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In the figure above, segments RS and TU represent two

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In the figure above, segments RS and TU represent two [#permalink] New post 15 Oct 2012, 09:17
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Gmat_Bible_DS_122.png
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In the figure above, segments RS and TU represent two positions of the same support beam leaning against the side SV of a structure. The length TV is how much greater than the length RV ?

(1) The length of SU is 2\sqrt{2} - 4 meters.
(2) The ratio of the length of VR to VT is 6 : 2.
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Oct 2012, 04:15, edited 1 time in total.
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Re: In the figure above, segments RS and T U represent two [#permalink] New post 15 Oct 2012, 13:23
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Dear joylive,

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

What's interesting about this problem is --- almost everything is determined by the prompt. We already know we have the two special right triangles ---- the 30-60-90 triangle and the 45-45-90 triangle --- so all those ratios are determined. Further, we know the hypotenuses, the length of the beam, are equal.
Here's a blog about these two triangles:
http://magoosh.com/gmat/2012/the-gmats- ... triangles/
For the GMAT Quant section, you need to know these two triangles cold.

Basically, all the ratios are completely determined by the prompt, so the only thing we need to determine the distance RT is one length.

(A) gives us a length ---it's sufficient
(B) gives us another ratio, which doesn't help us find a length -- not sufficient.

Remember, you need a length to find a length! That's a very basic geometry idea.

Answer = A

Does all this make sense?

Mike :-)
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Re: In the figure above, segments RS and T U represent two [#permalink] New post 15 Oct 2012, 18:04
I understand the triangles but I do not get how the hypotenuse can be equal for the two triangles.

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Re: In the figure above, segments RS and T U represent two [#permalink] New post 15 Oct 2012, 18:45
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joylive wrote:
I understand the triangles but I do not get how the hypotenuse can be equal for the two triangles.

Dear joylive,

The text of the problem says: "In the figure above, segments RS and TU represent two positions of the same support beam leaning against the side SV of a structure."

Thus, the same beam, in one position is RS, and then it slides down to a slightly lower angle, and it's at TU --- but it's the same beam, so it's the same length in both places. Does this make sense?

Mike :-)
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Re: In the figure above, segments RS and T U represent two [#permalink] New post 15 Oct 2012, 23:28
mikemcgarry wrote:
joylive wrote:
I understand the triangles but I do not get how the hypotenuse can be equal for the two triangles.

Dear joylive,

The text of the problem says: "In the figure above, segments RS and TU represent two positions of the same support beam leaning against the side SV of a structure."

Thus, the same beam, in one position is RS, and then it slides down to a slightly lower angle, and it's at TU --- but it's the same beam, so it's the same length in both places. Does this make sense?

Mike :-)


Ooooh! I completely missed it :x :roll: , now I got the idea.
Thanks mate! :)
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 16 Oct 2012, 04:17
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joylive wrote:
Attachment:
Gmat_Bible_DS_122.png
In the figure above, segments RS and TU represent two positions of the same support beam leaning against the side SV of a structure. The length TV is how much greater than the length RV ?

(1) The length of SU is 2\sqrt{2} - 4 meters.
(2) The ratio of the length of VR to VT is 6 : 2.


Similar questions to practice:
in-the-figure-above-segment-rs-and-tu-represent-two-position-126441.html
a-ladder-25-feet-long-is-leaning-against-a-wall-that-is-130364.html

Hope it helps.
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 16 Oct 2012, 20:59
Bunuel wrote:
joylive wrote:
Attachment:
Gmat_Bible_DS_122.png
In the figure above, segments RS and TU represent two positions of the same support beam leaning against the side SV of a structure. The length TV is how much greater than the length RV ?

(1) The length of SU is 2\sqrt{2} - 4 meters.
(2) The ratio of the length of VR to VT is 6 : 2.


Similar questions to practice:
in-the-figure-above-segment-rs-and-tu-represent-two-position-126441.html
a-ladder-25-feet-long-is-leaning-against-a-wall-that-is-130364.html

Hope it helps.


@Bunuel, thanks mate!
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 29 Oct 2012, 05:12
Hello Mike,

How will SU help is determining the required measurement? Can you please explain?
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 29 Oct 2012, 05:13
How can the ratio of VR/VT be more than 1 as given is statement 2? Isn't VR<VT?
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 29 Oct 2012, 19:17
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teal wrote:
Hello Mike,
How will SU help is determining the required measurement? Can you please explain?

Yes. This is a tricky thing about geometry DS problems. The basic idea is: to get a length, you need to be given a length. That Big Idea #1 in this problem. When you understand the implications of that statement, it's pure gold. If all you are given is angular relationships, as this prompt gives us, then the figures could be any size. This is the basic geometry idea of similarity. If you make scaled-up or scaled-down versions of any shape (square, pentagon, equilateral triangle, etc.), then all the absolute lengths are different, but both the angles and the ratio of the lengths are the same.

What this prompt gives us allows us to see that the figure consists of a 30-60-90 triangle and a 45-45-90 triangle, with hypotenuses of the same length. That means all the angles are determined, and all the ratios are determined, but we still don't know in absolute terms how big or small anything is. From the prompt alone, any length --- say SR --- could be 1 inch long or 1 mile long. It could be arbitrarily small or big.

When we are given the length of SU, that fixes one length, which has the consequence of fixing all the lengths. Now that we know one length, the figure cannot be arbitrarily big or small. It has to have exactly one size --- only one size of the shape will result in SU being exactly [2/sqrt(2) - 4] meters. So, knowing SU, we could calculate all the other lengths.

Here's the other big idea ---- in DS, don't do unnecessary calculations. You are not actually be asked to find the numerical answer for the prompt question. All you are being asked to to figure out whether the statements provide information sufficient to answer the prompt question. That's Big Idea #2. Here's a blog about DS on the GMAT.
http://magoosh.com/gmat/2012/introducti ... fficiency/

In geometry questions in particular, it's often the case that just one piece of information --- here, the length of SU --- is enough to determine everything. Actually doing that calculation, going through all the steps to figure out the difference (TV - RV) given that SU = [2/sqrt(2) - 4] ---- that would be a royal pain in the tuckus, and it is absolutely unnecessarily for answering this particular question.

If this were a PS problem --- given the prompt, and given SU = [2/sqrt(2) - 4], which of the following equals (TV - RV) --- that would be a 800+ level question, very difficult, because you would actually have to work through that solution. That is a highly unlikely question to see on the GMAT, unless you are getting absolutely everything else in the Quant section correct. By contrast, this question, the DS question in which you just have to judge whether it is sufficient or not, that's probably a 600 level question, not particular hard by GMAT standards. You really need to see how to use these geometric shortcuts to your advantage on the DS section.

Here's a blog that talks about another set of these geometric shortcuts.
http://magoosh.com/gmat/2012/gmat-data- ... nce-rules/

Does all this make sense?

Mike :-)
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 29 Oct 2012, 19:25
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teal wrote:
How can the ratio of VR/VT be more than 1 as given is statement 2? Isn't VR<VT?

Yes, you're quite right. Something is funky about that second statement. First of all, as you correctly point out, VR is part of VT, so how can the ratio be greater than1? Even if we take the reciprocal, that's not right either. In fact, the correct ratio is

VR/VT = \frac{sqrt(2)}{2}

In other words, if VR were the side of a square, VT would be the length of the diagonal of the square.

There's something sketchy about the question. It has only a sloppy hand-drawn diagram, and that second statement, as is, is faulty. I dare say, the question is not reproduced here at its original level of quality.

Cheers,
Mike :-)
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 24 May 2014, 12:11
mikemcgarry wrote:
teal wrote:
Hello Mike,
How will SU help is determining the required measurement? Can you please explain?

Yes. This is a tricky thing about geometry DS problems. The basic idea is: to get a length, you need to be given a length. That Big Idea #1 in this problem. When you understand the implications of that statement, it's pure gold. If all you are given is angular relationships, as this prompt gives us, then the figures could be any size. This is the basic geometry idea of similarity. If you make scaled-up or scaled-down versions of any shape (square, pentagon, equilateral triangle, etc.), then all the absolute lengths are different, but both the angles and the ratio of the lengths are the same.

What this prompt gives us allows us to see that the figure consists of a 30-60-90 triangle and a 45-45-90 triangle, with hypotenuses of the same length. That means all the angles are determined, and all the ratios are determined, but we still don't know in absolute terms how big or small anything is. From the prompt alone, any length --- say SR --- could be 1 inch long or 1 mile long. It could be arbitrarily small or big.

When we are given the length of SU, that fixes one length, which has the consequence of fixing all the lengths. Now that we know one length, the figure cannot be arbitrarily big or small. It has to have exactly one size --- only one size of the shape will result in SU being exactly [2/sqrt(2) - 4] meters. So, knowing SU, we could calculate all the other lengths.

Here's the other big idea ---- in DS, don't do unnecessary calculations. You are not actually be asked to find the numerical answer for the prompt question. All you are being asked to to figure out whether the statements provide information sufficient to answer the prompt question. That's Big Idea #2. Here's a blog about DS on the GMAT.
http://magoosh.com/gmat/2012/introducti ... fficiency/

In geometry questions in particular, it's often the case that just one piece of information --- here, the length of SU --- is enough to determine everything. Actually doing that calculation, going through all the steps to figure out the difference (TV - RV) given that SU = [2/sqrt(2) - 4] ---- that would be a royal pain in the tuckus, and it is absolutely unnecessarily for answering this particular question.

If this were a PS problem --- given the prompt, and given SU = [2/sqrt(2) - 4], which of the following equals (TV - RV) --- that would be a 800+ level question, very difficult, because you would actually have to work through that solution. That is a highly unlikely question to see on the GMAT, unless you are getting absolutely everything else in the Quant section correct. By contrast, this question, the DS question in which you just have to judge whether it is sufficient or not, that's probably a 600 level question, not particular hard by GMAT standards. You really need to see how to use these geometric shortcuts to your advantage on the DS section.

Here's a blog that talks about another set of these geometric shortcuts.
http://magoosh.com/gmat/2012/gmat-data- ... nce-rules/

Does all this make sense?

Mike :-)


Hi Mike,

Thanks for the breakdown. For arguments sake(i'm trying to bridge the gap) -- i can see the ratios of the angles and the sides. I originally thought that we would need to know a FULL length of the sides in question to determine the value, meaning, we would need to know SV instead of just SU. Are you saying we don't need to know that because at some point (if we were to solve it), we could find a single value that will make this SU true?

Thanks
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 24 May 2014, 16:28
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russ9 wrote:
Hi Mike,

Thanks for the breakdown. For arguments sake(i'm trying to bridge the gap) -- i can see the ratios of the angles and the sides. I originally thought that we would need to know a FULL length of the sides in question to determine the value, meaning, we would need to know SV instead of just SU. Are you saying we don't need to know that because at some point (if we were to solve it), we could find a single value that will make this SU true?

Thanks

Dear russ9,
I'm happy to respond. :-)

To some extent, this is just the magic of Geometry. If all the angles are fixed, then any length is enough to determine all the other lengths --- it doesn't matter whether the length we are given is a full side or some odd chunk between intersections. One length is enough. Now, it's also true --- if this were a PS problem, and we actually had to calculate the numerical value of the answer, it probably would be much easier if we were given a number for SV rather than for SU, but that's just a matter of convenience. In terms of DS, we don't have to worry about ease of calculation --- in fact, we don't need to calculate at all! All we need to know is that, given one length, we could find any other length. That's it. That's enough for sufficiency.

Does this make sense?
Mike :-)
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Re: In the figure above, segments RS and TU represent two [#permalink] New post 27 May 2014, 14:45
mikemcgarry wrote:
russ9 wrote:
Hi Mike,

Thanks for the breakdown. For arguments sake(i'm trying to bridge the gap) -- i can see the ratios of the angles and the sides. I originally thought that we would need to know a FULL length of the sides in question to determine the value, meaning, we would need to know SV instead of just SU. Are you saying we don't need to know that because at some point (if we were to solve it), we could find a single value that will make this SU true?

Thanks

Dear russ9,
I'm happy to respond. :-)

To some extent, this is just the magic of Geometry. If all the angles are fixed, then any length is enough to determine all the other lengths --- it doesn't matter whether the length we are given is a full side or some odd chunk between intersections. One length is enough. Now, it's also true --- if this were a PS problem, and we actually had to calculate the numerical value of the answer, it probably would be much easier if we were given a number for SV rather than for SU, but that's just a matter of convenience. In terms of DS, we don't have to worry about ease of calculation --- in fact, we don't need to calculate at all! All we need to know is that, given one length, we could find any other length. That's it. That's enough for sufficiency.

Does this make sense?
Mike :-)


Absolutely. Thanks for clarifying the doubt!
Re: In the figure above, segments RS and TU represent two   [#permalink] 27 May 2014, 14:45
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