Management Aptitude Test Sep 2013 Paper

(a) For first cube, probability of getting red face = $\frac{5}{6}$
and probability of getting blue late = $\frac{4}{6}$
Now, let the number of faces which are coloured red in colour = x
∴ Number of blue faces = 6 - x
Similarly, for second cube,
Probability of getting red face = $\frac{\mathit{x}}{6}$
and probability of getting blue face = $\frac{6-\mathit{x}}{6}$
Then, by given condition,
$\frac{5}{6}\times \frac{\mathit{x}}{6}$ + $\frac{1}{6}\times \frac{6-\mathit{x}}{6}=\frac{1}{2}$
$\frac{5}{36}\mathit{x}+\frac{6-\mathit{x}}{36}=\frac{1}{2}$
$\frac{5\mathit{x}+6-\mathit{x}}{36}=\frac{1}{2}$
4x + 6 = 18
4x = 18 - 6 = 12
x = $\frac{12}{4}$ = 3
∴ Number of red faces on second cube = 3