Difference between revisions of "1992 AJHSME Problems/Problem 14"

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Four gallons is <math> \frac{1}{2}-\frac{1}{3}= \frac{1}{6} </math>. Thus, capacity of the tank in gallons is <math> 6 \times 4= \boxed{\text{(D)}\ 24} </math>.
 
Four gallons is <math> \frac{1}{2}-\frac{1}{3}= \frac{1}{6} </math>. Thus, capacity of the tank in gallons is <math> 6 \times 4= \boxed{\text{(D)}\ 24} </math>.
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==See Also==
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{{AJHSME box|year=1992|num-b=13|num-a=15}}

Revision as of 21:36, 22 December 2012

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

$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 48$

Solution

Four gallons is $\frac{1}{2}-\frac{1}{3}= \frac{1}{6}$. Thus, capacity of the tank in gallons is $6 \times 4= \boxed{\text{(D)}\ 24}$.

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
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