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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 14 Sep 2013, 22:23
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Hello Bunuel,
Request you to please provide your comments on the doubt posted here-

Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect.
So, going by that approach;

x-y=1/2 ---(1)
x/y>1 --(2)
Substituting the value of x in equation(2)

(y+1/2)/y>1

Lets assume that y is positive-

(y+1/2) > y

1/2>0 --This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0

Now, Lets assume that y is negative-

Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality.
I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "-y" in the whole equation. Please clarify. Which of the following would be correct then

a) y+1/2 <y
b) y+1/2 < -y
c) -y+1/2 < -y

Please help.
Thanks
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 14 Sep 2013, 22:39
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imhimanshu
Hello Bunuel,
Request you to please provide your comments on the doubt posted here-

Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect.
So, going by that approach;

x-y=1/2 ---(1)
x/y>1 --(2)
Substituting the value of x in equation(2)

(y+1/2)/y>1

Lets assume that y is positive-

(y+1/2) > y

1/2>0 --This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0

Now, Lets assume that y is negative-

Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality.
I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "-y" in the whole equation. Please clarify. Which of the following would be correct then

a) y+1/2 <y
b) y+1/2 < -y
c) -y+1/2 < -y

Please help.
Thanks
Show more

Refer to the highlighted portion : Actually you don't have to take 2 cases at this point: The expression you have is : \(\frac{y+0.5}{y}>1 \to 1+\frac{0.5}{y}>1 \to \frac{1}{y}>0\)--> Hence, y>0.

As for your doubt, if y is negative, we cross-multiply it and get : \(y+0.5<y \to 0>0.5\), which is absurd.

If y is negative, then -y would be positive, and for multiplying a positive quantity, you don't need to flip signs. So , yes expression a is correct.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 03 Dec 2013, 23:07
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St 1) 2x-2y = 1 => 2 (x-y) = 1 => x-y =1/2 => all this tells us is that x > y (could be positive or negative) == hence INSUFF

St 2) x/y > 1 => we don't know if y is (+) or (-) . So we have two cases:

if y positive, then x>y; if y negative, then x<y (again INSUFF)

Combining 1) and 2) we get x>y (from 1) ...which means y is positive (from 2)

Hence, if y is positive, and x >y, then x is also positive. SUFF!!

Hope this was reasoned properly.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 23 Dec 2013, 03:53
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Manbehindthecurtain
Are x and y both positive?

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

Plug in approach that can be used without thinking much and very likely arrive at the correct answer.

Values to be taken: x positive and negative and find the corresponding values for y based on the statements
Note: x and y cannot be of different signs and also x cannot be zero as they will not satisfy (ii)

(i) x=10, we have y =9.5 .Both positive satisfied And now x=-10, we have y=-9.5. Both negative also satisfied .Different results. So (i) alone not sufficient

(ii) x=10, y can be positive. Both positive satisfied . And now x=-10, y can be negative. Both negative also satisfied. So (ii) alone not sufficient

(i) + (ii) x=10, y=9.5 satisfies both the statements . Both positive satisfied . And now x=-10. Value of y is found from (i) and is negative , but we see it does not satisfy (ii). So both cannot be negative .

So we can answer the question using (i) and (ii) together
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 05 Aug 2015, 06:37
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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.
Show more


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.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 05 Aug 2015, 07:07
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Matthias15



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.
Show more

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!
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 24 Sep 2015, 16:00
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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'
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 24 Sep 2015, 16:21
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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'.
'
Show more
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
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 22 Jun 2016, 21:17
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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.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 22 Jun 2016, 22:03
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Manbehindthecurtain
Are x and y both positive?

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


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.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 22 Jan 2017, 12:47
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What about the example of y=0? Then x/y is surely greater than 1, hence I chose answer E.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 22 Jan 2017, 13:50
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vgmatv
What about the example of y=0? Then x/y is surely greater than 1, hence I chose answer E.
Show more

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
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 20 Jun 2017, 22:27
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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.

Discussed here: https://gmatclub.com/forum/ds1-93964.htm ... approaches and also here along with other hard inequality problems: https://gmatclub.com/forum/inequality-an ... 86939.html

Hope it helps.
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
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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.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 21 Jun 2017, 13:36
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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
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 15 Jul 2017, 22:46
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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!
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 30 Apr 2018, 07:42
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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
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 30 Apr 2018, 07:54
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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
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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 \(\frac{1}{2y}=\frac{1}{negative}=negative<0\).
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 04 Aug 2018, 01:51
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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.
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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.
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 08 Aug 2018, 08:02
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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.

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r.baldawa
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 29 Sep 2018, 09:26
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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.
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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.
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