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Guys I didn't forget your request, just was collecting good questions to post.

So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers.

1. If \(6*x*y = x^2*y + 9*y\), what is the value of xy? (1) \(y – x = 3\) (2) \(x^3< 0\)

Re: Inequality and absolute value questions from my collection [#permalink]

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14 Oct 2013, 20:29

6. If x and y are integer, is y > 0? (1) x +1 > 0 (2) xy > 0

If y>0 then the question becomes does y represent a +ve integer?

From (1) x+1>0 x>-1

Now take to number line -----------------(-1)------0------1------2-------3----------- We DO NOT know anything about y.INSUFFICIENT.

From (2) xy>0 This means (i)-ve integer *-ve integer >0 (or) (ii)+ve integer *+ve integer >0 Also it means that neither x nor y is 0 Here in (i) y CAN be -ve (or) in (ii) y can be +ve. INSUFFICIENT.

NOW do (1+2) -----------------(-1)------0------1------2-------3----------- This becomes -----------------(-1)------0------1------2-------3----------- i.e., 0 is eliminated from solution set of x. WHY? Because, neither x nor y is 0 (stated earlier) Now look at below two statements, (i)-ve integer *-ve integer >0 (or) (ii)+ve integer *+ve integer >0 From number line it is established that x is >=1 Hence, case (i) is ELIMINATED We are left with case (ii) which states both are +ve====> y is a +ve integer. Answer: C

Re: Inequality and absolute value questions from my collection [#permalink]

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24 Oct 2013, 20:11

Bunuel wrote:

4. 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.

Answer: C.

Hi Bunuel

Can you please explain this solution again . I am finding it hard to understand why x = y + 1/2 is insufficient and also why c is the answer

4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

Hi Bunuel

Can you please explain this solution again . I am finding it hard to understand why x = y + 1/2 is insufficient and also why c is the answer

Regards

In the equation \(x = y + \frac{1}{2}\), there are two variables. Here

Case I :- if Y is positive then x would also be positive.

but if Y is negative then we can not say for sure what sign would X take.

Case II :- if \(y = -\frac{1}{4}\) then x would be \(\frac{1}{4}\)(positive)

Case III :- y = -100 then x would be -99.5.

We got three different answers so this statement is obviously insufficient. It is useful to remember important property of equations that 1) to get the value of 2 variables, we need 2 equations 2) To get the value of 3 variables, we need 3 equations 3) To get the value of n variables, we need n equations.

Re: Inequality and absolute value questions from my collection [#permalink]

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19 Feb 2014, 04:00

I get super confused in solving the below kind of inequalities. Can some expert help?

Are the below working's correct?

1/| n | >n • Case 1: n > 0… n 2 < 1… - 1 < n < 1 • Case 2: n < 0… - n 2 < 1… n 2 > -1…

x/| x | < x • Case 1: x > 0… x^2 > x… as x > 0, divide by x without changing signs… x > 1 • Case 2: x < 0… - x^2 > x… as x < 0, divide and only flip sign… - x < 1… x > -1

Re: Inequality and absolute value questions from my collection [#permalink]

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21 Feb 2014, 08:00

Bunuel wrote:

11. Is |x+y|>|x-y|? (1) |x| > |y| (2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2 OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.

Hi,

I didnt quite understand this part "(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin."

what in the second equation tells us that they both have to be the same sign?

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2 OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.

Hi,

I didnt quite understand this part "(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin."

what in the second equation tells us that they both have to be the same sign?

Thanks

If x and y have the opposite signs: ---x---0---y---- the distance between the origin and x will always be less than x and y.

Re: Inequality and absolute value questions from my collection [#permalink]

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04 Mar 2014, 05:24

For question 2 Option D. From S1:x<0 or x=-ve If x is -ve,y=|x|+x will always be zero.Hence suff.

From S2:y<1. y can NEVER BE -ve because if x is -ve y=0 as proved above.And if x is +ve y=2x which is +ve.We are given y<1 so only possibility is y=0.Suff.

I get super confused in solving the below kind of inequalities. Can some expert help?

Are the below working's correct?

1/| n | >n • Case 1: n > 0… n 2 < 1… - 1 < n < 1 • Case 2: n < 0… - n 2 < 1… n 2 > -1…

x/| x | < x • Case 1: x > 0… x^2 > x… as x > 0, divide by x without changing signs… x > 1 • Case 2: x < 0… - x^2 > x… as x < 0, divide and only flip sign… - x < 1… x > -1

1. \(\frac{1}{| n |}>n\) --> multiply by \(|n|\) (we can safely do that since |n|>0): \(n*|n| < 1\).

If \(n>0\), then we'll have \(n^2<1\) --> \(-1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\). If \(n<0\), then we'll have \(-n^2<1\) --> \(n^2>-1\). Which is true for any n from the range we consider. So, \(n*|n| < 1\) holds true for any negative value of n.

Thus \(\frac{1}{| n |}>n\) holds true if \(n<0\) and \(0<n<1\).

2. \(\frac{x}{|x|} < x\).

If \(x>0\), then we'll have \(\frac{x}{x} < x\) --> \(1<x\). If \(x<0\), then we'll have \(\frac{x}{-x} < x\) --> \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

Thus \(\frac{x}{|x|} < x\) holds true if \(-1<x<0\) or \(x>1\).

Re: Inequality and absolute value questions from my collection [#permalink]

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23 Apr 2014, 06:04

Bunuel wrote:

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Bunuel, the second statement signifies that n is true for all values less than "1". For example, n=0.5 --> 1/.5>.5 --> 2> .5 (True) Am i correct here.. Anyways , the answer still remains the same A

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Bunuel, the second statement signifies that n is true for all values less than "1". For example, n=0.5 --> 1/.5>.5 --> 2> .5 (True) Am i correct here.. Anyways , the answer still remains the same A

It holds true for any number less than 1 except 0.

Re: Inequality and absolute value questions from my collection [#permalink]

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08 May 2014, 23:16

Bunuel wrote:

7. |x+2|=|y+2| what is the value of x+y? (1) xy<0 (2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0. B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

Dear Bunnel

I am unable to understand the red highlighted portion. how i understood as was |x+2|=|y+2| means that both side have to be positive..

case 1: x=y (same sign for both or 0) case 2: x+y = -4 (1 has to be positive or even when both are negative)

my problem is i get confused here... please explain some Mod inequalities..
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

7. |x+2|=|y+2| what is the value of x+y? (1) xy<0 (2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0. B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

Dear Bunnel

I am unable to understand the red highlighted portion. how i understood as was |x+2|=|y+2| means that both side have to be positive..

case 1: x=y (same sign for both or 0) case 2: x+y = -4 (1 has to be positive or even when both are negative)

my problem is i get confused here... please explain some Mod inequalities..

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Is this also a right way to solve this: x^2+2xy+y^2=9a --> If we have to find x^2 + y^2 > 4a we should know whether 2xy>5a or 2xy<5a...No info on this.Hence Insuff x^2-2xy+y^2=a---> If we have to find x^2 + y^2 > 4a we should know whether 2xy>3a or 2xy<3a...No info..Hence Insuff

Combining both..we still dont know as both the cases are possible and hence E..

Bunuel Please Confirm..
_________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Is this also a right way to solve this: x^2+2xy+y^2=9a --> If we have to find x^2 + y^2 > 4a we should know whether 2xy>5a or 2xy<5a...No info on this.Hence Insuff x^2-2xy+y^2=a---> If we have to find x^2 + y^2 > 4a we should know whether 2xy>3a or 2xy<3a...No info..Hence Insuff

Combining both..we still dont know as both the cases are possible and hence E..

Bunuel Please Confirm..

Yes, the question from (1) becomes: is 2xy < 5a and from (2) it becomes: is 2xy > 3a. So, when we combine the question becomes is 3a < 2xy < 5a. It's unclear how you deduced that we cannot answer that question without plugging values.
_________________

Re: Inequality and absolute value questions from my collection [#permalink]

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20 May 2014, 02:05

Bunuel wrote:

9. Is n<0? (1) -n=|-n| (2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.

I see that all my incorrect answers were because of not considering the zero case..Is this really so frequently tested on the GMAT..Or is just to make tricky practice questions
_________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Is this also a right way to solve this: x^2+2xy+y^2=9a --> If we have to find x^2 + y^2 > 4a we should know whether 2xy>5a or 2xy<5a...No info on this.Hence Insuff x^2-2xy+y^2=a---> If we have to find x^2 + y^2 > 4a we should know whether 2xy>3a or 2xy<3a...No info..Hence Insuff

Combining both..we still dont know as both the cases are possible and hence E..

Bunuel Please Confirm..

Yes, the question from (1) becomes: is 2xy < 5a and from (2) it becomes: is 2xy > 3a. So, when we combine the question becomes is 3a < 2xy < 5a. It's unclear how you deduced that we cannot answer that question without plugging values.

I mean.. you do not have any restriction on x,y,a being a integer..So it seems that multiple values of x & y such that 2xy is both greater than and less than 4a are possible Is this the right approach..or should I stick to yours..?
_________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.

I see that all my incorrect answers were because of not considering the zero case..Is this really so frequently tested on the GMAT..Or is just to make tricky practice questions