Question

Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. The edge of the new cube formed is

(a) 12 cm

(b) 14 cm

(c) 16 cm

(d) 18 cm

Answer 1

Sides (edges)of 3 cubes are 6 cm, 8 cm, and 10 cm respectivley

$\therefore$ Volume of first cube $=(6{)}^3=216c{m}^3$

Volume of second cube $=(8{)}^3=512c{m}^3$

and volume of third cube

$=(10{)}^3=1000c{m}^3$

Sum of volumes of 3 cubes

$=216+512+1000=1728c{m}^3$

$\therefore$ Volume of new single cube $=1728c{m}^3$

$\therefore$ Edge $=\sqrt[3]{31728}=\sqrt[3]{3(12{)}^3}]$

=12 cm