A vessel having `30 m^(3)`of water is emptied through two openings, one small and the other large. Water flows out through the smaller opening at the rate of U m3/h and through the larger one at the rate of `V m^(3)//h`. Given that 3U + 2V = 70 and that the vessel gets fully emptied in 1 hour, what is V?

Options

A. `10m^(3)//h`
B. `20 m^(3)//h`
C. `30m^(3)//h`
D. `50m^(3)//h`

Answer

To solve the problem, we need to find the value of \( V \) given the conditions of the water flow through two openings in a vessel. Let's break it down step by step.

### Step 1: Set up the equations
We know that the total volume of water in the vessel is \( 30 \, m^3 \) and it gets emptied in 1 hour. The water flows out through two openings: one small opening at a rate of \( U \, m^3/h \) and one large opening at a rate of \( V \, m^3/h \).

From this information, we can write our first equation based on the total outflow:
\[ U + V = 30 \quad \text{(1)} \]

We are also given another equation:
\[ 3U + 2V = 70 \quad \text{(2)} \]

### Step 2: Solve for one variable
From equation (1), we can express \( U \) in terms of \( V \):
\[ U = 30 - V \quad \text{(3)} \]

### Step 3: Substitute into the second equation
Now, we substitute equation (3) into equation (2):
\[ 3(30 - V) + 2V = 70 \]
Expanding this gives:
\[ 90 - 3V + 2V = 70 \]
Combining like terms results in:
\[ 90 - V = 70 \]

### Step 4: Solve for \( V \)
Now, isolate \( V \):
\[ -V = 70 - 90 \]
\[ -V = -20 \]
\[ V = 20 \, m^3/h \]

### Step 5: Conclusion
Thus, the value of \( V \) is:
\[ \boxed{20 \, m^3/h} \]