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22 Nov 2018, 10:57
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C
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Difficulty:
15%
(low)
Question Stats:
83% (01:54) correct
17%
(02:32)
wrong
based on 218
sessions
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Not Attempted Yet
Sarah and James are running around a circular track. Sarah runs at a constant speed of 800 feet/minute and James runs at a constant speed of 1200 feet/minute. James is currently 80 feet behind Sarah, how many seconds will it take James to catch up to Sarah and run 40 feet ahead of her?
A. 5 Seconds
B. 10 Seconds
C. 18 Seconds
D. 40 Seconds
E. 50 Seconds
A. 5 Seconds
B. 10 Seconds
C. 18 Seconds
D. 40 Seconds
E. 50 Seconds
Jazzmin
Right now, James is 80 feet behind. He needs to be 40 feet ahead. So he needs to cover a relative distance of 80 + 40 = 120 feet.
Time taken = Relative distance/Relative Speed = 120/(1200 - 800) = 120/400 mins = 120/400 * 60 sec = 18 sec
Answer (C)
Note that the fact that they are running in a circle doesn't have much to do here.
Check this video for when to use relative speed: https://youtu.be/wrYxeZ2WsEM
General Discussion
22 Nov 2018, 20:18
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What is the source of this question?
Have to be careful with the information. Not only does he start 80ft behind, he also needs to get 40ft in front.
80+40=120. Rewriting the question: How many seconds does it take for James to gain 120ft on Sarah?
Speed·time=distance
Since there are two speeds and we are trying to find the difference between the speeds
(\(Speed_J\) - \(Speed_S\))·time=distance
\(\frac{1200ft}{min}- \frac{800ft}{min}·x = 120ft\)
\(x(\frac{400ft}{min}) = 120ft\)
\(\frac{x400ft}{min} = 120ft\)
We want to eliminate the feet units, so swap the denominator and the distance
\(\frac{x400ft}{120ft} = 1 min\)
Can now cancel units
\(\frac{x400}{120}=60 seconds\)
\(\frac{10x}{3}=60 seconds\)
10x=180 seconds
x=18 seconds
C
Have to be careful with the information. Not only does he start 80ft behind, he also needs to get 40ft in front.
80+40=120. Rewriting the question: How many seconds does it take for James to gain 120ft on Sarah?
Speed·time=distance
Since there are two speeds and we are trying to find the difference between the speeds
(\(Speed_J\) - \(Speed_S\))·time=distance
\(\frac{1200ft}{min}- \frac{800ft}{min}·x = 120ft\)
\(x(\frac{400ft}{min}) = 120ft\)
\(\frac{x400ft}{min} = 120ft\)
We want to eliminate the feet units, so swap the denominator and the distance
\(\frac{x400ft}{120ft} = 1 min\)
Can now cancel units
\(\frac{x400}{120}=60 seconds\)
\(\frac{10x}{3}=60 seconds\)
10x=180 seconds
x=18 seconds
C
