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work-rate problem - Manhattan test material

Expert replies
by waltz2salsa » Mon Oct 25, 2010 1:11 am
Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Mon Oct 25, 2010 2:34 am
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Quickest approach is to ballpark. Pi ≈ 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi ≈ 60 miles.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 ≈ 12 + 8.4 = 20.4.

The correct answer is B
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by waltz2salsa » Mon Oct 25, 2010 9:24 am
Many thanks man, similar explanation but quicker :)
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by gmatapril » Wed Mar 23, 2011 1:41 pm
why are we using combined rate why can't we just divide 52 by 2. since we just have to find car B's hours. please
clear my doubt

GMATGuruNY wrote:
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Quickest approach is to ballpark. Pi ≈ 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi ≈ 60 miles.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 ≈ 12 + 8.4 = 20.4.

The correct answer is B
Join the discussion

by GMATGuruNY » Thu Mar 24, 2011 12:28 pm
gmatapril wrote:why are we using combined rate why can't we just divide 52 by 2. since we just have to find car B's hours. please
clear my doubt

GMATGuruNY wrote:
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Quickest approach is to ballpark. Pi ≈ 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi ≈ 60 miles.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 ≈ 12 + 8.4 = 20.4.

The correct answer is B
After B has traveled for 10 hours alone, A starts to travel as well.
At this point we have to combine A's rate with B's rate because the two cars are traveling toward each other.
Since B's rate is 2 mph, each hour B travels 2 miles toward A.
Since A's rate is 3 mph, each hour A travels 3 miles toward B.
Thus, every hour A and B travel 2+3 = 5 miles toward each other.
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by gmatapril » Fri Mar 25, 2011 10:18 am
thanks a lot for clearing my doubt. just one more question if both the cars were moving in same direction then we would had subtracted their Rates.
thank you.

GMATGuruNY wrote:
gmatapril wrote:why are we using combined rate why can't we just divide 52 by 2. since we just have to find car B's hours. please
clear my doubt

GMATGuruNY wrote:
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Quickest approach is to ballpark. Pi ≈ 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi ≈ 60 miles.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 ≈ 12 + 8.4 = 20.4.

The correct answer is B
After B has traveled for 10 hours alone, A starts to travel as well.
At this point we have to combine A's rate with B's rate because the two cars are traveling toward each other.
Since B's rate is 2 mph, each hour B travels 2 miles toward A.
Since A's rate is 3 mph, each hour A travels 3 miles toward B.
Thus, every hour A and B travel 2+3 = 5 miles toward each other.
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by GMATGuruNY » Fri Mar 25, 2011 1:03 pm
gmatapril wrote:thanks a lot for clearing my doubt. just one more question if both the cars were moving in same direction then we would had subtracted their Rates.
thank you.

GMATGuruNY wrote:
gmatapril wrote:why are we using combined rate why can't we just divide 52 by 2. since we just have to find car B's hours. please
clear my doubt

GMATGuruNY wrote:
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Quickest approach is to ballpark. Pi ≈ 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi ≈ 60 miles.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 ≈ 12 + 8.4 = 20.4.

The correct answer is B
After B has traveled for 10 hours alone, A starts to travel as well.
At this point we have to combine A's rate with B's rate because the two cars are traveling toward each other.
Since B's rate is 2 mph, each hour B travels 2 miles toward A.
Since A's rate is 3 mph, each hour A travels 3 miles toward B.
Thus, every hour A and B travel 2+3 = 5 miles toward each other.
Yes, if A and B were moving in the same direction, we would subtract the slower rate from the faster rate.
Since A = 3 mph and B = 2 mph, in one hour A will travel 3-2 = 1 more mile than B.
Thus, if A and B leave from the same location at the same time, the rate at which A will move ahead of B = 3-2 = 1 mph.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by minkathebest » Sat Nov 10, 2012 2:51 pm
It is a pretty good question. The lesson that I learned here were couple:
1) look for what the question is asking. I stopped solving when I found the total time it takes for them to travel the asked distance, when the question was asking how much time B kept travelling.
2) This is a collision problem, we ADD the rates, not subtract.

It is helpful to draw a diagram on this one:

Draw a circle with an arrow going clock-wise, indicating that B is traveling in that direction.

Now, RTD chart

R T D
2mph x 10 h = 20 miles

so, B has been traveling for 20 miles when A starts.

The entire track is 2 pi r = 20 pi (This means that B is somewhere on the track, I drew another picture to just keep that in mind)

Now, the question is asking how long will it take for them to get past each other + add 12 miles.

So, we have total distance = 20pi - 20 + 12 = 20 pi - 8

Therefore,

20 pi 8 = (5) T
T = 4 pi - 1.6

Now, the fun has not stopped yet, we need to find the total time B has been traveling.

B had already traveled 10 hours so we add this in:

(4 pi - 1.6) + 10 = 4pi - 8.4 Answer.
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by Ankur87 » Sat Apr 20, 2013 9:59 am
Hi,

I guess we can't include 10 hrs which B already covered as the ques says for how many hours will car B have been travelling WHEN the cars have passed each other for the first time and put another 12 miles between them.
It is not asking the total time B traveled.

i.e. (20pie- 20) + 12 = (20pie - 8) miles / 5

Please clear my doubt.
GMATGuruNY wrote:
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Quickest approach is to ballpark. Pi ≈ 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi ≈ 60 miles.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 ≈ 12 + 8.4 = 20.4.

The correct answer is B
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by mevicks » Sat Sep 21, 2013 8:13 am
Ankur87 wrote:Hi,

I guess we can't include 10 hrs which B already covered as the ques says for how many hours will car B have been travelling WHEN the cars have passed each other for the first time and put another 12 miles between them.
It is not asking the total time B traveled.

i.e. (20pie- 20) + 12 = (20pie - 8) miles / 5

Please clear my doubt.
Hi Ankur,

If you carefully review what that question is asking is then you will come to a conclusion that 10 hours have to be indeed added into the final answer for B.
"for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them "

Have been traveling = Continuous tense, the action started in the past and still continues... B started long back and the poor chap is still driving!
and = +
So essentially its asking "How much distance traveled UNTIL NOW" + "How much for another 12 miles"

Hope that clears the doubt.

Regards,
Vivek
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by GMATinsight » Thu Oct 22, 2015 6:32 am
waltz2salsa wrote:Ques: Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4pi- 1.6
b) 4pi+ 8.4
c) 4pi+ 10.4
d) 2pi- 1.6
e) 2pi- 0.8

OA:B

I am not satisfied with the explanation provided. Can someone please help!

Regards,
Waltz2Salsa
Check the solution...

Answer: Option B
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by Brent@GMATPrepNow » Thu Oct 22, 2015 9:13 am
For practice, here's an easier question dealing with people on a circular track: https://www.beatthegmat.com/frank-and-ed ... 29750.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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