Question:
The price of a laptop is first increased by \(x\%\), then decreased by \(x\%\). The new price becomes \(\frac{7}{16}\) of the original price. Find the value of \(x\).
The price of a laptop is first increased by \(x\%\), then decreased by \(x\%\). The new price becomes \(\frac{7}{16}\) of the original price. Find the value of \(x\).
Show Hint
Increase by \(x\%\) and decrease by \(x\%\) never cancel each other. The net percentage change is
\[
-\frac{x^2}{100}\%.
\]
Updated On: Jun 8, 2026
- \(50\%\)
- \(25\%\)
- \(75\%\)
- \(62.5\%\)
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The Correct Option is C
Solution and Explanation
Concept:
When a quantity is increased by \(x\%\) and then decreased by the same \(x\%\), the net effect is always a decrease.
The formula for successive percentage changes is:
\[
\text{Final Value}
=
\text{Original Value}
\left(1+\frac{x}{100}\right)
\left(1-\frac{x}{100}\right)
\]
Using the identity
\[
(1+a)(1-a)=1-a^2,
\]
we obtain
\[
\text{Final Value}
=
\text{Original Value}
\left(1-\frac{x^2}{10000}\right).
\]
Step 1: Form the equation using the given information.
Let the original price be \(P\). After increase and decrease, \[ P\left(1-\frac{x^2}{10000}\right) = \frac{7}{16}P. \] Cancelling \(P\), \[ 1-\frac{x^2}{10000} = \frac{7}{16}. \]
Step 2: Solve for \(x^2\).
\[ \frac{x^2}{10000} = 1-\frac{7}{16} = \frac{9}{16}. \] Therefore, \[ x^2 = 10000\times\frac{9}{16} = 625\times9 = 5625. \]
Step 3: Find the value of \(x\).
\[ x=\sqrt{5625}=75. \] Hence, \[ \boxed{x=75\%} \]
Step 1: Form the equation using the given information.
Let the original price be \(P\). After increase and decrease, \[ P\left(1-\frac{x^2}{10000}\right) = \frac{7}{16}P. \] Cancelling \(P\), \[ 1-\frac{x^2}{10000} = \frac{7}{16}. \]
Step 2: Solve for \(x^2\).
\[ \frac{x^2}{10000} = 1-\frac{7}{16} = \frac{9}{16}. \] Therefore, \[ x^2 = 10000\times\frac{9}{16} = 625\times9 = 5625. \]
Step 3: Find the value of \(x\).
\[ x=\sqrt{5625}=75. \] Hence, \[ \boxed{x=75\%} \]
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