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In the figure above, segments AC and BC are each parallel

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In the figure above, segments AC and BC are each parallel  [#permalink]

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New post Updated on: 13 Aug 2014, 03:03
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In the figure above, segments AC and BC are each parallel to one of the rectangular coordinate axes. Is the length of AC greater than the length of BC?

(1) m = 1 and k = 3
(2) The slope of segment AB is 4/5

Originally posted by mahendru1992 on 13 Aug 2014, 02:50.
Last edited by Bunuel on 13 Aug 2014, 03:03, edited 2 times in total.
Edited the question
Magoosh GMAT Instructor
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Re: In the figure above, segments AC and BC are each parallel  [#permalink]

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New post 13 Aug 2014, 15:55
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mahendru1992 wrote:
Attachment:
The attachment Untitled.png is no longer available
In the figure above, segments AC and BC are each parallel to one of the rectangular coordinate axes. Is the length of AC greater than the length of BC?

(1) m = 1 and k = 3
(2) The slope of segment AB is 4/5

Dear mahendru1992,
I'm happy to help with this. :-)

Statement #1: well, with this, we know the x-coordinate of the first point is 1. Because it's drawn in Quadrant III, I'm going to assume you actually meant -1, but it doesn't really matter. The y-coordinate of the second point is 3. Well, that leaves the y-coordinate of the first point and the x-coordinate of the second point up for grabs. We could have:
Attachment:
two possibilities for ABC.JPG
two possibilities for ABC.JPG [ 21.83 KiB | Viewed 1732 times ]

With this information, the inequality could go either way. This statement, alone and by itself, is insufficient.

Statement #2: slope is a powerful piece of information!! See:
http://magoosh.com/gmat/2012/gmat-math- ... x-y-plane/
This means
4/5 = rise/run = AC/BC
BC*(4/5) = AC
Since all these numbers are positive, we know that BC must be bigger.
This statement, alone and by itself, is sufficient.

Answer = (B)

Does all this make sense?
Mike :-)
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Re: In the figure above, segments AC and BC are each parallel  [#permalink]

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New post 17 Aug 2014, 12:42
mikemcgarry wrote:
mahendru1992 wrote:
Attachment:
Untitled.png
In the figure above, segments AC and BC are each parallel to one of the rectangular coordinate axes. Is the length of AC greater than the length of BC?

(1) m = 1 and k = 3
(2) The slope of segment AB is 4/5

Dear mahendru1992,
I'm happy to help with this. :-)

Statement #1: well, with this, we know the x-coordinate of the first point is 1. Because it's drawn in Quadrant III, I'm going to assume you actually meant -1, but it doesn't really matter. The y-coordinate of the second point is 3. Well, that leaves the y-coordinate of the first point and the x-coordinate of the second point up for grabs. We could have:
Attachment:
two possibilities for ABC.JPG

With this information, the inequality could go either way. This statement, alone and by itself, is insufficient.

Statement #2: slope is a powerful piece of information!! See:
http://magoosh.com/gmat/2012/gmat-math- ... x-y-plane/
This means
4/5 = rise/run = AC/BC
BC*(4/5) = AC
Since all these numbers are positive, we know that BC must be bigger.
This statement, alone and by itself, is sufficient.

Answer = (B)

Does all this make sense?
Mike :-)


Hello.. you have switched B & C in your diagram. Will it affect the solution?
Also, as per MGMAT Geometry book, slope = rise/run = (y2-y1)/(x2-x1)
How does that relate to AC & BC. Please explain.
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Re: In the figure above, segments AC and BC are each parallel  [#permalink]

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New post 17 Aug 2014, 13:25
apsForGmat wrote:
Hello.. you have switched B & C in your diagram. Will it affect the solution?
Also, as per MGMAT Geometry book, slope = rise/run = (y2-y1)/(x2-x1)
How does that relate to AC & BC. Please explain.

Yes, I did make that mistake in the diagram, but you will notice that all my equations and conclusions are based on the original diagram.

Do you understand the statement "slope = rise/run"? Do you understand what "rise" and "run" mean? The rise is the vertical distance between the two points, the distance from one y-height to the other. In the original diagram, the rise = AC. The run is horizontal separation between the two points, the horizontal distance between the x-line of one point and the x-line of the other. In the original diagram, run = BC. Thus, rise/run = AC/BC.

I suggest you read the blog article on slope I gave in the previous post.

Does this make sense?
Mike :-)
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Mike McGarry
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Re: In the figure above, segments AC and BC are each parallel  [#permalink]

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New post 17 Aug 2014, 15:07
mikemcgarry wrote:
mahendru1992 wrote:
Attachment:
Untitled.png
In the figure above, segments AC and BC are each parallel to one of the rectangular coordinate axes. Is the length of AC greater than the length of BC?

(1) m = 1 and k = 3
(2) The slope of segment AB is 4/5

Dear mahendru1992,
I'm happy to help with this. :-)

Statement #1: well, with this, we know the x-coordinate of the first point is 1. Because it's drawn in Quadrant III, I'm going to assume you actually meant -1, but it doesn't really matter. The y-coordinate of the second point is 3. Well, that leaves the y-coordinate of the first point and the x-coordinate of the second point up for grabs. We could have:
Attachment:
two possibilities for ABC.JPG

With this information, the inequality could go either way. This statement, alone and by itself, is insufficient.

Statement #2: slope is a powerful piece of information!! See:
http://magoosh.com/gmat/2012/gmat-math- ... x-y-plane/
This means
4/5 = rise/run = AC/BC
BC*(4/5) = AC
Since all these numbers are positive, we know that BC must be bigger.
This statement, alone and by itself, is sufficient.

Answer = (B)

Does all this make sense?
Mike :-)


I though the answer must be C before I read your exp :( . Thanks a lot
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Re: In the figure above, segments AC and BC are each parallel  [#permalink]

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New post 30 Dec 2018, 12:55
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Re: In the figure above, segments AC and BC are each parallel   [#permalink] 30 Dec 2018, 12:55
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