QUESTION

CSAT

Hard

Maths

Prelims 2014

A gardener increased the area of his rectangular garden by increasing its length by 40% and decreasing its width by 20%. The area of the new garden

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Explanation

Let the original length and width of the garden be denoted by LL and WW. The original area = L×WL \times W.

The length of the garden is increased by 40%, so the new length = L×1.40L \times 1.40. The width of the garden is decreased by 20%, so the new width = W×0.80W \times 0.80.

The new area of the garden = New length ×\times New width = (L×1.40)×(W×0.80)(L \times 1.40) \times (W \times 0.80).

Simplifying, the new area = 1.40×0.80×L×W=1.12×L×W1.40 \times 0.80 \times L \times W = 1.12 \times L \times W.

The percentage increase in the area = New area - Original areaOriginal area×100\frac{\text{New area - Original area}}{\text{Original area}} \times 100.

So, the percentage increase = (1.12×L×WL×W)(L×W)×100=0.12×100=12%\frac{(1.12 \times L \times W - L \times W)}{(L \times W)} \times 100 = 0.12 \times 100 = 12\%.

Therefore, the area of the new garden has increased by 12%.

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