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' (say) Initial area = final area \[l\times b=l\times \frac{160}{100}\times b'\Rightarrow \,b'\,=\frac{100}{160}\,b=\frac{5}{8}b\] Percentage decriment in width \[=\frac{b-b'}{b'}\,\times 100=\frac{b-\frac{5}{8}b}{b}\times 100%\] \[=37.5%\]
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