Difference between revisions of "1986 AJHSME Problems/Problem 5"

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==See Also==
 
==See Also==
  
[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=4|num-a=6}}
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[[Category:Introductory Number Theory Problems]]

Revision as of 18:57, 20 May 2009

Problem

A contest began at noon one day and ended $1000$ minutes later. At what time did the contest end?

$\text{(A)}\ \text{10:00 p.m.} \qquad \text{(B)}\ \text{midnight} \qquad \text{(C)}\ \text{2:30 a.m.} \qquad \text{(D)}\ \text{4:40 a.m.} \qquad \text{(E)}\ \text{6:40 a.m.}$

Solution

There are $60$ minutes in an hour. So, we can easily eliminate some of the choices by noting that noon is exactly $720$ minutes away from midnight. Since $720 < 1000$, we know that it cannot be A or B. Because midnight is $720$ minutes away, we know that the contest ended $1000 - 720 = 280$ minutes after midnight. The highest multiple of 60 that will fit into $280$ is $240$, which is $4 \times 60$, and the remainder is $40$ minutes, meaning that the contest ended at $4:40 \text{a.m.}$

$4:40$ is $\boxed{\text{D}}$

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
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