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CAT 2024 Slot 3 — QA Question 18

Permutation & CombinationEasy

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

Answer & solution

  • A

    30

  • 25

  • C

    27.5

  • D

    20

Solution

Medium

Let the first increment be x%x\%; the second is 2x%2x\%. Apply the two successive multipliers and set the result equal to 187.5%187.5\% of the original. Solve the resulting quadratic in xx.

1

Set up the multiplier equation. First increase x%x\%, then 2x%2x\%.

(1+x100) ⁣(1+2x100)=1.875⇒ (1+x100) ⁣(1+2x100)=158\begin{aligned} &\left(1+\frac{x}{100}\right)\!\left(1+\frac{2x}{100}\right)=1.875\\ &\Rightarrow\ \left(1+\frac{x}{100}\right)\!\left(1+\frac{2x}{100}\right)=\frac{15}{8} \end{aligned}
2

Expand into a quadratic. Let u=x100u=\dfrac{x}{100}.

(1+u)(1+2u)=158⇒ 2u2+3u+1=158(expand)⇒ 2u2+3u−78=0⇒ 16u2+24u−7=0(×8)\begin{aligned} &(1+u)(1+2u)=\frac{15}{8}\\ &\Rightarrow\ 2u^2+3u+1=\frac{15}{8}\quad\text{(expand)}\\ &\Rightarrow\ 2u^2+3u-\frac{7}{8}=0\\ &\Rightarrow\ 16u^2+24u-7=0\quad\text{(}\times8\text{)} \end{aligned}
3

Solve and take the positive root.

u=−24±242+4⋅16⋅72⋅16=−24±102432⇒ u=−24±3232(1024=32)⇒ u=832=14(positive root)⇒ x=100u=25\begin{aligned} &u=\frac{-24\pm\sqrt{24^2+4\cdot16\cdot7}}{2\cdot16}=\frac{-24\pm\sqrt{1024}}{32}\\ &\Rightarrow\ u=\frac{-24\pm 32}{32}\quad\text{(}\sqrt{1024}=32\text{)}\\ &\Rightarrow\ u=\frac{8}{32}=\frac14\quad\text{(positive root)}\\ &\Rightarrow\ x=100u=25 \end{aligned}
25%25\%
CAT 2024 Slot 3 QA Q18: After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of sala — Solution | TheCATExam