BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

another one...is xy>0?

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 24
Joined: Thu Mar 07, 2013 5:30 am

another one...is xy>0?

by Mormuse » Sat Mar 30, 2013 11:30 am
1. x-y>-2
2. x-2y<-6

For xy>0, x and y must have same sign,

statement 1 works with:
both >0 x=1 y=4
both <0 x=-1 y=-4
or different signs but only y can be negative:
x=1 y=-4

statement 2 works with almost everything:
both >0 x=1 y=4
both <0 x=-10 y=-1
or different signs but only x can be negative:
x=-4 y=2

combining both statements:
x and y can only be both positive or negative, thus combining statements is sufficient. can someone confirm that its the only way to solve this, it seems long and if there is a faster way to solve it i would be grateful to have your feedback.
thx
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 73
Joined: Sun May 06, 2012 2:19 am
Location: Cape Town
Thanked: 6 times

by rintoo22 » Sat Mar 30, 2013 11:51 am
Hi claudayst,

Working out different combination for x and y based on Statements can be cumbersome. However the moment you realise that there can be +/- to determine result resort to the following method.

keep the inequaity sign in the same direction. Tweak one of the equation to make both eq. unidirectional. In this case I have tweaked statement 2.
x-y>-2 ....1)
-x+2y>6 ....2)

Now resolve the equation like simultaneous eq.

y > 4 and x > 2

Hope this helps.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Mar 30, 2013 12:09 pm
claudayst wrote: Is xy>0?

1. x-y>-2
2. x-2y<-6
Target question: Is xy > 0?

Statement 1: x-y > -2
There are several pairs of numbers that meet this condition. Here are two:
Case a: x = 5 and y = 1, in which case xy is greater than 0
Case b: x = 5 and y = -1, in which case xy is not greater than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x - 2y < -6
There are several pairs of numbers that meet this condition. Here are two:
Case a: x = 1 and y = 5, in which case xy is greater than 0
Case b: x = -1 and y = 5, in which case xy is not greater than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
Here's what we know:
x-y > -2
x-2y < -6

Since both inequalities have an x, let's isolate x in both of them to get:
y-2 < x
x < 2y-6

Aside: Notice that I rewrote them so that the 2 inequality symbols are pointing in the same direction.

Now we can combine these inequalities to get: y-2 < x < 2y-6
Next, remove the x to get: y-2 < 2y-6
Then subtract y from both sides and add 6 to both sides to get: 4 < y
Great, we now know that y is positive.
Also, if y-2 < x (and y>4), then we know that x must also be positive
Since we now know that x and y are positive, we can be certain that xy is greater than 0

So, the answer is C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 212
Joined: Mon Dec 06, 2010 12:52 am
Location: India
Thanked: 5 times
Followed by:1 members

by sachin_yadav » Mon Apr 08, 2013 11:56 pm
Brent@GMATPrepNow wrote: Great, we now know that y is positive.
Also, if y-2 < x (and y>4), then we know that x must also be positive
Since we now know that x and y are positive, we can be certain that xy is greater than 0
One of the best questions i have come across.
Last part is a bit confusing.

Brent, as you have taken statement 1 to get x > 0. Is it okay if we take statement 2 ?
Because if we take statement 2, then

x < 2y-6
y > 4
suppose y = 5
x < 2(5) - 6
x < 10 - 6
x < 4
So, x can be positive and negative both.

This part is confusing me.

Regards
Sachin
Never surrender
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 308
Joined: Thu Mar 29, 2012 12:51 am
Thanked: 16 times
Followed by:3 members

by Lifetron » Tue Apr 09, 2013 1:25 am
sachin_yadav wrote:
Brent@GMATPrepNow wrote: Great, we now know that y is positive.
Also, if y-2 < x (and y>4), then we know that x must also be positive
Since we now know that x and y are positive, we can be certain that xy is greater than 0
One of the best questions i have come across.
Last part is a bit confusing.

Brent, as you have taken statement 1 to get x > 0. Is it okay if we take statement 2 ?
Because if we take statement 2, then

x < 2y-6
y > 4
suppose y = 5
x < 2(5) - 6
x < 10 - 6
x < 4
So, x can be positive and negative both.

This part is confusing me.

Regards
Sachin
We can't answer with each statements individually. That's the reason for taking both the statements together.

The result is y-2 < x < 2y-6

There is no point in taking this result and substituting in one statement, which won't give the actual limits.
Top
Master | Next Rank: 500 Posts
Posts: 298
Joined: Tue Feb 16, 2010 1:09 am
Thanked: 2 times
Followed by:1 members

by Deepthi Subbu » Wed Apr 10, 2013 3:02 am
Hi Brent@GmatprepNow, I am a little doubful abt one of the statements

Now we can combine these inequalities to get: y-2 < x < 2y-6
Next, remove the x to get: y-2 < 2y-6

If at all we are going to remove soemthing from an inequality, we are supposed to do so to other parts also rite?
y-2 < x < 2y-6
y-2-x<0<2y-6-x

Where am I going wrong?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 212
Joined: Mon Dec 06, 2010 12:52 am
Location: India
Thanked: 5 times
Followed by:1 members

by sachin_yadav » Wed Apr 10, 2013 4:36 am
Deepthi Subbu wrote:
If at all we are going to remove soemthing from an inequality, we are supposed to do so to other parts also rite?
y-2 < x < 2y-6
y-2-x<0<2y-6-x

Where am I going wrong?
Deepthi,

This is transitive property:- if a < b < c, then a < c
Therefore, y-2 < 2y-6

Sachin
Never surrender
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 212
Joined: Mon Dec 06, 2010 12:52 am
Location: India
Thanked: 5 times
Followed by:1 members

by sachin_yadav » Wed Apr 10, 2013 4:50 am
gughanbose wrote:
We can't answer with each statements individually. That's the reason for taking both the statements together.

The result is y-2 < x < 2y-6

There is no point in taking this result and substituting in one statement, which won't give the actual limits.
Hi gughanbose,

Let me understand this clearly.
After taking both the statements together, the following inequality is y-2 < x < 2y-6.
After removing x, the following inequality is y-2 < 2y-6

After solving we get y > 4
Now here comes the doubtful part:- if from the above combined inequality we can take y-2 < x, then why can't we take x < 2y - 6, and after solving i get x < 4.
I wasn't taking the second statement. I was just working on the combined inequality.

Please let me know what i am missing.

Sachin
Never surrender
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 511
Joined: Wed Aug 11, 2010 9:47 am
Location: Delhi, India
Thanked: 344 times
Followed by:86 members

by Anju@Gurome » Wed Apr 10, 2013 5:12 am
sachin_yadav wrote:After taking both the statements together, the following inequality is y-2 < x < 2y-6.
After removing x, the following inequality is y-2 < 2y-6

After solving we get y > 4
Now here comes the doubtful part:- if from the above combined inequality we can take y-2 < x, then why can't we take x < 2y - 6, and after solving i get x < 4.
What's your logic behind that solution?
Anyway, if any inequality forks down to two or more independent inequalities, then our final solution is that one which satisfies the original inequality, not the forked ones.

Here, if we take the second part, x < 2y - 6
And we already know, y > 4.
So, (2y - 6) > (2*4 - 6) = 2
--> (2y - 6) > 2

Now, (2y - 6) is greater than both x and 2.
But as we don't know the relative position of x and 2 on the number line we cannot comment on the values of x.

For example, y = 10 ---> x < (2*10 - 6) ---> x < 14 ---> x can be -20 or 0 or 4 or 10 etc.
So x can have any values (positive, zero or negative) depending upon the value of y.

But if we take the first part, x > y - 2
And we already know, y > 4.
So, (y - 2) > (4 - 2) = 2
--> (y - 2) > 2

Now, x > (y - 2) > 2 > 0
Therefore, we have a definite range of values of x independent of the value of y.

So, our final solution is x > 2 as that will satisfy both parts of the inequality.

Hope that helps.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 212
Joined: Mon Dec 06, 2010 12:52 am
Location: India
Thanked: 5 times
Followed by:1 members

by sachin_yadav » Wed Apr 10, 2013 7:24 am
Anju,

Thank you so much and really appreciate for your reply.
I got your point on what makes x > 2. But i am not able to understand the below part.
Anju@Gurome wrote: Here, if we take the second part, x < 2y - 6
And we already know, y > 4.
So, (2y - 6) > (2*4 - 6) = 2
--> (2y - 6) > 2

But if we take the first part, x > y - 2
And we already know, y > 4.
So, (y - 2) > (4 - 2) = 2
--> (y - 2) > 2
What makes (2y - 6) > 2 and (y - 2) > 2 ? After all we are just plugging the values
(2*4 - 6) = 2
so, (2y - 6) = 2
or
(2*5 - 6) = 4
so, (2y - 6) = 4.

Sorry, just want to clarify the doubt.

Regards
Sachin
Never surrender
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 511
Joined: Wed Aug 11, 2010 9:47 am
Location: Delhi, India
Thanked: 344 times
Followed by:86 members

by Anju@Gurome » Wed Apr 10, 2013 7:58 am
sachin_yadav wrote:What makes (2y - 6) > 2 and (y - 2) > 2 ? After all we are just plugging the values
(2*4 - 6) = 2
so, (2y - 6) = 2
or
(2*5 - 6) = 4
so, (2y - 6) = 4.
Let's start over.
y - 2 < 2y - 6 ---> y > 4

For the first part,
  • --> y > 4
    --> y - 2 > 4 - 2 (Subtract 2 from both sides)
    --> (y - 2) > 2
For the second part,
  • --> y > 4
    --> 2y > 8 (Multiply both sides by 2)
    --> 2y - 6 > 8 - 6 (Subtract 2 from both sides)
    --> (2y - 6) > 2
Note that, I'm not plugging any random number for y and I'm not using the equality sign.
I'm starting with the result in our hand, i.e. y > 4 and manipulating it to get the bounds of (y - 2) and (2y - 6).

While solving inequality we should never plug numbers.
But we can plug numbers to check whether an inequality holds true or not.

For example, you are plugging y = 5 as y > 4.
Now, y - 2 = 5 - 2 = 3
As, x > y - 2, x > 3
But as I have shown the actual range of values for x is x > 2.
So, your solution is missing the range 2 < x < 3

Hope that helps.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 212
Joined: Mon Dec 06, 2010 12:52 am
Location: India
Thanked: 5 times
Followed by:1 members

by sachin_yadav » Wed Apr 10, 2013 9:13 pm
Thanks. I got your point. :D

Regards
Sachin
Anju@Gurome wrote:
Let's start over.
y - 2 < 2y - 6 ---> y > 4

For the first part,
  • --> y > 4
    --> y - 2 > 4 - 2 (Subtract 2 from both sides)
    --> (y - 2) > 2
For the second part,
  • --> y > 4
    --> 2y > 8 (Multiply both sides by 2)
    --> 2y - 6 > 8 - 6 (Subtract 2 from both sides)
    --> (2y - 6) > 2
Note that, I'm not plugging any random number for y and I'm not using the equality sign.
I'm starting with the result in our hand, i.e. y > 4 and manipulating it to get the bounds of (y - 2) and (2y - 6).

While solving inequality we should never plug numbers.
But we can plug numbers to check whether an inequality holds true or not.

For example, you are plugging y = 5 as y > 4.
Now, y - 2 = 5 - 2 = 3
As, x > y - 2, x > 3
But as I have shown the actual range of values for x is x > 2.
So, your solution is missing the range 2 < x < 3

Hope that helps.
Never surrender
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 308
Joined: Thu Mar 29, 2012 12:51 am
Thanked: 16 times
Followed by:3 members

by Lifetron » Wed Apr 10, 2013 10:02 pm
I have a very bad doubt :

a)y > 4
y-4 > 0 [Subtracting across inequalities]

b)y > 4
(y/4) > 1 [Dividing with the number of known sign]

Both of 'em are correct ?
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 511
Joined: Wed Aug 11, 2010 9:47 am
Location: Delhi, India
Thanked: 344 times
Followed by:86 members

by Anju@Gurome » Wed Apr 10, 2013 10:09 pm
gughanbose wrote:I have a very bad doubt :

a)y > 4
y-4 > 0 [Subtracting across inequalities]

b)y > 4
(y/4) > 1 [Dividing with the number of known sign]

Both of 'em are correct ?
Yes. Both of them are correct.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Top
Post new topic Post Reply
• Page 1 of 1