# Two water taps together can fill a tank in 9⅜ hours.The tap of larger diameter takes 10 hrs less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

The question is a real-life application of linear equations in two variables.

## Answer: The tap of smaller diameter can separately fill the tank in 25 hours whereas the larger tap can separately fill the tank in 15 hours.

Let's frame the equations suitable to the given data.

**Explanation:**

Let the time taken by:

- tap with a small diameter alone to fill the tank is t hours
- tap with a large diameter alone to fill the tank is (t - 10) hours.

So, in an hour:

- 1/t portion of the tank can be filled by the tap with a smaller diameter.
- 1/(t – 10) portion of the tank can be filled by the tap with a larger diameter.

Given that, the tank can be filled in 9⅜ hours = 75/8 hours if both the taps are left open.

Therefore, in an hour, both the taps can fill 8/75 portions of the tank.

1/t + 1/(t - 10) = 8/75

{(t-10) + t } / t(t - 10) = 8/75

{2t - 10} / t(t - 10) = 8/75

{2t - 10} / { t^{2} - 10t } = 8/75

{2t - 10} × 75 = { t^{2} - 10t } × 8

150t - 750 = 8t^{2} - 80t

8t^{2} - 80t - 150t + 750 = 0

8t^{2} - 230t + 750 = 0

2(4t^{2} - 115t + 375) = 0

4t^{2} - 100t -15t + 375 = 0 (splitting the middle-term)

4t(t - 25) - 15(t - 25) = 0

(4t - 15) (t - 25) = 0

t = 15/4 or t = 25

Here, t = 15/4 is not a feasible solution as t -10 value comes out to be negative.

So,

- the tap with a smaller diameter takes t hours i.e., 25 hours
- the tap with a larger diameter takes (t -10) = (25 -10) i.e., 15 hours.