Question

The length of rectangle is increased by 60%. The percentage by which the width has to be decreased to maintain the same area is

Answer 1

assume initially l=100 and b=100

so area =l*b

160(100-b)=100*100

b=37(1/2)%

Answer 2

$\huge\mathfrak{Answer}$

Let the length 100 metre and breadth be 100m, then

New length = 160m, new breadth = x,

$\sf \: Then \: \: \: \: \: \: \: \: 160 \times x = 100 \times 100 \\ \\ \sf \implies \: x = \frac{100 \times 100}{160} \implies \: x = \frac{125}{2}$

$\therefore \: \rm \: decrease \: in \: breadth \: = 100 - \frac{125}{2} \% \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf \: 37 \: \frac{1}{2} \%$

So the answer is 37 and 1/2 %