Difference between revisions of "1994 AJHSME Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | There are <math>2 \times 90 = 180</math> minutes of total playing time. Divided equally among the five children, each child gets <math>180/5 = \boxed{\text{(E)}\ 36}</math> minutes. Note that the answer key says C, but the answer is actually <math>\boxed{\text{(E)} | + | There are <math>2 \times 90 = 180</math> minutes of total playing time. Divided equally among the five children, each child gets <math>180/5 = \boxed{\text{(E)}\ 36}</math> minutes. Note that the answer key says C, but the answer is actually <math>\boxed{\text{(E)}}</math>, please notice this. |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|num-b=13|num-a=15}} | {{AJHSME box|year=1994|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:28, 30 October 2016
Problem
Two children at a time can play pairball. For 
minutes, with only two children playing at time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is
Solution
There are 
minutes of total playing time. Divided equally among the five children, each child gets 
minutes. Note that the answer key says C, but the answer is actually 
, please notice this.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
