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 - 2y) = 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`

Substituting values of y in (ii), we get

`x = 3/4 - 2`

`x = (-5)/4`

∴ `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.