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Re: Problem solving [#permalink]
26 Apr 2010, 05:55
what is the OA.
every option seem to be correct, but when I wanted to evalute all the options after putting the infinity as the value for left out variable, I wasn't sure. Only Option A, says it would be 0 all the time.
lets try B
(2). x=1 and y=1
xy + z =x (y +z) 1.1 + z = 1.(1+z) 1 + z = 1 +z
what if Z is infinity, does equality hold good in that case.
Re: Problem solving [#permalink]
26 Apr 2010, 09:12
Hi,
Only option 5 holds true. In other cases barring case 1, only one of the values holds true. In case 1, if we put x=0 we get z=0, i.e. we get a solution but nothing to tell us that the two sides match.
If XY + Z = X(Y+Z), then which of the following must be true?
A. X=0 AND Z=0 B. X=1 AND Y=1 C. Y=1 AND Z=0 D. X=1 OR Y=0 E. X=1 OR Z=0
Please explain the answer and also why certain choices are wrong? For example what's wrong with the choice X=1 and Y=1?
\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).
Answer: E.
To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.
Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.
Re: Problem solving [#permalink]
28 Apr 2010, 06:29
ghanala wrote:
If xy + z =x (y +z), then which of the following must be true?
(1). x=0 and z=0 (2). x=1 and y=1 (3). y=1 and z=0 (4). x=1 and y=0 (5) x=1 and z=0
Given xy+z= x (y + z) xy + z = xy + xz z = xz.
This means: 1] if x=0, then z can take any value, including 0 but not necessarily 0. Eliminate option A.
2] if z=0, then x can take any value, including 1 but not necessarily 1; moreover if we assume z not equal to 0 and divide by z on both sides. We get x=1; this means that 'z' can take non-zero values when x=1 Eliminate option E.
Now we need to examine options B, C & D.
Examining options B & D together as both suggest that x=1 when y=1 or when y=0 in the equation xy + z = xy + xz if x=1 then the eq becomes y + z = y + z. This is possible for all values of y & z not necessarily 1 or 0. Eliminate both B & D.
Examining option C: y=1 and z=0 in equation xy + z = xy + xz If y=1 then the eq becomes x + z = x + xz x + z = x(1+z) Here if z is any value other than 0 then the equation does not hold good. Hence z must be equal to 0.
Re: xy + z = x(y + z) [#permalink]
01 Aug 2010, 11:42
I see. Where did you come up with the z*(x-1) = 0 from? Thanks. Just wanted to clarify that. It's already making a lot more sense. I should have caught this but I guess I didn't =( _________________
Re: Problem solving [#permalink]
01 Apr 2012, 20:54
Hi Bunuel, As per your solution, after cancelling the xy terms, you subtracted z from xz i.e \(xz-z=0\) instead of cancelling the variable "z". Could you please explain that part. Is it because we run the risk of forgetting z=0 value? Thanks H
Bunuel wrote:
LM wrote:
Please explain the answer and also why certain choices are wrong? For example what's wrong with the choice X=1 and Y=1?
\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).
To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.
Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.
Re: Problem solving [#permalink]
01 Apr 2012, 21:06
imhimanshu wrote:
Hi Bunuel, As per your solution, after cancelling the xy terms, you subtracted z from xz i.e \(xz-z=0\) instead of cancelling the variable "z". Could you please explain that part. Is it because we run the risk of forgetting z=0 value? Thanks H
i am not Bunuel but may try to explain your doubt:
when you cancel terms like that you miss some solutions for the equation; the thing is you should never ever cancel terms like that.
Re: Problem solving [#permalink]
01 Apr 2012, 21:11
Expert's post
imhimanshu wrote:
Hi Bunuel, As per your solution, after cancelling the xy terms, you subtracted z from xz i.e \(xz-z=0\) instead of cancelling the variable "z". Could you please explain that part. Is it because we run the risk of forgetting z=0 value? Thanks H
Bunuel wrote:
LM wrote:
Please explain the answer and also why certain choices are wrong? For example what's wrong with the choice X=1 and Y=1?
\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).
To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.
Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.
Hope it helps.
We cannot reduce (divide) \(z(x-1)=0\) by \(z\) since it can be zero and division by zero is not allowed. Also if we do that we exclude the possible solution of the equation: \(z=0\). _________________
Re: GMAT PREP (PS) [#permalink]
01 Aug 2012, 17:41
Bunuel wrote:
LM wrote:
If XY + Z = X(Y+Z), then which of the following must be true?
A. X=0 AND Z=0 B. X=1 AND Y=1 C. Y=1 AND Z=0 D. X=1 OR Y=0 E. X=1 OR Z=0
Please explain the answer and also why certain choices are wrong? For example what's wrong with the choice X=1 and Y=1?
\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).
To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.
Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.
Hope it helps.
I am confused because answer choice (A) also satisfies the equation. X=0 AND Z=0 as it is one of many possible solutions. If Z=0 ==> (0)Y=(0)Y. X can take an an infinity amount of values, as well as Y. Very puzzled, it seems like we have two correct answer choices in the question.
Re: GMAT PREP (PS) [#permalink]
01 Aug 2012, 23:28
Expert's post
alphabeta1234 wrote:
Bunuel wrote:
LM wrote:
If XY + Z = X(Y+Z), then which of the following must be true?
A. X=0 AND Z=0 B. X=1 AND Y=1 C. Y=1 AND Z=0 D. X=1 OR Y=0 E. X=1 OR Z=0
Please explain the answer and also why certain choices are wrong? For example what's wrong with the choice X=1 and Y=1?
\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).
To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.
Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.
Hope it helps.
I am confused because answer choice (A) also satisfies the equation. X=0 AND Z=0 as it is one of many possible solutions. If Z=0 ==> (0)Y=(0)Y. X can take an an infinity amount of values, as well as Y. Very puzzled, it seems like we have two correct answer choices in the question.
We are asked "which of the following MUST be true".
Now, if \(z=0\) then \(x\) can take ANY value. So, \(x=0\) AND \(z=0\) is not necessarily true. _________________
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