Difference between revisions of "2018 AMC 8 Problems/Problem 12"
m (→Solution 2) |
|||
(20 intermediate revisions by 14 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==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? | + | 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, because we could just eyeball it to see that the watch would have passed <math>6</math> hours. But this method is better when the numbers are a bit more complex, which makes it both easier and reliable. Either way, our answer is <math>\boxed{\textbf{(B) }6:00}</math>. | ||
− | == Solution == | + | --BakedPotato66 |
− | + | --Rishi09 | |
− | {{AMC8 box|year=2018|num-b=11|num-a= | + | |
+ | == 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>. | ||
+ | (Note that this is because the ratio of the car clock to the watch clock is a ratio of 35:30 or 7:6 due to the clock being ahead by 5 minutes when the watch time has been 30 minutes since the "start time".) | ||
+ | |||
+ | ~edited by SHREYU.MEDATATI | ||
+ | |||
+ | == 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 <math>\boxed{\textbf{(B) }6:00}</math>. | ||
+ | |||
+ | ~SaxStreak | ||
+ | |||
+ | == Solution 4 (a version of solution 3) == | ||
+ | |||
+ | 7 hours have passed since 12:00 pm and 7 hours = 420 minutes because there are 60 minutes in an hour. Because every 35 minutes, the clock is ahead by 5 minutes, you need to divide 420 by 35 to find out how many times it happens. 420 divided by 35 is 12. Then you would multiply 12 by 5 because the clock is ahead by 5 minutes. 12 times 5 is 60, so that means that the clock is ahead by 60 minutes. In order to find the watch's time, you must find what was 60 minutes earlier than 7:00 which is <math>\boxed{\textbf{(B) }6:00}</math>. | ||
+ | |||
+ | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
+ | https://youtu.be/Klm4SbXQoL0 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/-1-frlYXgCU | ||
+ | |||
+ | https://youtu.be/WUjEqOZsAhg | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2018|num-b=11|num-a=13}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 16:13, 18 August 2024
Contents
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. Either way, our answer is .
--BakedPotato66 --Rishi09
Solution 2
When the car clock passes hours, the watch has passed hours, meaning that the time would be . (Note that this is because the ratio of the car clock to the watch clock is a ratio of 35:30 or 7:6 due to the clock being ahead by 5 minutes when the watch time has been 30 minutes since the "start time".)
~edited by SHREYU.MEDATATI
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
Solution 4 (a version of solution 3)
7 hours have passed since 12:00 pm and 7 hours = 420 minutes because there are 60 minutes in an hour. Because every 35 minutes, the clock is ahead by 5 minutes, you need to divide 420 by 35 to find out how many times it happens. 420 divided by 35 is 12. Then you would multiply 12 by 5 because the clock is ahead by 5 minutes. 12 times 5 is 60, so that means that the clock is ahead by 60 minutes. In order to find the watch's time, you must find what was 60 minutes earlier than 7:00 which is .
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
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.