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Difference between revisions of "2018 AMC 8 Problems/Problem 12"
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== Solution 1 ==
 
== 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.  
 
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.
+
<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>.
  
 
--BakedPotato66
 
--BakedPotato66
 +
--Rishi09
  
 
== Solution 2 ==
 
== 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>.
+
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 ==
 
== 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. In addition, 6 hours added to 12:00 pm (noon) makes the answer <math>\boxed{\textbf{(B) }6:00}</math>.
+
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
 +
 +
== 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 ==
 
== Video Solution ==

Latest revision as of 16:13, 18 August 2024

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, because we could just eyeball it to see that the watch would have passed $6$ 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 $\boxed{\textbf{(B) }6:00}$.

--BakedPotato66 --Rishi09

Solution 2

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

~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 $\boxed{\textbf{(B) }6:00}$.

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

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. AMC logo.png

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