The length of a rectangle is increased by 60%. By what per cent would the width have to be decreased to maintain the same area ?
[amp_mcq option1=”37.5 %” option2=”60%” option3=”75%” option4=”120%” correct=”option1″]
This question was previously asked in
UPSC CAPF – 2017
– The length is increased by 60%, so the new length $L’$ is $L + 0.60L = 1.60L$.
– Let the new width be $W’$. The new area $A’ = L’ \times W’ = 1.60L \times W’$.
– To maintain the same area, $A’ = A$, so $1.60L \times W’ = L \times W$.
– We can solve for $W’$: $W’ = \frac{L \times W}{1.60L} = \frac{W}{1.60} = \frac{W}{8/5} = \frac{5}{8}W$.
– The decrease in width is $W – W’ = W – \frac{5}{8}W = \frac{3}{8}W$.
– The percentage decrease in width is $\frac{\text{Decrease in width}}{\text{Original width}} \times 100\% = \frac{(3/8)W}{W} \times 100\% = \frac{3}{8} \times 100\%$.
– $\frac{3}{8} = 0.375$, so the percentage decrease is $0.375 \times 100\% = 37.5\%$.