If xyz 0 and w/x = x/y = y/z, then w/z must be equal to which of the

Question

If  xyz ≠ 0  and  \frac{w}{x} = \frac{x}{y} = \frac{y}{z} , then  \frac{w}{z}  must be equal to which of the following?

(A)  \frac{w}{x}

(B)  \frac{w}{x^3}

(C)  \frac{w^2}{x^2}

(D)  \frac{w^2}{x^3}

(E)  \frac{w^3}{x^3}

gmatophobia · Accepted Answer

\frac{w}{x} = \frac{x}{y}

w = \frac{x^2}{y}   -- (1)

\frac{x}{y} = \frac{y}{z}

z = \frac{y^2}{x}   -- (2)

\frac{(1) }{ (2) } = \frac{w }{ z }= \frac{\frac{x^2}{y}}{\frac{y^2}{x}} = (\frac{x}{y})^3

\frac{w }{ z }= (\frac{x}{y})^3

From the first equality we get -

\frac{w }{ z }= (\frac{w}{x})^3

Option E

andreagonzalez2k · Answer

In the problem statement it says that:

\frac{w}{x} = \frac{x}{y}

So you only have to substitute both values.

A faster way to solve it is to play with values:

w=8, x=4, y=2, z=1

\frac{8}{4} = \frac{4}{2} = \frac{2}{1}

\frac{w}{z} = \frac{8}{1} = 8

A)  \frac{w}{x} = \frac{8}{4} ≠ 8

B)  \frac{w}{x^3} = \frac{8}{4^3} = \frac{8}{64} ≠ 8

C)  \frac{w^2}{x^2} = \frac{8^2}{4^2} = \frac{64}{16} = 4 ≠ 8

D)  \frac{w^2}{x^3} = \frac{8^2}{4^3} = \frac{64}{64} = 1 ≠ 8

E)  \frac{w^3}{x^3} = \frac{8^3}{4^3} = \frac{2^9}{2^6} = 8

floridastudent · Answer

Hi can someone elaborate how you made that final jump from (w/z) = (x/y)^3 into (w/z) = (w/x)^3

MrWhite · Answer

Given in the question:  \frac{w }{ x} = \frac{x }{ y} => w = \frac{x^2}{y}

\frac{w}{z }= (\frac{x }{ y})^3 = (\frac{x^2 }{ yx})^3

Hence

\frac{w}{z } = (\frac{w }{ x})^3

Hope this helps.

chetan2u · Answer

\frac{w}{x} = \frac{x}{y} = \frac{y}{z} 
Multiply all three and equate to  \frac{w}{x} 
 \frac{w}{x} * \frac{x}{y} * \frac{y}{z} = \frac{w}{x} * \frac{w}{x} * \frac{w}{x} 
 \frac{w}{z} = (\frac{w}{x})^3

E

Or
substitute w=8 and x=4
 \frac{8}{4} = \frac{4}{2} = \frac{2}{1} 
So,  \frac{w}{z} = \frac{8}{1}=8 
A) \frac{w}{x}=\frac{8}{4}=2 
B) \frac{w}{x^3}=\frac{8}{4^3}=\frac{1}{8} 
C) \frac{w^2}{x^2}=\frac{8^2}{4^2}=\frac{4}{1}=4 
D) \frac{w^2}{x^3}=\frac{8^2}{4^3}=1 
E) \frac{w^3}{x^3}=\frac{8^3}{4^3}=8

E

alphab · Answer

problem is if you start with fractions it wont work e.g. w/x = 1/2, x/y = 2/4, y/z = 4/8

Neodymium · Answer

If we re-write the given equations, it appears as:

i. wy = x^2 => w = x^2 / y
ii. xz = y^2
iii. wz = x.y => z = x.y / w

So, substituting the above equations for w and z in the form of w / z we get:

w / z = (x^2/y)/(y^2/x) = (x/y)3, now we can see from the given set of equations that (x/y = w/x), so we back-substitute and get = (w/x)^3.

andreagonzalez2k · Answer

It works too.
In your example, you choose the values w=1, x=2, y=4. z=8.
w/z=1/8
And B, D or E could be the answers.

So, before choosing values, you must note that choosing w=1 is a bad idea because in the answers you have w, w^2 and w^3 and 1^anything is 1. So you have to choose the values carefully.

saynchalk · Answer

I have it till (x/y)^3 but can't get past that

SAKSHI2907 · Answer

this is the approach i used we know that the ratio of w/x=x/y=y/z is equal assume X/Y= 2/4 then w/x=1/2 and y/z=4/8 so, w/z=1/8 =1^3/2^3 =w^3/x^3 Hope this helps

saynchalk · Answer

Love the approach, but how did you figure this out?

SAKSHI2907 · Answer

Hey, as soon as i saw the ratios are equal i assumed all of their ratios to be 1/2 and plugged in the values accordingly. w/x=1/2 x/y=2/4 (which further simplifies to1/2) y/z=4/8 so now that i know the values i substituted them in w/x=1/8 1/8 can be further simplified as 1^3/2^3= w^3/x^3 Sorry if my approach and explanation still seems to be confusing or messy. I just played with the number. After all, everyone can be a good student, but not everyone can be a good teacher

GMATinsight · Answer

SHORTCUT

Let,  \frac{w}{x} = \frac{x}{y} = \frac{y}{z} = a

Let's multiply them all

ie.  \frac{w}{x} * \frac{x}{y} * \frac{y}{z} = a*a*a

i.e. \frac{w}{z} = a^3

Only Option E is a^3 form hence

Answer: Option E

bumpbot · Answer

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