# 2018 AMC 8 Problems/Problem 12

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem 12

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

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.

--BakedPotato66

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 