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 ?

Options

A. 32
B. 25
C. 29
D. 49

Answer

To solve the problem step-by-step, we will first establish the relationships between the capacities of the large, medium, and small ships, then calculate the total water carried by the first set of ships, and finally determine how many journeys the second set of ships would need to make to carry the same amount of water.

### Step 1: Establish Relationships
From the problem, we know:
- 4 large ships (L) = 7 small ships (S)
- 3 medium ships (M) = 2 large ships (L) + 1 small ship (S)

Let's denote:
- The capacity of a large ship as \( L \)
- The capacity of a medium ship as \( M \)
- The capacity of a small ship as \( S \)

From the first relationship:
\[ 4L = 7S \implies L = \frac{7}{4}S \]

From the second relationship:
\[ 3M = 2L + S \]
Substituting \( L \) from the first relationship:
\[ 3M = 2\left(\frac{7}{4}S\right) + S = \frac{14}{4}S + S = \frac{14}{4}S + \frac{4}{4}S = \frac{18}{4}S \]
Thus,
\[ M = \frac{18}{12}S = \frac{3}{2}S \]

### Step 2: Calculate Total Water Carried by the First Set of Ships
Now we have:
- \( L = \frac{7}{4}S \)
- \( M = \frac{3}{2}S \)

Next, we calculate the total water carried by 15 large ships, 7 medium ships, and 14 small ships, each making 36 journeys:
\[ \text{Total water} = \text{(Number of ships)} \times \text{(Capacity)} \times \text{(Journeys)} \]
Calculating for each type of ship:
- Water from large ships:
\[ 15L \times 36 = 15 \times \frac{7}{4}S \times 36 = \frac{105}{4}S \times 36 = \frac{3780}{4}S \]
- Water from medium ships:
\[ 7M \times 36 = 7 \times \frac{3}{2}S \times 36 = \frac{21}{2}S \times 36 = \frac{756}{2}S = 378S \]
- Water from small ships:
\[ 14S \times 36 = 14S \times 36 = 504S \]

Now, summing these amounts:
\[ \text{Total water} = \frac{3780}{4}S + 378S + 504S \]
Converting \( 378S \) and \( 504S \) to have a common denominator:
\[ 378S = \frac{1512}{4}S \quad \text{and} \quad 504S = \frac{2016}{4}S \]
Thus,
\[ \text{Total water} = \frac{3780 + 1512 + 2016}{4}S = \frac{7308}{4}S \]

### Step 3: Calculate Journeys for the Second Set of Ships
Now we need to find out how many journeys 12 large ships, 14 medium ships, and 21 small ships need to make to carry the same total amount of water:
\[ \text{Water from 12 large ships} = 12L \times J = 12 \times \frac{7}{4}S \times J = \frac{84}{4}S \times J = 21S \times J \]
\[ \text{Water from 14 medium ships} = 14M \times J = 14 \times \frac{3}{2}S \times J = 21S \times J \]
\[ \text{Water from 21 small ships} = 21S \times J \]

Total water from the second set:
\[ \text{Total water} = (21S + 21S + 21S) \times J = 63S \times J \]

### Step 4: Set the Total Water Equal
Now we set the total water from both sets equal:
\[ 63S \times J = \frac{7308}{4}S \]
Dividing both sides by \( S \) (assuming \( S \neq 0 \)):
\[ 63J = \frac{7308}{4} \]
Solving for \( J \):
\[ J = \frac{7308}{4 \times 63} = \frac{7308}{252} = 29 \]

### Final Answer
Thus, the number of journeys required for 12 large, 14 medium, and 21 small ships to bring the same quantity of water is **29 journeys**.

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