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I am studying the QR section of my Kaplan book.  So far I am doing okay... but I have come across this problem, and the explanation that the book gives is a little vague, or I am just not seeing it.  (I haven't had a math class in over 10 years... so I apologize if it's a little silly...)

q,r, and s are positive numbers; qrs > 12

Quantity A              Quantity B

qr                            3

5                             s

A. Quantity A is greater

B. Quantity B is greater

C. The two quantities are equal

D. The relationship cannot be determine from the information given   (correct answer)

If anyone is willing to help me learn how to determine this answer, I would appreciate it.

Thank you.

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Quantity A        Quantity B

qr                      3

5                       s

qr x s                3 x s

5                      s

qrs x 5             3 x 5

5

qrs                  15

Compare these values with qrs > 12

qrs could be more than 15 and still be more than 12 e.g. qrs = 16, 17, 18...

qrs could be less than 15 and still be more than 12 e.g. qrs = 14, 13

Hence the relationship cannot be determined from the information given. Hope this helps.

Edited by obviousbicycle
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Quantity A                            Quantity B

qr/5                                        3/s

As all are positive numbers, we can cross multiply;

so we get,

qrs                                      15

As obviousbicycle explained qrs could have different values, so  the relationship cannot be determine from the information given.

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Thank you!!  I have no idea why I was having such trouble with that one.  That makes sense now just by looking at it... I feel silly knowing how simple it was.

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Im struggling with a question and was hoping for some help.

"In the xy- coordinate plane, triangle RST is equilateral. Points R and T have coordinates (0,2) and (1,0), resepctively"

Quantity A= The perimeter of Triangle RST.

Quantity B= 3√5

I had answered that Quantity A was larger, but the correct answer is "C", they are of equal value..

My logic must be off, so I will not even explain what I did. Thanks for your help!

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using the distance formula you know that each side of the triangle equals the square root of 5 and as there are 3 equal sides  A = B.

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Equilateral means that all sides are equal

=> P = 3*s (P is perimeter, s is each side)

You have two coordinates of the triangle... so you can use the Pythagorean theorem to solve for s by considering the triangle formed (not the original triangle) when the origin is the third side of this new right triangle.

a^2 +b^2 = c^2  (c in this case is equivalent to s, and the two sides of this smaller triangle are the x and y axes... a and b are the distances along the X/Y axes)

if we consider a the x-axis distance, and b the y-axis distance, then we get

a = 2, b=1

substituting, we get

2^2+1^2=s^2

4+1=s^2

sqrt(5)=s

Plugging back into our perimeter formula, 3*sqrt(5) = P

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Yep there it is, thank you so much for your responses.. My mistake was just a simple and careless error.

I did:   2^2 + 1^2 = C^2   and some how managed to think that C was √9 instead of √5...  This translated into me assuming that each side of the triangle was 3, for a perimeter of 9..

Ugggh, these simple errors are holding me back from that elusive 170 With any luck and plenty of optimism, I will have made every simple error possible in practice to have gotten it out of my system before the real test next week.

Thanks again!

SP30

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Yep there it is, thank you so much for your responses.. My mistake was just a simple and careless error.

I did:   2^2 + 1^2 = C^2   and some how managed to think that C was √9 instead of √5...  This translated into me assuming that each side of the triangle was 3, for a perimeter of 9..

Ugggh, these simple errors are holding me back from that elusive 170 With any luck and plenty of optimism, I will have made every simple error possible in practice to have gotten it out of my system before the real test next week.

Thanks again!

SP30

When possible, slow down.  Most of the math is not that complex, and (and least in my experience), errors are generated from going too quickly and simply being worn down.  If you know what you're doing math-wise, those two battles will be your biggest fight.

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NBTR, I totally agree. In the past I've experienced some issues with the timing of my second math set. I nearly always go 20/20 on my initial quant set, but with the second (most difficult tier) quant set I get so frustrated because I KNOW I could figure out every single problem if I wasn't timed, but I don't always go about the advanced problems in the most time efficient manor. Essentially, I make stupid errors on easier problems because I'm too concerned with saving my time for a problem that will take me 3+ minutes to solve..

Thanks for sharing your experience, it's always nice to not feel all alone out here in the GRE-world.

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I just did a practice test, and the last 4 questions were all probability-related, with high level of difficulty! I did well in the first section so the second section was ridiculously hard. And we know - or at least me - that probability questions require a lot of time to be solved, so I bombed all four and that really ruined my score. I wonder if this scene is possible in the real test?

Edited by obviousbicycle
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I had an obscene amount of probability-related problems when I took the GRE. In all seriousness, even with my math background if I hadn't had as much experience tutoring students in introductory statistics classes, those sections would have taken me much longer.

Edited by 33andathirdRPM
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ah that's not fun. I can do probability, but definitely not 5 advanced level questions in like what, 7 minutes? That's just insane. Basic probability is okay, but when it comes to complex stuff like "find the probability of winning \$5 by throwing a dice 3 times - and each number on the dice has different amount of dollars etc etc" um, how about no ##### Share on other sites

Mine were nowhere near that tricky. One just needs to be able to compute a mean and a variance, or do problems like the following.

Suppose P( A ) = 2/5 and P( B ) = 2/3. Choose the correct option:

i ) 0 <= P(AB) < 0.2

ii ) .2 <= P(AB) < .4

iii ) .4 <= P(AB) < .6

iv ) .6 <= P(AB) < .8

v ) .8 <= P(AB) <= 1

Edited by 33andathirdRPM

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