2011 AMC 8 Problems/Problem 5

Problem

What time was it $2011$ minutes after midnight on January 1, 2011?

$\textbf{(A)}\ \text{January 1 at 9:31PM}$

$\textbf{(B)}\ \text{January 1 at 11:51PM}$

$\textbf{(C)}\ \text{January 2 at 3:11AM}$

$\textbf{(D)}\ \text{January 2 at 9:31AM}$

$\textbf{(E)}\ \text{January 2 at 6:01PM}$

Solution 1

There are $60$ minutes in an hour. $2011/60=33\text{r}31,$ or $33$ hours and $31$ minutes. There are $24$ hours in a day, so the time is $9$ hours and $31$ minutes after midnight on January 2, 2011. $\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$

Solution 2

Like the previous solution, $2011$ minutes is $33$ hours and $31$ minutes. That means it has to be January 1 at 9:31PM or January 2 at 9:31AM because they are the only ones that end with 31. But, the answer can’t be in January 1 because there are 24 hours in a day and 33 is greater than 24. So, the answer is $\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$ .

-RealCXY

Solution 3 (Estimate)

The answer choices are relatively far apart, so we can use estimation to solve this problem. $2100/60$ is around $35$ hours, or about $1.5$ days. The only answer choice around $1.5$ days from the midnight of January 1, 2011 is $\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$ .

MegaBoy6679 :D 23:31, 26 December 2022 (EST)

Video Solution by WhyMath

https://youtu.be/G4TaNyvDUOQ

2011 AMC 8 (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.

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