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The values of x and y vary with the value of z so that each

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The values of x and y vary with the value of z so that each  [#permalink]

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New post 07 Apr 2009, 15:58
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The values of x and y vary with the value of z so that each additive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 and the value of y increasing by a factor of 3. If x and y are positive for each z>0, what is the value of x/(x+y) when z=12?

(1) When z=6, x=5y
(2) z=0, x =y+1

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Re: x and y vary with the value of z  [#permalink]

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New post 07 Apr 2009, 23:00
botirvoy wrote:
The values of x and y vary with the value of z so that each additive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 and the value of y increasing by a factor of 3. If x and y are positive for each z>0, what is the value of x/(x+y) when z=12?

(1) When z=6, x=5y
(2) z=0, x =y+1


I've seen this in GMATFocus, but among the several thousand real GMAT questions I've seen, I've yet to see anything all that similar to this question, so there are probably more important problems to study. Still, the idea here is this:

-According to the stem, if we add 2 to z, x will double, and y will triple. So, to take an example, suppose when z = 6 that x = a, and y = b. Then when z = 8 (we add two to z), x = 2a (x doubles), and y = 3b (y triples). Similarly, if we add two again to z -- that is, when z = 10 -- then again x doubles and y triples; we have x = 2^2 * a, and y = 3^2 * b. Similarly, when z = 12, x = 2^3 * a and y = 3^3 * b, and so on.

-Using Statement 1, if z = 6, we know that x = 5y. If y = b when z = 6, then x = 5b. So working as above, when z = 12, x = 2^3 * 5b, and y = 3^3 * b. From there, you can work out x/(x+y), because the b will cancel.

-Using Statement 2, we know that x = y + 1 when z = 0. If y = d when z = 0, then x = d+1. Using the same logic as above, you can determine that when z = 12, x will be equal to 2^6 * (d + 1), and y will be equal to 3^6 * d. Now if you try to calculate x/(x+y), d will not cancel; we would need to know the value of d to calculate x/(x+y) here.

So the first statement is sufficient, the second not. The question is really just a convoluted way of testing whether you understand that ratios are based on multiplication and division, and not on addition, and uses what are known as 'logarithmic scales' to test that idea. It's a much more complicated version of question 99 in the PS section of OG11, which tests 'logarithmic scales' in a much more straightforward way - the question in the OG is in a much more common format than the above, and is worth understanding.
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Re: x and y vary with the value of z  [#permalink]

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New post 07 Apr 2009, 23:10
Ian has explained it all. It took me a minute to figure out the ratio but once that's figured out the solution is quite easy. A nice problem though. Thanks for posting.

-pradeep
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Re: x and y vary with the value of z  [#permalink]

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New post 08 Apr 2009, 08:12
IanStewart wrote:
botirvoy wrote:
The values of x and y vary with the value of z so that each additive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 and the value of y increasing by a factor of 3. If x and y are positive for each z>0, what is the value of x/(x+y) when z=12?

(1) When z=6, x=5y
(2) z=0, x =y+1


I've seen this in GMATFocus, but among the several thousand real GMAT questions I've seen, I've yet to see anything all that similar to this question, so there are probably more important problems to study. Still, the idea here is this:

-According to the stem, if we add 2 to z, x will double, and y will triple. So, to take an example, suppose when z = 6 that x = a, and y = b. Then when z = 8 (we add two to z), x = 2a (x doubles), and y = 3b (y triples). Similarly, if we add two again to z -- that is, when z = 10 -- then again x doubles and y triples; we have x = 2^2 * a, and y = 3^2 * b. Similarly, when z = 12, x = 2^3 * a and y = 3^3 * b, and so on.

-Using Statement 1, if z = 6, we know that x = 5y. If y = b when z = 6, then x = 5b. So working as above, when z = 12, x = 2^3 * 5b, and y = 3^3 * b. From there, you can work out x/(x+y), because the b will cancel.

-Using Statement 2, we know that x = y + 1 when z = 0. If y = d when z = 0, then x = d+1. Using the same logic as above, you can determine that when z = 12, x will be equal to 2^6 * (d + 1), and y will be equal to 3^6 * d. Now if you try to calculate x/(x+y), d will not cancel; we would need to know the value of d to calculate x/(x+y) here.

So the first statement is sufficient, the second not. The question is really just a convoluted way of testing whether you understand that ratios are based on multiplication and division, and not on addition, and uses what are known as 'logarithmic scales' to test that idea. It's a much more complicated version of question 99 in the PS section of OG11, which tests 'logarithmic scales' in a much more straightforward way - the question in the OG is in a much more common format than the above, and is worth understanding.



Yeah very different Q from what we have seen so far.

If x and y are positive for each z>0, what is the value of x/(x+y) when z=12?

What relevance does this have? B gives us a value of X for Z=0.
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Re: x and y vary with the value of z  [#permalink]

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New post 09 Apr 2009, 14:28
Thank you everyone!
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Re: x and y vary with the value of z  [#permalink]

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New post 13 Apr 2009, 19:01
I liked this question.
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Re: The values of x and y vary with the value of z so that each  [#permalink]

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Re: The values of x and y vary with the value of z so that each   [#permalink] 04 Jul 2017, 11:20
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