IPMAT Indore 2024 · Quantitative Ability · Question 7

Step 1 — write the successive changes. Starting from price P:

P\overset{+x\%}{\to}P\left( 1+x/100 \right)\overset{-x\%}{\to}P\left( 1+x/100 \right)\left( 1-x/100 \right)

Step 2 — use the difference of squares.

P\left( 1-\frac{{x}^{2}}{10000}\right)=0.9676\,P

Step 3 — solve.

\frac{{x}^{2}}{10000}=1-0.9676=0.0324 \Rightarrow{x}^{2}=324 \Rightarrow x=18

(The negative root is rejected since x is a percentage increase.)

Answer: 18

Shortcut. The net-change identity. An x\% rise followed by an x\% fall always leaves

-\frac{{x}^{2}}{100}\% \text{net change.}

A drop to 96.76\% is a net loss of 3.24\%, so \frac{{x}^{2}}{100}=3.24\Rightarrow{x}^{2}=324\Rightarrow x=18. Recognising 324={18}^{2} ends it in seconds.