Question

$A^{\prime }s$ income is $60$% more than that of $B$. By what percent is $B^{\prime }s$ income less than $A^{\prime }s$?

Answer 1

Let the salaries of $A,B$ be $x,y$.

But it is given that $A^{\prime }s$ salaray is $60\%$ more than that of $B$.

$x=y(1+\frac{60}{100})$

$x=y\left(\frac{160}{100}\right)$

Let $B^{\prime }s$ salary is $a\%$ less than that of $A$. So,

$\begin{array}{cc}l\left(\frac{160}{100}y\right)\left(1-\frac{a}{100}\right)& =y\\ 1-\frac{a}{100}& =\frac{100}{160}\\ 1-\frac{100}{160}& =\frac{a}{100}\\ \frac{60}{160}& =\frac{a}{100}\\ a& =37.5\%\end{array}$

Hence, $B^{\prime }s$ income is $37.5\%$ less than that of $A^{\prime }s$ salary.