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:

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) cannot be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

04 Sep 2010, 07:19

10

This post received KUDOS

1

This post was BOOKMARKED

Substituting (r,s) in the original equation: y=3x+2 => s=3r+2 => 3r-s+2=0. Question : is 3r-s-2=0. ?? statement 1 - (3r+2-s)(4r+9-s)=0 => (3r+2-s)=0 or (4r+9-s)=0 Not sufficient

Statement 2 - (4r-6-s)(3r+2-s)=0 => (3r+2-s)=0 or (4r-6-s)=0 Not sufficient Combining 1&2 -> common result (3r+2-s)=0 Sufficient. Answer - C

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

10 Jan 2012, 08:17

2

This post received KUDOS

Ok.. So lets look at both the statements individually:

First Statement 1:

1. (3r+2-s)(4r+9-s)=0

Look at the statement. It could mean two things. Either (3r+2-s) = 0 or (4r+9-s) = 0

if (3r+2-s) = 0, then yes 3r+2-s = 0 and hence s=3r+2 which is exactly like saying y=3x+2

But if 3r+2-s is not equal to 0 then we don't know for sure.

Now lets look at statement 2:

2. (4r-6-s)(3r+2-s)=0

Same implies for statement 2.

now lets assume that in the first statement 4r+9-s=0. If that is true, then from the second statement it is not possible for 4r-6-s to be equal to 0. Hence 3r+2-s = 0 and yes the point r,s which is the same as x,y lies on it !

Hence C
_________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

10 Jan 2012, 13:06

Easy one. GMAT loves these type of patterns.

NOTE: Notice the pattern here. When you have similar type of information in both statements, the answer is usually C, D or E.

1. rephrasing the equations gives us s=3r+2 and s=4r+9. obviously s=4r+9 does not lie, but s=3r+2 does 2. rephrasing the equations gives us s=3r+2 and s=4r-5. obviously s=4r-5 does not lie, but s=3r+2 does

Together, s=3r+2 and it does lie. C.
_________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Concentration: Entrepreneurship, General Management

GMAT 1: 730 Q49 V42

GPA: 3.44

WE: General Management (Entertainment and Sports)

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

20 Apr 2012, 06:25

Bunuel wrote:

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) can not be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Answer: C.

Bunuel,

Again, sorry for re-opening an old thread but I have a small doubt regarding this question. I got this question on my GMAT Prep today, had one quick look at the options and immediately saw that for the point to lie on the line, 3r+2-s has to be equal to zero. And since this information was given to us in two separate quadratic equations, it was necessary for us to have both the equations in order to discard (4r+9-s) and (4r-6-s) as solutions to the given equations.

However, at this point, I thought to myself that C would be too easy an answer and solved the equations to see if GMAT was upto its usual trickery. And this is where I went wrong.

Take for example the first equation. One solution gives us what we want, i.e., 3r+2-s = 0. However, if we solve the second solution 4r+9-s = 0, we get s = 4r+9. Now, if we plug this into the original equation y = 3x+2, we get:

y = 3x+2 Plugging the values of r and s into x and y:

4r+9 = 3r +2; Solving this gives us r = -7 and s = -19 and plugging these values of r and s in the original equation satisfies the equation:

-19 = 3(-7) + 2; -19 = -19

Since, this equation too satisfies the equation (and similarly, so does the second equation) I marked the answer as D.

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) can not be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Answer: C.

Bunuel,

Again, sorry for re-opening an old thread but I have a small doubt regarding this question. I got this question on my GMAT Prep today, had one quick look at the options and immediately saw that for the point to lie on the line, 3r+2-s has to be equal to zero. And since this information was given to us in two separate quadratic equations, it was necessary for us to have both the equations in order to discard (4r+9-s) and (4r-6-s) as solutions to the given equations.

However, at this point, I thought to myself that C would be too easy an answer and solved the equations to see if GMAT was upto its usual trickery. And this is where I went wrong.

Take for example the first equation. One solution gives us what we want, i.e., 3r+2-s = 0. However, if we solve the second solution 4r+9-s = 0, we get s = 4r+9. Now, if we plug this into the original equation y = 3x+2, we get:

y = 3x+2 Plugging the values of r and s into x and y:

4r+9 = 3r +2; Solving this gives us r = -7 and s = -19 and plugging these values of r and s in the original equation satisfies the equation:

-19 = 3(-7) + 2; -19 = -19

Since, this equation too satisfies the equation (and similarly, so does the second equation) I marked the answer as D.

Can you please tell me where I am going wrong?

Many thanks in advance.

The question asks whether \(3r+2-s=0\). Now, statement (1) says: \((3r+2-s)(4r+9-s)=0\), which means that either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both.

The above DOES NOT mean that we have a system of equations: \(3r+2-s=0\) and \(4r+9-s=0\). Because if it were so then you have an answer right away: \(3r+2-s=0\). No wonder that when you solve it as a system you got the values of \(s\) and \(r\) which make \(3r+2-s\) equal to zero: your whole starting point was that it does.

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

08 Jul 2013, 03:59

1

This post received KUDOS

gaurav2k101 wrote:

In the xy-plane , does the line with equation y=3x+2 contains the point(r,s)?

(1) (3r+2-s)(4r+9-s)=0 (2) (4r-6-s)(3r+2-s)=0

A*B=0==>THEN EITHER A/B IS ZERO OR BOTH ARE ZERO...this is a rule. now if r,s lies on point y=3x+2==>then 3r+2-s=0 now lets say 3r+2-s=A.. 4r+9-s=B 4r-6-s=C

now statement 1 SAYS that ...A*B=0===>Clearly we are not sure that A=0 or not similarly statement 2 says that ...A*C===>AGAIN WE CANNOT GAURANTEE THAT A=0..

NOW combining both we cam clearly say that A=0.. HENCE SUFFICIENT.
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabulary-list-for-gmat-reading-comprehension-155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment : http://www.youtube.com/watch?v=APt9ITygGss

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

30 Aug 2013, 15:41

Bunuel,

In statement (1), you have said that we have these possibilities: (3r + 2 - s) = 0 OR (4r + 9 - s) = 0 OR both.

But, when we have something like this: (x + 3)(x+5) = 0

In this case, there are only two possibilities, right? (x+3)=0 OR (x+5)=0. Both cannot be zero at the same time because "x" represents a single and unique value. Please confirm.

In statement (1), you have said that we have these possibilities: (3r + 2 - s) = 0 OR (4r + 9 - s) = 0 OR both.

But, when we have something like this: (x + 3)(x+5) = 0

In this case, there are only two possibilities, right? (x+3)=0 OR (x+5)=0. Both cannot be zero at the same time because "x" represents a single and unique value. Please confirm.

Yes.

(3r + 2 - s) = 0 and (4r + 9 - s) = 0 can both be true, for r=-7 and s=-19.

But for (x+3)=0 and (x+5)=0 both cannot be true simultaneously. Either x=-3 or x=-5.

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

Show Tags

07 Jun 2014, 12:40

could we also think of it as:

the following two equations define two non-identical hyperbola. 1) (3r+2-s) (4r+9-s)=0 2) (4r-6-s) (3r+2-s)=0

each equation represents a set of points on different hyperbola since the question states that (r,s) is one point, it must be a single intersection of the two hyperbola which cannot be found without both equations.

therefore: you need both equations to define one point (r,s) and then you can easily solve for whether (r,s) satisfies the first equation y = 3x+2