Difference between revisions of "1994 AJHSME Problems/Problem 3"
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==Problem== | ==Problem== | ||
| − | Each day Maria must work <math>8</math> hours | + | Each day Maria must work <math>8</math> hours. This does not include the <math>45</math> minutes she takes for lunch. If she begins working at <math>\text{7:25 A.M.}</math> and takes her lunch break at noon, then her working day will end at |
<math>\text{(A)}\ \text{3:40 P.M.} \qquad \text{(B)}\ \text{3:55 P.M.} \qquad \text{(C)}\ \text{4:10 P.M.} \qquad \text{(D)}\ \text{4:25 P.M.} \qquad \text{(E)}\ \text{4:40 P.M.}</math> | <math>\text{(A)}\ \text{3:40 P.M.} \qquad \text{(B)}\ \text{3:55 P.M.} \qquad \text{(C)}\ \text{4:10 P.M.} \qquad \text{(D)}\ \text{4:25 P.M.} \qquad \text{(E)}\ \text{4:40 P.M.}</math> | ||
Revision as of 12:15, 30 October 2016
Contents
Problem
Each day Maria must work 
hours. This does not include the 
minutes she takes for lunch. If she begins working at 
and takes her lunch break at noon, then her working day will end at
Solution
8 hours from 7:25 AM is 15:25 or 3:25 PM. 45 minutes from 25 minutes is 10 minutes after the hour, so her working day ends at 
-The Hacker
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
