Question

A rectangle has dimensions $15 \times 20$ cm$^2$. Its width is decreased by 5% while its length is increased such that its area remains unchanged. Then the percentage increase in its length is?

• A. $5 \frac{5}{19}$
• B. $5 \frac{6}{19}$
• C. $5 \frac{7}{19}$
• D. $5 \frac{8}{19}$

Hint:

If one dimension decreases, the other must increase proportionally to keep area constant.

Answer 1

The Correct Option is A

Concept: Area remains constant.
Step 1: Initial dimensions.
\[ {Length} = 20,\quad {Width} = 15 \]
Step 2: Width decreased by 5%.
\[ {New width} = 15 \times 0.95 \]
Step 3: Area remains same.
\[ 20 \times 15 = L \times (15 \times 0.95) \] \[ 300 = 14.25L \Rightarrow L = \frac{300}{14.25} \approx 21.0526 \]
Step 4: Percentage increase.
\[ \frac{21.0526 - 20}{20} \times 100 \approx 5.26% \] \[ = 5 \frac{5}{19}% \]
Hence, the increase is $5 \frac{5{19}%$.