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 that of the first increment, then the percentage of salary increase in the first increment was:

Question

Accepted Answer

Let Gopal's initial salary be $$100$$. Let the percentage increase in the first increment be $$x\%$$. Then, the percentage increase in the second increment is $$2x\%$$.

After two successive increments, his final salary becomes:  
$$
100 	imes (1 + \frac{x}{100}) 	imes (1 + \frac{2x}{100}) = 187.5
$$

Simplifying:  
$$
(1 + \frac{x}{100}) 	imes (1 + \frac{2x}{100}) = 1.875
$$

Expanding:  
$$
1 + \frac{x}{100} + \frac{2x}{100} + \frac{x \cdot 2x}{100^2} = 1.875
$$  
$$
1 + \frac{3x}{100} + \frac{2x^2}{10000} = 1.875
$$

Removing $$1$$ from both sides and multiplying through by $$10000$$:  
$$
300x + 2x^2 = 8750
$$  
Dividing by $$2$$:  
$$
x^2 + 150x - 4375 = 0
$$

Solving this quadratic equation using factorization or approximation gives $$x = 25$$. Thus, the percentage increase in the first increment is **25%**.

Answer

Answer

Answer