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 "2011 AMC 8 Problems/Problem 5"

(→‎Solution)
m
Line 19: Line 19:
  
 
Like the previous solution, <math>2011</math> minutes is <math>33</math> hours and <math>31</math> 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 <math>\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}</math>.
 
Like the previous solution, <math>2011</math> minutes is <math>33</math> hours and <math>31</math> 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 <math>\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}</math>.
 +
 
-RealCXY
 
-RealCXY
 +
 +
==Solution 3 (Estimate)==
 +
 +
The answer choices are relatively far apart, so we can use estimation to solve this problem. <math>2100/60</math> is around <math>35</math> hours, or about <math>1.5</math> days. The only answer choice around <math>1.5</math> days from the midnight of January 1, 2011 is <math>\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}</math>.
 +
MegaBoy6679 :D 23:31, 26 December 2022 (EST)
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=4|num-a=6}}
 
{{AMC8 box|year=2011|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:31, 26 December 2022

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)

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_5&oldid=184964"