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If x does not equal -y is (x-y)/(x+y)>1? [#permalink]

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07 Jul 2013, 16:47

Hey Guys,

I really need help on the thought process with DS not necessary how to solve the problem. I have a hypothetical DS below. The goal of my question isn't to solve the problem but to find what necessary information is need to begin with to solve the problem.

===============================Question 1================================ If x does not equal -y is (x-y)/(x+y)>1?

(1) y<0 (2) x+y>0

Lets take the two scenarios:

a) If x+y>0 which implies (x-y)>(x+y) ---> y>0 b) If x+y<0 which implies (x-y)<(x+y) ----> y<0

Now Statement (1) in the DS meets scenario (b)'s criteria. So does it make it sufficient? If we know y<0 does this imply x+y<0 and therefore (x-y)/(x+y)>1?

When we use multiple "if-then" scenarios does the questionmark on the statement still extend? In other words should statements (a) & (b) correctly be

(x-y)/(x+y)>1 ?? a) If x+y>0 which implies (x-y)>(x+y) ?? ---> y>0 ?? b) If x+y<0 which implies (x-y)<(x+y) ?? ----> y<0 ??

========================================Question 2========================= To illustrate the point further he is another example: Is x^2-y^2<0?

(1) |x|<|y| (2) x+y>0

Is x^2-y^2<0? can be reworded as x^2<y^2 ?? sqrt(x^2)<sqrt(y^2) ?? |x|<|y| ??

Notice how the question mark extended all the way down as we manipulated the question to get the new question is |x|<|y| ??

In our first statement (1) |x|<|y| matches our question. Hence sufficient.

Is this logic of extending the question mark the same for "if-then" scenarios?

Re: If x does not equal -y is (x-y)/(x+y)>1? [#permalink]

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08 Jul 2013, 01:37

Lets just consider the left hand side of the inequality:

(x-y)/(x+y),

1) y < 0 , the left hand side of the numerator can now be rewritten as : (x+ |y|) / (x-|y|). We can still not say that the numerator is larger than the denominator as we still do not know the value of x e.g. , if x=-4 , y = -1

then , (x-y)/(x+y) = -3/-4 = 3/4 (<1)

if x =4 , y = -1

then , (x-y)/(x+y) = 5/3 (>1)

So this is insufficient

2) x+y > 0

=> x > -y this is insufficient too

but if consider 1 and 2 together

from the second statement we have , x > -y ans since y < 0

x > |y|

So both statements are needed to prove the inequality

Let me first present how this problem is intended to be solved by the test writers and then I will examine your approach.

When you see a question like Is (x-y)/(x+y)>1?, then one should always try to simplify or rewrite this by bringing all the terms to one side to make > 0 or < 0. It is always easier to examine if an expression is positive or negative as opposed to comparing two quantities, as is the case in the original question. This is a standard set up on the GMAT.

If we rewrite the above expression as: (x-y)/(x+y) - 1 >0 and clean it up, we end up with Is 2y/(x+y) < 0 ? This is a lot easier to deal with.

Statement 1: (1) y<0 Just use simple examples: x = -2 and y = -3, Answer is No, x = 3 and y = -2, Answer is Yes. Insufficient.

Statement 2: (2) x+y>0 Again use examples: x = 2 and y = 3, Answer is No, x = 4 and y = -2, Answer is Yes. Insufficient.

Statement 1 and 2: This is easy because now we know that the numerator is negative and denominator is positive. Answer is a definite Yes. Sufficient.

One could do this problem in the original form using examples, but that is typically harder. The GMAT is testing your ability to rephrase the question and simplify everything before using examples. This is a classic GMAT set up.

===============================Question 1================================ If x does not equal -y is (x-y)/(x+y)>1?

(1) y<0 (2) x+y>0

Here you are rephrasing the original question under two cases, the first where x+y is positive and the second when x+y is negative. In both cases the resulting simplification is still a question. I have modified it below.

a) If x+y>0 then Is y>0? (this is the new question) or b) If x+y<0 then Is y<0? (this is the new question)

Statement 1 tells us that y<0, the original question is still Is y>0 if x+y>0 and Is y<0 if x+y<0? This is of course getting convoluted but we will pursue this further.

Based on your approach the fact given in statement 1 does not make it sufficient, because if y<0, then I can choose x to be sufficiently positive such that x+y>0 and the question is still Is y>0, but y<0 and the answer is a No.

However, if I choose x to be negative as well and we have y<0 which then tells us that x+y<0, and now the question is Is y<0 and the answer is Yes. Therefore, the statement 1 is insufficient. This is mentally a very difficult process, at least for me.

Now Statement (1) in the DS meets scenario (b)'s criteria. So does it make it sufficient? If we know y<0 does this imply x+y<0 and therefore (x-y)/(x+y)>1?

When we use multiple "if-then" scenarios does the question mark on the statement still extend?

In summary, the answer to your question is Yes.

(x-y)/(x+y)>1 ?? a) If x+y>0 which implies (x-y)>(x+y) ?? ---> y>0 ?? b) If x+y<0 which implies (x-y)<(x+y) ?? ----> y<0 ??

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