Question

Two water taps together can fill a tank in `8 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.

Answer 1

Let the tap of smaller diameter takes ‘x’ hours to fill the tank alone. 

Then, the tap of larger diameter takes ‘x – 4’ hours to fill the tank alone.

In 1 hour, part of tank filled by tap of smaller diameter = `1/x`

In 1 hour, part of tank filled by tap of larger diameter = `1/(x - 4)`

1n 1 hour, part of tank filled by both will be = `1/x + 1/(x - 4)`

In `8 8/9` hours, part of tank filled by both which is completely filled

⇒ `8 8/9 (1/x + 1/(x - 4)) = 1`

⇒ `80/9 (1/x + 1/(x - 4)) = 1`

⇒ `((x - 4) + x)/(x(x - 4)) = 9/80`

⇒ 80(2x – 4) = 9(x² – 4x)

⇒ 160x – 320 = 9x² – 36x

⇒ 9x² – 196x + 320 = 0

`x = (196 ± sqrt((196)^2 - 4 xx 9 xx 320))/(2 xx 9)`

`x = (196 ± sqrt(49^2 xx 4^2 - 4 xx 9 xx 4 xx 80))/18`

`x = (196 ± 4sqrt(49^2 - 720))/18`

`x = (196 ± 4 xx sqrt(1681))/18`

`x = (196 ± 4 xx 41)/18`

`x = (196 ± 164)/18`

⇒ `x = 16/9, 20`

But `x = 16/9` is not possible

Time taken to separately fill the tank by

⇒ Tap of smaller diameter = 20 hours.

⇒ Tap of larger diameter = 20 – 4 = 16 hours.