Question

If A's income is $10\%$ more than B's income, by how much percent B's income is less than A's income?

• A. $9\frac{1}{11}\%$
• B. $10\%$
• C. $20\%$
• D. $90\frac{10}{11}\%$

Answer 1

The correct option is A

$9\frac{1}{11}\%$

Let the B's salary be $\mathrm{x}$

A's income$=$$10\%\mathrm{more}\mathrm{than}\mathrm{x}$

Step 1:- Calculate the A's income

$10\%\mathrm{of}\mathrm{x}=$$\frac{10}{100}\times \text{x}$

$=\frac{\mathrm{x}}{10}$

A's income$=\mathrm{x}+\frac{\mathrm{x}}{10}$

$=\frac{11\mathrm{x}}{10}$

Step 2:- Calculate how much percent B's income is less than A's income

Difference in both's income$=\frac{11\mathrm{x}}{10}-\mathrm{x}$

$=\frac{\mathrm{x}}{10}$

Required Percentage$=\left(\frac{\mathrm{Difference}\mathrm{in}\mathrm{income}}{\mathrm{A}\text{'}\mathrm{s}\mathrm{income}}\times 100\right)\%$

$=\left(\frac{{\displaystyle \frac{\mathrm{x}}{10}}}{{\displaystyle \frac{11\mathrm{x}}{10}}}\times 100\right)\%$

$$\begin{gathered} =\left(\frac{100}{11}\right)\% \\ =9\frac{1}{11}\% \end{gathered}$$

B's income is $9\frac{1}{11}\%$less than A's income hence option A is correct.