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Help with a Maths Question

Old 23rd Sep 2008, 17:20
  #1 (permalink)  
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Question Help with a Maths Question

Hello All,

I came across this question on the internet whilst revising maths for COMPASS. I'd very much appreciate it if anyone could please enlighten me with the most direct approach to answering the it:

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 travelling 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 hours
C. 5 hours
D. 6 hours
E. 7 hours

Thank you!
Ouseburn is offline  
Old 23rd Sep 2008, 17:42
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Join Date: Jan 2008
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I just did it in my head.

The answer is 4 hours. By the time the second cyclist sets of the first cyclist is 18 miles ahead (6 mph x 3 hours = 18 miles). So, from the point when the second cyclist sets off he is matching the first cyclists speed AND exceeding it at a rate of 4 mph (10mph - 6mph = 4mph).

So travelling for 4 hours, catching the 2nd cyclist at a rate of 4mph reduces the 18 miles deficit to 0 (4 hours x 4mph = 18).

I'm sure someone may be able to post an easy formula for you to remember.
Shiver me timbers! is offline  
Old 25th Sep 2008, 16:37
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Thanks mate
Ouseburn is offline  
Old 26th Sep 2008, 10:02
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Is it me or does the question not make it clear whether they are looking for total time elapsed or time elapsed since the second bike started?

One gives B, the other gives E
EGHH is offline  
Old 26th Sep 2008, 10:59
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The right answer is B, because if the first one starts at 2 o'clock, in 7 1/2 hrs he will make 6*7 1/2 iaw 45 miles.The second one will starts at 5 o'clock so he will make in 4 1/2 * 10 45 miles too.So the second one will catch the first one at 9 1/2 o'clock

so first have v1 and the second one v2, the distance is the same, we have to find out the time t

d=v2*(t-3) so v1*t=v2*t-3*v2 so v2*t -v1*t=3*v2 so t(v2-v1)=3v2 so t=3v2/ v2-v1 so 30/4 so 7,5 iaw 7 hrs and 30 minutes

so the time is 7 1/2 - 3 = 4 1/2
airman13 is offline  
Old 26th Sep 2008, 11:57
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Time to Catch Up = (Vslowest x d) / (Vfastest - Vslowest)

might be easier to remember
ChrisLKKB is offline  
Old 26th Sep 2008, 13:50
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EGHH - it says from the time the SECOND cyclists started - so it's only catch up time. B.

Had it been FIRST cyclist - then yes. E.

I did it the same way David UK did.

airman13 - say that again - v e r y slowly?
Keygrip is offline  
Old 26th Sep 2008, 14:36
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Lol, thanks KeyGrip, you helped me complete my thoroughly planned and totally intended demonstration of why it's important to read the question properly. Um, yep, that'll do it.

I swear that "from the time the second cyclist started biking" wasn't there when I read it the first time
EGHH is offline  

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