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!
rates and circles
This topic has expert replies
- ssmiles08
- Master | Next Rank: 500 Posts
- Posts: 472
- Joined: Sun Mar 29, 2009 6:54 pm
- Thanked: 56 times
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)
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)
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.
-
- Master | Next Rank: 500 Posts
- Posts: 399
- Joined: Wed Apr 15, 2009 3:48 am
- Location: india
- Thanked: 39 times
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
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
It does not matter how many times you get knocked down , but how many times you get up
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!
I have to read the qs stem properly I guess!
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
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
-
- GMAT Instructor
- Posts: 1302
- Joined: Mon Oct 19, 2009 2:13 pm
- Location: Toronto
- Thanked: 539 times
- Followed by:164 members
- GMAT Score:800
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
https://www.beatthegmat.com/hollyyy-diff ... tml#201896
Kaplan Teacher in Toronto
- GMAT680
- Senior | Next Rank: 100 Posts
- Posts: 57
- Joined: Wed Feb 17, 2010 6:15 pm
- Location: MD
- Thanked: 2 times
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
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