(a) An amount of ₹ 5,000 is to be deposited in three different bonds bearing 6%, 7% and 8% per year respectively. Total annual income is ₹ 358. If the income from first two investments is ₹ 70 more than the income from the third, then find the amount of investment in each bond by rank method. OR (b) A continuous random variable X has the following probability function. | Value of $X = x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | | --- | --- | --- | --- | --- | --- | --- | --- | --- | | $P(x)$ | $0$ | $k$ | $2k$ | $2k$ | $3k$ | $k^2$ | $2k^2$ | $7k^2 + k$ | (i) Find k. (ii) Evaluate $P(x \frac{1}{2}$, then find the minimum value of $x$.

(a) An amount of ₹ 5,000 is to be deposited in three different bonds bearing 6%, 7% and 8% per year respective

Question 21 of 44

An amount of ₹ 5,000 is to be deposited in three different bonds bearing 6%, 7% and 8% per year respectively. Total annual income is ₹ 358. If the income from first two investments is ₹ 70 more than the income from the third, then find the amount of investment in each bond by rank method. OR
A continuous random variable X has the following probability function.

Value of $X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

Find k.
Evaluate $P(x < 6)$ , $P(x \geq 6)$ and $P(0 < x < 5)$ .
If $P(X \leq x) > \frac{1}{2}$ , then find the minimum value of $x$ .

Puducherry TnboardTamil Nadu HSC (DGE) Commerce Board 2022Subjective· 5mImportance★★★★★

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(a) Solve the consistent $3\times3$ system for the three bond amounts: $x=1000,\ y=2200,\ z=1800$ . (b) $k=0.1$ ; $P(x<6)=0.81$ , $P(x\ge6)=0.19$ , $P(0<x<5)=0.8$ ; minimum $x=4$ .

(a) Investment in each bond (rank / matrix method).

Let $x,y,z$ be the amounts (in Rs) invested at 6%, 7%, 8%. Then

$x+y+z=5000,$

$0.06x+0.07y+0.08z=358,$

$0.06x+0.07y-0.08z=70\quad(\text{income of first two is }70\text{ more than third}).$

Writing $AX=B$ with

$A=\begin{pmatrix}1&1&1\\ 6&7&8\\ 6&7&-8\end{pmatrix},\; X=\begin{pmatrix}x\\y\\z\end{pmatrix},\; B=\begin{pmatrix}5000\\35800\\7000\end{pmatrix}$

(the last two equations multiplied by 100). Here $\rho(A)=\rho([A|B])=3=$ number of unknowns, so the system is consistent with a unique solution.

Subtract the 3rd equation from the 2nd: $(0.06x+0.07y+0.08z)-(0.06x+0.07y-0.08z)=358-70$ , i.e. $0.16z=288\Rightarrow z=1800.$

Then $0.06x+0.07y=358-0.08(1800)=214$ and $x+y=5000-1800=3200.$

From $x=3200-y$ : $0.06(3200-y)+0.07y=214\Rightarrow192+0.01y=214\Rightarrow y=2200,$ so $x=1000.$

Check: incomes $60+154+144=358$ ; first two $214=144+70$ . Correct.

(b) Probability function.

The probabilities must sum to $1$ :

$0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1$

$\Rightarrow 9k+10k^2=1\Rightarrow 10k^2+9k-1=0\Rightarrow(10k-1)(k+1)=0.$ …

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