Difference between revisions of "1992 AJHSME Problems/Problem 13"

 
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==See Also==
 
==See Also==
 
{{AJHSME box|year=1992|num-b=12|num-a=14}}
 
{{AJHSME box|year=1992|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 00:09, 5 July 2013

Problem

Five test scores have a mean (average score) of $90$, a median (middle score) of $91$ and a mode (most frequent score) of $94$. The sum of the two lowest test scores is

$\text{(A)}\ 170 \qquad \text{(B)}\ 171 \qquad \text{(C)}\ 176 \qquad \text{(D)}\ 177 \qquad \text{(E)}\ \text{not determined by the information given}$

Solution

Because there was an odd number of scores, $91$ must be the middle score. Since there are two scores above $91$ and $94$ appears the most frequent (so at least twice) and $94>91$, $94$ appears twice. Also, the sum of the five numbers is $90 \times 5 =450$. Thus, the sum of the lowest two scores is $450-91-94-94= \boxed{\text{(B)}\ 171}$.

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
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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