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{6 x}{100} $
Income from second investment $=₹ \frac{7 y}{100}$
Income from third investment $=₹ \frac{8 z}{100}$
$ \therefore $ Total annual income $=₹\left(\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}\right)$
$\Rightarrow \frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}=358 $
$\Rightarrow 6 x+7 y+8 z=35800$
It is given that the combined income from the first two investments is $₹ 70$ more than the income from the third.
$\therefore \frac{6 x}{100}+\frac{7 y}{100}=70+\frac{8 z}{100} $
$\Rightarrow 6 x+7 y-8 z=7000$
Thus, the system of equations is
$x+y+z =5000$
$6 x+7 y+8 z =35800$
$6 x+7 y-8 z=7000 \text { or } A X=B$
where $ A=\begin{bmatrix}1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8\end{bmatrix}, X=\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and $B=\begin{bmatrix} 5000 \\ 35800 \\ 7000\end{bmatrix}$
Now, $|A|=\begin{vmatrix}1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8\end{vmatrix}$
$=1(-56-56)-1(-48-48)+1(42-42)$
$=-16 \neq 0$
$\therefore$ The solution of the system $A X=B$ 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 \operatorname{adj}(A) =\begin{bmatrix} -112 & 96 & 0 \\ 15 & -14 & -1 \\1 & -2 & 1\end{bmatrix}^{\prime}=\begin{bmatrix}-112 & 15 & 1 \\96 & -14 & -2 \\0 & -1 & 1 \end{bmatrix} $
$\therefore A^{-1} =\frac{1}{|A|}(\operatorname{adj} A)$
$ =\frac{-1}{16} \begin{bmatrix} -112 & 15 & 1 \\96 & -14 & -2 \\ 0 & -1 & 1\end{bmatrix}$
Hence, solution is given by $X=A^{-1} B$
$=-\frac{1}{16}\begin{bmatrix} -112 & 15 & 1 \\96 & -14 & -2 \\0 & -1 & 1\end{bmatrix}\begin{bmatrix} 5000 \\35800 \\7000\end{bmatrix}$
$ =-\frac{1}{16} \begin{bmatrix} -560000+537000+7000 \\480000-501200-14000 \\0-35800+7000 \end{bmatrix} $
$\Rightarrow \begin{bmatrix} x \\y \\z\end{bmatrix} =\begin{bmatrix} 1000 \\2200 \\1800\end{bmatrix} $
$\Rightarrow x =1000, y=2200 \text { and } z=1800$
Hence, three investments are of ₹ 1000 , ₹ 2200 and ₹ 1800 , respectively.