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Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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06 Sep 2015, 00:52

Bunuel wrote:

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.

(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.

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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06 Sep 2015, 05:13

reto wrote:

Bunuel wrote:

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.

You get from statement 1, 2x-2y=1 ---> \(x=\frac{1+2y}{2}\).....(1)

From statement 2, \(\frac{x}{y} >1\)---> \(\frac{x}{y}-1>0\) --->\(\frac{x-y}{y} > 0\) , now substitute for x from (1) above, you get,

\(\frac{\frac{1+2y}{2} -y}{y} > 0\) ---> \(\frac{\frac{1+2y-2y}{2}}{y} > 0\) ----> \(\frac{1}{2y} >0\) ---> \(\frac{1}{y} >0\) ---> \(y >0\)and as from (1),

\(x=\frac{1+2y}{2}\), for y>0, x>0 as well.

Alternately, you can view it as:

\(\frac{x}{y} >1\) ---> x and y both have the same sign (i.e. both of them are either <0 or >0). x>0 ---> y >0 and when x<0 ---> y <0 and vice versa. This is true when \(\frac{x}{y} > 0\)

\(\frac{x}{y} <1\) ---> x and y both have different sign i.e. x>0 ---> y <0 and when x<0 ---> y >0 and vice versa. This is true when \(\frac{x}{y} < 0\) Hope this helps.
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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06 Sep 2015, 20:31

Hi Bunuel I understoof your line of solving this question. But just wanted to say in GMAT two statements cannot contradict each other. Statement 1 says that slope of line is 1, which means x and y co-ordinates will always have same value, which means the ratio x/y = 1 always. However statement 2 says that x/y>1, how is this possible. This contradicts statement 1. Correct me if I am wrong.

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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06 Sep 2015, 20:48

Shreeya24 wrote:

Hi Bunuel I understoof your line of solving this question. But just wanted to say in GMAT two statements cannot contradict each other. Statement 1 says that slope of line is 1, which means x and y co-ordinates will always have same value, which means the ratio x/y = 1 always. However statement 2 says that x/y>1, how is this possible. This contradicts statement 1. Correct me if I am wrong.

You are correct in saying that 2x-2y=1 has slope of 1 but it is not a must that this statement means that x/y = 1. Having slope of 1 does not mean that the x and y coordinates will be equal (to make x/y =1).

You can confirm this by looking at the set of values satisfying the above equation:

(1.5,1) (2.5,2) (3,2.5)

So you see that none of these sets mean x/y > 1.

But your understanding that an official DS question will never have contradictory statements is correct. Though this is not the case for this question.

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

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

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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24 Sep 2015, 17:05

jitendra31 wrote:

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+1/2 x/y>1 --> x−y/y>0 --> substitute x --> 1/y>0 --> y is positive, and as x=y+1/2, 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'

As a rule for GMAT quant, a/0 = NOT DEFINED and as such you MUST NOT CONSIDER this case. BY definition, infinity can be both + or - . So your logic of using 'infinity' does not stand. Also , as mentioned before by me you should never consider any case with 0 in the denominator. If there is a case that is for example, a/(b-2) = 3 then for GMAT, you must have b \(\neq\) 2

This is GMAT's rules and hence you have to follow them.
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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22 Jun 2016, 10:05

As per my solution i am getting E. Kindly help me understand where i am going wrong as the OA is 'C'

1. First statement: 2(x)-2(y)=1 --> x-y=0.5

Now two cases:

(a) X=2.5 , y= 2 , difference is 0.5 --> Both are positives (b) X= -2, y= -2.5 , difference is 0.5 --> both are negatives

Hence insufficient

2. Second statement: (x/y) >1 --> x>y

Both the numbers can be positive or negative hence insuff

1+2 : Both the statements combined

x-y= 0.5 & x>y

Now two cases:

(a) X=2.5 , y= 2 , difference is 0.5 --> Both are positives ( x>y) (b) X= -2, y= -2.5 --> -2 -(-2.5) is equal to -2 + 2.5 which results into 0.5. Also -2 > -2.5

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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22 Jun 2016, 10:15

1

This post received KUDOS

sgrover18 wrote:

As per my solution i am getting E. Kindly help me understand where i am going wrong as the OA is 'C'

1. First statement: 2(x)-2(y)=1 --> x-y=0.5

Now two cases:

(a) X=2.5 , y= 2 , difference is 0.5 --> Both are positives (b) X= -2, y= -2.5 , difference is 0.5 --> both are negatives

Hence insufficient

2. Second statement: (x/y) >1 --> x>y

Both the numbers can be positive or negative hence insuff

1+2 : Both the statements combined

x-y= 0.5 & x>y

Now two cases:

(a) X=2.5 , y= 2 , difference is 0.5 --> Both are positives ( x>y) (b) X= -2, y= -2.5 --> -2 -(-2.5) is equal to -2 + 2.5 which results into 0.5. Also -2 > -2.5

difference is 0.5 --> both are negatives

This is insuff, hence 'E' should be the answer

Couple of things:

1. x/y >1 DOES NOT mean that x>y (think x=-3 and y=-2). It only means that both x and y will be of same sign.

2. When x=-2, y= -2.5, you are going against S2 that x/y >1 , your case makes it the opposite, x/y < 1. This is where you are making a mistake.

By combining the 2 statements, you will get C as the correct answer.
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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22 Jun 2016, 11:20

Engr2012 wrote:

sgrover18 wrote:

As per my solution i am getting E. Kindly help me understand where i am going wrong as the OA is 'C'

1. First statement: 2(x)-2(y)=1 --> x-y=0.5

Now two cases:

(a) X=2.5 , y= 2 , difference is 0.5 --> Both are positives (b) X= -2, y= -2.5 , difference is 0.5 --> both are negatives

Hence insufficient

2. Second statement: (x/y) >1 --> x>y

Both the numbers can be positive or negative hence insuff

1+2 : Both the statements combined

x-y= 0.5 & x>y

Now two cases:

(a) X=2.5 , y= 2 , difference is 0.5 --> Both are positives ( x>y) (b) X= -2, y= -2.5 --> -2 -(-2.5) is equal to -2 + 2.5 which results into 0.5. Also -2 > -2.5

difference is 0.5 --> both are negatives

This is insuff, hence 'E' should be the answer

Couple of things:

1. x/y >1 DOES NOT mean that x>y (think x=-3 and y=-2). It only means that both x and y will be of same sign.

2. When x=-2, y= -2.5, you are going against S2 that x/y >1 , your case makes it the opposite, x/y < 1. This is where you are making a mistake.

By combining the 2 statements, you will get C as the correct answer.

Thank you so much ! That makes perfect sense. I guess I forgot the basic rule that if numbers are negative on the either side of the inequality then the sign of the inequality reverses (as you quoted -3<-2)

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.

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 example, if x = 1 and y = ½, x and y are both positive; however, if x = -1/2 and y = -1, x and y are both negative. 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. Since x = ½ + y, x also must be greater than zero.

Answer: C
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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11 Nov 2016, 21:56

I solved this using a combination of logic and testing general cases.

For (1), think of x-y = 1/2 as the difference between two points on the number line. This difference can occur at if 1. x and y = + (AND x>y) 2. x and y = - (AND y>x) 3. x = +, y= - (AND x>y)

Clearly not sufficient.

(2) tells us that x and y have the same sign AND y>x. Also not sufficient on it's own.

Finally, using (2) we can rule out the first and third options for (1) and you end up with C as the answer.

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