Difference between revisions of "1993 AJHSME Problems/Problem 15"

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<math>\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3</math>
 
<math>\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3</math>
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==Solution==
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Say that the four numbers are <math>a, b, c,</math> & <math>97</math>, with <math>a \le b \le c \le 97</math>. Then <math>\frac{a+b+c+97}{4} = 85</math>. What we are trying to find is <math>\frac{a+b+c}{3}</math>. Solving, <cmath>\frac{a+b+c+97}{4} = 85</cmath> <cmath>a+b+c+97 = 340</cmath> <cmath>a+b+c = 243</cmath> <cmath>\frac{a+b+c}{3} = 81 \Rightarrow \mathrm{(A)}</cmath>

Revision as of 13:26, 1 May 2012

Problem

The arithmetic mean (average) of four numbers is $85$. If the largest of these numbers is $97$, then the mean of the remaining three numbers is

$\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3$

Solution

Say that the four numbers are $a, b, c,$ & $97$, with $a \le b \le c \le 97$. Then $\frac{a+b+c+97}{4} = 85$. What we are trying to find is $\frac{a+b+c}{3}$. Solving, \[\frac{a+b+c+97}{4} = 85\] \[a+b+c+97 = 340\] \[a+b+c = 243\] \[\frac{a+b+c}{3} = 81 \Rightarrow \mathrm{(A)}\]

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