A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:*

A. 6 hours

B. 10 hours

C. 15 hours

D. 30 hours

Let's denote the time taken by the first pipe to fill the tank as \( x \) hours. According to the problem, the second pipe takes 5 hours less than the first pipe, so it takes \( x - 5 \) hours. The third pipe takes 4 hours less than the second pipe, so it takes \( (x - 5) - 4 = x - 9 \) hours.

### Step-by-Step Solution:

1. **Work Done by Each Pipe:**
   - First pipe: \(\text{Work rate} = \frac{1}{x}\)
   - Second pipe: \(\text{Work rate} = \frac{1}{x - 5}\)
   - Third pipe: \(\text{Work rate} = \frac{1}{x - 9}\)

2. **Combined Work Rate of the First and Second Pipes:**
   Since the first two pipes operating together take the same time to fill the tank as the third pipe alone:
   \[
   \frac{1}{x} + \frac{1}{x - 5} = \frac{1}{x - 9}
   \]

3. **Solve the Equation:**
   To solve this equation, we'll first find a common denominator on the left-hand side:
   \[
   \frac{(x - 5) + x}{x(x - 5)} = \frac{1}{x - 9}
   \]
   Simplifying:
   \[
   \frac{2x - 5}{x(x - 5)} = \frac{1}{x - 9}
   \]

   Cross-multiplying to eliminate the denominators:
   \[
   (2x - 5)(x - 9) = x(x - 5)
   \]

   Expand both sides:
   \[
   2x^2 - 18x - 5x + 45 = x^2 - 5x
   \]
   Simplify:
   \[
   2x^2 - 23x + 45 = x^2 - 5x
   \]
   Move all terms to one side:
   \[
   x^2 - 18x + 45 = 0
   \]

4. **Solve the Quadratic Equation:**
   We can solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \( a = 1 \), \( b = -18 \), and \( c = 45 \).
   \[
   x = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(45)}}{2(1)}
   \]
   Simplify:
   \[
   x = \frac{18 \pm \sqrt{324 - 180}}{2} = \frac{18 \pm \sqrt{144}}{2} = \frac{18 \pm 12}{2}
   \]
   This gives two possible solutions:
   \[
   x = \frac{18 + 12}{2} = \frac{30}{2} = 15 \quad \text{or} \quad x = \frac{18 - 12}{2} = \frac{6}{2} = 3
   \]

5. **Check the Validity:**
   Since the second pipe takes \( x - 5 \) hours and \( x = 15 \) gives a positive time, \( x = 15 \) is the correct answer (because \( x = 3 \) would give a negative or unrealistic time for the second pipe).

The correct answer is **C. 15 hours**.

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you determine the combined work rate of two or more pipes working together?
2. What is the quadratic formula, and how is it used to solve equations?
3. How can this type of problem be solved using algebraic substitution?
4. Can you solve similar problems involving three or more variables?
5. What are some practical applications of work-rate problems in real life?

**Tip:** For work-rate problems, always express the work rates of each individual component and combine them accordingly. This makes setting up and solving equations easier.

Learn how to determine the time required by the first pipe to fill a tank based on work rate problems. This problem involves algebraic manipulation and the quadratic formula, focusing on calculating individual work rates and their combined effect. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation