If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?

before attempting this question simplify it:
\(\frac{z*(2x+y)}{(2xz-y)}\)

We can see that we do not get value of z from (1)

We can see that we do not get value of x and y from (2)

Also after combining we get the value of numerator but not denominator.

E is the answer.

We can start by simplifying the question:

(2xz + yz)/(2xz – y) = ?

z(2x + y)/(2xz – y) = ?

Statement One Alone:

2x + y = 3

Although statement one provides a value for 2x + y, we still do not have a value for z or 2xz –y. Thus, statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

z = 2

Although statement two provides a value for z, we still do not have a value for 2xz – y or 2x + y. Thus, statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using statements one and two, we have values for z and 2x + y. However, since we still do not have a value for 2xz - y, we do not have enough information to answer the question.

Answer: E
_________________

I read this question as Y = 2xz and I ended up with undefined.

\(\frac{(2xz + yz)}{(2xz – y)}\)=\(\frac{z(2x+y)}{(2xz-y)}\)

1) 2x + y = 3
No information about\(\frac{z}{(2xz – y)}\)

2) z = 2
No information about \(\frac{(2x+y)}{(2xz-y)}\)

1) & 2) combined
No information about denominator 2xz-y

E :D
_________________

Hi Jeff,

I understand that statement 1 & 2 alone are not sufficient but I opted for C in this way

From statement 2 we get z=2
denominator becomes 4x-y
=> 2x^2 -1y^2
=> (2x+y) (2x-y)

And numerator becomes 2(2x+y)

2x+y cancels out from both numerator and denominator

now we are left with 2/2x-y

from statement 1 we get 2x+y = 3
=> y=3-2x

putting value of y in 2/2x-y

we get 2/2x-3-2x
which comes up to 2/-3

I know this must be one of the silliest question you have encountered but can`t help myself with this.

your help will be highly appreciated.

Thanks in advance!

Hi

The highlighted part. I think what you have done here is not correct. 4x - y cannot be written as (2x)^2 - y^2
Instead 4x = (2√x)^2 and y = (√y)^2, so we can write 4x - y as (2√x + √y)(2√x - √y)

So the answer will be E only

ok here is my reasoning

statement one: x could be 0, or 1 and y could be 3 or 1

statement two: clrealy is not enough as we dont know values of x and y

combing the two statements is not enough because we still dont know values of x and y

is my reasong correct ? can anyone comment

Hello Dave

Generally speaking, in these type of questions, its not a good idea to go by this approach (that we need all values to answer this question), as sometimes we can get an answer without getting values of each and every variable. Its about manipulating the algebraic expressions.

Eg,, lets consider that we have to find this: (2xy - yz)/(4z - 8x)
Do we need to find values of each of x, y, z to solve the above. No. Because we can manipulate the above expression by taking out 'y' common from the numerator and '-4' common from the denominator.

So, it becomes y*(2x-z)/-4*(2x-z). Now 2x-z cancels out from both numerator and denominator and we are left with: y/-4 or -y/4. So what I am trying to say is that to find (2xy - yz)/(4z - 8x), we just need the value of y, values of x and z are NOT required.

amanvermagmat many thanks for explanation. one question how from this (2xz + yz)/(2xz – y) you got this (2xy - yz)/(4z - 8x) ?

many thanks for explanation. one question how from this (2xz + yz)/(2xz – y) you got this (2xy - yz)/(4z - 8x) ?[/quote]

Hello Dave

I didnt get (2xy - yz)/(4z - 8x) from (2xz + yz)/(2xz – y).. I was just giving you a different example to explain my point.

Hello,

I opted for option C, because I got values of x and y by combining two equations.

From statement 1, we can deduce (2x+y) = 3. And from statement 2, z = 2.
Inserting both these values in the equation, z(2x + y) / 2xz - y , we get 6 / 4x - y.

6 / 4x - y, can be rewritten as 4x - y = 6 ------- (1)
From statement 1, we have the equation, 2x + y = 3 ------ (2)
Combining Eq. (1) and (2), we can get x = 3/2, and y = 0. These values also hold true the condition in question stem i.e y not equal to 2xz.

By these unique values, we can get a unique value of the equation. Therefore, I chose C.

Can anyone please explain whether I am correct in my reasoning.

Thanks !

Hello

Till this step everything is fine: that we need to get 6/(4x-y)

This is what we NEED to find. I think you have already equated this to '1' (and thats why you get 4x-y = 6). We cannot assume 6/(4x-y) to be anything, because thats what we need to find in the first place.

MathRevolution can we use your trick here?
_________________

Answer: Option E

Video solution by GMATinsight



Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK HERE.
Two MUST join YouTube channels : GMATinsight (1000+ FREE Videos) and GMATclub

Hello, everyone. I came across this question today and noticed that, as of this writing, 13 percent of people have opted for (C). I wanted to discuss why this cannot be the answer without resorting to self-evident logic. By factoring and substituting, you can indeed work out quite a bit of the expression:

\(\frac{z(2x+y)}{(2xz-y)}\)

\(\frac{(2)(3)}{(2x(2)-y)}\)

\(\frac{6}{4x-y}\)

Now, from statement (1), you can isolate the y and substitute into the above expression:

\(2x+y=3\)

\(y=3-2x\)

But you have to be careful when substituting, or you could flip a sign, a mistake that a few people may be making. (I have highlighted such an error in the quoted post above.)

\(\frac{6}{4x-(3-2x)}\)

\(\frac{6}{4x-3+2x}\)

\(\frac{6}{6x-3}\)

Or, if you prefer the reduced fraction,

\(\frac{2}{2x-1}\)

That is it. Since we cannot determine the value of x, we will not be able to determine the value of the quotient even with the two statements together. Thus, the answer must be (E).

I hope this post may prove helpful to those who may have gone wayward but not understood why.

- Andrew
_________________

Jeff, my thought process here was, 3 variables - 2 equations, hence, it will not be sufficient to get the value. Did I jump the gun or can that thought process be used here?