Difference between revisions of "2018 AMC 8 Problems/Problem 12"
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== Solution 3 == | == Solution 3 == | ||
| − | From 12:00 pm (noon) to 7:00 pm, the car clock has passed 12 35-minute cycles (12 X 35 = 420) because 7 hours = 420 minutes. So, 12 30-minute cycles (12 X 30 = 360) for the watch time are 360 minutes, which is 6 hours. | + | From 12:00 pm (noon) to 7:00 pm, the car clock has passed 12 35-minute cycles (12 X 35 = 420) because 7 hours = 420 minutes. So, 12 30-minute cycles (12 X 30 = 360) for the watch time are 360 minutes, which is 6 hours. Therefore, 6 hours added to 12:00 pm (noon) makes the answer <math>\boxed{\textbf{(B) }6:00}</math>. |
~SaxStreak | ~SaxStreak | ||
Revision as of 05:15, 1 January 2023
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?
Solution 1
We see that every 
minutes the clock passes, the watch passes 
minutes. That means that the clock is 
as fast the watch, so we can set up proportions.
. Cross-multiplying we get 
. Now, this is obviously redundant, because we could just eyeball it to see that the watch would have passed 
hours. But this method is better when the numbers are a bit more complex, which makes it both easier and reliable.
--BakedPotato66
Solution 2
When the car clock passes 
hours, the watch has passed hours, meaning that the time would be 
.
Solution 3
From 12:00 pm (noon) to 7:00 pm, the car clock has passed 12 35-minute cycles (12 X 35 = 420) because 7 hours = 420 minutes. So, 12 30-minute cycles (12 X 30 = 360) for the watch time are 360 minutes, which is 6 hours. Therefore, 6 hours added to 12:00 pm (noon) makes the answer .
~SaxStreak
Video Solution
~savannahsolver
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
