Are x and y both positive? (1) 2x - 2y = 1 (2) \frac{x}{y} > 1

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.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Back to Forum

Create Topic

Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1

1 2 3 4

Matthias15

Joined: 05 Aug 2015

Last visit: 26 Feb 2016

Posts: 18

Own Kudos:

[1]

Given Kudos: 21

Posts: 18

Kudos: 38

[1]

05 Aug 2015, 06:37

Kudos

Bookmarks

Bunuel

arijitb1980

Bunuel

Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
\(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hello Bunuel,

I am preparing for GMAT and you are a master in quantitatives,I salute you. However this problem I did not understand why it is C.I got E by plugging in numbers.
Considerin Stmt (1)
2x-2y=1
Means x-y=1/2
Now i used 2 numbers x=2,y=1.5 and x= -1.5 and y= -2
So (1) is insufficient as both x and y can be both +ve or -ve.
Stmt(2)
x/y > 1
Means x >y
If i use the above numbers x=2,y=1.5 and x= -1.5 and y= -2
then this holds true but cannot determine if x and y are both positive or negetive.
Combining (1) and (2) I still cannot determine if x and y are both positive or negetive .
So for me answer is E .
Please let me know what am I missing here.

Thanks

As stated above, x= -1.5 and y= -2 does not satisfy the second statement, hence you cannot use these values when plugging numbers.

Hi,

While I see that translating (x/y)>1 into x>y must be wrong, as clearly it lets me use a different range of numbers (and is the reason I chose E instead of C), I don't understand what is wrong with my algebra.

Why can't I multiply (x/y) * y > 1 * y and get x>y?

Thank you.

GMATinsight

Major Poster

Joined: 08 Jul 2010

Last visit: 14 Sep 2026

Posts: 7,352

Own Kudos:

17,520

[1]

Given Kudos: 128

Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator

Location: India

GMAT: QUANT+DI EXPERT

Schools: IIM (A) ISB '24

GMAT 1: 750 Q51 V41 Verified score

WE:Education (Education)

Products:

Expert

Expert reply

Schools: IIM (A) ISB '24

GMAT 1: 750 Q51 V41 Verified score

Posts: 7,352

Kudos: 17,520

[1]

05 Aug 2015, 07:07

Kudos

Bookmarks

Matthias15

Hi Matthias15

If (x/y) > 1

Case-1: If y >0 then x > y

Case-2: If y <0 then x < y

Please note: Multiplying a Negative Number on both sides require the Inequation sign to flip [as mentioned in Case-2]

That's where you are wrong.

I hope it helps!

jitendra31

Joined: 04 Dec 2014

Last visit: 08 Nov 2016

Posts: 8

Own Kudos:

Given Kudos: 37

Posts: 8

Kudos: 65

24 Sep 2015, 16:00

Kudos

Bookmarks

What happens when y = o?

X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'.

Even in this case -

"One of the approaches:
2x−2y=1 --> x=y+12
xy>1 --> x−yy>0 --> substitute x --> 1y>0 --> y is positive, and as x=y+12, x is positive too. Sufficient."

The moment you assume y = 0, you can easily see that 1/y is infinity that is greater than 0 and hence, satisfies the equation.

Am i missing something. My answer would be 'E'

BrentGMATPrepNow

Major Poster

Joined: 12 Sep 2015

Last visit: 31 Oct 2025

Posts: 6,727

Own Kudos:

37,344

[10]

Given Kudos: 799

Location: Canada

Expert

Expert reply

Posts: 6,727

Kudos: 37,344

[10]

24 Sep 2015, 16:16

Kudos

Bookmarks

Manbehindthecurtain

Are x and y both positive?

(1) 2x-2y = 1
(2) x/y > 1

Target question: Are x and y both positive?

Statement 1: 2x - 2y = 1
There are several pairs of numbers that satisfy this condition. Here are two:
Case a: x = 1 and y = 0.5, in which case x and y are both positive
Case b: x = -0.5 and y = -1, in which case x and y are not both positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x/y > 1
This tells us that x/y is positive. This means that either x and y are both positive or x and y are both negative. Here are two possible cases:
Case a: x = 4 and y = 2, in which case x and y are both positive
Case b: x = -4 and y = -2, in which case x and y are not both positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2
Statement 1 tells us that 2x - 2y = 1.
Divide both sides by 2 to get: x - y = 1/2
Solve for x to get x = y + 1/2

Now take the statement 2 inequality (x/y > 1) and replace x with y + 1/2 to get:
(y + 1/2)/y > 1
Rewrite as: y/y + (1/2)/y > 1
Simplify: 1 + 1/(2y) > 1
Subtract 1 from both sides: 1/(2y) > 0
If 1/(2y) is positive, then y must be positive.

Statement 2 tells us that either x and y are both positive or x and y are both negative.
Now that we know that y is positive, it must be the case that x and y are both positive
Since we can now answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

BrentGMATPrepNow

Major Poster

Joined: 12 Sep 2015

Last visit: 31 Oct 2025

Posts: 6,727

Own Kudos:

37,344

[2]

Given Kudos: 799

Location: Canada

Expert

Expert reply

Posts: 6,727

Kudos: 37,344

[2]

24 Sep 2015, 16:21

Kudos

Bookmarks

jitendra31

What happens when y = o?
X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'.
'

It may seem that 0.5/0 = infinity, but this is not the case.
If we approach 0 from the positive side, then it looks like 0.5/0 is a REALLY BIG POSITIVE NUMBER
0.5/0.1 = 5
0.5/0.01 = 50
0.5/0.001 = 500
0.5/0.0001 = 5000
0.5/0.00001 = 50000
etc.

But what if we approach 0 from the NEGATIVE side:
0.5/(-0.1) = -5
0.5/(-0.01) = -50
0.5/(-0.001) = -500
0.5/(-0.0001) = -5000
0.5/(-0.00001) = -50000
Here it looks like 0.5/0 will be a REALLY BIG NEGATIVE NUMBER

This is why we say that x/0 is undefined.

Cheers,
Brent

JeffTargetTestPrep

Target Test Prep Representative

Joined: 04 Mar 2011

Last visit: 05 Jan 2024

Posts: 2,973

Own Kudos:

8,970

[5]

Given Kudos: 1,646

Status:Head GMAT Instructor

Affiliations: Target Test Prep

Expert

Expert reply

Posts: 2,973

Kudos: 8,970

[5]

06 May 2016, 06:42

Kudos

Bookmarks

Are x and y both positive?

(1) 2x - 2y = 1
(2) x/y > 1

Solution:

We need to determine whether x and y are both positive.

Statement One Alone:

2x – 2y = 1

Simplifying statement one we have:

2(x – y) = 1

x – y = ½

The information in statement one is not sufficient to determine whether x and y are both positive. For instance if x = 1 and y = ½, x and y are both positive; however if x = -1/2 and y = -1, x and y are not both positive. We can eliminate answer choices A and D.

Statement Two Alone:

x/y > 1

Using the information in statement two, we see that x and y can both be positive or both be negative. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using the information in statements one and two we know that x – y = ½ and that x/y > 1. Isolating x in the equation we have: x = ½ + y. We can now substitute ½ + y for x in the inequality x/y > 1 and we have:

(1/2 + y)/y > 1

(1/2)/y + y/y > 1

1/2y + 1 > 1

1/2y > 0

Thus, y must be greater than zero. Also, since x/y is greater than one, x also must be greater than zero.

Answer: C

MathRevolution

Math Revolution GMAT Instructor

Joined: 16 Aug 2015

Last visit: 27 Sep 2022

Posts: 10,057

Own Kudos:

20,422

Given Kudos: 4

GMAT 1: 760 Q51 V42 Verified score

GPA: 3.82

Expert

Expert reply

GMAT 1: 760 Q51 V42 Verified score

Posts: 10,057

Kudos: 20,422

22 Jun 2016, 21:17

Kudos

Bookmarks

There are 2 variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that the correct answer is C. Using 1) and 2), from 2(x-y)=1>0 we get 2(x-y)>0, x>y. Using 2), if we multiply both sides by y^2, we get xy>y^2, xy-y^2>0, y(x-y)>0. Since x-y>0, we get y>0. From x>y>0, x>0. The answer is always yes and the condition is sufficient. Hence, the correct answer is C.



For cases where we need 2 more equations, 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.

Attachments

variable approach's answer probability.jpg [ 219.74 KiB | Viewed 16402 times ]

adiagr

Joined: 18 Jan 2010

Last visit: 05 Oct 2019

Posts: 202

Own Kudos:

1,172

Given Kudos: 9

GMAT 1: 710 Q48 V40 Verified score

Posts: 202

Kudos: 1,172

22 Jun 2016, 22:03

Kudos

Bookmarks

Manbehindthecurtain

Are x and y both positive?

(1) 2x-2y = 1
(2) x/y > 1

Statement 1
2x-2y = 1

x - y = (1/2)

x = 3/2; y = 1 ; both x and y positive

x = 1/4, y = -1/4; x is positive, y is negative

so not sufficient

Statement 2

(x/y) > 1

x = 3/2; y = 1 ; (x/y) is more than 1. both x and y positive

x = -2; y = -1 ; (x/y) is more than 1. both x and y negative

so not sufficient

Combining Statement 1 and 2

x - y = 1/2

x = y+(1/2)

x/y > 1

{y+(1/2)} / y is more than 1

1+(1/2y) > 1

(1/2y) > 0

y has to be positive. because x = y+1/2, so x also has to be positive.

C is the answer.

vgmatv

Joined: 01 May 2016

Last visit: 26 Feb 2017

Posts: 2

Kudos: 0

22 Jan 2017, 12:47

Kudos

Bookmarks

What about the example of y=0? Then x/y is surely greater than 1, hence I chose answer E.

BrentGMATPrepNow

Major Poster

Joined: 12 Sep 2015

Last visit: 31 Oct 2025

Posts: 6,727

Own Kudos:

37,344

[1]

Given Kudos: 799

Location: Canada

Expert

Expert reply

Posts: 6,727

Kudos: 37,344

[1]

22 Jan 2017, 13:50

Kudos

Bookmarks

vgmatv

What about the example of y=0? Then x/y is surely greater than 1, hence I chose answer E.

x/0 can be really big OR really small...

From my earlier post:

If we approach 0 from the positive side, then it looks like 0.5/0 is a REALLY BIG POSITIVE NUMBER
0.5/0.1 = 5
0.5/0.01 = 50
0.5/0.001 = 500
0.5/0.0001 = 5000
0.5/0.00001 = 50000
etc.

But what if we approach 0 from the NEGATIVE side:
0.5/(-0.1) = -5
0.5/(-0.01) = -50
0.5/(-0.001) = -500
0.5/(-0.0001) = -5000
0.5/(-0.00001) = -50000
Here it looks like 0.5/0 will be a REALLY BIG NEGATIVE NUMBER

This is why we say that x/0 is undefined.

Cheers,
Brent

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 17 Sep 2026

Posts: 113,607

Own Kudos:

841,791

Given Kudos: 111,601

Products:

Expert

Expert reply

Posts: 113,607

Kudos: 841,791

20 Jun 2017, 22:27

Kudos

Bookmarks

TaN1213

Bunuel

Hi Buñuel,

If we put x=1/2 & y=0 , would you please explain how are both statements together sufficient? Both statements hold true for these values of x and y, yet 0 is not positive.

Regards.

chris558

Are x and y both positive?

1) 2x-2y=1
2(x-y)=1
x-y=1/2
-->3/4-1/4=1/2....YES
-->-1/4-(-3/4)=1/2...NO
INSUFFICIENT

2) x/y>1
This just means that x and y have the same sign. They're either both positive or both negative.
INSUFFICIENT

1&2)
x=1/2+y

(1/2+y)/y>1
y/2 + 1 > 1
y/2 > 0 which means that Y is greater than 0. And since both x and y have the same sign, both x and y are Positive. YES.

Answer is C.

Sent from my Redmi Note 4 using GMAT Club Forum mobile app

y cannot be 0 because in this case x/y will not be defined (division by 0 is not allowed) and not greater than 1 as given in the second statement.

ydmuley

Retired Moderator

Joined: 19 Mar 2014

Last visit: 01 Dec 2019

Posts: 807

Own Kudos:

929

Given Kudos: 199

Location: India

Concentration: Finance, Entrepreneurship

Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20

GPA: 3.5

Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20

Posts: 807

Kudos: 929

21 Jun 2017, 13:36

Kudos

Bookmarks

Are x and y both positive?

(1) 2x-2y = 1

2(x-y) = 1

x-y = 1/2 = 05

Lets substitute some values and check:

x = 0.1 & y = -0.4

0.1 - (-0.4) = 0.5 ==============> Answer to the question is NO

x = 1 & y = 0.5

1 - 05 = 0.5 =================> Answer to the questions is YES

As we are getting multiple answers, Statement (1) is Not Sufficient.

(2) x/y > 1

Which means both x & y have same sing, they can be either positive or both negative.

As we are getting multiple answers, statement (2) us Not Sufficient.

Now, lets combine both (1) and (2)

we get:

x = y + 1/2

also, we know x/y>1

which means: (y+1/2)/y>1

which means y is > 0

as y>0, x has to be > 0 as well. So we can come to the conclusion that both x & y are positive.

Hence, Answer is C

grassmonkey

Joined: 22 Jan 2017

Last visit: 08 Jun 2018

Posts: 24

Own Kudos:

Given Kudos: 5

Posts: 24

Kudos: 24

15 Jul 2017, 22:46

Kudos

Bookmarks

First of all, everyone should really read Bunel's earlier posts before just tagging him. He must have written the same explanation five times!

Let me try:

Statement 1:
2x - 2y =1
2*(x-y) = 1
x - y =0.5

Ok cool, but you can probably tell intuitively that a difference of 0.5 doesn't really help. X and Y could be 1 and 0.5 or 0.25 and -0.25.

So this is not really that helpful, however it was important to do this algebra because, often on DS questions, you need to see how the statements work together and simpler expressions are much easier to work with. So, always at least try and simplify stuff wherever possible as the GMAT really seems to reward that reflex.

Statement 2:
x/y > 1

This should be an immediate reaction: same sign. If you don't immediately think that, practice until you do.

Now, as Bunel mentioned, there are two approaches. Number picking and algebra. Whenever possible, I try to use algebra. It's really personal preference on a question like this, but I have a panic that goes off in my head where I am scared I am missing a potential extreme value. So I try to use algebra to avoid that feeling.

Whenever you have an equation and an inequality, you always want to substitute the equation into the inequality (the other way around is not really permitted).

Notice that it actually doesn't matter which value you isolate. Because you know x and y have the same sign, you are just trying to figure out the sign of either one of them.

equation 1: x = y + 0.5
equation 2: x/y > 1

Now you can clean up the second equation as Bunel did, or you can go straight for it.

(y + 0.5)/y > 1
y/y + 0.5/y >1
1 + 0.5/y > 1
0.5y > 0

So a positive number (0.5) multiplied by y is greater that 0. Ok so y has to be positive. And if y is positive......so is x (because statement 2 told us that).

(C)

Keep practicing everyone!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 17 Sep 2026

Posts: 113,607

Own Kudos:

841,791

[19]

Given Kudos: 111,601

Products:

Expert

Expert reply

Posts: 113,607

Kudos: 841,791

[19]

05 Apr 2018, 17:11

Kudos

Bookmarks

Are x and y both positive?

GRAPHIC APPROACH.

Notice that the question is basically asks whether the point (x, y) is in the first quadrant.

(1) \(2x - 2y = 1\). Draw line \(y=x-\frac{1}{2}\):

Not sufficient.

(2) \(\frac{x}{y} > 1\). Draw line \(\frac{x}{y}=1\). The solutions is the green region:

Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Answer: C.

Show Spoiler

Attachment:

graph.png [ 6.13 KiB | Viewed 98338 times ]

Attachment:

graph %282%29.png [ 5.95 KiB | Viewed 97671 times ]

Hungluu92vn

Joined: 10 Feb 2017

Last visit: 01 Mar 2024

Posts: 33

Own Kudos:

Given Kudos: 163

Location: Viet Nam

GPA: 3.5

WE:General Management (Education)

Posts: 33

Kudos: 87

30 Apr 2018, 07:42

Kudos

Bookmarks

Hi everyone,

Sorry but i dont understand the condition that 1/2y > 0 so y must be positive. But i think y can also be negative because when y is negative so 2y is still positive and 1/2y > 0 anyway. Where am i wrong? Please help. Thank you so much

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 17 Sep 2026

Posts: 113,607

Own Kudos:

841,791

Given Kudos: 111,601

Products:

Expert

Expert reply

Posts: 113,607

Kudos: 841,791

30 Apr 2018, 07:54

Kudos

Bookmarks

Hungluu92vn

For \(\frac{1}{2y}>0\) to be true y must be positive. It cannot be negative because in this case 2y = 2*negative = negative, so \(\frac{1}{2y}=\frac{1}{negative}=negative<0\).

tatz

Joined: 04 May 2014

Last visit: 24 Nov 2018

Posts: 45

Own Kudos:

Given Kudos: 39

Concentration: Strategy, Operations

Posts: 45

Kudos: 9

04 Aug 2018, 01:51

Kudos

Bookmarks

dauntingmcgee

I found this one easiest to solve by drawing a graph. Clearly 1) and 2) alone are not sufficient as discussed, so what remains to be seen is if 2) adds enough information to 1) to determine if both x and y are positive.

Drawing a quick graph of the line y=x-1/2 we find that the x-intercept of the line is (0.5,0) and the y-intercept is (0,-0.5). From this graph we can clearly see that we don't need to worry about anything in the 4th quadrant (+x/-y is not >1) or the 3rd quadrant (|x|<|y|, therefore x/y is not >1). All that is left is the 1st quadrant, in which x and y are both positive.

Sufficient.

Can you please explain your statement "3rd quadrant (|x|<|y|, therefore x/y is not >1)". Why we don't need to worry about IIIrd quadrant. ?
My confusion is - Since x/y >1 - this implies both have either -ve sign or +ve sign. If both are -ve, they must lie in 3rd quadrant.

Blackishmamba

Joined: 02 Aug 2017

Last visit: 19 Dec 2022

Posts: 34

Own Kudos:

228

Given Kudos: 144

Concentration: Strategy, Nonprofit

Schools: ISB '20

GPA: 3.71

Schools: ISB '20

Posts: 34

Kudos: 228

08 Aug 2018, 08:02

Kudos

Bookmarks

Much simpler approach is here:
1) x-y=0.5
Clearly, insufficient as difference betn 2 +ve numbers as well as 2 -ve numbers can be 0.5
2) x/y>1
Insufficient, as x,y both can be either +ve or -ve

Considering both together,
x/y>1
If both are -ve, then x<y and hence x-y<0. So,there is no way that x-y=0.5 hence both cannot be zero.
If both are positive, x>y and hence x-y=0.5 can hold true.
Sufficient.

Posted from my mobile device

r.baldawa

Current Student

Joined: 16 Apr 2017

Last visit: 12 Apr 2021

Posts: 5

Own Kudos:

Given Kudos: 270

Schools: Desautels '23 (A)

Posts: 5

Kudos: 3

29 Sep 2018, 09:26

Kudos

Bookmarks

Bunuel

Are x and y both positive?

(1) \(2x-2y=1\). Well this one is clearly insufficient. You can do it with number plugging OR consider the following: both x and y are positive means that point (x,y) is in the I quadrant. \(2x-2y=1\) --> \(y=x-\frac{1}{2}\). We know it's an equation of a line and basically the question asks whether this line (all (x,y) points of this line) is only in I quadrant. This line for sure passes I quadrant (for example, x = 1.5 and y = 1) but it cannot entirely be only in I quadrant, so there must be some (x, y) points whose coordinates are not both positive. Not sufficient.

(2) \(\frac{x}{y}>1\) --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally. One of the approaches:

From (1): \(2x-2y=1\), so \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\);

Substitute x into (2): \(\frac{y + \frac{1}{2}}{y}>1\);

\(1 + \frac{1}{2y}>1\);

\(\frac{1}{2y}>0\);

\(y > 0\). \(y\) is positive. Thus, \(x=y+\frac{1}{2}=positive+positive=positive\). So, \(x\) is positive too. Sufficient.

Answer: C.

You can also check GRAPHIC APPROACH below.

Hello,

1/2y > 0 means y could be 0 and 1/2y could be infinity which is greater than 0.
So we can't actually remove the point of y not equal to 0.

Regards,
Ronak B.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 17 Sep 2026

Posts: 113,607

Own Kudos:

841,791

Given Kudos: 111,601

Products:

Expert

Expert reply

Posts: 113,607

Kudos: 841,791

29 Sep 2018, 09:51

Kudos

Bookmarks

r.baldawa

Bunuel

Hello,

1/2y > 0 means y could be 0 and 1/2y could be infinity which is greater than 0.
So we can't actually remove the point of y not equal to 0.

Regards,
Ronak B.

Dividing by 0 is not allowed. anything/0 is undefined and not infinity.