Difference between revisions of "1990 AJHSME Problems/Problem 20"

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==Solution==
 
==Solution==
  
Let <math>S</math> be the sum of all the incomes but the largest one.  For the actual data, the mean is <math>\frac{S+98000}{1000}</math>, and for the incorrect data the mean is <math>\frac{S+980000}{1000}</math>.  The difference is <math>882\rightarrow \boxed{\text{The answer is A}}</math>
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Let <math>S</math> be the sum of all the incomes but the largest one.  For the actual data, the mean is <math>\frac{S+98000}{1000}</math>, and for the incorrect data the mean is <math>\frac{S+980000}{1000}</math>.  The difference is <math>882, or \rightarrow \boxed{\text{A}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 17:27, 7 May 2021

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 $882, or \rightarrow \boxed{\text{A}}$

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions
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