2018 AMC 8 Problems/Problem 12

Revision as of 23:47, 14 January 2022 by Ethan0619 (talk | contribs) (Problem)


The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?

$\textbf{ (A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10$

Solution 1

We see that every $35$ minutes the clock passes, the watch passes $30$ minutes. That means that the clock is $\frac{7}{6}$ as fast the watch, so we can set up proportions. $\dfrac{\text{car clock}}{\text{watch}}=\dfrac{7}{6}=\dfrac{7 \text{ hours}}{x \text{ hours}}$. Cross-multiplying we get $x=6$. Now this is obviously redundant, we could just eyeball it and see that the watch would have passed $6$ hours, but this method better when the numbers are a bit more complex, which makes it both easier and reliable.


Solution 2

When the car clock passes $7$ hours, the watch has passed $6$ hours, meaning that the time would be $\boxed{\textbf{(B) }6:00}$

Video Solution


See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

Invalid username
Login to AoPS