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Difference between revisions of "1986 AJHSME Problems/Problem 5"
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==Solution==
 
==Solution==
  
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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>
+
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 D
+
<math>4:40</math> is <math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 17:53, 24 January 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 Problems

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