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# The values of x and y vary with the value of z

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

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19 Dec 2015, 15:22
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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: The values of x and y vary with the value of z  [#permalink]

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19 Dec 2015, 22:58
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I first rewrote x/(x+y) to (x+y)/x -> 1+y/x? -> y/x?

(1) When z=6, x=5y so at z=6 if y=p, x will be 5p
Since we know for every 2 in z, x grow by factor 2 and y by factor 3, at z=12, y= 3^3p and x=2^3*5*p. we know we can calculate y/x. SUFF

(2) z=0, x =y+1. Let y=a and so x=a+1. So at z=12, y=3^6*a and x=2^6(a+1)
Now y/x = (3^6*a) / (2^6(a+1)). No solution unless we know a. INSUF

Ans: A
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Re: The values of x and y vary with the value of z  [#permalink]

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20 Dec 2015, 06:44
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NoHalfMeasures 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

For the relations given, you can assume that x=pz and y=qz. For the conditions wherein z --> z+2 we get, 2x=p(z+2) and 3y=q(z+2)

We need to find, x/(x+y) for z=12 --> 7p/(7p+14q/3). So we will get a unique value if we can have a relation such as p=kq or vice versa.

Per statement 1, when z=6, x=5y. This is exactly what we need and will give us p=10q/3, giving us a unique answer ---> sufficient.

Per statement 2, when z=0, x=y+1, you will end up getting p=2q/3 + 1but substituting this into the expression of x/(x+y), you will not be able to get a unique value. Not sufficient.

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

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21 Dec 2015, 02:21
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NoHalfMeasures 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

1- value of x/x+y when x = 5y is 5y/6y = 5/6 suff
2- value of x/x+y when x = y+1 is y+1/2y+1 insuff
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Re: The values of x and y vary with the value of z  [#permalink]

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14 Mar 2016, 23:57
NoHalfMeasures 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

(1) value of x=5y when z=6
so when z=6 it increases by 6 values.
but it is given for every increase in 2 values of z x will be a factor of 2
so when z=12 we have x=3*(5y)
so with new value of x we get a unique value as (15y)/(15y+y)---->15/16
suff...
(2) z=0, x =y+1
new value of x when z=12----->x=6*(y+1)
substituting this new value in x/(x+y)
we get (6y+6)/(6y+6+y)------>6y+6)/(7y+6)
thus we dont know the exact value of y ,we can't get any unique value
thus insuff...

Ans A
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Re: The values of x and y vary with the value of z  [#permalink]

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30 Apr 2016, 03:33
NoHalfMeasures 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

Hi Bunuel,

Can you please explain this question.
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Re: The values of x and y vary with the value of z  [#permalink]

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01 May 2016, 01:01
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NoHalfMeasures wrote:
I first rewrote x/(x+y) to (x+y)/x -> 1+y/x? -> y/x?

(1) When z=6, x=5y so at z=6 if y=p, x will be 5p
Since we know for every 2 in z, x grow by factor 2 and y by factor 3, at z=12, y= 3^3p and x=2^3*5*p. we know we can calculate y/x. SUFF

(2) z=0, x =y+1. Let y=a and so x=a+1. So at z=12, y=3^6*a and x=2^6(a+1)
Now y/x = (3^6*a) / (2^6(a+1)). No solution unless we know a. INSUF

Ans: A

NoHalfMeasures is the one who has explained this question correctly so far, so I will try to clarify by expanding on NoHalfMeasures' explanation.

First, we have to correctly interpret the question. What the question means is that, every time z increases by 2, x is multiplied by a factor of 2, and y is multiplied by a factor of 3. So, if x = m and y = n when z = 0, we get the following sequences of values of x, y, and z:

When $$z = 0: x = m, y = n$$
When $$z = 2: x = 2*m, y = 3*n$$
When $$z = 4: x = 2^2*m, y = 3^2*n$$
When $$z = 6: x = 2^3*m, y = 3^3*n$$
When $$z = 8: x = 2^4*m, y = 3^4*n$$
When $$z = 10: x = 2^5*m, y = 3^5*n$$
When $$z = 12: x = 2^6*m, y = 3^6*n$$

(1) We are told that when z = 6, x = 5y. As NoHalfMeasures suggests, let's call this value of y = p, so x = 5*p when z = 6. Writing out the sequences of values of x, y, and z from z = 6 to z = 12 looks like this:

When $$z = 6: x = 5*p, y = p$$
When $$z = 8: x = 2*5*p, y = 3*p$$
When $$z = 10: x = 2^2*5*p, y = 3^2*p$$
When $$z = 12: x = 2^3*5*p, y = 3^3*p$$

So, based on (1), when z = 12:

$$\frac{x}{x+y} = \frac{2^3*5*p}{2^3*5*p + 3^3*p}$$

We can factor p out of the numerator and denominator to get:

$$\frac{x}{x+y} = \frac{2^3*5}{2^3*5 + 3^3}$$

$$\frac{x}{x+y} = \frac{40}{40+27} = \frac{40}{67}$$

SUFFICIENT

Having seen this, we could realize that, if we know the ratio of x:y for some value of z, this ratio will be multiplied by 2/3 every time z increases by 2. This means that, for any specific value of z, x and y will both be specific multiples of some value like p, which can then be factored out of $$\frac{x}{x+y}$$ to give us a single value of $$\frac{x}{x+y}$$ when z = 12. Thus, any statement that tells us the ratio x:y for a specific value of z will be sufficient.

(2) We are told that when z = 0, x = y+1. As NoHalfMeasures suggests, let's call this value of y = a, so x = a+1 when z = 0. Writing out the sequences of values of x, y, and z from z = 0 to z = 12 now looks like this:

When $$z = 0: x = a+1, y = a$$
When $$z = 2: x = 2*(a+1), y = 3*a$$
When $$z = 4: x = 2^2*(a+1), y = 3^2*a$$
When $$z = 6: x = 2^3*(a+1), y = 3^3*a$$
When $$z = 8: x = 2^4*(a+1), y = 3^4*a$$
When $$z = 10: x = 2^5*(a+1), y = 3^5*a$$
When $$z = 12: x = 2^6*(a+1), y = 3^6*a$$

So, based on (2), when z = 12:

$$\frac{x}{x+y} = \frac{2^6*(a+1)}{2^6*(a+1) + 3^6*a}$$

$$\frac{x}{x+y} = \frac{64a + 64}{64a + 64 + 729a}$$

$$\frac{x}{x+y} = \frac{64a + 64}{793a + 64}$$

NOT SUFFICIENT

We cannot factor a out of this expression, so we cannot get a specific value for $$\frac{x}{x+y}$$ when z = 12 without knowing the value of a. Unlike with statement (1), we don't know the ratio x:y for any value of z. In general, if we are given a statement that does not allow us to find the ratio x:y for a specific value of z, we will not have sufficient information to find a single value for $$\frac{x}{x+y}$$ when z = 12.

I hope this longer explanation clarifies this problem for more of you. Please let me know if you have any questions!
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Re: The values of x and y vary with the value of z  [#permalink]

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29 Sep 2018, 09:43
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