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:

I don't understand your explanation, nitya34. Why a must be positive?

IMO from the first part of the question we can extract that a & b must be of the same sign, either positive or negative:

\(a=2, b=3\) \((-2, 3), (-3, 2)\) -> same quadrant

\(a=-2, b=-3\) \((2, -3), (3, -2)\) -> same quadrant

\(a=-2, b=3\) \((2, 3), (-3, -2)\) -> different quadrant

From (1) we know that x & y must be of the same sign too, but we don't know whether positive or negative and therefore it is insufficient. From (2) we know that a & x must be of the same sign but we don't know anything about Y, therefore it is insufficient. From (1) & (2) we know that a,b,x,y must be of the same sign and therefore are in the same quadrant = sufficient.

157) If ab ≠ 0 and (-a, b) and (-b, a) are in the same quadrant, is (-x, y) in this quadrant? a. xy > 0 b. ax > 0

OA:

Given that (-a,b) and (-b,a) are in the same quadrant. Hence we can deduce that the points are either in the I or III quadrant.

We need to determine whether (-x,y) also belongs to either I or III quadrant.

Statement A: xy>0. With this, we can deduce that either both x and y is +ve or both -ve.

However we cannot tie this information with the two given points (-a,b) and (-b,a). Insufficient.

Statement B: ax>0. With this again, we can deduce that either both a and x is +ve or both -ve. But we do not have sufficient information about y hence we cannot make out if (-x,y) is in the I or III quadrant. Insufficient.

With both the statement put together, we know that the sign of a and x is the same and the sign of x and y is the same. Hence we can deduce that the sign of x and y is the same. If the sign of x and y is the same then the point (-x,y) should also be in the I or III quadrant.

Sufficient. Hence answer is C. _________________

Support GMAT Club by putting a GMAT Club badge on your blog

157) If ab ≠ 0 and (-a, b) and (-b, a) are in the same quadrant, is (-x, y) in this quadrant? a. xy > 0 b. ax > 0

OA:

Given that (-a,b) and (-b,a) are in the same quadrant. Hence we can deduce that the points are either in the I or III quadrant.

We need to determine whether (-x,y) also belongs to either I or III quadrant.

Statement A: xy>0. With this, we can deduce that either both x and y is +ve or both -ve.

However we cannot tie this information with the two given points (-a,b) and (-b,a). Insufficient.

Statement B: ax>0. With this again, we can deduce that either both a and x is +ve or both -ve. But we do not have sufficient information about y hence we cannot make out if (-x,y) is in the I or III quadrant. Insufficient.

With both the statement put together, we know that the sign of a and x is the same and the sign of x and y is the same. Hence we can deduce that the sign of x and y is the same. If the sign of x and y is the same then the point (-x,y) should also be in the I or III quadrant.

Sufficient. Hence answer is C.

How can you deduce - "Hence we can deduce that the points are either in the I or III quadrant." ??? Also whatever you have proved using statement 1 and 2 can be proved by the statement 1 alone.

take a = 2 and b = 1 => -a,b = -2,1 and -b,a = -1,2 -------> both lies in 2nd quadrant. Also if you interchange the signs it can be proved that they lie in the 4th quadrant.

=> either both a and b are +ve or -ve i.e. both should have the same sign.

If it is not true, suppose a =1 and b = -2 => -a,b = -1,-2 and -b,a = 2,1 => do they lie in same quadrant? NO

Statement 1. xy>0 => either both are -ve or both are +ve => either (-x, y) is in 2nd or 4th quadrant. Hence not sufficient.

Statement 2. ax>0 => a and x have same sign. Since we do not know anything of b and y we can not predict. No sufficient.

Statement 1 and 2 taken together: x and y have same sign. a and x have same sign => a,b,x,y all have the same sign. => if a and b have -ve sign then x and y also have -ve sign and vice versa. => the required point lies in the quadrant in which the given two lie.

Hence C

Note: If we prove (-x, y) either lie in 2 or 4 quadrant, then it is wrong. it is possible that the two points lie in 2nd and the given point (-x, y) lies in the 4th.

I must say - very good question. What is the source? _________________

"With both the statement put together, we know that the sign of a and x is the same and the sign of x and y is the same. Hence we can deduce that the sign of x and y is the same. If the sign of x and y is the same then the point (-x,y) should also be in the I or III quadrant."

I dont understand your reasoning. Just from (1) you know sign of x & y is the same. Why do you need to deduce from (1) & (2) ?

Given : (-a,b) & (-b,a) are in the same quadrant ==> a & b should have same sign. from (i) x,y have same sign from (ii) a,x have same sign Together from (i) , (ii) and given information Sign of a=b=x=y . Hence (-x,y) will be in the same quadrant as (-a,b) & (-b,a)

157) If ab ≠ 0 and (-a, b) and (-b, a) are in the same quadrant, is (-x, y) in this quadrant? a. xy > 0 b. ax > 0

OA:

Given that (-a,b) and (-b,a) are in the same quadrant. Hence we can deduce that the points are either in the I or III quadrant.

We need to determine whether (-x,y) also belongs to either I or III quadrant.

Statement A: xy>0. With this, we can deduce that either both x and y is +ve or both -ve.

However we cannot tie this information with the two given points (-a,b) and (-b,a). Insufficient.

Statement B: ax>0. With this again, we can deduce that either both a and x is +ve or both -ve. But we do not have sufficient information about y hence we cannot make out if (-x,y) is in the I or III quadrant. Insufficient.

With both the statement put together, we know that the sign of a and x is the same and the sign of x and y is the same. Hence we can deduce that the sign of x and y is the same. If the sign of x and y is the same then the point (-x,y) should also be in the I or III quadrant.

Sufficient. Hence answer is C.

How can you deduce - "Hence we can deduce that the points are either in the I or III quadrant." ??? Also whatever you have proved using statement 1 and 2 can be proved by the statement 1 alone.

take a = 2 and b = 1 => -a,b = -2,1 and -b,a = -1,2 -------> both lies in 2nd quadrant. Also if you interchange the signs it can be proved that they lie in the 4th quadrant.

=> either both a and b are +ve or -ve i.e. both should have the same sign.

If it is not true, suppose a =1 and b = -2 => -a,b = -1,-2 and -b,a = 2,1 => do they lie in same quadrant? NO

Statement 1. xy>0 => either both are -ve or both are +ve => either (-x, y) is in 2nd or 4th quadrant. Hence not sufficient.

Statement 2. ax>0 => a and x have same sign. Since we do not know anything of b and y we can not predict. No sufficient.

Statement 1 and 2 taken together: x and y have same sign. a and x have same sign => a,b,x,y all have the same sign. => if a and b have -ve sign then x and y also have -ve sign and vice versa. => the required point lies in the quadrant in which the given two lie.

Hence C

Note: If we prove (-x, y) either lie in 2 or 4 quadrant, then it is wrong. it is possible that the two points lie in 2nd and the given point (-x, y) lies in the 4th.

I must say - very good question. What is the source?

Yes Gurpreet, you are correct. I madly rushed through the problem after incorrectly assuming that (-a,b) and (-b,a) are in the same quadrant. Thanks for pointing it out and also for the solution. +1 to you. _________________

Support GMAT Club by putting a GMAT Club badge on your blog

If ab is not equal to 0 , and points (-a,b) and (-b,a) are in the same quadrant of the xy – plane , is point (-x,y) in this same quadrant? 1) xy > 0 2) ax > 0

Hi everyone..would appreciate if you could solution to the question below...thanks in advance.

If axb (a times b) is different than zero and points (-a, b) and (-b,a) are in the same quadrant of the xy-plane is point (-x,y) in this same quadrant?

Re: Data Sufficiency Question GMAT Prep [#permalink]

Show Tags

15 Aug 2011, 07:26

Question tells us that A and B have the same sign. (1) Not sufficient because xy could both be negative and ab could both be positive or vice-versa (2) NS because ax having the same sign does not mean by do as well Together sufficient because we know xy have the same sign, and that sign is the same as a

one question..``Stem says, both points are in 2nd or 4th quad `` which is assuming +, + or -,- for a and b, since a times b is different than zero..how come we dont evaluate +,- and -,+ options for a and b? thanks

gmatclubot

Re: DS - points on quadrant
[#permalink]
15 Aug 2011, 12:20

Post your Blog on GMATClub We would like to invite all applicants who are applying to BSchools this year and are documenting their application experiences on their blogs to...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...