Question

After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was

• A. 27.5
• B. 30
• C. 25
• D. 20

Answer 1

The Correct Option is C

Let Gopal's initial salary be $S$.
After the first increment, his salary becomes:
\[ S_1 = S \times \left(1 + \frac{x}{100}\right) \]
where $x$ is the percentage increase in the first increment.
After the second increment, his salary becomes:
\[ S_2 = S_1 \times \left(1 + \frac{2x}{100}\right) \]
We are given that his final salary is 187.5. $S_2 = S \times 1.875$
Substituting $S_2 = S_1 \times \left(1 + \frac{2x}{100}\right)$:
\[ S \times \left(1 + \frac{x}{100}\right) \times \left(1 + \frac{2x}{100}\right) = S \times 1.875 \]
Canceling out $S$ and solving the equation:
\[ \left(1 + \frac{x}{100}\right) \times \left(1 + \frac{2x}{100}\right) = 1.875 \]
Expanding the terms:
\[ 1 + \frac{x}{100} + \frac{2x}{100} + \frac{2x^2}{10000} = 1.875 \]
Simplifying:
\[ 1 + \frac{3x}{100} + \frac{2x^2}{10000} = 1.875 \]
Subtract 1 from both sides:
\[ \frac{3x}{100} + \frac{2x^2}{10000} = 0.875 \]
Multiply the entire equation by 10000 to eliminate the denominators:
\[ 300x + 2x^2 = 87500 \]
Rearrange:
\[ 2x^2 + 300x - 87500 = 0 \]
Solving this quadratic equation using the quadratic formula:
\[ x = \frac{-300 \pm \sqrt{300^2 - 4 \times 2 \times (-87500)}}{4} \]
\[ x = \frac{-300 \pm \sqrt{790000}}{4} \implies x = \frac{-300 \pm 890}{4} = \frac{590}{4} = 25 \]
Thus, the percentage increase in the first increment is 25.