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# Help with a Maths Question

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# Help with a Maths Question

23rd Sep 2008, 17:20

Join Date: Apr 2008
Location: UK
Posts: 20
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 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

Thank you!
23rd Sep 2008, 17:42

Join Date: Jan 2008
Posts: 88
I just did it in my head.

The answer is 4 ½ 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 ½ hours, catching the 2nd cyclist at a rate of 4mph reduces the 18 miles deficit to 0 (4 ½ hours x 4mph = 18).

I'm sure someone may be able to post an easy formula for you to remember.
25th Sep 2008, 16:37

Join Date: Apr 2008
Location: UK
Posts: 20
Thanks mate
26th Sep 2008, 10:02

Join Date: Oct 2007
Location: _... .. ._
Posts: 27
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
26th Sep 2008, 10:59

Join Date: Sep 2007
Location: inside of a pretty bustard
Age: 49
Posts: 263
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

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

d=v1*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
26th Sep 2008, 11:57

Join Date: Aug 2008
Location: between a rock and a hard place
Posts: 261
Time to Catch Up = (Vslowest x d) / (Vfastest - Vslowest)

might be easier to remember
26th Sep 2008, 13:50

Moderator

Join Date: Mar 2003
Location: Orlando, Florida
Age: 65
Posts: 2,586
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?
26th Sep 2008, 14:36

Join Date: Oct 2007
Location: _... .. ._
Posts: 27
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

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