1993 AJHSME Problems/Problem 15

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 $a, b, c,$ & $97$ . Then $\frac{a+b+c+97}{4} = 85$ . What we are trying to find is $\frac{a+b+c}{3}$ . Solving, $\frac{a+b+c+97}{4} = 85$ $a+b+c+97 = 340$ $a+b+c = 243$ $\frac{a+b+c}{3} = \boxed{\mathrm{(A)}\ 81}$

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1993 AJHSME Problems/Problem 15

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