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 Work Concept:
Instead of directly forming the rate equation, we first assume the capacity of the tank as LCM of the denominators to simplify calculation.
Combined time = 8 8/9 hours = 80/9 hours.

Let total work (tank capacity) = 80 units.
Then combined rate = 9 units per hour.

Step 2: Let Individual Times be:
Let time taken by smaller tap = x hours.
Then time taken by larger tap = (x − 4) hours.

Their individual rates:
Smaller tap = 80/x units per hour.
Larger tap = 80/(x − 4) units per hour.

Since together they fill 9 units per hour:
80/x + 80/(x − 4) = 9

Step 3: Forming the Equation:
Take LCM:
80(x − 4) + 80x / x(x − 4) = 9

80(2x − 4) = 9x(x − 4)

160x − 320 = 9x² − 36x

Rearranging:
9x² − 36x − 160x + 320 = 0
9x² − 196x + 320 = 0

Step 4: Solving the Quadratic:
Using quadratic formula:
x = [196 ± √(196² − 4×9×320)] / 18

= [196 ± √(38416 − 11520)] / 18
= [196 ± √26896] / 18
= [196 ± 164] / 18

Possible values:
x = 360/18 = 20
x = 32/18 ≈ 1.77 (Rejected, as x − 4 becomes negative)

Step 5: Final Values:
Time taken by smaller tap = 20 hours.
Time taken by larger tap = 20 − 4 = 16 hours.

Final Answer:
The smaller tap fills the tank in 20 hours and the larger tap fills it in 16 hours.