Question

Two numbers are less than the third number by $30\%$ and $37\%$ respectively. How much percent is the second number less than by the first?

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

Answer: (C)

$10$

Let the third number is $x$.
Then first number $=\left(\begin{array}{c}100-30\end{array}\right)\%$ of $x$.
$=70\%$ of $x$ $=\frac{7x}{10}$
Second number is $\frac{63x}{100}$
Difference $=\frac{7x}{10}-\frac{63x}{00}=\frac{7x}{100}$
So, required percentage is. difference is what percent of first number
$\left(\begin{array}{c}\frac{7x}{100}\times \frac{10}{7x}\times 100\end{array}\right)\%=10\%$