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# If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0

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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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Hi amitjash,

I guess we need to consider the case where X=Y then |x-y|= 0.

The statements are required to establish whether X=Y.
Actually the question is whether X=Y. From question stem xy+z=z => XY = 0 => X=0 / Y=0 / X=Y=0

xy+z=z, is |x-y|>0?

1. x# 0 (Not equal to)

2. Y= 0
From question stem xy+z=z => XY = 0

Stmt 1: x# 0 (Not equal to) ; combined with question stem Y = 0. => X#Y hence |x-y|> 0 always ; sufficient
Stmt 2: Y=0 gives 2 possibilities Y=X = 0 => |x-y|= 0 & Y=0 X#0 => X#Y hence |x-y|> 0; not sufficient.

Answer A
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
Ian / Karishma

I may be loosing it all But this is how I think

If xy = 0. This certainly is no brainer x = 0 or y = 0

But the real problem is going to happen if x and y are both zero. Since x = 0 / y and since y and x are both zero we can infer that x is 0/0 or undefined. However we have assumed x=0. Hence I think we cannot assume that x and y are both zero. Can we ?

Not sure. Thoughts??
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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gmat1220 wrote:
Ian / Karishma

I may be loosing it all But this is how I think

If xy = 0. This certainly is no brainer x = 0 or y = 0

But the real problem is going to happen if x and y are both zero. Since x = 0 / y and since y and x are both zero we can infer that x is 0/0 or undefined. However we have assumed x=0. Hence I think we cannot assume that x and y are both zero. Can we ?

Not sure. Thoughts??

gmat1220: As per your request, let me provide the logic here.

Both, gurpreetsingh and your high school, are correct.

Given that x*y = 0.
Think of it in this way: I have two numbers. I don't know their values. But when I multiply them, I get 0. So at least one of the numbers have to be 0. Both can also be 0 since 0*0 = 0. Nothing says that they cannot be equal.

Why were you taught 'Either x or y is 0 in high school?'
Because 'Either or' implies 'At least one'. It is counter intuitive to how we think about 'Either or'. In language, we generally think 'Either or' means either A or B but not both. This is incorrect implication in logic. 'Either or' means at least one of A and B. Both are also possible. (If you do not agree, check out https://en.wikipedia.org/wiki/Logical_disjunction)

Next, how do you explain 0 = 0/0? If you remember, in many DS questions where you have equations with denominators, you are specifically given that the denominator is not 0. e.g.
Given x = y/(x - z), x not equal to z,...
You can only divide something by x if you know that it is not zero.
xy = 0 cannot be re-written as x = 0/y because you do not know whether y is 0 or not. You do not divide an equation by a variable until and unless you know that the variable is not 0.

Hope that clears up the doubts.
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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If xy + z = z, is |x - y| > 0? or given xy = 0 (either x or y or both 0), is |x-y|> 0? Squaring, is x^2 + y^2 -2xy > 0 ?

(1) x # 0 (Not equal to): x^2 is +ve. y = 0. So, x^2 + y^2 -2xy > 0
(2) y = 0: if x = 0, NO. If x = 1, x^2 + y^2 -2xy > 0. YES.

A.
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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zaarathelab wrote:
Is xy + z = z, is |x-y| > 0 ?

(1) x ≠ 0
(2) y = 0

KINDLY EXPLAIN STEP BY STEP
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Is xy + z = z, is |x-y| > 0 ? [#permalink]
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shakticnb wrote:
zaarathelab wrote:
Is xy + z = z, is |x-y| > 0 ?

(1) x ≠ 0
(2) y = 0

KINDLY EXPLAIN STEP BY STEP

If xy + z = z, is |x-y| > 0 ?

xy + z = z;
Cancel z: xy = 0. This implies that either x or y (or both) is 0.

(1) x ≠ 0. Since x is not 0, then y = 0. Thus the question becomes is |x| > 0. The absolute value of a non-zero number (x) is always positive. So, we have a definite YES answer to the question. Sufficient.

(2) y = 0. The question becomes is |x| > 0. If x = 0, then the answer is NO but if x ≠ 0, then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

Is xy + z = z, is |x-y| > 0 ?

(1) x ≠ 0
(2) y = 0

in the original condition we have xy+z=z, xy=0 and the question asks for |x-y|>0?
Since we have 2 variables and 1 equation, in order to match the number of variables and equations we need 1 more equation. Since there is 1 each in 1) and 2), D is likely the answer.

In case of 1), y = 0 thus the answer is always yes. Therefore it is sufficient.
In case of 2), x = 2 and y = 0 thus the answer is yes, but if x = 0 and y = 0 the answer is no. Therefore it is not sufficient.
The answer is A
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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amitjash wrote:
If xy + z = z, is |x - y| > 0?

(1) x ≠ 0
(2) y = 0

We can start by simplifying the given equation:

xy + z = z

xy = 0

We see that either x or y or both x and y must be zero.

We also see that the question is asking whether the absolute value of x - y is greater than zero. Since the absolute value of any quantity will be positive except when the quantity is 0, the only way in which absolute value of x - y won’t be greater than zero is if it’s equal to zero, and that will only occur if x = y. Let’s keep this in mind as we analyze our statements.

Statement One Alone:

x ≠ 0

Since xy = 0, if x does not equal zero, then y MUST equal zero. Thus, since x cannot equal y, the absolute value of x - y will always be greater than zero. Statement one alone is sufficient to answer question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

y = 0

Since y = 0, x could or could not equal zero, and thus we do not have enough information to answer the question. For instance, if y = 0 and x = 1, then |x - y| is greater than zero; however, if y = 0 and x = 0, then |x - y| is not greater than zero.

Answer: A
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Re: If xy + z = z, is |x - y| > 0 ? (1) x ≠ 0 (2) y = 0 [#permalink]
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In case of 1)
, y = 0 thus the answer is always yes.
Therefore it is sufficient.

In case of 2),
x = 2 and y = 0
thus the answer is yes,
but if x = 0 and y = 0
the answer is no.

Therefore it is not sufficient.
The answer is A
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Re: If xy + z = z, is |x - y| > 0 ? (1) x 0 (2) y = 0 [#permalink]
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