Difference between revisions of "2018 AMC 8 Problems/Problem 12"
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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? | 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? | ||
| − | <math>\textbf{(A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10</math> | + | <math>\textbf{ (A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10</math> |
| + | |||
| + | == Solution 1 == | ||
| + | We see that every <math>35</math> minutes the clock passes, the watch passes <math>30</math> minutes. That means that the clock is <math>\frac{7}{6}</math> as fast the watch, so we can set up proportions. | ||
| + | <math>\dfrac{\text{car clock}}{\text{watch}}=\dfrac{7}{6}=\dfrac{7 \text{ hours}}{x \text{ hours}}</math>. Cross-multiplying we get <math>x=6</math>. Now this is obviously redundant, we could just eyeball it and see that the watch would have passed <math>6</math> hours, but this method better when the numbers are a bit more complex, which makes it both easier and reliable. | ||
| + | |||
| + | --BakedPotato66 | ||
| + | |||
| + | == Solution 2 == | ||
| + | When the car clock passes <math>7</math> hours, the watch has passed <math>6</math> hours, meaning that the time would be <math>\boxed{\textbf{(B) }6:00}</math> | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/-1-frlYXgCU | ||
| + | |||
| + | ==See Also== | ||
{{AMC8 box|year=2018|num-b=11|num-a=13}} | {{AMC8 box|year=2018|num-b=11|num-a=13}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 23:47, 14 January 2022
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, we could just eyeball it and see that the watch would have passed 
hours, but this method 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 
Video Solution
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. 
