A) \[11\frac{1}{9}%\]
B) \[9\frac{1}{11}%\]
C) 11%
D) 10%
Correct Answer: B
Solution :
Let the length = 100 m and height = x m Area = 100 x New length = 110 m and let new height \[=(x-y%\,\,\text{of}\,x)\] Then, \[110\times \,\left( x-\frac{y}{100}\cdot x \right)=100\times x\] \[\Rightarrow \] \[110\times \left( 1-\frac{y}{100} \right)=100\] \[\Rightarrow \] \[1-\frac{y}{100}=\frac{100}{110}\] \[\Rightarrow \] \[\frac{y}{100}=1-\frac{100}{110}=\frac{10}{110}=\frac{1}{11}\] \[\therefore \] \[y=\frac{1000}{11}=9\frac{1}{11}%\]
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