A) \[\frac{5}{9}\]
B) 10
C) \[\frac{9}{5}\]
D) 20
Correct Answer: C
Solution :
Let length \[=x.\] and the breadth\[=y\] The original area \[=xy\] New length = 150 % of \[x=\frac{150x}{100}=\frac{3x}{2}\] New breadth =120% of y\[=\frac{120y}{100}=\frac{6y}{5}\] New area \[=\frac{3x}{2}\times \frac{6y}{5}=\frac{9}{5}xy=\frac{9}{5}\] (Original area)You need to login to perform this action.
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