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Man Cycling

by sandipgumtya » Sun Feb 21, 2016 8:30 am
A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?

A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes

Sorry,as this may be a repetition. Pl help to understand the concept here.OA-B.
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by GMATGuruNY » Sun Feb 21, 2016 9:39 am
A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes
The time interval between consecutive buses is equal to how often the buses DEPART from the station: every 5 minutes, every 6 minutes, etc.
All of the buses -- in each direction -- travel at the same uniform speed.
The result is that the distance between consecutive buses is always the same.

The distance between consecutive buses can be any value.
Let the distance between consecutive buses = 24 units.
Let b = the rate of each bus and c = the rate of the cyclist.

SAME DIRECTION:
Here, the buses and the cyclist are COMPETING, so we SUBTRACT their rates.
The time needed for the next bus to CATCH UP to the cyclist is 12 minutes.
Thus:
b-c = d/t = 24/12 = 2 units per minute.

OPPOSITE DIRECTIONS:
Here, the buses and the cyclist are WORKING TOGETHER to cover the distance between them, so we ADD their rates.
The time needed for the cyclist and the next oncoming bus to PASS EACH OTHER is 4 minutes.
b+c = d/t = 24/4 = 6 units per minute.

Adding the two equations, we get:
(b-c) + (b+c) = 2+6
2b = 8
b=4 units per minute.

Since the rate of each bus is 4 units per minute and the distance between consecutive buses is 24 units:
The time interval between consecutive buses = d/r = 24/4 = 6 minutes.

The correct answer is B.
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by [email protected] » Sun Feb 21, 2016 12:27 pm
Hi sandipgumtya,

There was a lengthy discussion of this question here:

https://www.beatthegmat.com/speed-problem-t272446.html

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by sandipgumtya » Mon Feb 22, 2016 8:52 pm
Hi Mitch,
I could not get the underlying concept yet.Can u pl explain once more?
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by GMATGuruNY » Tue Feb 23, 2016 6:20 am
sandipgumtya wrote:Hi Mitch,
I could not get the underlying concept yet.Can u pl explain once more?
Here is the same approach applied to a different distance:

Let the distance between consecutive buses = 12 units.
Let b = the rate of each bus and c = the rate of the cyclist.

SAME DIRECTION:
Here, the buses and the cyclist are COMPETING, so we SUBTRACT their rates.
The time needed for the next bus to CATCH UP to the cyclist is 12 minutes.
Thus:
b-c = d/t = 12/12 = 1 unit per minute.

OPPOSITE DIRECTIONS:
Here, the buses and the cyclist are WORKING TOGETHER to cover the distance between them, so we ADD their rates.
The time needed for the cyclist and the next oncoming bus to PASS EACH OTHER is 4 minutes.
b+c = d/t = 12/4 = 3 units per minute.

Adding the two equations, we get:
(b-c) + (b+c) = 1+3
2b = 4
b = 2 units per minute.

Since the rate of each bus is 2 units per minute, and the distance between consecutive buses is 12 units:
The time interval between consecutive buses = d/r = 12/2 = 6 minutes.

Regardless of the distance between consecutive buses, the time interval between consecutive buses is THE SAME:
6 minutes.

The correct answer is B.
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by sandipgumtya » Thu Feb 25, 2016 7:57 pm
thanks a lot Mitch.Got it.
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by Matt@VeritasPrep » Tue Mar 01, 2016 11:29 pm
Another way of thinking about this:

If two people are traveling TOWARD each other, we can add their rates to determine when they'll meet.

If one person is trying to catch another, we can subtract the slower person's rate from the faster person's rate to determine when they'll meet.

Suppose that our bus has rate b, that the distance between buses is d, and (for simplicity's sake) that our cyclist's rate is 15 mph.

Let's pick as our starting point a time at which the cyclist is overtaken by one bus and meets another. At that point, the buses behind each of the two he has just seen are exactly d miles away, in either direction.

In twelve minutes, our cyclist travels 3 miles. In that time, the bus that overtakes him travels (d + 3) miles: the d miles between buses, plus the 3 miles required to catch the cyclist.

In four minutes, our cyclist travels 1 mile. In that time, the bus that he meets travels (d - 1) miles: the d miles between buses minus the 1 mile the cyclist has traveled.

Using D = RT, we now have

(d + 3) = b * 12

and

(d - 1) = b * 4

This gives us d = 3 and b = 1/2.

We want to solve for t in the equation

d = b * t, so

3 = (1/2)t

or

6 = t

This question seems about as hard as a GMAT rate problem could be, so don't let it make you hop in front of a bus! It's quite unlikely to appear on the exam, and if it does, you can be confident that you're doing very well.
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