A) 100/11%
B) 100/9%
C) 9%
D) 10%
Correct Answer: A
Solution :
Let the length of the rectangle \[=l\] and breadth of the rectangle \[=b\] \[\therefore \] Area of rectangle \[=lb\] New length of the rectangle \[=\frac{l\times 110}{100}=\frac{11l}{10}\] and let new breadth of the rectangle \[=B\] \[\therefore \] Area of new rectangles \[=\frac{11l}{10}B\] By given condition, \[lb=\frac{11l}{10}B\] \[\Rightarrow \] \[B=\frac{10}{11}b\] \[\therefore \] Decrease in breadth of the rectangle \[=b-\frac{10}{11}b=\frac{b}{11}\] Percentage decrease \[=\frac{b/11}{b}\times 100%=\frac{100}{11}%\]You need to login to perform this action.
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