Difference between revisions of "1994 AJHSME Problems/Problem 14"
Mrdavid445 (talk | contribs) (Created page with "==Problem== Two children at a time can play pairball. For <math>90</math> minutes, with only two children playing at time, five children take turns so that each one plays the s...") |
Lamboreghini (talk | contribs) (→Solution) |
||
| (8 intermediate revisions by 5 users not shown) | |||
| Line 4: | Line 4: | ||
<math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36</math> | <math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36</math> | ||
| + | |||
| + | ==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. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:06, 7 November 2018
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.
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. 
