Question

An amount of $₹5000$ is put into three investments at the rates of interest $6\%,7\%$ and $8\%$ per annum, respectively. The total annual income is $₹358$. If the combined income from the first two investments is ₹ 70 more than the income from the third, then the amount of each investment (in ₹) is

• A. $1000,2200,180$
• B. $1000,2200,1100$
• C. $1000,2200,1800$
• D. $2200,1800,1100$

Answer 1

Answer: (C)

Let $₹x,₹y$, and $₹z$ be the investments at the rates of interest of $6\%,7\%$ and $8\%$ per annum, respectively. Then,
Total investment $=₹5000$
$\Rightarrow x+y+z=5000$
Now, income from first investment $=₹\frac{6x}{100}$
Income from second investment $=₹\frac{7y}{100}$
Income from third investment $=₹\frac{8z}{100}$
$\therefore$ Total annual income $=₹\left(\frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}\right)$
$\Rightarrow \frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}=358$
$\Rightarrow 6x+7y+8z=35800$
It is given that the combined income from the first two investments is $₹70$ more than the income from the third.
$\therefore \frac{6x}{100}+\frac{7y}{100}=70+\frac{8z}{100}$
$\Rightarrow 6x+7y-8z=7000$
Thus, the system of equations is
$x+y+z=5000$
$6x+7y+8z=35800$
$6x+7y-8z=7000\text{ or }AX=B$
where $A=\left[\begin{array}{ccc}1& 1& 1\\ 6& 7& 8\\ 6& 7& -8\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}5000\\ 35800\\ 7000\end{array}\right]$
Now, $|A|=\left|\begin{array}{ccc}1& 1& 1\\ 6& 7& 8\\ 6& 7& -8\end{array}\right|$
$=1(-56-56)-1(-48-48)+1(42-42)$
$=-16\ne 0$
$\therefore$ The solution of the system $AX=B$ exists and is given by
$X=A^{-1}B$
Now, the cofactors of $A$ are
$A_{11}=-112;A_{12}=96;A_{13}=0;A_{21}=15;A_{22}=-14,$
$A_{23}=-1;A_{31}=1;A_{32}=-2;A_{33}=1$
$\therefore \mathrm{adj}(A)={\left[\begin{array}{ccc}-112& 96& 0\\ 15& -14& -1\\ 1& -2& 1\end{array}\right]}^{\prime }=\left[\begin{array}{ccc}-112& 15& 1\\ 96& -14& -2\\ 0& -1& 1\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}(\mathrm{adj}A)$
$=\frac{-1}{16}\left[\begin{array}{ccc}-112& 15& 1\\ 96& -14& -2\\ 0& -1& 1\end{array}\right]$
Hence, solution is given by $X=A^{-1}B$
$=-\frac{1}{16}\left[\begin{array}{ccc}-112& 15& 1\\ 96& -14& -2\\ 0& -1& 1\end{array}\right]\left[\begin{array}{c}5000\\ 35800\\ 7000\end{array}\right]$
$=-\frac{1}{16}\left[\begin{array}{c}-560000+537000+7000\\ 480000-501200-14000\\ 0-35800+7000\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1000\\ 2200\\ 1800\end{array}\right]$
$\Rightarrow x=1000,y=2200\text{ and }z=1800$
Hence, three investments are of ₹ 1000 , ₹ 2200 and ₹ 1800 , respectively.