The length of a rectangle is decreased by $20$ % , then the percentage width be increased to maintain the same area, is

25

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20

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30

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40

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Solution

The correct option is A $25$ %
Let the original length be $l$

New length $= l - \frac{20}{100} l = 0.8 l$

Original width $= b$

New width $= b + x$

Original Area=New Area

$l b = (0.8 l) (b + x)$

$\Rightarrow l b = 0.8 l b + 0.8 l b$

$\Rightarrow 0.2 l b = 0.8 l x$

$\Rightarrow x = \frac{b}{4}$

$% i n c r e a s e = \frac{x}{b} \times 100$

$= \frac{b}{4 \times b} \times 100 = 25 %$

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The length of a rectangle is decreased by 20 % , then the percentage width be increased to maintain the same area, is

The correct option is A 25 %Let the original length be lNew length =l−20100l=0.8lOriginal width =bNew width =b+xOriginal Area=New Arealb=(0.8l)(b+x)⇒lb=0.8lb+0.8lb⇒0.2lb=0.8lx⇒x=b4%increase=xb×100 =b4×b×100=25%

The length of a rectangle is decreased by $20$ % , then the percentage width be increased to maintain the same area, is