#### Question: Two water taps together can fill a tank in 9(\\frac{3}{8})) hours. The tap of larger dimensions 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.

#### Video Explanation

#### Explanatory Answer

Step 1: Frame equations based on the information about the time taken by the two taps to fill the tank.

Let the time taken by the tap of larger dimension be t hours.

∴ time taken by the tap of smaller dimension = (t + 10) hours.

The larger tap will fill \\frac{1}{t}) of the tank in 1 hour

The smaller tap will fill \\frac{1}{t + 10}) of the tank in 1 hour

Together the 2 taps will fill \\frac{1}{t}) + \\frac{1}{t + 10}) of the tank in 1 hour ----------(1)

Statement: Two water taps fill a tank in 9\\frac{3}{8}) hrs

i.e., in \\frac{75}{8}) hours

So, the taps will fill \\frac{1}{75/8})= \\frac{8}{75}) of the tank in 1 hour ----------(2)

Equate (1) and (2)

\\frac{1}{t}) + \\frac{1}{t + 10}) = \\frac{8}{75})

Because t ≠ 0 and t cannot be -10, multiply the entire equation by t(t + 10)

(t + 10) + t = \\frac{8}{75})(t (t + 10))

2t + 10 = \\frac{8}{75})(t^{2} + 10t)

75(2t + 10) = 8t^{2} + 80 t

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

8t^{2} - 70t - 750 = 0

Divide the equation by 2 :

4t^{2} - 35t - 375 = 0

##### Step 2: Factorize to find the values of ‘t’

4t^{2} – 60t + 25t – 375 = 0

4t(t – 15) + 25 (t – 15) = 0

(4t + 25) (t – 15) = 0

t = \\frac{-25}{4}) and t = 15

Time taken is positive

∴ t = 15 hrs and t + 10 = 15 + 10 = 25 hrs