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Circular Motion Problem

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Circular Motion Problem

by knight247 » Tue Nov 18, 2014 9:33 am
X, Y & Z are moving along a circular path of 500 yards and speeds 10, 20 & 45 yards/ minute respectively. All three of them start from the same point and move in the same direction.

(A) When and where will all three of them meet for the first time?
(B) At what time will all three of them meet at the original STARTING POINT?


No OA. Detailed explanations would be appreciated.
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by Mathsbuddy » Wed Nov 19, 2014 2:41 am
Although X,Y and Z travel in a circular path, this is not necessarily a circular motion problem.
If you think of it as a straight line of 500 yards that repeats indefinitely, it can be solved without worrying about the circle. Does this help?
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by Mathsbuddy » Wed Nov 19, 2014 3:14 am
Further to my last entry, I think the easiest way to answer this is to consider a graph.

See picture.

You can see that A and B will only coincide at distances that are multiples of 1000 yards and times that are
multiples of 100 minutes
Every 100 minutes, C will travel 4500 yards. So C will intercept A and B at multiples of 9000 yards.
So All 3 will first meet at 9000 yards, which is also the starting point (being a multiple of 500 yards).
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by GMATGuruNY » Wed Nov 19, 2014 4:08 am
knight247 wrote:X, Y & Z are moving along a circular path of 500 yards and speeds 10, 20 & 45 yards/ minute respectively. All three of them start from the same point and move in the same direction.

(A) When and where will all three of them meet for the first time?
(B) At what time will all three of them meet at the original STARTING POINT?


No OA. Detailed explanations would be appreciated.
X and Y:
For Y to meet X, Y must travel 1 circumference further than X.
Since Y's rate - X's rate = 20-10 = 10 yards per minute, every minute Y travels 10 more yards than X.
Since the circumference = 500 yards, the time for Y to travel 1 circumference further than X = 500/10 = 50 minutes.
Implication:
Since Y travels 1 circumference further than X every 50 minutes, Y meets X at every 50-minute interval.

Y and Z:
For Z to meet Y, Z must travel 1 circumference further than Y.
Since Z's rate - Y's rate = 45-20 = 25 yards per minute, every minute Z travels 25 more yards than Y.
Since the circumference = 500 yards, the time for Z to travel 1 circumference further than Y = 500/25 = 20 minutes.
Implication:
Since Z travels 1 circumference further than Y every 20 minutes, Z meets Y at every 20-minute interval.

X, Y and Z:
The first interval at which Z will meet Y and Y will meet X -- in other words, the time required for all 3 to meet -- is equal to the LCM of 50 and 20 = 100 minutes.
In 100 minutes, the distance traveled by X = 10*100 = 1000 yards.
In 100 minutes, the distance traveled by Y = 20*100 = 2000 yards.
In 100 minutes, the distance traveled by Z = 45*100 = 4500 yards.
Since the distance in each case is a multiple of 500 -- the length of the entire path -- after 100 minutes X, Y and Z will all meet at the starting point.

Correct answers:
A) 100 minutes, at the starting point
B) 100 minutes
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by Mathsbuddy » Wed Nov 19, 2014 4:33 am
I have just seen that my gradients are incorrect in the method below. I will re-post with them corrected.
Mathsbuddy wrote:Further to my last entry, I think the easiest way to answer this is to consider a graph.

See picture.

You can see that A and B will only coincide at distances that are multiples of 1000 yards and times that are
multiples of 100 minutes
Every 100 minutes, C will travel 4500 yards. So C will intercept A and B at multiples of 9000 yards.
So All 3 will first meet at 9000 yards, which is also the starting point (being a multiple of 500 yards).
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by Mathsbuddy » Wed Nov 19, 2014 5:10 am
Hi there GuruNY,

I totally agree with your solution (without any doubt).
However, I am totally perplexed where my mistake is in the attached new diagram (at the bottom, below your quote).
Clearly:
X travels 500 yards per 50 mins = 10 yards/min
Y travels 1000 yards per 50 mins = 20 yards/min
Z travels 2250 yards per 50 mins = 45 yards/min
I have tried to show this on the graphs to find the common intersection.
The graph clearly shows that X and Y will only meet every 1000 yards.
Hence we need to find where Z will first intercept them at a multiple of 1000 yards.
This will be at 9000 yards (=2250 x 4)
9000/45 = 200 mins

Therefore the first common interception will be at 200 mins at the starting point.

Could you please see why my response does not match yours? I'm scratching my head here!

Thanks,
Mathsbuddy



GMATGuruNY wrote:
knight247 wrote:X, Y & Z are moving along a circular path of 500 yards and speeds 10, 20 & 45 yards/ minute respectively. All three of them start from the same point and move in the same direction.

(A) When and where will all three of them meet for the first time?
(B) At what time will all three of them meet at the original STARTING POINT?


No OA. Detailed explanations would be appreciated.
X and Y:
For Y to meet X, Y must travel 1 circumference further than X.
Since Y's rate - X's rate = 20-10 = 10 yards per minute, every minute Y travels 10 more yards than X.
Since the circumference = 500 yards, the time for Y to travel 1 circumference further than X = 500/10 = 50 minutes.
Implication:
Since Y travels 1 circumference further than X every 50 minutes, Y meets X at every 50-minute interval.

Y and Z:
For Z to meet Y, Z must travel 1 circumference further than Y.
Since Z's rate - Y's rate = 45-20 = 25 yards per minute, every minute Z travels 25 more yards than Y.
Since the circumference = 500 yards, the time for Z to travel 1 circumference further than Y = 500/25 = 20 minutes.
Implication:
Since Z travels 1 circumference further than Y every 20 minutes, Z meets Y at every 20-minute interval.

X, Y and Z:
The first interval at which Z will meet Y and Y will meet X -- in other words, the time required for all 3 to meet -- is equal to the LCM of 50 and 20 = 100 minutes.
In 100 minutes, the distance traveled by X = 10*100 = 1000 yards.
In 100 minutes, the distance traveled by Y = 20*100 = 2000 yards.
In 100 minutes, the distance traveled by Z = 45*100 = 4500 yards.
Since the distance in each case is a multiple of 500 -- the length of the entire path -- after 100 minutes X, Y and Z will all meet at the starting point.

Correct answers:
A) 100 minutes, at the starting point
B) 100 minutes
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by Mathsbuddy » Wed Nov 19, 2014 5:27 am
At last I have reconciled my errors!

The wrong-doing was to plot repeated time rather than repeated distance!

Now I have achieved the objective of depicting the intersection of all 3 lines graphically.

Clearly they all meet at the start point (500 yards = 0 yards in a circle) at 100 minutes.
This fully complies with GuruNY's correct solution.
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by GMATGuruNY » Wed Nov 19, 2014 7:00 am
It should be noted that the GMAT would provide answer choices:
X, Y and Z are moving along a circular path of 500 yards at speeds of 10 yards per minute, 20 yards per minute, and 45 yards per minute, respectively. All three start from the same point and move in the same direction. After m minutes, X, Y and Z meet for the first time. What is the value of m?

20
50
100
150
200
When taking the GMAT, we could PLUG IN THE ANSWERS, which represent the value of m.
Since m is equal to the time required for X, Y and Z to meet for the FIRST TIME, we should start with the smallest answer choice.
When the correct answer choice is plugged in, X, Y and Z will all be the SAME DISTANCE FROM THE STARTING POINT.

A: 20
In 20 minutes, the distance traveled by X = r*t = 10*20 = 200 yards
In 20 minutes, the distance traveled by Y = r*t = 20*20 = 400 yards.
Since X and Y are not the same distance from the starting point, eliminate A.

B: 50
In 50 minutes, the distance traveled by X = r*t = 10*50 = 500 yards
In 50 minutes, the distance traveled by Y = r*t = 20*50 = 1000 yards.
In 50 minutes, the distance traveled by Z = r*t = 45*50 = 2250 yards.
Here, X and Y are each at the starting point (since each has traveled a multiple of 500 yards), while Z is 250 yards from the starting point (since Z has traveled a multiple of 500 yards + 250).
Since X, Y and Z are not all the same distance from the starting point, eliminate B.

C: 100
In 100 minutes, the distance traveled by X = r*t = 10*100 = 1000 yards.
In 100 minutes, the distance traveled by Y = r*t = 20*100 = 2000 yards.
In 100 minutes, the distance traveled by Z = r*t = 45*100 = 4500 yards.
Here, distance in each case is a multiple of 500 -- the length of the entire path -- implying that X, Y and Z will all be at the starting point.
Since X, Y and Z are all the same distance from the starting point -- 0 yards -- the correct answer is C.
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by Mathsbuddy » Wed Nov 19, 2014 7:27 am
Thank you GuruNY,

Certainly, 'plugging in' can save time. May I add an idea that could speed up the 'plugging in' process for selected questions.

If you systematically plug in answers in order A,B,C,D, then you could have up to 4 eliminations.
However, if you plug in the median answer first, then you might only have up to 2 eliminations before moving on (assuming you are of course correct!) In GMAT, time is precious, so this might be a good time saver, where there is a clear 'order' correlation between the answers and the question. (For example, a question that requires squaring a negative answer would undermine this principle). Nonetheless, there is no harm with starting with answer C as a matter of course.

(For computer programmers, this is based on binary elimination as used in "Bubble Sort").

Let me explain:

If you choose answer C first and it is too high, then you only have answers A and B remaining.
From these, pick the one with the 'easier' values. If wrong, then elimination would imply the other is correct. By easier values, I mean that working with x = 1 is far easier than x = 0.12674, to save time>

If you choose answer C first and it is too low, then you only have answers D and E remaining.
As explained above, a test of one of these will either solve or eliminate.

Therefore, I conclude that 'plugging in' answer C first could save half the time when eliminating answers in many cases. However, it is crucial to UNDERSTAND when such elimination would be appropriate. For example, I don't think it would work in this question. Would GuruNY agree?

GMATGuruNY wrote:It should be noted that the GMAT would provide answer choices:
X, Y and Z are moving along a circular path of 500 yards at speeds of 10 yards per minute, 20 yards per minute, and 45 yards per minute, respectively. All three start from the same point and move in the same direction. After m minutes, X, Y and Z meet for the first time. What is the value of m?

20
50
100
150
200
When taking the GMAT, we could PLUG IN THE ANSWERS, which represent the value of m.
Since m is equal to the time required for X, Y and Z to meet for the FIRST TIME, we should start with the smallest answer choice.
When the correct answer choice is plugged in, X, Y and Z will all be the SAME DISTANCE FROM THE STARTING POINT.

A: 20
In 20 minutes, the distance traveled by X = r*t = 10*20 = 200 yards
In 20 minutes, the distance traveled by Y = r*t = 20*20 = 400 yards.
Since X and Y are not the same distance from the starting point, eliminate A.

B: 50
In 50 minutes, the distance traveled by X = r*t = 10*50 = 500 yards
In 50 minutes, the distance traveled by Y = r*t = 20*50 = 1000 yards.
In 50 minutes, the distance traveled by Z = r*t = 45*50 = 2250 yards.
Here, X and Y are each at the starting point (since each has traveled a multiple of 500 yards), while Z is 250 yards from the starting point (since Z has traveled a multiple of 500 yards + 250).
Since X, Y and Z are not all the same distance from the starting point, eliminate B.

C: 100
In 100 minutes, the distance traveled by X = r*t = 10*100 = 1000 yards.
In 100 minutes, the distance traveled by Y = r*t = 20*100 = 2000 yards.
In 100 minutes, the distance traveled by Z = r*t = 45*100 = 4500 yards.
Here, distance in each case is a multiple of 500 -- the length of the entire path -- implying that X, Y and Z will all be at the starting point.
Since X, Y and Z are all the same distance from the starting point -- 0 yards -- the correct answer is C.
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