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If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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15 Jun 2016, 03:30
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If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)? 1) 2x + y = 3 2) z = 2
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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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15 Jun 2016, 04:31
AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 before attempting this question simplify it: \(\frac{z*(2x+y)}{(2xzy)}\) 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.



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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06 Dec 2016, 18:19
AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 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
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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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12 Dec 2016, 14:06
I read this question as Y = 2xz and I ended up with undefined.



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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22 Dec 2016, 05:42
AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 \(\frac{(2xz + yz)}{(2xz – y)}\)=\(\frac{z(2x+y)}{(2xzy)}\) 1) 2x + y = 3 No information about\(\frac{z}{(2xz – y)}\) 2) z = 2 No information about \(\frac{(2x+y)}{(2xzy)}\) 1) & 2) combined No information about denominator 2xzy E :D
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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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21 Jan 2018, 02:59
JeffTargetTestPrep wrote: AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 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 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 4xy => 2x^2 1y^2 => (2x+y) (2xy) And numerator becomes 2(2x+y) 2x+y cancels out from both numerator and denominator now we are left with 2/2xy from statement 1 we get 2x+y = 3 => y=32x putting value of y in 2/2xy we get 2/2x32x 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!



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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21 Jan 2018, 23:30
marwahshubham wrote: JeffTargetTestPrep wrote: AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 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 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 4xy => 2x^2 1y^2 => (2x+y) (2xy)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



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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22 Apr 2018, 06:20
AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 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



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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22 Apr 2018, 11:10
dave13 wrote: AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 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*(2xz)/4*(2xz). Now 2xz 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.



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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22 Apr 2018, 12:59
amanvermagmat wrote: dave13 wrote: AbdurRakib wrote: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3 2) z = 2 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*(2xz)/4*(2xz). Now 2xz 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) ?



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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22 Apr 2018, 21:53
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.



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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05 Nov 2018, 23:54
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 !



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Re: If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
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08 Nov 2018, 00:34
nihar98 wrote: 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/(4xy) This is what we NEED to find. I think you have already equated this to '1' (and thats why you get 4xy = 6). We cannot assume 6/(4xy) to be anything, because thats what we need to find in the first place.



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