Difference between revisions of "1971 AHSME Problems/Problem 16"
Coolmath34 (talk | contribs) (Created page with "== Problem == After finding the average of <math>35</math> scores, a student carelessly included the average with the <math>35</math> scores and found the average of these <...") |
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If we add <math>18</math> to the first <math>35</math> numbers, the new average is <math>\frac{648}{36} = 18</math> still. | If we add <math>18</math> to the first <math>35</math> numbers, the new average is <math>\frac{648}{36} = 18</math> still. | ||
| − | The two averages are the same, therefore the answer is <math>\textbf{(A) }1:1.</math> | + | The two averages are the same, therefore the answer is <math>\boxed{\textbf{(A) }1:1}.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=15|num-a=17}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:00, 5 August 2024
Problem
After finding the average of 
scores, a student carelessly included the average with the scores and found the
average of these 
numbers. The ratio of the second average to the true average was
Solution
Assume the scores are the first natural numbers: 
The average of the scores is 
If we add 
to the first numbers, the new average is 
still.
The two averages are the same, therefore the answer is 
-edited by coolmath34
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
