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07 Aug 2020, 02:14
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:03) correct
31%
(02:45)
wrong
based on 231
sessions
History
Date
Time
Result
Not Attempted Yet
Car A is 20 miles behind Car B, which is traveling in the opposite direction along the same route as Car A. Car A is traveling at a constant speed of 65 miles per hour and Car B is traveling at a constant speed of 55 miles per hour. Approximately how many minutes will it take for Car A and B be on a 20 miles distance from each other again?
(a) 10
(b) 200
(c) 20
(d) 45
(e) 36
(a) 10
(b) 200
(c) 20
(d) 45
(e) 36
07 Aug 2020, 02:35
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Given
To Find
Approach and Working Out
Correct Answer: Option D
- • Car A is 20 miles behind Car B, which is traveling in the opposite direction along the same route as Car A.
- o The behind word is creating confusion here. If you look at the image, then this statement indicates that this is case 2.
o In that case, they will never be 20 miles apart unless they are moving in a circular direction.
• Car A is traveling at a constant speed of 65 miles per hour and Car B is traveling at a constant speed of 55 miles per hour.
To Find
- • Minutes it will take for Car A and B be on a 20 miles distance from each other again.
- o The only possible way to understand this problem when they are initially in the position as case 1 and then they are in the position as in case 2.
Approach and Working Out
- • It is given that both the distances are 20 miles.
- o Relatively, they are traveling 40 miles.
- o They will take \(\frac{40}{120}\) hours = \(\frac{1}{3} \)hours or 20 minutes.
Correct Answer: Option D
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07 Aug 2020, 07:46
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BrendaMSL
HiBrendaMSL,
Welcome to GMAT club. Here is the solution to your question.
Now cars are 20 miles apart. For them to be 20 miles apart again, first cars must meet and then travel away from each other, until they are 20 miles apart again.
Let \(t\) be the time in which they meet then we can form this equation \(\hspace{1mm}\)\(55t + 65t = 20 \Rightarrow \hspace{2mm} t = \frac{1}{6}\) or 10 minutes
Now let \(x\) be the time in which they are 20 Miles apart traveling away from each other, again we can form the equation \(55x + 65x = 20 \Rightarrow x = \frac{1}{6}\) or 10 minutes.
So total time to be 20 Miles apart again 10 +10 = 20 Minutes
Hope it's clear.
