Circular Track

[kean.allison]

From MGMAT CAT #6
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?
A) 4- 1.6
B) 4 + 8.4
C) 4 + 10.4
D) 2 - 1.6
E) 2 - 0.8

Can somebody help? I'm getting lost in the MGMAT official explanation. I'm bad at these problems anyway, but I think the circular track and the opposite directions are throwing me and I don't know how to think about this.

[nitin_prakash_khanna]

I think your answers are missing the ¶ (pie) , lets call it P.

So answer choices should be
A) 4P- 1.6
B) 4 P+ 8.4
C) 4P + 10.4
D) 2P - 1.6
E) 2P - 0.8
Now to the solution....

It will be beneficial to quickly draw a circle and keep track of A & B's movement.

We need to calculate time when A & B are 12 miles apart given all the conditions and constraint given in Question.

step 1.
B has a lead of 10 hours.

SInce its a circular track, the total circumference is 2*10*P = 20P (remember P is pie and 10 is radius of track , given)

B's speed is 2 mph, in 10 hrs it will travel 2*10 = 20 miles.

Step 2.

Now A starts from the same point in and in opposite direction. Now how far are A & B on the track......if you have the figure handy, you can see from the starting point if B has travelled 20 miles on a track of 20P length. the distance to original point is 20P-20, where A is standing currently.

A's speed is 3 mph.

Now when will they cross , will be given by the below equation, assuming they cross after t hours.

20P-20-2t = 3t
5t = 20P-20
t = 4P-4 hours

Step 3.
After corssing each other they continue to move in opposite direction, A is moving at 3 mph, B is moving at 2 mph. After how many hours distance between them will 12 miles.

3t+2t = 12
t = 2.4 hours

Final step
how long B has travelled
intial 10 hrs + the time both crossed + the time it took for the distance betn them to be 12 miles

10+4P-4+2.4
4P+8.4
Ans B.

:)

[esledge]

I do this one much the same as Nitin. In fact, exactly the same up to this step:

the distance to original point is 20P-20, where A is standing currently.

We can save a bit of time if you use the "working together shortcut," in which we add the rates of the two cars.

We have a distance of 20pi-20 plus another 12 after the cars pass each other.

We have a combined speed of 2mph + 3mph (i.e. the cars approach each other on the track at 5mph).

Time traveling together= Distance/Rate = (20pi-20+12)miles/5mph = (20pi-8)miles/5mph = (4pi-1.6) hours

Note that (A) is the trap answer! We must remember that this is just the time since car A started, and we need to add 10 hours head start that car B had.

Time is therefore 4pi-1.6+10 = 4pi - 8.4.

[rkim81]

Hi two questions, I got this same question on my MGMAT CAT #3, is there a reason why the first poster got it on the 6th and I got it on the 3rd? This was one of the last questions, that I guessed to my benefit correctly.

Now for the mathematics. I understand that you have to make an RTD chart and come up with the equation 3t+2t+20=20pie. But why do you add 10 again after you've already solved? You've already added 10 to the time of Car B, isn't this adding the time twice?

[esledge]

Hi two questions, I got this same question on my MGMAT CAT #3, is there a reason why the first poster got it on the 6th and I got it on the 3rd? This was one of the last questions, that I guessed to my benefit correctly.

The GMAT exam is adaptive. That means that there is a big pool of questions and the algorithm selects questions for you based on your demonstrated ability (i.e. how you've answered previous questions), the mix of PS/DS you've seen on a given exam, the mix of topics you've already seen on a given exam, and even (within these and other parameters) randomly. You just happened to see it on your 3rd exam, but anyone can see it at different points or maybe not at all.

By the way, this is why we want people to post the first few words of a question as the subject, rather than posting the exam/question # it was for them.

Now for the mathematics. I understand that you have to make an RTD chart and come up with the equation 3t+2t+20=20pie. But why do you add 10 again after you've already solved? You've already added 10 to the time of Car B, isn't this adding the time twice?

The first time Nitin used the 10 hours for Car B, it was to figure out where on the track B was at the time when A started driving. He was essentially "resetting" the question so he could start from the time when Car A began, as if both cars were starting simultaneously.

He then calculates the time from the simultaneous beginning to the meeting time to get the total time that passed while both were in motion. This is Car A's total time, but 10 hours less than Car B's time, since it ignors the 10 hour reset we did earlier.

If it were Car A's distance that mattered, we would just use this time. But since it is Car B's distance, we need to add 10 back to it's time.

I think you are thinking of it as an advance of 10 hours for B and then another advance of 10 hours. If fact, we took 10 hours away from B at the beginning (by assuming the distance it travelled was already done), so we have to give those 10 hours (and that already-travelled distance) back.

[tienvunguyen]

This method, I believe, only works when the circular perimeter is longer than the total distance that Car B travels. What if we have a shorter track and during those 10 hours, Car B travels around the circle more than once? In that case, 2t+20+3t = 2*r*pi*k (where k is the number of laps car B already covered during 10 hours + 1). And this problem will get more complicated.

Don't you agree?

[RonPurewal]

This method, I believe, only works when the circular perimeter is longer than the total distance that Car B travels. What if we have a shorter track and during those 10 hours, Car B travels around the circle more than once? In that case, 2t+20+3t = 2*r*pi*k (where k is the number of laps car B already covered during 10 hours + 1). And this problem will get more complicated.

Don't you agree?

well, sure. on pretty much any problem in the world, there are going to be additional factors you could introduce that would complicate the problem further; this is no exception.

if you introduce this extra issue (a shorter track that will allow the cars to "lap" each other), then the problem will become far too "tricky" for the typical standards of this test.

on this problem, though, there is enough information to guarantee that this situation will not occur.

[chidam_baram_m]

Hi, I have a further question on this: we have calculated t as 4pi-4. When we are doing the final step in the calculation, we again calculate t as 2.4 hrs. Should this be some other variable such as t1..cause t=4pi-4 gives 8.56 as the value, assuming pi as 3.14. This way, 5t would be 40+hrs, which is ridiculous. So should some other variable be used here..?please explain.

[mschwrtz]

If I understand nitin_prakash_khanna's answer, then yes, it does seem to represent a couple of different time values as t. So I think that you followed it correctly.

[nehajadoo]

thanks all
i had this ques too on the MGMAT prac test and was wrong so the explanations above def helped :)

However, side concern. I did not take any notice Which prac test this came on. Are you guys saying that the test themselves are adaptive? as in, if i did well on test1 overall, test2 would be harder?
i thought the adaptive algorithm is Within a test and test 1,2,3 etc all start with same level difficulty and adapt in the test itself?

[mschwrtz]

nehajadoo, the tests are adaptive ONE AT A TIME, just as you understood. Our tests happen to share a pool of questions, so that the one person might see a question on CAT 1 that another sees on CAT 4. Until you reset your exams, we'll remember which questions you've seen, just so that you don't see them again, but you'll start every CAT with an otherwise perfectly clean slate.

[vijaykumar_ntpc]

perimeter of the track 20pie
in 10 hours B travelled 20
distance between A and B when A starts 20p-20
relative speed =2+3=5
time taken to complete 20p-20; when both meet each other=(20p-20)/5
time taken to travel 12 after meeting when both are going in opposite direction=12/5
total time of A=(20p-20)/5+12/5=4p-1.6
total time of B=4p-1.6+10=4p+8.4

[mschwrtz]

That looks right.