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

[venkhatapriya]

Nice !

[RonPurewal]

Nice !

indeed.

[zarak_khan]

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

This is how I solved this question:

Statement 1--> x - y = 0.5
Values for x, y and x-y
x = 1, y = 0.5, x-y = 0.5 --> x & y > 0
x = 2, y = 1.5, x-y = 0.5 --> x & y > 0
x = 0, y = -0.5, x- y = 0.5 --> x & y not >0 or <0
x = -0.5, y = -1, x - y = 0.5 --> x & y < 0
Insufficient

Statement 2 --> x /y > 1
Values for x, y, x/y
x = 2, y = 1, x/y = 2 --> x & y > 0
x = -1, y = -1/3, x/y = 3 --> x & y < 0
Insufficient

Statement 1 and 2 Combined --> x-y = 0.5 & x/y > 1
Values for x, y, x-y, x/y
x = 1, y = 0.5, x-y = 0.5, x/y>1 --> x & y > 0
x = -1, y = -1.5, x-y = 0.5, x/y<1 --> 2nd statement not satisfied
x = -2, y = -1.5, x-y = -0.5 --> 2nd statement not satisfied

Since Statement 1 and 2 combined give me valid answers for x & y > 0 only, I can conclude that C is the answer choice.

Tutors, is my strategy good enough or too time consuming for actual test conditions?

Thanks!

[RonPurewal]

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

This is how I solved this question:

Statement 1--> x - y = 0.5
Values for x, y and x-y
x = 1, y = 0.5, x-y = 0.5 --> x & y > 0
x = 2, y = 1.5, x-y = 0.5 --> x & y > 0
x = 0, y = -0.5, x- y = 0.5 --> x & y not >0 or <0
x = -0.5, y = -1, x - y = 0.5 --> x & y < 0
Insufficient

Statement 2 --> x /y > 1
Values for x, y, x/y
x = 2, y = 1, x/y = 2 --> x & y > 0
x = -1, y = -1/3, x/y = 3 --> x & y < 0
Insufficient

Statement 1 and 2 Combined --> x-y = 0.5 & x/y > 1
Values for x, y, x-y, x/y
x = 1, y = 0.5, x-y = 0.5, x/y>1 --> x & y > 0
x = -1, y = -1.5, x-y = 0.5, x/y<1 --> 2nd statement not satisfied
x = -2, y = -1.5, x-y = -0.5 --> 2nd statement not satisfied

Since Statement 1 and 2 combined give me valid answers for x & y > 0 only, I can conclude that C is the answer choice.

Tutors, is my strategy good enough or too time consuming for actual test conditions?

Thanks!

well, we can't really comment on whether it is too time-consuming; that's really a matter of how quickly you can do this sort of computation.
however, the approach that you have taken to the combined statements (1) and (2) is not efficient. from the looks of things, when you got to the point where you were considering the two statements together, you were still taking BOTH
(a) values that fail to satisfy statement 1
AND
(b) values that fail to satisfy statement 2
!!!

so, the question is this -- what is the system that you were using to generate these values? were they just random values?

if you are "plugging in" on a DS problem, and reach a situation in which you have the two statements together, do the following:
* take a bunch of values that satisfy ONE of the statements (note that you should already have such lists, from considering the individual statements)
* cross off the ones that DON'T satisfy the OTHER statement
* investigate the ones that remain

in your case, a nice jumping-off point would be the list that you have already compiled for statement 1:
x = 1, y = 0.5
x = 2, y = 1.5
x = 0, y = -0.5
x = -0.5, y = -1
the last two of these don't satisfy statement 2, so you can just forget about them.
the two choices that remain both give the same answer to the question (i.e., "yes").
if that's not enough evidence to convince you, investigate a few more choices that satisfy statement 1 -- you'll notice that the only ones that also pass the test of statement 2 consist of two positive numbers, with X greater than Y.

[sudaif]

Can someone please help? I get answer E when I solve this.

statement 1) x - y = 1/2 is insufficient because x=2, y=3/2 gives you x - y =1/2 and both x and y are positive. on the other hand, x=-3/2 and y=2, gives you x - y =1/2 and x is negative while y is positive. thus insuff.

statement 2) x/y >1
Implies that if y>0, then x>Y>0 and if y<0, then X<Y<0.
clearly insufficient.

statement 1 + 2
say, we take y>0, let's say x=2 and y=3/2...that satisfies x -y=1/2 from statement 1 and we know that x>Y>0 --> both +ive
say, we take y<0, let's say x=-2, and y=-5/2....that satisfies x - y =1/2 from statement and we know that X<Y<0 -->both negative

shouldn't the answer be E then?

[RonPurewal]

sudaif --

say, we take y<0, let's say x=-2, and y=-5/2....that satisfies x - y =1/2 from statement and we know that X<Y<0 -->both negative

nope.
in this case x < y < 0 is not true. (it's not true that -2 < -5/2.)

[tejkumar.m]

good one.. Hi Ron,

The figures I have taken are not going along with the OA. Please can you clarify

x=1, y=1/2 , both positive, x-y=1/2 (cond1) and x>y (Cond2)
x=-1/4, y=-3/4 , both negative, x-y=1/2 (cond1) and x>y (Cond2)

How can we say for sure that both are sufficient.

Thanks

[tim]

It is not true that both are sufficient. Can you clarify the question?

[tejkumar.m]

Thanks Tim for responding. The question is the same as mentioned in the previous posts in this thread.

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

This is a GMAT Prep question and the OA shows as Both statements together can solve the question. Also, I can see that the OA is acknowledged by everyone in the posts (in this thread) including the staff.

Please can you tell me if I am missing something here. Thanks

[ChrisB]

Hi,

The key to the problem is combining the statements together. I understand that you have come up with numbers that satisfy the two constraints from the statements, but that is not equivalent to combining the information in the statements.

To do so, you must attack the problem by combining the statements. This is accomplished by manipulating statement 1 to isolate the variable x:

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

x can now be substituted into the inequality from statement (2)

(1/2 + y) / y >1

You may be tempted to multiply both sides by y, but you can't do that here because we do not know whether y is positive or negative. If y were negative, then the inequality sign would have to flip. For this reason, we don't cross multiply. We can, however, simplify the fraction. Doing so yields:

(1/(2y)) + (y/y) = 1/(2y) + 1 > 1
Subtracting one from both sides simplifies this to
1/(2y) > 0
The only way that 1 / (2y) > 0 is if 2y > 0. This is because 1 is positive and a positive number must be divided by another positive number to yield a positive quotient.

Finally, if y is positive then we know x is positive according to statement (2). Thus we can answer YES that both x and y are definitely both positive.

Thanks,
Chris

[milinjc]

Statement I and II alone are clearly not sufficient on their own.

Statement I & II together:

2x - 2y = 1
2(x/y) - 2 = 1/y (Divide both side of equality by y)
2*GT1 - 2 = 1/y (leverage statement II)
GT2 - 2 = 1/y

left side of equality is clearly >0

Hence, 1/y > 0 and that makes y>0
If y>0, x in statement II (x/y>1) greater than 0 as well. x>0

So, both statements together are sufficient.

Thanks.
Milin

[RonPurewal]

Hi milinjc,
Do you have a question? If you do, it's not clear exactly what that question is.

Thanks.

[vishalc581]

Hi


I am getting OA as E.



Statement -1: x-y =1/2

Statement 2: x>y


From 1 and 2

When x = 1/3 & y =-1/6


We get 1/2 but both x and y are not positive.



When we take x = 1 & y = 1/2
Then x-y =1/2

Now both x and y are positive.


Can you please correct me.

[tim]

x/y > 1 does NOT mean that x > y. This is only true when y is positive. Take a look at your first example; it definitely violates the x/y > 1 constraint even though x > y in that example.

[RonPurewal]

Statement 2: x>y

think about the step that gets you from the original statement (x/y > 1) to this version.

now, think about when you can actually perform that step. (i.e., "I can only do that to both sides if I know that ______________").