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 | + | 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> |
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| + | 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. | ||
==See Also== | ==See Also== | ||
Latest revision as of 11:26, 13 June 2023
Problem
The annual incomes of 
families range from 
dollars to 
dollars. In error, the largest income was entered on the computer as 
dollars. The difference between the mean of the incorrect data and the mean of the actual data is
Solution
Let 
be the sum of all the incomes but the largest one. For the actual data, the mean is 
, and for the incorrect data the mean is 
. The difference is 
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.
See Also
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
