Difference between revisions of "1992 AJHSME Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | Four gallons is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math> of the tank. Thus, capacity of the tank in gallons is <math>6 cdot 4= \boxed{\text{(D)}\ 24}</math>. | + | Four gallons is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math> of the tank. Thus, capacity of the tank in gallons is <math>6 \cdot 4= \boxed{\text{(D)}\ 24}</math>. |
== See Also == | == See Also == | ||
{{AJHSME box|year=1992|num-b=13|num-a=15}} | {{AJHSME box|year=1992|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 09:36, 6 January 2025
Problem
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
Solution
Four gallons is 
of the tank. Thus, capacity of the tank in gallons is 
.
See Also
| 1992 AJHSME (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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
