Difference between revisions of "1986 AJHSME Problems/Problem 5"
5849206328x (talk | contribs) (New page: ==Problem== A contest began at noon one day and ended <math>1000</math> minutes later. At what time did the contest end? <math>\text{(A)}\ \text{10:00 p.m.} \qquad \text{(B)}\ \text{mid...) |
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==Solution== | ==Solution== | ||
| − | + | 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 a.m.</math> | |
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| + | <math>4:40</math> is D | ||
==See Also== | ==See Also== | ||
[[1986 AJHSME Problems]] | [[1986 AJHSME Problems]] | ||
Revision as of 18:51, 24 January 2009
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 D
