• question_answer A large cube is formed from the material obtained by melting three smaller cubes of side 3, 4 and 5 cm. What is the ratio of the total surface areas of the smaller cubes and the large cube? A) $2:1$                   B)       $3:2$C) $25:18$  D)                   $27:20$E) None of these

Explanation Option [c] is correct. Volume of the large cube $=({{3}^{3}}+{{4}^{3}}+{{5}^{3}})=216\,c{{m}^{3}}$ Let the edge of the large cube be a So, ${{a}^{3}}~=216\Rightarrow ~a=6\text{ }cm$ $\therefore$ Required ratio = $\left( \frac{6\times \left( {{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{6\times {{6}^{2}}} \right)=\frac{50}{36}=25:18$