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rates and circles

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by vkb16 » Wed Sep 30, 2009 3:58 am
Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?

(Pie = p)

4P - 1.6
4P + 8.4
4P + 10.4
2P - 1.6
2P - 0.8

OA is B
My qs is, why do we have to involve Pie in the ans?
I got ans as 24 by following the catchup speed method.
For A to catch up, it will take 20/ (3-2) = 20 hrs.
Now for a to pass 12 miles ahead of B, it will take 12/3 = 4 more hours.
Am I missing something?

thanks!
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by ssmiles08 » Wed Sep 30, 2009 5:57 am
The radius of the circle is 10, so the distance would be the circumference of the circle which is 20pi.

2(t + 10) + 3t - 12 = 20pi

5t + 8 = 20pi

t = 4pi - 1.6

the time for car B is t + 10 since its already been on the road for 10 more hours.

4pi - 1.6 + 10 = 4pi + 8.4 (B)
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by xcusemeplz2009 » Wed Sep 30, 2009 6:09 am
IMO B

in 10 hrs car B will cover 20 miles at a speed of 2mph

tot dis is 20pi miles

when A will start dis left for passing each other is 20pi-20 miles

relative speed is 3+2=5

time taken to cross is (20pi-20)/5=4pi-4

time taken for another 12 miles is 12/5=2.4

tot time= 4pi-4+2.4=4pi-1.6

now we need to add 10 hrs for tot time of B

4pi-1.6+10=4pi+8.4
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by vkb16 » Wed Sep 30, 2009 10:38 pm
thanks a lot guys! I realised my mistake.. I didnt notice that the 2 cars were going in Opposite directions, and hence we have to ADD the rates, not subtract them!

I have to read the qs stem properly I guess!
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by enniguy » Wed Jan 13, 2010 8:42 pm
I did the same mistake of subtracting both the speeds. Wonder why. Anyways, this method could be faster I guess.

Total Distance = 20pi - 20 (B has already traveled) + 12 (Extra 12 miles from Qn)
= 20pi - 8.

Total Speed = 3 + 2

time = Distance/Speed
= 4pi - 1.6
But B has already traveled for 10 hours, So
4pi+8.4
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by Testluv » Wed Jan 13, 2010 8:57 pm
I discussed a solution to this question, and provided some tips for multiple object rate problems here:

https://www.beatthegmat.com/hollyyy-diff ... tml#201896
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by GMAT680 » Sun Mar 07, 2010 10:59 am
here is my approach to solve this problem.

Car B: D(B) = 2(t+10)
Car A: D(A) = 2pi*r - D(B) + 12 = 3t

Substitute Statement one on 2:
2pi*r - 2(t+10) + 12 = 3t
20pi -2t -20 +12 = 3t
20pi -8= 5t
t = 4pi - 8/5
add 10 hours to find time for t(B) = 4pi -(8/5) + 10 = 4pi + 42/5
Hence t(B) = 4pi + 8.4
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