Question

Two water taps together can fill a tank in \(8\frac{8}{9}\) hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Hint:

When solving quadratic equations for time and work, always reject the value that makes any of the individual times negative or realistically impossible.

Answer 1

Step 1: Understanding the Concept:
The rate of filling is the reciprocal of the time taken. For two taps working together, the sum of their separate rates equals the combined rate.
Step 2: Key Formula or Approach:
If time taken is \(t\), work rate is \(1/t\).
Step 3: Detailed Explanation:
1. Let the time taken by the smaller tap to fill the tank be \(x\) hours.
2. Then, the time taken by the larger tap \(= (x - 4)\) hours.
3. Combined time \(= 8\frac{8}{9} = \frac{80}{9}\) hours.
4. Combined rate \(= \frac{9}{80}\) tank/hour.
5. According to the problem:
\[ \frac{1}{x} + \frac{1}{x - 4} = \frac{9}{80} \]
\[ \frac{(x - 4) + x}{x(x - 4)} = \frac{9}{80} \implies \frac{2x - 4}{x^2 - 4x} = \frac{9}{80} \]
\[ 80(2x - 4) = 9(x^2 - 4x) \implies 160x - 320 = 9x^2 - 36x \]
\[ 9x^2 - 36x - 160x + 320 = 0 \implies 9x^2 - 196x + 320 = 0 \]
6. Solving using quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{196 \pm \sqrt{(-196)^2 - 4(9)(320)}}{18} = \frac{196 \pm \sqrt{38416 - 11520}}{18} \]
\[ x = \frac{196 \pm \sqrt{26896}}{18} = \frac{196 \pm 164}{18} \]
- \(x = \frac{360}{18} = 20\)
- \(x = \frac{32}{18} \approx 1.77\) (Rejected as \(x-4\) would be negative).
7. Time for smaller tap \(= 20\) hours.
Time for larger tap \(= 20 - 4 = 16\) hours.
Step 4: Final Answer:
The smaller tap fills the tank in 20 hours and the larger tap fills it in 16 hours.