Two water taps together can fill a tank in `1(7)/(8)` hours. The tap with 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.

#### Solution

Let the first tap takes x hours to completely fill tank

⇒ Second tap will take 2 hours less

⇒ According to question

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

`("x"-2+"x")/("x"("x"-2)) = (8)/(15)`

`(2"x"-2)/("x"("x"-2)) = (18)/(15)`

`((2"x"-1))/("x"("x"-2)) = (18)/(15)`

`15 ("x"-1) = 4"x" ("x" -2)`

`15"x" - 15 = 4"x"^2 - 8"x"`

`4"x"^2 - 23"x" + 15 = 0`

`4"x"^2 - 20"x" - 3"x" + 15 = 0`

`4"x"("x" - 5) - 3 ("x" -5) = 0`

`("x" - 5) (4"x" - 3) = 0`

`"x" = 5 or (3)/(4)`

Since `(3)/(4) - 2 = "Negative time" (3)/(4)` is not possible.

Which is not possible

⇒ x = 5

Rate of 1^{st }pipe = 5 hours

Rate of 2^{nd} pipe = 5 - 2 = 3hours