# Difference between revisions of "2012 AMC 8 Problems/Problem 3"

## Problem

On February 13 The Oshkosh Northwester listed the length of daylight as 10 hours and 24 minutes, the sunrise was $6:57\textsc{am}$, and the sunset as $8:15\textsc{pm}$. The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?

$\textbf{(A)}\hspace{.05in}5:10\textsc{pm}\quad\textbf{(B)}\hspace{.05in}5:21\textsc{pm}\quad\textbf{(C)}\hspace{.05in}5:41\textsc{pm}\quad\textbf{(D)}\hspace{.05in}5:57\textsc{pm}\quad\textbf{(E)}\hspace{.05in}6:03\textsc{pm}$

## Solution 1

The problem wants us to find the time of sunset and gives us the length of daylight and time of sunrise. So all we have to do is add the length of daylight to the time of sunrise to obtain the answer. Convert 10 hours and 24 minutes into $10:24$ in order to add easier.

Adding, we find that the time of sunset is $6:57\textsc{am} + 10:24 \implies 17:21 \implies \boxed{\textbf{(B)}\ 5:21\textsc{pm}}$.

## Solution 2 (Answer choices)

The length of daylight is 10 hours and 24 minutes, but for now, we can just look at the 24 minutes. Add 24 minutes to 57 part of sunrise, and you see that the sunset time must end in a 21. We see that the only option that ends with 21 is $\implies \boxed{\textbf{(B)}\ 5:21\textsc{pm}}$.

## See Also

 2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 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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