If both length and breadth of a cuboid is increased by 50 percent, th…

Let the dimensions of the cuboid be l, b, h

50% = $\frac{1}{2}$

The changed dimensions $\frac{3}{2}$l, $\frac{3}{2}$b, h'

Since volume remains the same, hence

Volume ⇒ lbh = $\frac{3}{2}$l.$\frac{3}{2}$b.h'

                h' = $\frac{4}{9}$h

Height decreased by = 9R - 4R = 5R

Percentage decreased in height = $\frac{5}{9}$ x 100 = 55.55%

 

Alternate Solution:

Using Ratio method:

	Old		New
Length	2	:	3
Breath	2	:	3
Height	H	:	h
Area	1	:	1

 Now volume:

$\frac{2×2×H}{3×3×h}$ = $\frac{1}{1}$

$\frac{H}{h}$ = $\frac{9}{4}$

Height decreased by = 9R - 4R = 5R

Percentage decreased in height = $\frac{5}{9}$ x 100 = 55.55%