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

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!

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

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.

A is the correct answer.
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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

(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

Hi Bunuel,

Can you please explain this question.
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You can answer this question quite quickly with very little math (experts please let me know if this a valued approach)

We know that every time 2 is added to z; x increases by a factor of 2 (2x), and y increases by a factor of 3 (3y)

\(\frac{(2x)}{(2x+3y)}\) for any value of z
In order to find the value of x/(x+y) when z=12, x from the numerator and both x and y from the denominator have to cancel out or derive the value of x and y to a certain value of Z from the statements


(1) When z=6, x=5y
Substitute 5y for x, we can directly eliminate y from the numerator and denominator and obtain a value for z = 6, which we can figure out that same value for z =12
Sufficient

2.) (2) z=0, x =y+1
When you substitute y+1 for x, there's no way to factor out and cancel out Y from the equation, leaving you with 1 variable.
Not sufficient.

Answer is A

is this from gmatprep? pls, go in the right direction.

Z=6, x= 5y; y= x/5 --- (1)
when z=12 = 2^3x = 3^3 *x/5
x/(x+y) = 2^3x/2^3x + 3^3x/5 = 40/67

Can't do it with statement 2

so answer is A

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