the given system. Hence, the given system is consistent and it has infinitely many solutions given by x = $\frac{3}{5}(2+k), y =\frac{4}{5}(2+k), z = k$, where $k \in R$.

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Question 65 Marks

If $A=\left[\begin{array}{rrr}2 & 0 & -3 \\ 1 & 4 & 5\end{array}\right], B=\left[\begin{array}{rr}3 & 1 \\ -1 & 0 \\ 4 & 2\end{array}\right]$ and $C=\left[\begin{array}{rr}4 & 7 \\ 2 & 1 \\ 1 & -1\end{array}\right]$, verify that $A(B+C)=A B+A C$.

Answer

$A ( B + C )= A \left(\left[\begin{array}{rr}3 & 1 \\ -1 & 0 \\ 4 & 2\end{array}\right]+\left[\begin{array}{rr}4 & 7 \\ 2 & 1 \\ 1 & -1\end{array}\right]\right)= A \left[\begin{array}{rr}3+4 & 1+7 \\ -1+2 & 0+1 \\ 4+1 & 2+(-1)\end{array}\right]=\left[\begin{array}{rrr}2 & 0 & -3 \\ 1 & 4 & 5\end{array}\right]\left[\begin{array}{ll}7 & 8 \\ 1 & 1 \\ 5 & 1\end{array}\right]$
$=\left[\begin{array}{cc}2.7+0.1+(-3) .5 & 2.8+0.1+(-3) .1 \\ 1.7+4.1+5.5 & 1.8+4.1+5.1\end{array}\right]=\left[\begin{array}{cc}-1 & 13 \\ 36 & 17\end{array}\right] \ldots\text{(i)}$
$AB =\left[\begin{array}{rrr}2 & 0 & -3 \\ 1 & 4 & 5\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 0 \\ 4 & 2\end{array}\right]$
$=\left[\begin{array}{cc}2.3+0 \cdot(-1)+(-3) \cdot 4 & 2.1+0.0+(-3) \cdot 2 \\ 1.3+4 \cdot(-1)+5.4 & 1.1+4.0+5.2\end{array}\right]=\left[\begin{array}{rr}-6 & -4 \\ 19 & 11\end{array}\right]$
and $A C=\left[\begin{array}{rrr}2 & 0 & -3 \\ 1 & 4 & 5\end{array}\right]\left[\begin{array}{rr}4 & 7 \\ 2 & 1 \\ 1 & -1\end{array}\right]$
$=\left[\begin{array}{cc}2.4+0.2+(-3) \cdot 1 & 2.7+0.1+(-3) \cdot(-1) \\ 1.4+4.2+5.1 & 1.7+4.1+5 \cdot(-1)\end{array}\right]=\left[\begin{array}{rr}5 & 17 \\ 17 & 6\end{array}\right]$
$\therefore A B+A C=\left[\begin{array}{rr}-6 & -4 \\ 19 & 11\end{array}\right]+\left[\begin{array}{rr}5 & 17 \\ 17 & 6\end{array}\right]$
$=\left[\begin{array}{rr}-6+5 & -4+17 \\ 19+17 & 11+6\end{array}\right]=\left[\begin{array}{rr}-1 & 13 \\ 36 & 17\end{array}\right] \ldots\text{(ii)}$
From (i) and (ii), we get
A(B + C) = AB + AC

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Applied Maths 5 Marks Questions

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