Difference between revisions of "1986 AJHSME Problems/Problem 5"
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− | There are <math>60</math> minutes in an hour. So, we can easily eliminate some of the choices by noting that noon is exactly <math>720</math> minutes away from midnight. Since <math>720 < 1000</math>, we know that it cannot be A or B. Because midnight is <math>720</math> minutes away, we know that the contest ended <math>1000 - 720 = 280</math> minutes after midnight. The highest multiple of 60 that will fit into <math>280</math> is <math>240</math>, which is <math>4 \times 60</math>, and the remainder is <math>40</math> minutes, meaning that the contest ended at <math>4:40 \text{a.m.}</math> | + | There are <math>60</math> minutes in an hour. So, we can easily eliminate some of the choices by noting that noon is exactly <math>720</math> minutes away from midnight. Since <math>720 < 1000</math>, we know that it cannot be A or B. Because midnight is <math>720</math> minutes away, we know that the contest ended <math>1000 - 720 = 280</math> minutes after midnight. The highest multiple of 60 that will fit into <math>280</math> is <math>240</math>, which is <math>4 \times 60</math>, and the remainder is <math>40</math> minutes, meaning that the contest ended at <math>4:40 \text{ a.m.}</math> |
<math>4:40</math> is <math>\boxed{\text{D}}</math> | <math>4:40</math> is <math>\boxed{\text{D}}</math> |
Latest revision as of 12:47, 21 February 2019
Problem
A contest began at noon one day and ended minutes later. At what time did the contest end?
Solution
There are minutes in an hour. So, we can easily eliminate some of the choices by noting that noon is exactly minutes away from midnight. Since , we know that it cannot be A or B. Because midnight is minutes away, we know that the contest ended minutes after midnight. The highest multiple of 60 that will fit into is , which is , and the remainder is minutes, meaning that the contest ended at
is
See Also
1986 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.