2430.11 – Three Valves

Each valve, $A$ , $B$ , and $C$ , when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in 1 hour; with only valves $A$ and $C$ open, it takes 1.5 hours; and with only valves $B$ and $C$ open, it takes 2 hours. The number of hours required with only valves $A$ and $B$ open is:

$1.1$
$1.15$
$1.2$
$1.25$
$1.75$

Solution

Let $T$ be the volume of the tank. Let $a$ , $b$ , and $c$ be such that:

$A$ takes $a$ hours to fill the tank alone $\leadsto$ $A$ fills $\dfrac{T}{a}$ in 1 hour.
$B$ takes $b$ hours to fill the tank alone $\leadsto$ $B$ fills $\dfrac{T}{b}$ in 1 hour.
$C$ takes $c$ hours to fill the tank alone $\leadsto$ $C$ fills $\dfrac{T}{c}$ in 1 hour.

Now, all 3 together fill the tank in one hour:

$\dfrac{T}{a} + \dfrac{T}{b} + \dfrac{T}{c} = T$

$A$ and $C$ take $1.5$ hours to fill the tank, so in one hour:

$\dfrac{T}{a} + \dfrac{T}{c} = \dfrac{2T}{3}$

$B$ and $C$ take 2 hours, so in one hour:

$\dfrac{T}{b}+ \dfrac{T}{c} = \dfrac{T}{2}$

The first two equations give us $\dfrac{T}{b}= \dfrac{T}{3}$ , so $b = 3$ hours. The third equation gives us $\dfrac{T}{3} + \dfrac{T}{c} = \dfrac{T}{2}$ , so $\dfrac{T}{c} = \dfrac{T}{6}$ and $c = 6$ hours.

Finally, plug in $b$ and $c$ to get $\dfrac{T}{a} + \dfrac{T}{6} = \dfrac{2T}{3}$ , so $\dfrac{T}{a} = \dfrac{T}{2}$ and $a = 2$ hours.

Interim check: $\dfrac{T}{2} + \dfrac{T}{3} + \dfrac{T}{6} = T$ .

In one hour, $A$ and $B$ fill $\dfrac{T}{2} + \dfrac{T}{3} = \dfrac{5T}{6}$ , so they need $\dfrac{6}{5} = 1.2$ hours to fill the tank. The answer is (c).