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Let's break down and solve the problem given in the image:

**Problem Statement:**

Two water taps together can fill a tank in \( 9 \frac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

---

### Solution:

Let the time taken by the smaller tap to fill the tank be \( x \) hours.

Then, the larger tap takes \( x - 10 \) hours to fill the tank.

The combined rate of both taps filling the tank is:

- Smaller tap's rate = \( \frac{1}{x} \) (fraction of the tank filled per hour)
- Larger tap's rate = \( \frac{1}{x - 10} \)

Together, they fill the tank in \( 9 \frac{3}{8} \) hours, which is \( \frac{75}{8} \) hours. Their combined rate is the reciprocal of this, i.e.,

\[
\frac{1}{\frac{75}{8}} = \frac{8}{75}
\]

So, the equation for their combined rates is:

\[
\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}
\]

---

### Step-by-Step Solution:

1. **Set up the equation**:
\[
\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}
\]

2. **Find a common denominator on the left-hand side**:
\[
\frac{(x - 10) + x}{x(x - 10)} = \frac{8}{75}
\]
This simplifies to:
\[
\frac{2x - 10}{x(x - 10)} = \frac{8}{75}
\]

3. **Cross multiply** to eliminate the fractions:
\[
75(2x - 10) = 8x(x - 10)
\]

4. **Expand both sides**:
\[
150x - 750 = 8x^2 - 80x
\]

5. **Move all terms to one side** to form a quadratic equation:
\[
8x^2 - 230x + 750 = 0
\]

6. **Solve the quadratic equation** using the quadratic formula:
\[
x = \frac{-(-230) \pm \sqrt{(-230)^2 - 4(8)(750)}}{2(8)}
\]
\[
x = \frac{230 \pm \sqrt{52900 - 24000}}{16}
\]
\[
x = \frac{230 \pm \sqrt{28900}}{16}
\]
\[
x = \frac{230 \pm 170}{16}
\]

7. **Find the two possible solutions**:
\[
x = \frac{230 + 170}{16} = \frac{400}{16} = 25
\]
or
\[
x = \frac{230 - 170}{16} = \frac{60}{16} = 3.75
\]

Since \( x \) represents the time taken by the smaller tap to fill the tank, and it cannot be 3.75 (because the larger tap would take negative time), we conclude:

\[
x = 25 \, \text{hours}
\]

Thus, the smaller tap takes **25 hours** to fill the tank, and the larger tap, taking 10 hours less, takes:

\[
25 - 10 = 15 \, \text{hours}
\]

---

### Final Answer:

- The smaller tap takes **25 hours**.
- The larger tap takes **15 hours**.

---

Would you like further clarification or details on any part of this solution? Here are 5 related questions to explore:

1. How would the solution change if the taps could fill the tank in \( 8 \) hours together?
2. Can you generalize this method for three taps instead of two?
3. How can you verify the solution using unit rates of filling the tank?
4. How would you solve if the taps' difference in filling time was unknown?
5. What happens if the taps work at a constant but variable rate (e.g., faster or slower over time)?

**Tip:** Always convert mixed numbers (like \( 9 \frac{3}{8} \)) into improper fractions to make the calculations easier.

Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Learn how to solve a work and time problem involving two taps filling a tank, one of which takes 10 hours less than the other. The solution demonstrates the use of quadratic equations and work rate concepts, making it ideal for Grades 9-12 students.

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