Solve the following problems using two variables :
Two taps A and B can together fill a swimming pool in 15 days . Taps A and B are kept open for 12 days and then tap B is closed . It takes another 8 days for the pool to be filled . How many days does each tap require to fill the pool ?

Let the rate at which tap A fills the pool be \(\frac{1}{x}\) pools per day, and the rate at which tap B fills the pool be \(\frac{1}{y}\) pools per day, where \(x\) and \(y\) are the number of days it takes for taps A and B, respectively, to fill the pool individually.

### Step 1: Using the information that both taps together fill the pool in 15 days:
The combined rate of both taps is \(\frac{1}{x} + \frac{1}{y}\), and they fill the pool in 15 days, so:
\[
\frac{1}{x} + \frac{1}{y} = \frac{1}{15}
\]
This is the first equation.

### Step 2: Using the information that both taps are open for 12 days and then only tap A is open for another 8 days:
- For the first 12 days, both taps are open, so they fill \(\left( \frac{1}{x} + \frac{1}{y} \right) \times 12\) of the pool.
- For the remaining 8 days, only tap A is open, so it fills \(\frac{1}{x} \times 8\) of the pool.

Together, the pool is fully filled, so:
\[
12\left(\frac{1}{x} + \frac{1}{y}\right) + 8\left(\frac{1}{x}\right) = 1
\]
This is the second equation.

### Step 3: Solving the system of equations:
From the first equation:
\[
\frac{1}{x} + \frac{1}{y} = \frac{1}{15}
\]
Substitute \(\frac{1}{x} + \frac{1}{y} = \frac{1}{15}\) into the second equation:
\[
12 \times \frac{1}{15} + 8 \times \frac{1}{x} = 1
\]
Simplifying:
\[
\frac{12}{15} + \frac{8}{x} = 1
\]
\[
\frac{4}{5} + \frac{8}{x} = 1
\]
Subtract \(\frac{4}{5}\) from both sides:
\[
\frac{8}{x} = 1 - \frac{4}{5} = \frac{1}{5}
\]
Solve for \(x\):
\[
x = \frac{8}{\frac{1}{5}} = 40
\]

### Step 4: Solving for \(y\):
Substitute \(x = 40\) into the first equation:
\[
\frac{1}{40} + \frac{1}{y} = \frac{1}{15}
\]
Subtract \(\frac{1}{40}\) from both sides:
\[
\frac{1}{y} = \frac{1}{15} - \frac{1}{40}
\]
Find a common denominator:
\[
\frac{1}{y} = \frac{8}{120} - \frac{3}{120} = \frac{5}{120} = \frac{1}{24}
\]
Solve for \(y\):
\[
y = 24
\]

### Final Answer:
- Tap A can fill the pool in 40 days.
- Tap B can fill the pool in 24 days.

Would you like a more detailed explanation or any clarifications?

Here are 5 related questions to expand on this problem:
1. What happens if the rates of taps A and B are altered? How will it affect the solution?
2. How would the solution change if the taps were left open for different time intervals?
3. What is the general formula for solving such problems with two variables?
4. How can we modify this problem to include a third tap?
5. How does solving with two variables compare to solving similar problems with just one variable?

**Tip:** For rate problems, always break down the work done by each part and use algebraic equations to track their contributions over time.

Two taps A and B can together fill a swimming pool in 15 days. Taps A and B are kept open for 12 days and then tap B is closed. It takes another 8 days for the pool to be filled. How many days does each tap require to fill the pool?

Learn how to solve a work problem involving two taps filling a pool. Taps A and B together fill the pool in 15 days, but tap B is closed after 12 days. Find out how many days each tap takes to fill the pool individually with a step-by-step algebraic solution. Suitable for Grades 8-10.

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