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: What is the absolute difference between the cubes of two [#permalink]

Show Tags

16 Mar 2012, 05:03

8

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

16 Mar 2012, 05:04

IMO the correct answer is C first the requirement is to find the value of |x^3 - y^3| or |(x-y)( x^2 + y^2 + xy)| 1) x = y + 2 or x -y = 2, Not sufficient 2) (x+y)^2 = 49 + xy, as we simplify we get x^2 + y^2 + xy = 49 insufficient because we don't know the value of x - y from this statement combine 1 and 2, we get the value of (x-y) and ( x^2 + y^2 + xy) hence sufficient. _________________

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

16 Mar 2012, 05:08

Hey Bunnel, thanx for a nice explanation . I have learned something new today with ur explanation. Every time i read ur posts, it further reimposes my faith that ur a true legend. Thanx a ton. _________________

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.

Could you please help me to understand where is the flaw in my reasoning: I chose the answer D for this problem, since the first statement gives us the relationship between two variables : if the second item is two more than the first one, doesn't it mean x=x+2, where x+2 =y

Attachments

Screen Shot 2012-04-04 at 10.57.07 PM.png [ 31.45 KiB | Viewed 3196 times ]

Could you please help me to understand where is the flaw in my reasoning: I chose the answer D for this problem, since the first statement gives us the relationship between two variables : if the second item is two more than the first one, doesn't it mean x=x+2, where x+2 =y

Merging similar topics. Please refer to the solutions above and ask if anything remains unclear. _________________

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

04 Apr 2012, 12:21

But, Bunuel, why can't i set x=x+2, where x+2=y? Since the first statement tells us "one of the integers is 2 greater than the other". The only thing confused me - "ONE of the integers", what means i don't know the exact order May be my question is stupid one- but i just need to grasp it fully

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

04 Apr 2012, 12:35

Expert's post

Galiya wrote:

But, Bunuel, why can't i set x=x+2, where x+2=y? Since the first statement tells us "one of the integers is 2 greater than the other". The only thing confused me - "ONE of the integers", what means i don't know the exact order May be my question is stupid one- but i just need to grasp it fully

First of all x=x+2 doesn't make any sense, because x's cancel out and we have that 0=2, which is not true.

Next, "one of the integers is 2 greater than the other integer" CAN be expressed as x+2=y, (assuming that y is larger), but even in this case we cannot get the single numerical value of y^3-x^3, for example consider: y=2, x=0 and y=3, x=1.

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

05 Apr 2012, 04:25

Expert's post

vdbhamare wrote:

IMO C..

nice work bunnel.....

But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?

You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3-y^3=(x-y) (x^2+x y+y^2)\) --> (1) says that \(x-y=2\) and (2) says that \(x^2+x y+y^2=49\) --> \(x^3-y^3=(x-y) (x^2+x y+y^2)=2*49=98\).

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

29 Nov 2013, 08:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

09 Jul 2014, 02:57

Bunuel wrote:

vdbhamare wrote:

IMO C..

nice work bunnel.....

But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?

You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3-y^3=(x-y) (x^2+x y+y^2)\) --> (1) says that \(x-y=2\) and (2) says that \(x^2+x y+y^2=49\) --> \(x^3-y^3=(x-y) (x^2+x y+y^2)=2*49=98\).

Hope it's clear.

It was difficult to see that \(x^3-y^3\) can we written as \((x-y) (x^2+x y+y^2)\)

is there an easier way to understand this or must it be learnt by heart.

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

09 Jul 2014, 03:17

Expert's post

qlx wrote:

Bunuel wrote:

vdbhamare wrote:

IMO C..

nice work bunnel.....

But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?

You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3-y^3=(x-y) (x^2+x y+y^2)\) --> (1) says that \(x-y=2\) and (2) says that \(x^2+x y+y^2=49\) --> \(x^3-y^3=(x-y) (x^2+x y+y^2)=2*49=98\).

Hope it's clear.

It was difficult to see that \(x^3-y^3\) can we written as \((x-y) (x^2+x y+y^2)\)

is there an easier way to understand this or must it be learnt by heart.

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

11 Jul 2014, 06:05

Bunuel wrote:

What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Answer: C.

Hope it's clear.

Hello Bunuel . After solving stem 2 and coming to the point of \(x^2+xy+y^2=49\) , I deduced that since it is an equation in 2 variables it will not be solvable. I substituted from 1 ( y=x+2) and then realized that it transforms to a simple quadratic and then with the mentioned restriction , came to C. Would this be a right approach ?

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

11 Jul 2014, 11:38

Expert's post

himanshujovi wrote:

Bunuel wrote:

What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Answer: C.

Hope it's clear.

Hello Bunuel . After solving stem 2 and coming to the point of \(x^2+xy+y^2=49\) , I deduced that since it is an equation in 2 variables it will not be solvable. I substituted from 1 ( y=x+2) and then realized that it transforms to a simple quadratic and then with the mentioned restriction , came to C. Would this be a right approach ?

Have you read this: Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3). _________________

Re: What is the absolute difference between the cubes of two [#permalink]

Show Tags

16 Sep 2015, 04:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...