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

Mrdavid445 (talk | contribs) (Created page with "==Problem== The arithmetic mean (average) of four numbers is <math>85</math>. If the largest of these numbers is <math>97</math>, then the mean of the remaining three numbers i...") |
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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 . If the largest of these numbers is , then the mean of the remaining three numbers is

## Solution

Say that the four numbers are & , with . Then . What we are trying to find is . Solving,