Two numbers are respectively 20% and 25% lower than a third number. By how much percentage is the second number lower than the first? (R.R.B., 2006)
(a)5%
(b)$6\frac{1}{4}\%$
(c)8%
(d)$10\frac{1}{2}\%$
Answer
Answer (as printed): B
Explanation
Let the third number be $x$. Then, first number $=80\%$ of $x=\frac{80}{100} \times x=\frac{4x}{5}$. Second number $=75\%$ of $x=\frac{75}{100} \times x=\frac{3x}{4}$. Difference $=\frac{4x}{5}-\frac{3x}{4}=\frac{x}{20}$. $\therefore$ Required percentage $=\left(\frac{x}{20} \times \frac{5}{4x} \times 100\right)\%=\frac{25}{4}\%=6\frac{1}{4}\%$.
Explanation as extracted from the printed page; notation may be imperfect.