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An amount of ₹ 5000 is put into three investments at 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹358. If the total annual income from first two investments is ₹70 more than the income from the third, find the amount of each investment by the matrix method. HINT: Let these investments be ₹x, ₹y and ₹z, respectively. Then, $x+y+z=5000, \ldots$ (i) $\begin{array}{l} \frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}=358 \Rightarrow \\ 6 x+7 y+8 z=35800 \ldots (ii) \end{array}$ And, $\frac{6 x}{100}+\frac{7 y}{100}=\frac{8 z}{100}+70$ $\Rightarrow 6 x+7 y-8 z=7000 . \ldots \text { (iii) }$
An amount of ₹ 5000 is put into three investments at 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹358. If the total annual income from first two investments is ₹70 more than the income from the third, find the amount of each investment by the matrix method. HINT: Let these investments be ₹x, ₹y and ₹z, respectively. Then, $x+y+z=5000, \ldots$ (i) $\begin{array}{l} \frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}=358 \Rightarrow \\ 6 x+7 y+8 z=35800 \ldots (ii) \end{array}$ And, $\frac{6 x}{100}+\frac{7 y}{100}=\frac{8 z}{100}+70$ $\Rightarrow 6 x+7 y-8 z=7000 . \ldots \text { (iii) }$
CBSE | Class 12 | Excercise 8A | Maths | RS Aggarwal | System of Linear Equations
An amount of ₹ 5000 is put into three investments at 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹358. If the total annual income from first two investments is ₹70 more than the income from the third, find the amount of each investment by the matrix method. HINT: Let these investments be ₹x, ₹y and ₹z, respectively. Then, $x+y+z=5000, \ldots$ (i) $\begin{array}{l} \frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}=358 \Rightarrow \\ 6 x+7 y+8 z=35800 \ldots (ii) \end{array}$ And, $\frac{6 x}{100}+\frac{7 y}{100}=\frac{8 z}{100}+70$ $\Rightarrow 6 x+7 y-8 z=7000 . \ldots \text { (iii) }$

Solution:

Suppose the investments are $\mathrm{x} \mathrm{x}$, Fy and $\mathrm{F} \mathrm{z}$, respectively.
Therefore, $x+y+z=5000$
$\begin{array}{l}
\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}=358 \\
6 x+7 y+8 z=35800
\end{array}$
And, $\frac{6 x}{100}+\frac{7 y}{100}=\frac{8 z}{100}+70$
$6 x+7 y-8 z=7000$
Now converting in the matrix form,
$\begin{array}{l}
A X=B \\
{\left[\begin{array}{ccc}
1 & 1 & 1 \\
6 & 7 & 8 \\
6 & 7 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
5000 \\
35800 \\
7000
\end{array}\right]} \\
R_{3}-R_{2} \\
{\left[\begin{array}{ccc}
1 & 1 & 1 \\
6 & 7 & 8 \\
0 & 0 & -16
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
5000 \\
35800 \\
-28800
\end{array}\right]} \\
R_{2}-6 R_{1} \\
{\left[\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & -16
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
5000 \\
5800 \\
-28800
\end{array}\right]}
\end{array}$
Again converting into the equations we get
$\begin{array}{l}
x+y+z=5000 \\
Y+2 z=5800 \\
-16 z=-28800 \\
Z=1800 \\
Y+2 \times 1800=5800 \\
Y=5800-3600 \\
Y=2200 x+2200+1800=5000 \\
X=5000-4000 \\
X=1000
\end{array}$
Amount of $1000,2200,1800$ were invested in the investments of $6 \%, 7 \%, 8 \%$ respectively.

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