Question

Large (L), medium (M) and small (S) ships bring water. Given: \(4L = 7S\); and \(3M\) carry the same as \(2L + 1S\). A fleet of \(15L, 7M, 14S\) made 36 journeys and brought a certain quantity of water. How many journeys would \(12L, 14M, 21S\) need to bring the same quantity?

• A. 32
• B. 25
• C. 29
• D. 49

Hint:

When multiple "types" work together, convert them to one common unit using the given equivalences; this avoids rounding and gives exact integer answers.

Answer 1

The Correct Option is C

Step 1: Express all capacities in "small-ship equivalents".
From \(4L=7S \Rightarrow L=\dfrac{7}{4}S=1.75S\).
From \(3M = 2L + 1S = 2\cdot\dfrac{7}{4}S + 1S = \dfrac{9}{2}S \Rightarrow M=\dfrac{3}{2}S=1.5S\).
Step 2: Compute first fleet's capacity per journey.
\(15L + 7M + 14S = 15\cdot\dfrac{7}{4}S + 7\cdot\dfrac{3}{2}S + 14S = 26.25S + 10.5S + 14S = 50.75S.\) Step 3: Total work done.
Total water \(W = 36 \times 50.75S = 1827S\) (exact). Step 4: Second fleet's capacity per journey.
\(12L + 14M + 21S = 12\cdot\dfrac{7}{4}S + 14\cdot\dfrac{3}{2}S + 21S = 21S + 21S + 21S = 63S.\) Step 5: Required journeys.
Journeys \(= \dfrac{W}{63S} = \dfrac{1827S}{63S} = \boxed{29}.\)