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1986 AJHSME Problems/Problem 5: Difference between revisions

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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 \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 $1000$ 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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