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Re: Inequalities Help - OG Question [#permalink]
28 Jul 2011, 03:20

I originally came up with B - but I'm obviously wrong

I think I've violated a rule of inequalities here, so, at the risk of looking stupid, I'll write out my train of thought:

I rephrased the given into:

1- is x - y > x + y ? (I think that is where I am fundamentally wrong - are you allowed to cross-multiply inequalities?)

then

2- is x > x + 2y ?

then

3- is 0 > 2y or is Y < 0 ?

I immediately spotted "b" as answering the given, so I selected it...

I'm gonna go have a look in the MGMAT inequalities guide, but I thought I'd post the question up here anyway for other peoples' benefit _________________

Re: Inequalities Help - OG Question [#permalink]
28 Jul 2011, 04:21

Its E. My method was to multiply top and bottom with (x+y)/(x+y) = 1 You get (x^2 - y^2) / (x+y)^2

In this case it doesn't matter if you know x is positive and y is negative because it all squares out to positive. So you can't tell is top or bottom is bigger so maybe less than 1 or bigger than 1.

Re: Inequalities Help - OG Question [#permalink]
28 Jul 2011, 08:26

MackyCee wrote:

I originally came up with B - but I'm obviously wrong

I think I've violated a rule of inequalities here, so, at the risk of looking stupid, I'll write out my train of thought:

I rephrased the given into:

1- is x - y > x + y ? (I think that is where I am fundamentally wrong - are you allowed to cross-multiply inequalities?)

then

2- is x > x + 2y ?

then

3- is 0 > 2y or is Y < 0 ?

I immediately spotted "b" as answering the given, so I selected it...

I'm gonna go have a look in the MGMAT inequalities guide, but I thought I'd post the question up here anyway for other peoples' benefit

The most important thing to remember about inequalities is that you have to reverse the inequality if you multiply or divide by a negative. In this case you can't tell what the sign of x + y will be, so cross multiplying is dangerous - you have to consider two scenarios.

Q139) if "x" is not equal to "-y", is x-y / x+y > 1?

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

We know that we cannot multiply an inequality by an unknown if we don't know whether the unknown is a positive value or a negative value. The inequality behaves differently in the two cases. If the unknown is positive, the inequality stays as it is. If the unknown is negative, the inequality flips sign. Therefore, x-y > x+y is not an option. Next, '>1' is much more complicated that '>0' where we just need to consider whether the variables are positive or negative. Therefore, bring whatever is on the right hand side to the left hand side.

Question: Is \frac{(x - y)}{(x + y)} > 1 Is \frac{(x - y)}{(x + y)} - 1 > 0 Is \frac{-2y}{(x + y)} > 0

For this expression to be positive, y/(x+y) should be negative. Therefore, either 'y should be negative and (x+y) should be positive' or 'y should be positive and (x+y) should be negative'.

Statement 1: X > 0 No info about y so not sufficient. Statement 2: Y < 0 No info about x so not sufficient.

Both together, we know that y is negative. We now need to know the sign of (x+y). Just knowing that x is positive doesn't tell us the sign of (x+y) because we don't know which one out of x and y has greater absolute value. If absolute value of x is greater than that of y, x+y is positive. If absolute value of x is less than that of y, x+y is negative. We do not know the sign of x+y so we still cannot say whether \frac{-2y}{(x + y)} is positive. Not sufficient. Answer E. _________________

Statement 2: |x + 2| > 5 This translates to x < -7 or x > 3. x could be positive or negative. Not sufficient.

Both together, we get that 3 < x < 8. These will be the only values satisfying both conditions. If this is not intuitive, draw the two inequalities on the number line. x should lie between -2 and 8 and it should be either less than -7 or greater than 3. The only overlapping region is from 3 to 8. x will be positive so sufficient.

Re: Inequalities Help - OG Question [#permalink]
31 Jul 2011, 11:21

hey,

i simplified to is 0>y^2+xy is 0>y(y+x)

Statement 1: X > 0 X positive but Y negative gives 1 solution and Y positive gives different solution so NS

Statement 2: Y < 0

Y big negative and X small positive gives NO. Y little negative and X big positive gives yes.

combining: from S2 you can conclude that the answer varies on the values of X and Y. again X big positive and Y little negative the answer is 0>y(y+x) i.e 0> -1(-1+8)

X little positive and Y big negative answer is 0<y(y+x) i.e 0< -8(-8+3).

Re: Inequalities Help - OG Question [#permalink]
01 Aug 2011, 01:05

Expert's post

dimri10 wrote:

can someone help please:

in the second Q.(the modulus) i recieved 2 solutions for S1:

3<x<8 and -2<x

1. do i have to combine the solution of S1 to -2<x<8.

How did you get this? Even if you want to use the two step algebra: Step 1: Assume x >= 3 (x - 3) < 5 x < 8 So 3 <= x < 8

Step 2: Assume x < 3 and -(x - 3) < 5 gives x > -2 -2 < x < 3

So basically x can lie from -2 to 3 and from 3 to 8 i.e. it can lie from -2 to 8. OR Just say that if |x - 3| < 5, then -5 < x - 3 < 5. Adding 3 to the inequality, -2 < x < 8.

dimri10 wrote:

for S2 I got: x>3 [highlight]and[/highlight] -7>x

How do i combine both statements to 3<x<8?

The highlighted part has to be 'OR'. When you have |x + 2| > 5, then x+2 > 5 OR x+2 < -5 i.e. x > 3 OR x < -7 To satisfy the following two conditions: 1. -2 < x < 8 2. x>3 OR -7>x

Attachment:

Ques3.jpg [ 4.45 KiB | Viewed 968 times ]

From the figure, which are the only values of x satisfying both the conditions? Between 3 and 8, right? So 3 < x < 8 _________________

Re: Inequalities Help - OG Question [#permalink]
04 Aug 2011, 07:34

i am sorry karishma. just got confused. the number system approach is way better to solve such questions. i have seen some other posts in that issue (yours as well as Bunuel) and now it's a cake. 11 out of 10.

gmatclubot

Re: Inequalities Help - OG Question
[#permalink]
04 Aug 2011, 07:34

It’s been a long time, since I posted. A busy schedule at office and the GMAT preparation, fully tied up with all my free hours. Anyways, now I’m back...

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