Two cyclists start biking from a trail's start 3 hours

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Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 1/2 hours
C. 5 3/4 hours
D. 6 hours
E. 7 1/2 hours

The OA is B.

Please, can someone explain this PS question? I appreciate your help. Thanks!

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by Shahrukh@mbabreakspace » Fri Jul 06, 2018 5:45 am
Total distance travelled will be the same when the cyclist will catch up.
Let it catch up in 't' j=hours
So, 6*(t+3)= 10*t
=>t= 4.5 Hours

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by [email protected] » Fri Jul 06, 2018 5:52 am
Hi All,

We're told that two cyclists start biking from a trail's start 3 hours apart; he second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. We're asked how much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking. This is an example of a 'Chase Down' question and can be solved with a bit of Arithmetic and the Distance Formula.

The first cyclist has a 3 hour "head start" - and at a rate of 6 miles/hour will be 18 miles AHEAD of the second cyclist when that second cyclist starts moving. The difference in their two speeds is 4 miles/hour, meaning that for every hour that they are BOTH cycling, the second cyclist will 'catch up' 4 miles. Thus, it's fairly easy to calculate how long it will take for the second cyclist to 'chase down' the first...

(18 mile head start)/(4 miles/hour) = 4.5 hours

Final Answer: B

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by Scott@TargetTestPrep » Mon Jul 09, 2018 6:57 pm
BTGmoderatorLU wrote:Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 1/2 hours
C. 5 3/4 hours
D. 6 hours
E. 7 1/2 hours
We can let the time of the first cyclist = t + 3 and the time of the second cyclist = t, and thus:

6(t + 3) = 10t

6t + 18 = 10t

18 = 4t

18/4 = t

4.5 = t

Thus, 4.5 hours from the time the second cyclist started biking will pass before the second cyclist catches the first.

Answer: B

Scott Woodbury-Stewart
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