Question

The length of a rectangle is reduced by 30%. By what percent would the width have to be increased to maintain the original area? *

Answer 1

Answer:

333%

Step-by

Let W = any  number , eg 9cm

Let L = any number, eg 4 cm

A = LW

A = (9×4)cm²

A = 36cm²

A1  = (9 - 30%)×4x

     = 2.7×4x

     = 10.8xcm²

A =A1

36cm² = (10.8x)cm²

x = (36cm²)/(10.8cm²)

x = 3.33

x = 333%

∴ The width must be increased 3.33 or 333% times

Answer 2

Answer:

Therefore, to maintain the original area, the width has to be increased by 42.85%

Step-by-step explanation:

$area \ of \ rectangle = length(breadth) = lb$

Given, in case 2,  length of the rectangle is reduced by 30%

Let 'x' be the % of width to be increased

Area of both the rectangles should be same

$l.b = l(\frac{100-30}{100}) \ b (\frac{100+x}{100})$

100 × 100 = 70 × (100+x)

10000 = 7000 + 70 x

70x = 3000

$x = \frac{3000}{70}$

x = 42.85%  

Therefore, to maintain the original area, the width has to be increased by 42.85%