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 \(x\).

• A. \(50\%\)
• B. \(25\%\)
• C. \(75\%\)
• D. \(62.5\%\)

Hint:

Increase by \(x\%\) and decrease by \(x\%\) never cancel each other. The net percentage change is \[ -\frac{x^2}{100}\%. \]

Answer 1

The Correct Option is C

Step 1: Understand what happens with the two changes.
The price first goes up by $x\%$ and then comes down by the same $x\%$. When you raise a value and then lower it by the same percent, you do not return to the start. The net result is always a small drop.

Step 2: Write the multiplying factors.
An increase of $x\%$ multiplies the price by $\left(1+\frac{x}{100}\right)$. A decrease of $x\%$ multiplies it by $\left(1-\frac{x}{100}\right)$. Let the original price be $P$.

Step 3: Combine the two factors.
The final price is \[ P\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right). \] Using the simple identity $(1+a)(1-a)=1-a^2$, this becomes \[ P\left(1-\frac{x^2}{10000}\right). \]
Step 4: Use the given final price.
We are told the final price is $\frac{7}{16}$ of $P$. So \[ P\left(1-\frac{x^2}{10000}\right)=\frac{7}{16}P. \] Cancel $P$ from both sides: \[ 1-\frac{x^2}{10000}=\frac{7}{16}. \]
Step 5: Solve for $x^2$.
Move terms to find the fraction lost: \[ \frac{x^2}{10000}=1-\frac{7}{16}=\frac{9}{16}. \] Multiply both sides by $10000$: \[ x^2=10000\times\frac{9}{16}=625\times9=5625. \]
Step 6: Take the square root to get $x$.
\[ x=\sqrt{5625}=75. \] So the percentage is $75\%$. \[ \boxed{x=75\%} \]