1990 AJHSME Problems/Problem 20

Problem

The annual incomes of $1,000$ families range from $8200$ dollars to $98,000$ dollars. In error, the largest income was entered on the computer as $980,000$ dollars. The difference between the mean of the incorrect data and the mean of the actual data is

$\text{(A)}\ \text{882 dollars} \qquad \text{(B)}\ \text{980 dollars} \qquad \text{(C)}\ \text{1078 dollars} \qquad \text{(D)}\ \text{482,000 dollars} \qquad \text{(E)}\ \text{882,000 dollars}$

Solution

Let $S$ be the sum of all the incomes but the largest one. For the actual data, the mean is $\frac{S+98000}{1000}$ , and for the incorrect data the mean is $\frac{S+980000}{1000}$ . The difference is $882, or \rightarrow \boxed{\text{A}}$

This assumes that incomes of all the families are unique. Otherwise, more than one family could have $98000 as income and the above reasoning will not work.

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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
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1990 AJHSME Problems/Problem 20

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