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1. Large, medium and small ships are used to bring water. 4 large ships carry as much water as 7 small ships, 3 medium ships carry the same amount of water as 2 large ships and 1 small ship. 15 large, 7 medium and 14 small ships, each made 36 journeys and brought a certain quantity of water. In how many journeys would 12 large, 14 medium and 21 small ships bring the same quantity ?

A. 25

B. 29

C. 32

D. 49

Solution:

Let, L = large ships, M = medium ships, S = small ships
Now from the question,
4L ≡ 7S

\begin{aligned} 15L = (\frac{7}{4} \times 15) S = \frac{105}{4} S \\ Also, 2L = (\frac{7}{4} \times 2) S = \frac{7}{2} S \\ 3M \equiv 2 L + 1S \equiv (\frac{7}{2} + 1) S = \frac{9}{2} S \\ \Leftrightarrow 7M = (\frac{9}{2} \times \frac{1}{3} \times 7) S = \frac{21}{2} S \\ ∴ (15L + 7M + 14S) ships \\ \equiv (\frac{105}{4} + \frac{21}{2} + 14) S \\ = \frac{203}{4} S \\ (12 large + 14 medium + 2 1 small) ships \\ \equiv [(\frac{7}{4} \times 12) + (\frac{21}{2} \times 2) + 21] S \\ = (21 + 21 + 21) S \\ = 63 S \end{aligned}

Let the required number of journeys be x
More ships, Less journeys (Indirect proportion)

\begin{aligned} ∴ 63 : \frac{203}{4} :: 36 : x \\ \Leftrightarrow 63 x = \frac{203}{4} \times 36 = 1827 \\ \Leftrightarrow x = \frac{1827}{63} \\ \Leftrightarrow x = 29 \end{aligned}

Alternate:
Now from the question,

\begin{aligned} 4 L = 7 S \\ \Rightarrow \frac{L}{S} = \frac{7}{4} \\ \Rightarrow L = 7 x; S = 4 x \\ 3 M = 2 L + S \\ \Rightarrow M = \frac{2 \times 7 x + 4 x}{3} \\ \Rightarrow M = 6 x \end{aligned}

Thus, ratio of large, medium and small = 7 : 6 : 4
So, numbers of journeys:

\begin{aligned} = \frac{(15 \times 7 + 7 \times 6 + 14 \times 4) 36}{12 \times 7 + 14 \times 6 + 21 \times 4} \\ = \frac{7308}{252} \\ = 29 \end{aligned}

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