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In the figure above, segments RS and TU represent two [#permalink]

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15 Oct 2012, 09:17

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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.

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.

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 thesame support beamleaning 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?

Re: In the figure above, segments RS and T U represent two [#permalink]

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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 thesame support beamleaning 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 , now I got the idea. Thanks mate!

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.

Re: In the figure above, segments RS and TU represent two [#permalink]

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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.

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.

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.

Re: In the figure above, segments RS and TU represent two [#permalink]

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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.

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 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.

Re: In the figure above, segments RS and TU represent two [#permalink]

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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.

I dont understand how the difference between two sides i.e. SU can help ascertain the length of the sides.

Please explain.

Thank You

Dear ds11, First of all, please read this blog about asking good questions. http://magoosh.com/gmat/2014/asking-exc ... questions/ Students often underestimate how much the habit of asking excellent questions can move their studies forward.

Now, as to your specific question: do you realize that this very question has been answered repeated on this page already? What part of the previous explanations do you not understand? Make very clear --- what did you understand from what you read, and what connections did you not understand? If you make that clear, I and the other experts will be happy to help you.

Re: In the figure above, segments RS and TU represent two [#permalink]

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23 May 2015, 09:24

I understand the rationale to this question. Just to check though, how can the difference between two measurements be negative? (e.g. surely the value in satement 1 is negative?)

You're correct - the length of that line segment CANNOT be negative. The likely explanations for this error are:

1) Whoever originally wrote the question erred somehow (either in the math or in typing up the question). 2) The original poster mis-typed the prompt when submitting it.

Either way, the logic behind the correct answer still holds. The GMAT is far more meticulous with its question-writing and vetting process, so you won't deal with this type of error on Test Day.

Re: In the figure above, segments RS and TU represent two [#permalink]

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12 Sep 2016, 18:32

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