Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(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...)
 
(→‎Solution)
 
(4 intermediate revisions by 3 users not shown)
Line 7: Line 7:
 
==Solution==
 
==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 \text{ a.m.}</math>
 +
 
 +
<math>4:40</math> is <math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
  
−
[[1986 AJHSME Problems]]
+
{{AJHSME box|year=1986|num-b=4|num-a=6}}
 +
[[Category:Introductory Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:47, 21 February 2019

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_5&oldid=103703"