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Bunuel
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Bunuel
If x, y, and z are positive integers, is (x + y)z greater than (y + z)x ?

Is (x + y)z > (y + z)x?
Is xz + yz > xy + xz?
Cancel xz on both sides: is yz > xy?
Reduce by y: is z > z?

(1) y > x. Not sufficient.
(2) z > x. Sufficient.

Answer: B.
Bunuel
A bit typo. It is better if you edit it.
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Bunuel
If x, y, and z are positive integers, is (x + y)z greater than (y + z)x ?

Is (x + y)z > (y + z)x?
Is xz + yz > xy + xz?
Cancel xz on both sides: is yz > xy?
Reduce by y: is z > z?

(1) y > x. Not sufficient.
(2) z > x. Sufficient.

Answer: B.
Bunuel
A bit typo. It is better if you edit it.
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Edited. Thank you.
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GMAT 1: 760 Q51 V42
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Asad
If x, y, and z are positive integers, is (x + y)z greater than (y + z)x ?

(1) y > x
(2) z > x

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The question (x+y)z > (y+z)x is equivalent to z > x for the following reason

(x+y)z > (y+z)x
⇔ xz+yz > xy+zx
⇔ yz > xy
⇔ z > x, since y > 0

It is equivalent to condition 2) and it is sufficient.


Condition 1)

Since we don’t have any information about z, condition 2) is not sufficient.

Therefore, B is the answer.
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Asad
If x, y, and z are positive integers, is (x + y)z greater than (y + z)x ?

(1) y > x
(2) z > x

target (x + y)z greater than (y + z)x
xz+yz>yx+zx
yz>yx
z>x
#1
y>x
insufficient
#2
z>x
sufficient clearly
OPTION B
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@experts here, we could cancel xz and reduce by y only cus the question stem mentioned them to be positive integers, correct? if this weren't mentioned we couldn't have done the cancellation, right?
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