A Gardener Increases The Area Of His Rectangular Garden By Increasing

1.	A Gardener Increases The Area Of His Rectangular Garden By Increasing Its Length By 40% And Decreasing By 20%. The Area Of The New Garden?
Answer» LET the length and breadth of garden be L and b respectively THEREFORE, AREA = l x b NEW length after 40 % increase = l x 140/100 = 1.4% New breadth after 20% decrease = b x 80/100 =0.8b New area = 1.41 x 0.8b = 1.12lb Therefore, change in area percentage =1.12ib-ib/ib100%= 12% increase (New area – original area = 0.12> 0 i.e.., Increase). Let the length and breadth of garden be l and b respectively Therefore, Area = l x b New length after 40 % increase = l x 140/100 = 1.4% New breadth after 20% decrease = b x 80/100 =0.8b New area = 1.41 x 0.8b = 1.12lb Therefore, change in area percentage =1.12ib-ib/ib100%= 12% increase (New area – original area = 0.12> 0 i.e.., Increase).

A Gardener Increases The Area Of His Rectangular Garden By Increasing Its Length By 40% And Decreasing By 20%. The Area Of The New Garden?

Answer»

LET the length and breadth of garden be L and b respectively

NEW length after 40 % increase = l x 140/100 = 1.4%

New breadth after 20% decrease = b x 80/100 =0.8b

New area = 1.41 x 0.8b = 1.12lb

Therefore, change in area percentage =1.12ib-ib/ib*100%= 12% increase (New area – original area = 0.12> 0 i.e.., Increase).

Let the length and breadth of garden be l and b respectively

Therefore, Area = l x b

New length after 40 % increase = l x 140/100 = 1.4%

New breadth after 20% decrease = b x 80/100 =0.8b

New area = 1.41 x 0.8b = 1.12lb

Therefore, change in area percentage =1.12ib-ib/ib*100%= 12% increase (New area – original area = 0.12> 0 i.e.., Increase).