Difference between revisions of "2018 AMC 8 Problems/Problem 12"

(Created page with "==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...")
 
(20 intermediate revisions by 13 users not shown)
Line 1: Line 1:
==Problem 12==
+
==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>.
 +
 
 +
--BakedPotato66
 +
--Rishi09
 +
 
 +
== 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>.
 +
 
 +
== 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 ==
 +
https://youtu.be/-1-frlYXgCU
 +
 
 +
https://youtu.be/WUjEqOZsAhg
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2018|num-b=11|num-a=13}}
 +
 
 +
{{MAA Notice}}

Revision as of 10:17, 15 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?

$\textbf{ (A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10$

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

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

https://youtu.be/-1-frlYXgCU

https://youtu.be/WUjEqOZsAhg

~savannahsolver

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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