Questions & Answers

Question

Answers

A) \[2000\]

B) \[2400\]

C) \[2600\]

D) \[2800\]

Answer
Verified

Given: Total income of \[A\] and $B$\[=\text{ }6000\], $A$spend \[60%\] of his income and \[B\] spends \[80%\] of his income and both have equal savings.

Let income of\[A\]is \[x\]

Then income of \[B\]will be \[6000-x\][ As, \[income\text{ }of\text{ }A\text{ }+\text{ }income\text{ }of\text{ }B\text{ }=6000\]]

\[A\]spends \[60%\]of his income.

So, \[A's\] Saving is \[x\text{ }-\text{ }60%\text{ }of\text{ }x\]

Convert it in the mathematical form and simplify.

$\begin{align}

& =x-\dfrac{60}{100}of\text{ }x \\

& =x-\dfrac{6x}{10} \\

& =\dfrac{10x-6x}{10} \\

& =\dfrac{4x}{10}.............(1) \\

\end{align}$

Similarly Now, \[B\] spends \[80%\] of his income

So, \[B's\] saving is \[\left( 6000\text{ }-\text{ }x \right)\text{ }-\text{ }80%\text{ }of\text{ }\left( 6000\text{ }\text{ }x \right)\]

Simplify the above equations -

\[\begin{align}

& \Rightarrow \left( 6000-x \right)-\dfrac{80}{100}of(6000-x) \\

& =\left( 6000-x \right)-\dfrac{8(6000-x)}{10} \\

& =\dfrac{10\left( 6000-x \right)-8(6000-x)}{10} \\

& =\dfrac{60000-10x-48000+8x}{10} \\

& =\dfrac{12000-2x}{10}.................(2) \\

\end{align}\]

According to question eq(1) is equal to eq(2) as both \[A\] and \[B\] has equal saving.

$\dfrac{4x}{10}=\dfrac{12000-2x}{10}$

[Same denominator from both sides of the equations cancels each other]

$\begin{align}

& \Rightarrow 4x=12000-2x \\

& \Rightarrow 4x+2x=12000 \\

& \Rightarrow 6x=12000 \\

& \therefore x=\dfrac{12000}{6} \\

& \therefore x=2000 \\

\end{align}$

Therefore, the required solution – The income of A is $=Rs.\ \text{2000}$