Difference between revisions of "1973 AHSME Problems/Problem 14"

Latest revision as of 12:59, 20 February 2020

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 $\frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5} = \boxed{\textbf{(C)}\ 1.2}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 19: / Line 19: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=13|num-a=15}}
+{{AHSME 30p box|year=1973|num-b=13|num-a=15}}
 [[Category:Introductory Algebra Problems]]

Page

Toolbox

Search

Difference between revisions of "1973 AHSME Problems/Problem 14"

Latest revision as of 12:59, 20 February 2020

Problem

Solution

See Also