A) \[37\frac{1}{2}%\]
B) \[60%\]
C) \[75%\]
D) \[120%\]
Correct Answer: A
Solution :
Let the initial length and width are \[l\] and b respectively. Final length\[=l\times \frac{160}{100}\] Final width\[={{b}_{1}}\](say) Initial area = final area \[l\times b=l\times \frac{160}{100}\times {{b}_{1}}\Rightarrow {{b}_{1}}=\frac{100}{160}=\frac{5}{8}b\] Percentage decrement in width \[=\frac{b-{{b}_{1}}}{{{b}_{1}}}\times 100=\frac{b-\frac{5}{8}b}{b}\times 100%\] \[=37.5%\]
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