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Difference between revisions of "1994 AJHSME Problems/Problem 13"

(Created page with "==Problem== The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is <math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(...")
 
(→‎Uh Oh Unluky 13!)
 
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==Problem==
 
==Problem==
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The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is
 
The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is
  
 
<math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}</math>
 
<math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}</math>
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==Solution==
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The number halfway between is the average.
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<cmath>\frac{\frac16 + \frac14}{2} = \frac{\frac{2}{12} + \frac{3}{12}}{2} = \boxed{\text{(C)}\ \frac{5}{24}}</cmath>
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==See Also==
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{{AJHSME box|year=1994|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 12:24, 30 October 2016

Problem

The number halfway between $\dfrac{1}{6}$ and $\dfrac{1}{4}$ is

$\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}$

Solution

The number halfway between is the average.

\[\frac{\frac16 + \frac14}{2} = \frac{\frac{2}{12} + \frac{3}{12}}{2} = \boxed{\text{(C)}\ \frac{5}{24}}\]

See Also

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

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