Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If xy+z=z, is |x-y|>0? 1. x# 0 (Not equal to) 2. Y= 0

I dont understand... here why do we need any statement at all?? For whatever value of x-y, |x-y| will remain positive isnt it? Can some one explain?

First of all: xy+z=z --> xy=0 --> x=0 or y=0 or both, x=y=0.

Next: absolute value is always non-negative |some \ expression|\geq{0}, so |x-y|\geq{0}. Question asks whether |x-y|>0, so basically question ask whether x=y=0, as only case when |x-y| is not more than zero is when it equals to zero, and this happens when x=y=0.

(1) x\neq{0} --> then y=0, so x\neq{y} --> |x-y|>0. Sufficient.

(2) y=0 --> if x=0 then |x-y|=0 and the answer to the question is NO, but if x\neq{0} then |x-y|>0 and the answer to the question is YES. Two different answers, hence not sufficient.

Re: Confustion modulus problem. [#permalink]
18 May 2011, 22:33

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 ?

Re: Confustion modulus problem. [#permalink]
20 May 2011, 03:42

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??

If x*y=z

and you consider y = z/x on condition that x is not equal to 0.

hence if x*y=0 then you can not say x=0/y unless u assume y is not equal to zero.

Re: Confustion modulus problem. [#permalink]
20 May 2011, 03:47

garimavyas wrote:

the answer is A . very simple , the question has stated xy=0 so ,either x=0 or y=0 , now A states x not = 0. hence y=0 is clear from A itself.

now |x-y| >0 always, provided (x-y) is not equal to 0.

This question is simple indeed, but since the user has posted the question it may not be that easy for him to understand. Sometimes we miss simple things, either because we don't know the concept or because we overlook them.

I have followed many of your posts in which you have stated - very simple straight fwd answer. I m glad it was easy for you to answer the question. But look from other users' prospective; they wouldn't have posted if the question had been easy for them.

The reason why I have pointed out this small thing is to let you know that such words could discourage the person. He would simply think he is not smart enough to answer such a simple and straight fwd question.

Re: Confustion modulus problem. [#permalink]
20 May 2011, 19:37

Exactly ! Statement 2) says y = 0 and we know x y = 0 then it means x is non zero to avoid 0/0 form for y

S1 is sufficient and S2 is sufficient.

Hence the answer should be D Not sure why people are saying A

gurpreetsingh wrote:

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??

If x*y=z

and you consider y = z/x on condition that x is not equal to 0.

hence if x*y=0 then you can not say x=0/y unless u assume y is not equal to zero.

Re: Confustion modulus problem. [#permalink]
20 May 2011, 20:56

Your argument: "xy = 0 S2: y = 0 Therefore, it means x is non zero to avoid 0/0 form for y Therefore, x > 0"

Think about this for a second, it's deeply flawed. If you assume your argument, you are essentially suggesting that we can never multiply two zeros together, as we need to "avoid 0/0 form for y", yet clearly:

Let x = 0, y = 0 0 * 0 = 0

The problem with your line of thinking is that you are treating division-by-zero as a possible operation that we "avoid". It's not avoided, its simply undefined.

Exactly ! Statement 2) says y = 0 and we know x y = 0 then it means x is non zero to avoid 0/0 form for y

S1 is sufficient and S2 is sufficient.

Hence the answer should be D Not sure why people are saying A

gurpreetsingh wrote:

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??

If x*y=z

and you consider y = z/x on condition that x is not equal to 0.

hence if x*y=0 then you can not say x=0/y unless u assume y is not equal to zero.

Re: Confustion modulus problem. [#permalink]
20 May 2011, 21:37

this is interesting. Challenge this if x(x-1) =0 can you assume that both the factors are zero at the same time. It will result in paradox. If x=0 and x-1 =0 then 1=0 which is nonsense! So I still standby my argument. I wonder if Karishma or Ian will voice their opinion.

Re: Confustion modulus problem. [#permalink]
21 May 2011, 03:30

gmat1220 wrote:

:-) this is interesting. Challenge this if x(x-1) =0 can you assume that both the factors are zero at the same time. It will result in paradox. If x=0 and x-1 =0 then 1=0 which is nonsense! So I still standby my argument. I wonder if Karishma or Ian will voice their opinion.

Posted from my mobile device

if x(x-1) =0 then either x=0 or x=1 not both

in any equation (x-a)(x-b)(x-c)=0 the possible answers are either x=a, x=b, x=c or any combination of two or all three. eg a=b or a=c. If a,b,c are different then only either of them could be true at one time, but we will have three possible answers.

Re: Confustion modulus problem. [#permalink]
22 May 2011, 01:09

gmat1220 wrote:

:-) this is interesting. Challenge this if x(x-1) =0 can you assume that both the factors are zero at the same time. It will result in paradox. If x=0 and x-1 =0 then 1=0 which is nonsense! So I still standby my argument. I wonder if Karishma or Ian will voice their opinion.

Posted from my mobile device

You are correct, in that example, you can not assume both are zero at the same time. What you are missing here is that these both contain the same variable, x. This not what is presented in the question.

I recognize you do not accept the answer to this question, but step back and consider the implication of your argument. You are suggesting that two independent variables, x and y, can not be zero, because when multiplied together they equal zero. This is a non-sensical conclusion to reach, much like if someone reach 1 = 0 in your above example. I am sure you accept this conclusion makes no sense, yet you are not questioning your own argument to find out what has led you to reach this conclusion.

More than happy for either of those people to weigh in on this, you will note that Bunnel already has.

Re: Confustion modulus problem. [#permalink]
22 May 2011, 06:35

1

This post received KUDOS

Expert's post

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 http://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.

Re: Confustion modulus problem. [#permalink]
22 May 2011, 19:32

Ohh yes ! That clears up the cloud. Thanks so much...... I was taking the literal meaning of "either". I don't think I ever interpreted "either" as atleast one.

While Karishma's posts are awesome, gurpreetsingh, pike and fluke thanks to you!

gmatclubot

Re: Confustion modulus problem.
[#permalink]
22 May 2011, 19:32