Question

The percentage increase in the surface area of a cube when each side is increased to 3/2
times the original length is

(A) 225 (B) 200 (C) 175 (D) 125 correct answer plz​

Answer 1

To find:-

$\sf Increase \: in \: percentage\: of \: surface\: area \: when \\ {\sf{each \: side \: of \: a \: cube \: is \: increased \: to {\dfrac{3}{2}}}}$

Assumption:-

Let the side of 1st cube = x

$\sf Side \: of \: cube \: increase \: by{\dfrac{3}{2}} = \dfrac{3x}{2}$

Solution:-

1st case,

Side of first cube = x

$\sf{Surface\: Area \: of \: cube = {6(a)}^{2}}$

$\sf{Surface \: area \: of \: 1st \: cube = {6(x)}^{2}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: = {6x}^{2} square\: units}$

2nd case,

$\sf{Side\: of \: 2nd \: cube = \dfrac{3x}{2}}$

$\sf{Surface\: area \: of \: 2nd \: cube = 6 \times \bigg({\dfrac{3x}{2}\bigg)}^{2}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: =6\times \bigg(\dfrac{{9x}^{2}}{4}\bigg)square\:units}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: = \dfrac{{54x}^{2}}{4} square\:units}$

Increase in percentage of surface area:-

$\sf{\dfrac{Surface\: area\: of \: 2nd \: cube - Surface\: area \: of \: 1st\: cube}{Surface\:Area\:of\:1st\:cube}\times100}$

= $\sf{\dfrac{{54x}^{2}}{4} - {{6x}^{2}}\times \dfrac{1}{{6x}^{2}} \times 100}$

= $\sf{\dfrac{{54x}^{2} - {24x}^{2}}{4} \times \dfrac{1}{{6x}^{2}}\times100}$

= $\sf{\dfrac{{30x}^{2}}{4} \times \dfrac{1}{{6x}^{2}} \times 100}$

= 125 %

Therefore, there is 125% increase in surface area of the cube.