Question

b) Two pipes together can fill a tank in 6
hours. If the pipe of smaller
diameter takes 3 hours more than the pipe of larger diameter to fill the
tank separately, find the time taken by each pipe to fill the tank
separately?​

Answer 1

Two pipes take 6 hours to fill the tank.

$1 \text{ hour} = \dfrac{1}{6} \text{ of the tank}$

Define x:

Let the smaller pipe takes x hours to fill the tank

$1 \text{ hour} = \dfrac{1}{x} \text{ of the tank}$

The bigger pipe takes (x - 3) hours to fill the tank

$1 \text{ hour} = \dfrac{1}{x- 3} \text{ of the tank}$

Together:

$1 \text{ hour} = \dfrac{1}{x} + \dfrac{1}{x - 3}$

Solve x:

$\dfrac{1}{x} + \dfrac{1}{x - 3} = \dfrac{1}{6}$

$\dfrac{x - 3}{x(x - 3)} + \dfrac{x}{x(x - 3)} = \dfrac{1}{6}$

$\dfrac{x - 3 + x}{x(x - 3)} = \dfrac{1}{6}$

$\dfrac{2x - 3}{x(x - 3)} = \dfrac{1}{6}$

$6(2x - 3) = x(x - 3)$

$12x - 18 = x^2 - 3x$

$x^2 - 15x + 18 = 0$

$x = 1.315 \text{ or } x = 13.685$

⇒ x = 13.685 because x must be greater than 6

Find the time taken by the pipes:

$\text{Smaller Pipe} = x$

$\text{Smaller Pipe} = 13.7 \text { hours}$

$\text{Bigger Pipe} = x - 3$

$\text{Bigger Pipe} = 13.7 - 3$

$\text{Bigger Pipe} =10.7 \text { hours}$

Answer: smaller pipe takes 13.7 hours and the bigger pipe takes 10.7 hours