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A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& {\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}} \cr
& = \left( {{3^3} + {4^3} + {5^3}} \right) = 216\,c{m^3} \cr
& {\text{Let}}\,{\text{the}}\,{\text{edge}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}}\,{\text{be}}\,a \cr
& So,\,{a^3} = 216\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 6\,cm \cr
& \therefore {\text{Required}}\,{\text{ratio}} \cr
& = {\frac{{6 \times \left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}}} \cr
& = \frac{{50}}{{36}} \cr
& = 25:18 \cr} $$
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