Question

The length of a rectangle is increased by $60$%. By what percent should the width be decreased to maintain the same area?

• A. $33\frac{1}{2}$%
• B. $35\frac{1}{2}$%
• C. $37\frac{1}{2}$%
• D. $39\frac{1}{2}$%

Answer 1

The correct option is C $37\frac{1}{2}$%

Let ${}^{\prime }{l}^{\prime }$=length

${}^{\prime }{b}^{\prime }$=breadth

$A=lb$

Now $L=l+\frac{60}{100}\times \ell =1.6l$

$b^1=b-bx$

$A^1=L{b}^1=1.6l(b-bx)$

Given

$A=A^1$

$=lb=1.6l(b-bx)$]

$=b=1.6b(1-x)$

$=\frac{1}{1.6}=1-x$

$=x=11-\frac{10}{16}$

$x=\frac{6}{16}$

$\%x=\frac{6}{16}\times 100$

$=\frac{75}{2}$

$\%x=35\frac{1}{2}\%$