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

Question

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

NoHalfMeasures · Answer

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

ENGRTOMBA2018 · Answer

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.

jimwild · Answer

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

rohit8865 · Answer

(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

sairam595 · Answer

Hi  Bunuel ,

Can you please explain this question.

JeffYin · Answer

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!

Jsound996 · Answer

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

thangvietnam · Answer

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

hkkat · Answer

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

Rishab2409 · Answer

To solve this, let's first establish how x and y change as z increases:

Every time z increases by 2:
 
 x is multiplied by 2 
 y is multiplied by 3

Since z changes from 6 or 0 up to 12 in steps of 2, we can determine the factors by which x and y scale.

Statement (1):  When z = 6, x = 5y.

Let a and b be the values of x and y at z = 6. We are given a = 5b. To go from z = 6 to z = 12, z increases by 2 a total of 3 times (since  ).
 
 x = a * 2^3 = 8a 
 y = b * 3^3 = 27b

Now, substitute a = 5b into x:
 
 x = 8(5b) = 40b 
 y = 27b

Calculate the ratio   at z = 12 ---->   ------> 40/67.  Sufficient.

Statement (2):  When z = 0, x = y + 1.

Let a and b be the values at z = 0. We know a = b + 1. To go from z = 0 to z = 12, z increases by 2 a total of 6 times (since  ).
 
 x = a * 2^6 = 64a 
 y = b * 3^6 = 729b

Now, substitute a = b + 1 into x:
 
 x = 64 (b+1) = 64b + 64  
 y = 729b

Calculate the ratio   at z = 12 ---->   ------>  . The value of this ratio depends on the specific value of b (which we do not know). For different values of b, we get different ratios.  Not Sufficient.

Correct Answer (A)

Hope it was helpful!

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

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