Symbiosis National Aptitude Test SNAP 2012 Paper

Let the quantity carried by large, medium and small ships be L, M and S respectively.
4L = 7S, 3M = 2L + S
⇒ L = $\frac{7}{4}$ S and 3M = $2\times \frac{7}{2}$ S + S = $\frac{7}{2}$ S + S = $\frac{9}{2}$ S
⇒ M = $\frac{3}{2}$ S
Now 15 L + 7M + 14S = $\frac{7}{4}\times 15\mathit{S}+\frac{3}{2}\times 7\mathit{S}+14\mathit{S}$
= $\frac{105\mathit{S}}{4+\frac{21}{2}\mathit{S}+14\mathit{S}}$
= $\frac{105\mathit{S}+42\mathit{S}+56\mathit{S}}{4}$
= $\frac{203}{4}$ S
Total quantity carried = $\frac{203}{4}\mathit{S}$ × 36 = 1827 S
Again, 12L + 14M + 21S = $\frac{7}{4}$ × 12 S + $\frac{3}{2}$ × 14 X + 21 X
= 21S + 21S + 21S = 63S
∴ Number of journeys = $\frac{1827\mathit{S}}{63\mathit{S}}$ = 29