Question

Two water taps together can fill a tank in `1 7/8` hours. The tap with a longer diameter takes 2 hours less than the tap with a smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Answer 1

Let the time in which tap with a longer and smaller diameter can fill the tank separately be x hours and y hours respectively

`1/"x" + 1/"y" = 8/15`.....(i)

And x = y - 2.......(ii)

On substituting x = y - 2 from (ii) in (i), we get

`1/("y" - 2) + 1/"y" = 8/15`

`=> ("y + y" - 2)/("y"^2 - 2"y") = 8/15`

 `=>` 15(2y - 2) = 8(y² - 2y)

`=>` 8y² - 46y + 30 = 0

`=>` 4y² - 20y - 3y + 15 = 0

`=>` (4y - 3)(y - 5)= 0

`=> "y" = 3/4`, y = 5

Sunstituting values of y in (ii), we get

x = `3/4` - 2

x = `(-5)/4`

`therefore "x" ≠ (-5)/4`

(time cannot be negative)
Hence, the time taken by tap with longer diameter is 3 hours and the time taken by tap with smaller diameter is 5 hours, in order to fill the tank separately.