Data Sufficiency

If x != -y, is (x-y) / (x+y) > 1?

(1) X > 0
(2) Y < 0

Standard DS answers apply...

OA: E

I'm not sure how they got the right answer, so if you post, post your reasoning as well.

I am not sure how X>0 and x! = -y !!! Only true if x = a negative odd number.

I am interested to learn more.

Simplify to make your life easier.

Simplify this expression (x-y)/(x+y) - 1

-2y/(x+y) or -2y(x+y)/(x+y)^2

Since (x+y)^2 is +ve and -2 is -ve, just focus on whether y(x+y) > 0

y(x+y) > 0


1.x > 0, look at y(x+y). Useless
2. y < 0, same problem.

Using both, still inefficient, since we need to know whether x > -y

E.


Don't focus too much on x != -y, it is there to make the question error-free, as that expression can't be defined at x != -y.

cbenk121 wrote:If x != -y, is (x-y) / (x+y) > 1?

(1) X > 0
(2) Y < 0

Standard DS answers apply...

OA: E

I'm not sure how they got the right answer, so if you post, post your reasoning as well.

IMO E
simplifying
(x-y)/(x+y)>1
(x-y)/(x+y)-1>0
-2y/(x+y)>0
to get the ans we need to know 3 things x +ve or -ve
y + or - and x >< then y


s1) x>0 insuff as we have only one info
s2) insuff as we have only one inf

together insuff as we have only 2 inf

So, what is this for x! = -y ? does it tell us if x and way have the same sign ?

heshamelaziry wrote:So, what is this for x! = -y ? does it tell us if x and way have the same sign ?

when x = -y, x+y = 0 and (x-y)/(x+y) is undefined. It is there just for that purpose, making sure that the function is defined.

palvarez wrote:

heshamelaziry wrote:So, what is this for x! = -y ? does it tell us if x and way have the same sign ?

when x = -y, x+y = 0 and (x-y)/(x+y) is undefined. It is there just for that purpose, making sure that the function is defined.

But this is x !, x Factorial. Is this a typo ?

heshamelaziry wrote:

palvarez wrote:

heshamelaziry wrote:So, what is this for x! = -y ? does it tell us if x and way have the same sign ?

when x = -y, x+y = 0 and (x-y)/(x+y) is undefined. It is there just for that purpose, making sure that the function is defined.

But this is x !, x Factorial. Is this a typo ?

well, it is not typo. This sign "!=" is borrowed from programming where people use for "not equal to"

cbenk121 wrote:If x != -y, is (x-y) / (x+y) > 1?

(1) X > 0
(2) Y < 0

Standard DS answers apply...

OA: E

I'm not sure how they got the right answer, so if you post, post your reasoning as well.

Hi cbenk 121,

Most often, the question isn't difficult to think about but the statements are complex. Every so often, the question is complicated but the statements might not be so bad. Here, we get a complicated algebra question (and the statements provide readily digestible information). Whenever we get a complicated algebra question, we should always try to simply the algebra before going to the statements.

When you see inequality in a DS statement, the first thing you should ask is whether you have to worry about sign-flip. When you see inequality in a DS question stem, and you want to simplify the question, you should try to do it without sign-flipping.

With inequalities you can treat it EXACTLY like an equation except for just one instance: when multiplying or dividing by a negative. In DS when you see inequality and you have to cross-multiply with unknowns, watch it! Because they are unknowns, unless provided other information, we don't know their pos/neg signs, and if those signs vary, inequality signs start flipping. We don't know the sign of (x+y), so, if we were to cross multiply, we would have two inequalities, the one that is the subject of this thread and: Is x-y < -(x +y)?

BUT, we don't have to do that. Instead, we can subtract the 1 from both sides just as if that inequality sign were an equal sign (remember inequality is same as equal in all operations except multiplication and division).

Then:

Is x-y/x+y - 1> 0?

x-y/x+y - x+y/x+y >0?

Is -2y/x+y>0? (as xcusemplz and palvarez point out)

Now, we go to the statements. Here, we should pick numbers to help our reasoning. When picking numbers with inequality, be organized, and pick different numbers (+)(-) and (+)(+) or (-)(-).

When things are being divided, in order to be positive, they need to have to the same sign. Statement 1 tells us x is pos. But don't know about y so insufficient. Same problem with statement 2.

Combo:

X is positive and y is negative. We should pick one large positive and one "small" negative number, and then reverse it.

Large pos and "small" negative: Try x = 10 and y = -5. Then, the left hand side is positive, and the answer is yes.

Small pos and "large" negative: Try x = 5 and y = -10. Then, the left hand side is negative, and the answer is no.

Because we get both a yes and no answer, the answer is E.

Note: when we get to the point of combining the statements, either we can a) pick numbers to help our reasoning, b) rely on reasoning alone or c) use algebra alone. a and b are better than c.

Another high-end DS inequality issue is in dealing with systems of inequalities. In these cases, you can stack inequalities and add them (the same way you add equations) IF (and only if) the arrows are pointing in the same direction. If the arrows are pointing in opposite directions, you can subtract one inequality from another but you can't add them. However, here, it is usually easier to rewrite one of the inequalities, and add them.

Testluv wrote:

cbenk121 wrote:If x != -y, is (x-y) / (x+y) > 1?

(1) X > 0
(2) Y < 0

Standard DS answers apply...

OA: E

I'm not sure how they got the right answer, so if you post, post your reasoning as well.

Hi cbenk 121,

Most often, the question isn't difficult to think about but the statements are complex. Every so often, the question is complicated but the statements might not be so bad. Here, we get a complicated algebra question (and the statements provide readily digestible information). Whenever we get a complicated algebra question, we should always try to simply the algebra before going to the statements.

When you see inequality in a DS statement, the first thing you should ask is whether you have to worry about sign-flip. When you see inequality in a DS question stem, and you want to simplify the question, you should try to do it without sign-flipping.

With inequalities you can treat it EXACTLY like an equation except for just one instance: when multiplying or dividing by a negative. In DS when you see inequality and you have to cross-multiply with unknowns, watch it! Because they are unknowns, unless provided other information, we don't know their pos/neg signs, and if those signs vary, inequality signs start flipping. We don't know the sign of (x+y), so, if we were to cross multiply, we would have two inequalities, the one that is the subject of this thread and: Is x-y < -(x +y)?

BUT, we don't have to do that. Instead, we can subtract the 1 from both sides just as if that inequality sign were an equal sign (remember inequality is same as equal in all operations except multiplication and division).

Then:

Is x-y/x+y - 1> 0?

x-y/x+y - x+y/x+y >0?

Is -2y/x+y>0? (as xcusemplz and palvarez point out)

Now, we go to the statements. Here, we should pick numbers to help our reasoning. When picking numbers with inequality, be organized, and pick different numbers (+)(-) and (+)(+) or (-)(-).

When things are being divided, in order to be positive, they need to have to the same sign. Statement 1 tells us x is pos. But don't know about y so insufficient. Same problem with statement 2.

Combo:

X is positive and y is negative. We should pick one large positive and one "small" negative number, and then reverse it.

Large pos and "small" negative: Try x = 10 and y = -5. Then, the left hand side is positive, and the answer is yes.

Small pos and "large" negative: Try x = 5 and y = -10. Then, the left hand side is negative, and the answer is no.

Because we get both a yes and no answer, the answer is E.

Note: when we get to the point of combining the statements, either we can a) pick numbers to help our reasoning, b) rely on reasoning alone or c) use algebra alone. a and b are better than c.

Another high-end DS inequality issue is in dealing with systems of inequalities. In these cases, you can stack inequalities and add them (the same way you add equations) IF (and only if) the arrows are pointing in the same direction. If the arrows are pointing in opposite directions, you can subtract one inequality from another but you can't add them. However, here, it is usually easier to rewrite one of the inequalities, and add them.

I do not think this problem is an example of ""systems of inequalities". Could you give an example ?

That makes sense..the negative sign would flip the sign. Good to remember; thanks also for the tip on solving systems of equations with inequalities.

BTW: In that case, you could just plug and chug with original equation: no need to simplify. I saw that (1) and (2) on their own were insufficient, so then plugging in two different sets of value for X and Y would yield (C) is insufficient as well.

Thanks!

(BTW, yes x != -y is borrowed from programming...you wouldn't actually see this on GMAT, but until they put a key on my board of "=" with a slash through it..)

heshamelaziry wrote:

Testluv wrote:

cbenk121 wrote:If x != -y, is (x-y) / (x+y) > 1?

(1) X > 0
(2) Y < 0

Standard DS answers apply...

OA: E

I'm not sure how they got the right answer, so if you post, post your reasoning as well.

Hi cbenk 121,

Most often, the question isn't difficult to think about but the statements are complex. Every so often, the question is complicated but the statements might not be so bad. Here, we get a complicated algebra question (and the statements provide readily digestible information). Whenever we get a complicated algebra question, we should always try to simply the algebra before going to the statements.

When you see inequality in a DS statement, the first thing you should ask is whether you have to worry about sign-flip. When you see inequality in a DS question stem, and you want to simplify the question, you should try to do it without sign-flipping.

With inequalities you can treat it EXACTLY like an equation except for just one instance: when multiplying or dividing by a negative. In DS when you see inequality and you have to cross-multiply with unknowns, watch it! Because they are unknowns, unless provided other information, we don't know their pos/neg signs, and if those signs vary, inequality signs start flipping. We don't know the sign of (x+y), so, if we were to cross multiply, we would have two inequalities, the one that is the subject of this thread and: Is x-y < -(x +y)?

BUT, we don't have to do that. Instead, we can subtract the 1 from both sides just as if that inequality sign were an equal sign (remember inequality is same as equal in all operations except multiplication and division).

Then:

Is x-y/x+y - 1> 0?

x-y/x+y - x+y/x+y >0?

Is -2y/x+y>0? (as xcusemplz and palvarez point out)

Now, we go to the statements. Here, we should pick numbers to help our reasoning. When picking numbers with inequality, be organized, and pick different numbers (+)(-) and (+)(+) or (-)(-).

When things are being divided, in order to be positive, they need to have to the same sign. Statement 1 tells us x is pos. But don't know about y so insufficient. Same problem with statement 2.

Combo:

X is positive and y is negative. We should pick one large positive and one "small" negative number, and then reverse it.

Large pos and "small" negative: Try x = 10 and y = -5. Then, the left hand side is positive, and the answer is yes.

Small pos and "large" negative: Try x = 5 and y = -10. Then, the left hand side is negative, and the answer is no.

Because we get both a yes and no answer, the answer is E.

Note: when we get to the point of combining the statements, either we can a) pick numbers to help our reasoning, b) rely on reasoning alone or c) use algebra alone. a and b are better than c.

Another high-end DS inequality issue is in dealing with systems of inequalities. In these cases, you can stack inequalities and add them (the same way you add equations) IF (and only if) the arrows are pointing in the same direction. If the arrows are pointing in opposite directions, you can subtract one inequality from another but you can't add them. However, here, it is usually easier to rewrite one of the inequalities, and add them.

I do not think this problem is an example of ""systems of inequalities". Could you give an example ?

Is X positive?

(1) x + y > 5
(2) -x + y < -1

Using TestLuv's method, can rewrite as...

x + y > 5
x - y > 1

So 2x > 6...x > 3..x is positive, answer would be C.

Sucks to not be a programmer trying to do that problem.

I thought that read Xfactorial ... made the problem much harder.

cbenk121 wrote:That makes sense..the negative sign would flip the sign. Good to remember; thanks also for the tip on solving systems of equations with inequalities.

BTW: In that case, you could just plug and chug with original equation: no need to simplify. I saw that (1) and (2) on their own were insufficient, so then plugging in two different sets of value for X and Y would yield (C) is insufficient as well.

Thanks!

(BTW, yes x != -y is borrowed from programming...you wouldn't actually see this on GMAT, but until they put a key on my board of "=" with a slash through it..)

Hi cbenk121,

yes, it is true that there was no need for simplification, and that if we knew we actually had two inequalities in the question's equation, we could have just plugged numbers directly into them.

But, then again, if you know that multiplying is dangerous but that subtraction is safe, it may be quicker to quickly simplify, and then plug numbers (might not even have to) into the simplified inequality.

However, this narrow procedural issue most likely comes down to the test-takers's preference!