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If xy+z=x(y+z), which of the following must be true?

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If xy+z=x(y+z), which of the following must be true? [#permalink] New post 23 Sep 2010, 12:44
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If xy+z=x(y+z), which of the following must be true?

A. x=0 and y=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
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Mar 2012, 22:12, edited 3 times in total.
Corrected OA
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Re: must be true [#permalink] New post 23 Sep 2010, 12:46
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TomB wrote:
If xy+z=x(y+z), which of the following must be true?
A. x=0 and y=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


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.

Hope it helps.
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Re: must be true [#permalink] New post 27 Oct 2010, 18:17
Thanks again Bunuel . I somehow always got stucked at option C :roll: Your explanation is awesome :-D
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Re: must be true [#permalink] New post 05 Dec 2010, 08:19
Thanks Bunuel..The OA given is wrong its E thanks for proving it ..I was left wondering how answer could be D,Because I also got my answer as E
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Re: If xy + z = x(y + z) [#permalink] New post 08 Dec 2010, 09:00
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COULD or MUST be true questions:
ds-number-theory-101025.html?hilit=must%20true
number-properties-question-101150.html?hilit=must%20true
gmat-club-please-explain-83605.html?hilit=must%20true
must-be-true-101575.html?hilit=must%20true
gmat-prep-question-101282.html?hilit=must%20true
ab-2-c-is-even-101751.html?hilit=must%20true
mgmat-inequalities-101732.html?hilit=must%20true#p788920
division-and-inequalities-87707.html?hilit=could%20true%20following#p666131

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Re: must be true [#permalink] New post 08 Dec 2010, 09:18
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Thanks Bunuel for the links! :-D
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Re: must be true [#permalink] New post 19 Dec 2010, 21:26
Bunuel wrote:
TomB wrote:
xy+z=x(y+z). which of the following must be true?
a) x=0 and y=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


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.

Hope it helps.


Yeah the explanation was very helpful..
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Re: Which of the following must be true - Gmatprep [#permalink] New post 10 Jan 2012, 12:05
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Hi there! I'm happy to help with this. :)

So, the question:
If xy + z = x(y + z), then
(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

So, as a very general rough approximation rule, when variables are multiplied, the mathematical consequences will far more often involve the word "or" than the word "and." That's just a rough-and-ready rule to use if all else is lost.

Let's look at this question. We have: xy + z = x(y + z)

Distribute on the right side: xy + z = xy + xz

Subtract xy from both sides: z = xz

So first of all, we get this funny equation with x & z in it, without y, so that means: there's no restriction on y, only restrictions on x & z.

In solving the equation z = xz, you may be tempted to divide by z, but WHOA THERE! Before you divide by a variable, you have to ask yourself the deeply heartfelt question: could this variable equal zero? If so, then dividing by it would break all mathematical laws, and Santa Claus would find out you've been bad. So, what do we do?

We'll look at two cases: (i) z = 0, and (ii) z =/ 0
(i) if z = 0, then both sides of the equation x = xz equal zero, and the equation is true. So, one possibility that makes the equation true is z = 0
(ii) if z is not equal to zero, then we can divide by z, and get 1 = x. That's the other solution that makes the equation true.

The solutions are x = 1 or z = 0, answer choice E.

Here's another practice question for you.

http://gmat.magoosh.com/questions/110

The question at that link should be followed by a video explanation.

Please let me know if you have any questions about what I've said. Good luck tomorrow!! :)

Mike :)
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Re: Which of the following must be true - Gmatprep [#permalink] New post 10 Jan 2012, 19:46
Given xy+z = xy+xz => xz-z = 0

so z(x-1) = 0 => z=0, x=1 (E)
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Re: Which of the following must be true - Gmatprep [#permalink] New post 10 Jan 2012, 20:01
mikemcgarry has given an exhaustive explanation!

Simply put, it's basic algebra. Simplify the expression and equate to zero.
Watch out for the use of "and" and "or" in the answer choices.
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Re: Triple variable algebra problem [#permalink] New post 20 Mar 2012, 20:25
xy + z = x(y+z)
=> xy+z = xy + xz
=> z = xz
=> z=0 or x=1

Option (E)
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Re: Number properties [#permalink] New post 13 Apr 2013, 20:53
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singhmaharaj wrote:
If xy + z = x(y+z) 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 and y=0

A. X=1 and z=0




xy + z = x(y+z)-----------> xy + z = xy+xz ---------> z = xz --------> x=1 and Z= 0 or 1
Answer E.
Note :- Here the value of y does not matter since xy = xy going to be zero for any value of y
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Re: Number properties [#permalink] New post 13 Apr 2013, 21:49
singhmaharaj wrote:
If xy + z = x(y+z) 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 and y=0

A. X=1 and z=0


xy + z = x(y+z)
=> xy + z = xy + xz
=> z = xz

The statement is true for both options A and E.
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Re: Number properties [#permalink] New post 13 Apr 2013, 22:53
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rakeshd347 wrote:
singhmaharaj wrote:
If xy + z = x(y+z) 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 and y=0

E. X=1 and z=0



xy+z=xy+xz…cancel xy form both sides….xz=z….deduct z from both sides xz-z=0 or z(x-1)=0 so x=1 and z=0

Correct answer is E

Please KUDOS if my answer helps


z(x-1) = 0 .... either x-1 can be zero or z can be zero...

possible options can be z=0 and x-1 be anything or z can be anything and x-1 can be zero ....

Also the question didn't mention anything about x,y,z being positive
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Re: Number properties [#permalink] New post 13 Apr 2013, 23:23
Correct question is
If xy+z = x(y+z), 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 or y=0
5. x=1 or z=0

Solution :
xy + z = xy + xz
z = xz i.e. z(x-1) = 0
so, z = 0 or x = 1 or both
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Re: If xy+z=x(y+z), which of the following must be true? [#permalink] New post 20 Oct 2013, 10:33
When I substituted b the answer match up as well... Some one please explain.
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Re: If xy+z=x(y+z), which of the following must be true? [#permalink] New post 20 Oct 2013, 10:35
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haotian87 wrote:
When I substituted b the answer match up as well... Some one please explain.


Please read the following post carefully: if-xy-z-x-y-z-which-of-the-following-must-be-true-101575.html#p787274
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If xy+z=x(y+z), which of the following must be true?   [#permalink] 20 Oct 2013, 10:35
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