Question

Two numbers are $20\%$ and $25\%$ less than a third number respectively. How much $\%$ is the second number less than the first?

• A. $33\frac{1}{3}\%$
• B. $6\frac{2}{3}\%$
• C. $6\frac{1}{4}\%$
• D. $11\frac{1}{9}\%$

Answer 1

The correct option is C $6\frac{1}{4}\%$

Let the Numbers be $x,y$ and $z$

given that

$x$ is $20\%$ less than $z$

$\Rightarrow x=$ $80\%$ of $z$

$\Rightarrow x=\frac{80}{100}\times z=0.8z$

$y$ is $25\%$ less than $z$ given

$\Rightarrow y=75\%$ of $z$

$\Rightarrow y=\frac{75}{100}\times z=0.75z$

Now we have, $x-y=0.80z-0.75z$

$\Rightarrow x-y=0.05z$

$\%$ by which the second number is less than first is

$=\frac{x-y}{x}\times 100\%$

$=\frac{0.05z}{0.80z}\times 100\%$

$=\frac{5}{80}\times 100\%$

$=\frac{25}{4}\%=6\frac{1}{4}\%$