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Are x and y both positive?

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Source: — Data Sufficiency |
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by eagleeye » Wed May 16, 2012 5:31 pm
Hi PGMAT:

We are given :

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

Let's look at 2 first. We are given x/y>1 which means x/y is greater than zero as well. Therefore, either x and y are both positive or x and y are both negative. Hence this statement is not sufficient.

Case (I):
x and y are both negative.

Then we have 2x-2y = 1 which means x = y + 1/2 that is x > y.
but we have x/y > 1 ; multiplying both sides by y, the sign changes (whenever we multiply an inequality by a negative quantity the sign becomes opposite).
Then we have: y * x/y < y which becomes x < y , which contradicts x > y as we found from x = y + 1/2. Hence both x and y cannot be negative. Hence both x and y are positive and hence C is the correct answer.

Even though we got C as the answer just from the first case, I just want to show you that the other case is valid as I just claimed.

Case (II):
x and y are both positive. Then 1) still remains x = y + 1/2 implying x > y

But for 2) x/y > 1; multiplying both sides by y, the sign remains the same ( since y here is positive, whenever you multiply by a positive number sign of inequality remains unchanged).
Then 2) becomes : y * x/y > y which gives us x > y which is consistent. Hence we get a definite answer from using both conditions making the answer C.

Let me know if that helps :)
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by Stuart@KaplanGMAT » Wed May 16, 2012 9:36 pm
PGMAT wrote:Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

[spoiler]I think both statements together are insufficient and E, but OA is C. Can some one explain?[/spoiler]
Eagleeye gives a good math explanation; let's look at the question purely using number properties concepts.

Q: are x and y BOTH positive.

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

Let's think about what this means: on the number line, x is 1/2 to the right of y. Does that tell us anything about their signs? NO - insufficient.

(2) x/y > 1

This tells us 2 things:

1) x and y must have the same sign (since x/y is positive); and
2) the magnitude of x is greater than the magnitude of y (if not, we'd have a fraction).

However, x and y could be positive or negative, so insufficient.

TOGETHER:

From (1), we know that x is to the right of y on the number line.
From (2), we know that x and y have the same sign and that |x|>|y|

The only way for x and y to have the same sign, |x|>|y| AND x to be to the right of y on the number line is for them both to be positive. Sufficient, choose C.
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by aneesh.kg » Thu May 17, 2012 1:17 am
I love solving such a problem using co-ordinate geometry. This method is more visual, and thus more convincing.

Question: Is (x,y) in the 1st quadrant?

Statement(1) is a straight line with a slope = 1 and having a y-intercept = -1/2
Statement(2) is the region y < x when y is positive (the region below the line y = x above the y-axis), and y > x when y is negative (the region above the line y =x below the y-axis).

Image

As we see from the figure, the common solution (the Blue portion) lies in the first quadrant. So the answer to that is YES!

[spoiler](C)[/spoiler] is correct.

More problems solved using the help of co-ordinate geometry:
1) https://www.beatthegmat.com/question-pac ... tml#471307
2) https://www.beatthegmat.com/number-prope ... tml#470013
3) https://www.beatthegmat.com/if-b-1-and-2 ... tml#469327
4) https://www.beatthegmat.com/if-y-0-is-y- ... tml#469314
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by Anurag@Gurome » Thu May 17, 2012 7:43 pm
PGMAT wrote:Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

[spoiler]I think both statements together are insufficient and E, but OA is C. Can some one explain?[/spoiler]
(1) 2x - 2y = 1
x and y both positive means that point (x, y) is in the first quadrant.
2x - 2y = 1 implies y = x - 1/2, and it's an equation of a line and the question asks whether this line is only in first quadrant, which is 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.

Combining (1) and (2), 2x - 2y = 1 implies x = y + 1/2
x/y > 1 implies (x - y)/y > 0
Substituting the value of x, 1/y > 0 implies y is positive and since x = y + 1/2, so x is also positive; SUFFICIENT.

The correct answer is C.
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by i_have_no_cool_username » Fri May 18, 2012 4:49 am
Firstly - why do I feel like I've seen this problem somewhere before....PGMAT, can you quote the source of the question please? I would like to see if there is an explanation included in the source. Thanks!

Secondly, I think I'm going mad but I somehow keep getting to E for the answer using PIN. Can someone evaluate my solution for me, please?

Statement 1: x-y = 1/2
Insufficient since x and y can be any value and any sign.

Statement 2: x/y > 1
Indicates that x and y share same sign, and x > y.

Together: x and y must have same sign, x > y, and x - y = 1/2
Scenario 1 (Positive numbers: x = 1, y = 1/2; x > y)
1 - 1/2 = 1/2 (Yes, both numbers are positive)

Scenario 2 (Negative numbers: x = -3/2, y = -2; x > y)
-3/2-(-2) = 1/2 (No, both numbers are not positive)

Conflicting solutions, hence I chose E. So, what did I miss?
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by aneesh.kg » Fri May 18, 2012 5:04 am
i_have_no_cool_username wrote:Firstly - why do I feel like I've seen this problem somewhere before....PGMAT, can you quote the source of the question please? I would like to see if there is an explanation included in the source. Thanks!

Secondly, I think I'm going mad but I somehow keep getting to E for the answer using PIN. Can someone evaluate my solution for me, please?

Statement 1: x-y = 1/2
Insufficient since x and y can be any value and any sign.

Statement 2: x/y > 1
Indicates that x and y share same sign, and x > y.

Together: x and y must have same sign, x > y, and x - y = 1/2
Scenario 1 (Positive numbers: x = 1, y = 1/2; x > y)
1 - 1/2 = 1/2 (Yes, both numbers are positive)

Scenario 2 (Negative numbers: x = -3/2, y = -2; x > y)
-3/2-(-2) = 1/2 (No, both numbers are not positive)

Conflicting solutions, hence I chose E. So, what did I miss?
Hi,

You're goofing up the second Statement.

x/y > 1 DOES NOT necessarily mean that x > y.

when you have x/y > 1, and if y is positive, then you can multiply with 'y' on both sides and the sign of the inequality does change.
So, for y > 0, x > y.

BUT (and this is a big but)
if y is negative, when you multiply with 'y' on both sides, the sign of the inequality will change.
So, for y < 0, x < y.

No wonder you are getting x < y (for the y = - 2 case).

P.S.: This problem appears in one of the tests on the official GMAT site. You might've seen it in one of those tests.
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by Stuart@KaplanGMAT » Fri May 18, 2012 5:49 am
i_have_no_cool_username wrote:
Scenario 2 (Negative numbers: x = -3/2, y = -2; x > y)
-3/2-(-2) = 1/2 (No, both numbers are not positive)
[/color]
Conflicting solutions, hence I chose E. So, what did I miss?
Hi!

(2) says that x/y > 1

If you plug in x = -3/2 and y = -2, you get:

x/y = -(3/2)/-2 = -(3/2)*-(1/2) = 3/4 which is NOT greater than 1. Since your numbers don't satisfy the statement, they're impermissible and must be discarded.

As Aneesh aptly notes, you have to be VERY careful when manipulating inequalities with variables; remember the key rule for inequalities: if you multiply or divide both sides by a negative, you flip the direction of the inequality.

Accordingly, if you multiply or divide both sides by a variable with an unknown sign, you have to consider two cases - that the variable is positive (leave the inequality the way it is) and that the variable is negative (flip the inequality).
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by i_have_no_cool_username » Fri May 18, 2012 7:42 am
Stuart Kovinsky wrote:

Hi!

(2) says that x/y > 1

If you plug in x = -3/2 and y = -2, you get:

x/y = -(3/2)/-2 = -(3/2)*-(1/2) = 3/4 which is NOT greater than 1. Since your numbers don't satisfy the statement, they're impermissible and must be discarded.
Hi Stuart! Thanks for the explanation - I actually did come to that step of checking whether my PIN values of x/y > 1, and when it didn't, I thought "But everything else makes sense...", which was why I was wondering what went wrong.

So in case PIN gets confusing, as in my scenario, is this how the logic works without plugging in numbers:

Statements 1 and 2: x-y = 1/2; x and y has the same sign; AND -
either x < y (when both are negative in order to fulfill x/y > 1)
OR x > y (when both are positive in order to fulfill x/y > 1)

But since no negative numbers can fulfill both rules of x < y and x-y=1/2, x and y must be positive.

Makes sense? Or am I going mad again...:)

Also another avenue I was pondering: Statement 2 manipulation -
x/y > 1
x/y - 1 > 0
(x-y)/y > 0
Also to say (1/y)(x-y) > 0
Meaning y > 0 or x > y

And I'm stuck. How should I combine these two outcomes with Statement 1 to decide on final answer?

P.S. PGMAT, I'm sorry for latching onto your discussion thread with my questions. This was your question afterall! Apologies. I will carry my curiosity on to another thread if I'm still unsure. Sorry! But thanks for posting up this question!
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by GMATGuruNY » Fri May 18, 2012 8:11 am
i_have_no_cool_username wrote:
Stuart Kovinsky wrote:

Hi!

(2) says that x/y > 1

If you plug in x = -3/2 and y = -2, you get:

x/y = -(3/2)/-2 = -(3/2)*-(1/2) = 3/4 which is NOT greater than 1. Since your numbers don't satisfy the statement, they're impermissible and must be discarded.
Hi Stuart! Thanks for the explanation - I actually did come to that step of checking whether my PIN values of x/y > 1, and when it didn't, I thought "But everything else makes sense...", which was why I was wondering what went wrong.

So in case PIN gets confusing, as in my scenario, is this how the logic works without plugging in numbers:

Statements 1 and 2: x-y = 1/2; x and y has the same sign; AND -
either x < y (when both are negative in order to fulfill x/y > 1)
OR x > y (when both are positive in order to fulfill x/y > 1)

But since no negative numbers can fulfill both rules of x < y and x-y=1/2, x and y must be positive.

Makes sense? Or am I going mad again...:)

Also another avenue I was pondering: Statement 2 manipulation -
x/y > 1
x/y - 1 > 0
(x-y)/y > 0
Also to say (1/y)(x-y) > 0
Meaning y > 0 or x > y

And I'm stuck. How should I combine these two outcomes with Statement 1 to decide on final answer?

P.S. PGMAT, I'm sorry for latching onto your discussion thread with my questions. This was your question afterall! Apologies. I will carry my curiosity on to another thread if I'm still unsure. Sorry! But thanks for posting up this question!
Perhaps my solution here will help:

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by i_have_no_cool_username » Fri May 18, 2012 8:45 am
GMATGuruNY wrote:
Perhaps my solution here will help:

https://www.beatthegmat.com/are-x-and-y- ... 89906.html
Wow. Your approach is really clear! Thank you, GMATGuruNY!
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