# 1973 AHSME Problems/Problem 14

## Problem

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 $\textbf{(A)}\ 1.1\qquad\textbf{(B)}\ 1.15\qquad\textbf{(C)}\ 1.2\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 1.75$

## Solution

Let the rate of water flowing through valve $A$ be $a$, the rate of water flowing through valve $B$ be $b$, and the rate of water flowing through valve $C$ be $c$. WLOG, let the volume of the tank be 1 liter, and let the units for the rates be liters per hour. With this information, we can write three equations. $$\frac{1}{a+b+c} = 1$$ $$\frac{1}{a+c} = \frac{3}{2}$$ $$\frac{1}{b+c} = 2$$ Manipulate each equation to get $$1 = a+b+c$$ $$\frac{2}{3} = a+c$$ $$\frac{1}{2} = b+c$$ Solving for $a$ yields $a = \tfrac{1}{2}$, and solving for $b$ yields $b = \tfrac{1}{3}$. The number of hours to fill the tub with only valves $A$ and $B$ on is $\frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5} = \boxed{\textbf{(C)}\ 1.2}$.