Question

Given two positive integers $x$ and $y$ with $x<y$. By what percent is $x$ less than $y$?

• A. $\frac{100(y-x)}{x}$
• B. $\frac{100(x-y)}{x}$
• C. $\frac{100(y-x)}{y}$
• D. $100(x-y)$

Answer 1

Answer: (C)

$\frac{100(y-x)}{y}$

Given that, $x<y$

Let $x$ be $k\%$ less than $y$

$\Rightarrow$ $y-x=\frac{k}{100}\times y$

$\therefore$ $k=\frac{100(y-x)}{y}$