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# M10-33

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:43
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Difficulty:

45% (medium)

Question Stats:

71% (02:12) correct 29% (02:34) wrong based on 80 sessions

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If $$X$$, $$Y$$, and $$Z$$ are positive integers, is $$(X-Y) * (Y-Z) * (X-Z) \gt 0$$?

(1) $$X^2 + YZ = XY + XZ$$

(2) $$XY - Y^2 = XZ - YZ$$

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16 Sep 2014, 00:43
1
Official Solution:

Statement (1) by itself is sufficient. $$X^2 + YZ = XY + XZ$$ simplifies to $$(X-Y) * (X-Z) = 0$$.

Statement (2) by itself is sufficient. $$XY - Y^2 = XZ - YZ$$ simplifies to $$(X-Y) * (Y-Z) = 0$$.

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22 Nov 2014, 12:26
Simplification:

Stmt 1 becomes: (x^2)+yz-xy-xz=0
If you use foil backwards, you'll get (x-y)(x-z)=0.
Now we know that either (x-y) or (x-z) is 0.
Both of these are in the question stem. Replace one of the factors with a 0, and you'll see the entire left side must equal 0.
This is sufficient.

Question: How should we know that we need to foil?

Also, when I factor stmt 2, I get (-y+z)(y-x)=0. Are we allowed to multiply both factors on the left by -1 because doing the same to the 0 on the right would have no effect?

Thanks
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09 Dec 2014, 20:56
I dont know why, but at the moment I cannot figure out how from statement 1, you reached (x-y)(x-z)=0, maybe because it is already late and my brains refuse to work

X^2 + YZ = XY + XZ
XY+XZ-X^2 - YZ
X(X-Z)+Y(X-Z)
(X+Y)(X-Z)=0
or
from X^2 and -XZ we get X(X-Z)
and from YZ - XY = Y(Z-X) but Z-X is not the same as X-Z
where am I making the mistake?
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10 Dec 2014, 04:15
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1
mvictor wrote:
I dont know why, but at the moment I cannot figure out how from statement 1, you reached (x-y)(x-z)=0, maybe because it is already late and my brains refuse to work

X^2 + YZ = XY + XZ
XY+XZ-X^2 - YZ
X(X-Z)+Y(X-Z)
(X+Y)(X-Z)=0
or
from X^2 and -XZ we get X(X-Z)
and from YZ - XY = Y(Z-X) but Z-X is not the same as X-Z
where am I making the mistake?

$$X^2 + YZ = XY + XZ$$;

Group everything on one side: $$(X^2 -XY) + (YZ -XZ)=0$$;

Factor out X and -Z: $$X(X -Y) -Z(X-Y)=0$$;

Factor out X-Y: $$(X-Y)(X-Z)=0$$.

Does this make sense?
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01 Jun 2016, 07:18
JackSparr0w wrote:
Simplification:

Stmt 1 becomes: (x^2)+yz-xy-xz=0
If you use foil backwards, you'll get (x-y)(x-z)=0.
Now we know that either (x-y) or (x-z) is 0.
Both of these are in the question stem. Replace one of the factors with a 0, and you'll see the entire left side must equal 0.
This is sufficient.

Question: How should we know that we need to foil?

Also, when I factor stmt 2, I get (-y+z)(y-x)=0. Are we allowed to multiply both factors on the left by -1 because doing the same to the 0 on the right would have no effect?

Thanks

You don't really need to multiply stmt 2 by -1, the statement as you've factored still gives you enough to answer the question. You can multiply by -1, but you have it to do it to everything in the equation which might be messy after you've already factored it.

And incidentally, when you set the factor to zero to solve, it has the same result.
(-y+z)=0 --> z-y=0 -->z=y
(-z+y)=0 --> y-z=0 -->y=z

As to the question on how we know we need to FOIL - I don't know - anyone know of any clues that would indicate (reverse)FOIL-ing is the best way to go?
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13 Jan 2017, 08:06
I think this is a high-quality question.
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04 May 2017, 04:18
Bunuel wrote:
If $$X$$, $$Y$$, and $$Z$$ are positive integers, is $$(X-Y) * (Y-Z) * (X-Z) \gt 0$$?

(1) $$X^2 + YZ = XY + XZ$$

(2) $$XY - Y^2 = XZ - YZ$$

Here's what I did to solve this:

(X-Y) * (Y-Z) * (X-Z) > 0
i.e. neither (X-Y) nor (Y-Z) nor (X-Z) =0
or X!=Y and Y!=Z and X!=Z

(1) $$X^2 + YZ = XY + XZ$$
or $$X^2 - XZ = XY - YZ$$
or $$X(X - Z) = Y(X - Z)$$
or $$X = Y$$

Hence the (X-Y) * (Y-Z) * (X-Z) = 0 * (Y-Z) * (X-Z) = 0.
Sufficient.

(2) $$XY - Y^2 = XZ - YZ$$
or $$XY - XZ = Y^2 - YZ$$
or $$X(Y - Z) = Y(Y - Z)$$
or $$X = Y$$

Hence the (X-Y) * (Y-Z) * (X-Z) = 0 * (Y-Z) * (X-Z) = 0.
Sufficient.

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04 May 2017, 04:36
Considering statement (1) alone:

$$x^2$$ + yz = xy + xz
=> $$x^2$$ - xy - xz + yz = 0
=> x(x - y) - z(x - y) = 0
=> (x - y)(x - z) = 0
If 2 of the three brackets in the question are equal to 0, then the whole product is equal to 0.
SUFFICIENT. BCE goes out.

Considering statement (2) alone:

xy - $$y^2$$ = xz - yz
=> xy - $$y^2$$ - xz + yz = 0
=> y(x - y) - z(x - y) = 0
=> (x - y)(y - z) = 0
If 2 of the three brackets in the question are equal to 0, then the whole product is equal to 0.
SUFFICIENT. A goes out.
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31 Oct 2018, 20:48
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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31 Aug 2019, 06:57
I think this is a high-quality question and I agree with explanation.
Re M10-33   [#permalink] 31 Aug 2019, 06:57
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# M10-33

Moderators: chetan2u, Bunuel