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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
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:
Re: approach to these kind of qs?? [#permalink]
03 Aug 2010, 04:17
bibha wrote:
Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2
Please take the trouble of explaining one step at a time
To me I try to address (ii) first because it looks easiest
Pick x and y that meet the conditions given (x=5 y=3) |5-3| > |5+3| FALSE Try different x and y thinking we are using absolutes I picked two negatives (x=-3 y=-5) |-3-(-5)| > |-3+-5| 2 > 8 False So we see that ii alone is sufficient. The answer is B or D
To investigate i I am lost at the algebra, so I just try to think about which squares are 9 apart
3^2=9 4^2=16 5^2=25 luckily on the GMAT you typically work with smaller squares
So I pick X=5 and Y=4 because \(5^2-4^2=9\) Since we are squaring keep in mind that X can be 5 or -5 While Y can be 4 or -4
The test in the given equation |x-y| > |x+y| (5,4) gives us 1 > 9 False (5,-4) gives us 9 > 1 True (-5,4) gives us 9 > 1 True (-5,-4) gives us 1 > 9 False
So A alone doesn't work
The correct answer is B.
In proofreading my work I found a major error in my calculation with the absolutes. This changed my answer on this question. On the real GMAT I would have guessed probably E after 3-4 minutes of frustration!
Re: approach to these kind of qs?? [#permalink]
03 Aug 2010, 09:28
TallJTinChina wrote:
bibha wrote:
Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2
Please take the trouble of explaining one step at a time
To me I try to address (ii) first because it looks easiest
Pick x and y that meet the conditions given (x=5 y=3) |5-3| > |5+3| FALSE Try different x and y thinking we are using absolutes I picked two negatives (x=-3 y=-5) |-3-(-5)| > |-3+-5| 2 > 8 False So we see that ii alone is sufficient. The answer is B or D
To investigate i I am lost at the algebra, so I just try to think about which squares are 9 apart
3^2=9 4^2=16 5^2=25 luckily on the GMAT you typically work with smaller squares
So I pick X=5 and Y=4 because \(5^2-4^2=9\) Since we are squaring keep in mind that X can be 5 or -5 While Y can be 4 or -4
The test in the given equation |x-y| > |x+y| (5,4) gives us 1 > 9 False (5,-4) gives us 9 > 1 True (-5,4) gives us 9 > 1 True (-5,-4) gives us 1 > 9 False
So A alone doesn't work
The correct answer is B.
In proofreading my work I found a major error in my calculation with the absolutes. This changed my answer on this question. On the real GMAT I would have guessed probably E after 3-4 minutes of frustration!
In plugging in numbers approach - like here, (ii) works for 2 examples selected, but how do we universally say it is true? Is there not a more general approach? I think for speed we just select a few examples, but how do we guarantee it? I am always confused on this point. So I jump to prove it for good and I made a mistake here
I saw x^2-y^2 = 9, took it to mean (x+y)(x-y)=9 and said x+y,x-y have to be both 3 or -3 and jumped at A. Wrong, does not say that they have to be integers...
Please explain or point me to the strategy in dealing with these please! _________________
Re: approach to these kind of qs?? [#permalink]
04 Aug 2010, 00:34
2
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
bibha wrote:
Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2
Please take the trouble of explaining one step at a time
I'd start solving by analyzing the stem.
First approach:
Is \(|x-y|>|x+y|\)? In which cases \(|x-y|\) will be more than \(|x+y|\)?
If \(x\) and \(y\) have the same sign (both positive or both negative) then absolute value of their sum (\(|x+y|\)) will be more than absolute value of their difference (\(|x-y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x+y|\), for example \(|5+3|\) or \(|-5-3|\) and in \(|x-y|\) they will not.
If \(x\) and \(y\) have opposite signs then absolute value of their difference (\(|x-y|\)) will be more than absolute value of their sum (\(|x+y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x-y|\), for example \(|5-(-3)|\) or \(|-5-3|\) and in \(|x+y|\) they will not.
So basically the question is: do \(x\) and \(y\) have opposite signs?
Second approach:
Is \(|x-y|>|x+y|\)? As both sides of inequality are non-negative we can safely square them: is \((x-y)^2>(x+y)^2\)? --> is \(x^2-2xy+y^2>x^2+2xy+y^2\)? --> is \(xy<0\)? Again the question becomes: do \(x\) and \(y\) have opposite signs?
Next:
(1) x^2-y^2 = 9 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.
(2) x-y=2 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.
(1)+(2) \(x+y=4.5\) and \(x-y=2\) --> directly tells us that \(|x-y|=2<|x+y|=4.5\). Sufficient. (If you solve system of equations you'll see that \(x\) and \(y\) have the same sign: they are both positive).
Answer: C.
PLUGGING NUMBERS:
On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.
(1) \(x^2-y^2=9\) --> try \(x=5\) and \(y=4\) (\(x^2-y^2=25-16=9\)) --> \(|x-y|=1<|x+y|=9\), so we have answer YES. Now let's try another set of numbers to get NO: \(x=5\) and \(y=-4\) --> \(|x-y|=9>|x+y|=1\), so now we have answer NO. Thus this statement is not sufficient.
We can do the same with statement (2) as well.
For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient.
Re: approach to these kind of qs?? [#permalink]
07 Aug 2010, 03:37
Bunuel wrote:
bibha wrote:
Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2
Please take the trouble of explaining one step at a time
I'd start solving by analyzing the stem.
First approach:
Is \(|x-y|>|x+y|\)? In which cases \(|x-y|\) will be more than \(|x+y|\)?
If \(x\) and \(y\) have the same sign (both positive or both negative) then absolute value of their sum (\(|x+y|\)) will be more than absolute value of their difference (\(|x-y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x+y|\), for example \(|5+3|\) or \(|-5-3|\) and in \(|x-y|\) they will not.
If \(x\) and \(y\) have opposite signs then absolute value of their difference (\(|x-y|\)) will be more than absolute value of their sum (\(|x+y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x-y|\), for example \(|5-(-3)|\) or \(|-5-3|\) and in \(|x+y|\) they will not.
So basically the question is: do \(x\) and \(y\) have opposite signs?
Second approach:
Is \(|x-y|>|x+y|\)? As both sides of inequality are non-negative we can safely square them: is \((x-y)^2>(x+y)^2\)? --> is \(x^2-2xy+y^2>x^2+2xy+y^2\)? --> is \(xy<0\)? Again the question becomes: do \(x\) and \(y\) have opposite signs?
Next:
(1) x^2-y^2 = 9 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.
(2) x-y=2 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.
(1)+(2) \(x+y=4.5\) and \(x-y=2\) --> directly tells us that \(|x-y|=2<|x+y|=4.5\). Sufficient. (If you solve system of equations you'll see that \(x\) and \(y\) have the same sign: they are both positive).
Answer: C.
PLUGGING NUMBERS:
On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.
(1) \(x^2-y^2=9\) --> try \(x=5\) and \(y=4\) (\(x^2-y^2=25-16=9\)) --> \(|x-y|=1<|x+y|=9\), so we have answer YES. Now let's try another set of numbers to get NO: \(x=5\) and \(y=-4\) --> \(|x-y|=9>|x+y|=1\), so now we have answer NO. Thus this statement is not sufficient.
We can do the same with statement (2) as well.
For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient.
Hope it helps.
Excellent Bunuel!! Simply Excellent. When it is said that x-y=2 and x^2-y^2=9 why does it not jump at me that x+y must be 4.5!!! Got your point about plugging in numbers.. Can disprove but proving universally is difficult... _________________
Re: approach to these kind of qs?? [#permalink]
07 Aug 2010, 08:09
well I eliminated A and B out as in both case x and y can be both and negative in either case.... which will obvoiusly give dual case... so D also out....
now here was big problem C and E.... I was not able to decide here i was trying to establish a case where x must be + or - same with y if we can do that we will have 1 another yes or no as answer
Re: approach to these kind of qs?? [#permalink]
30 Jan 2014, 16:20
Bunuel wrote:
bibha wrote:
Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2
Please take the trouble of explaining one step at a time
I'd start solving by analyzing the stem.
First approach:
Is \(|x-y|>|x+y|\)? In which cases \(|x-y|\) will be more than \(|x+y|\)?
If \(x\) and \(y\) have the same sign (both positive or both negative) then absolute value of their sum (\(|x+y|\)) will be more than absolute value of their difference (\(|x-y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x+y|\), for example \(|5+3|\) or \(|-5-3|\) and in \(|x-y|\) they will not.
If \(x\) and \(y\) have opposite signs then absolute value of their difference (\(|x-y|\)) will be more than absolute value of their sum (\(|x+y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x-y|\), for example \(|5-(-3)|\) or \(|-5-3|\) and in \(|x+y|\) they will not.
So basically the question is: do \(x\) and \(y\) have opposite signs?
Second approach:
Is \(|x-y|>|x+y|\)? As both sides of inequality are non-negative we can safely square them: is \((x-y)^2>(x+y)^2\)? --> is \(x^2-2xy+y^2>x^2+2xy+y^2\)? --> is \(xy<0\)? Again the question becomes: do \(x\) and \(y\) have opposite signs?
Next:
(1) x^2-y^2 = 9 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.
(2) x-y=2 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.
(1)+(2) \(x+y=4.5\) and \(x-y=2\) --> directly tells us that \(|x-y|=2<|x+y|=4.5\). Sufficient. (If you solve system of equations you'll see that \(x\) and \(y\) have the same sign: they are both positive).
Answer: C.
PLUGGING NUMBERS:
On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.
(1) \(x^2-y^2=9\) --> try \(x=5\) and \(y=4\) (\(x^2-y^2=25-16=9\)) --> \(|x-y|=1<|x+y|=9\), so we have answer YES. Now let's try another set of numbers to get NO: \(x=5\) and \(y=-4\) --> \(|x-y|=9>|x+y|=1\), so now we have answer NO. Thus this statement is not sufficient.
We can do the same with statement (2) as well.
For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient.
Hope it helps.
I like second approach best, squaring both sides is definetely something you can do here in order to manipulate the problem more easily getting rid of the absolute values in both sides and saving time instead of plugging numbers
Re: Is |x - y| > |x + y|? [#permalink]
10 Nov 2015, 10:46
Expert's post
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 independent equations ensures a solution.
Is |x - y| > |x + y|?
(1) x^2 - y^2 = 9 (2) x - y = 2
If we modify the question, |x - y| > |x + y|?, or (|x - y|)^2 > (|x + y|)^2?, or (x - y)^2 > (x + y)^2?, or x^2-2xy+y^2>x^2+2xy+y^2?, or -2xy>2xy?, or 0>4xy?, or xy<0?. The conditions do not have xy<0, so there are 2 variables (X,y) and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer. Looking at the conditions together, x-y=2, x+y=9/2. This is sufficient and the answer becomes (C).
For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...