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!