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

Revision as of 20:07, 3 July 2013

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 $\boxed{\text{D}}$

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.

@@ Line 15: / Line 15: @@
 {{AJHSME box|year=1986|num-b=4|num-a=6}}
 [[Category:Introductory Number Theory Problems]]
+{{MAA Notice}}

Page

Toolbox

Search

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

Revision as of 20:07, 3 July 2013

Problem

Solution

See Also