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Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 10:33

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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]
02 Dec 2009, 01:05

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kaptain wrote:

Bunuel, you are correct. The key is understanding that the two equations are an 'OR' (either one is true depending on whether x^2-4 is positive or negative) and not an 'AND' (both are correct).

You mentioned that inequalities cannot be added 'the way' I did. I believe you are not saying that we cannot add inequalities. I saw an interesting discussion here - > http://www.beatthegmat.com/combining-in ... 21610.html (Sorry for the cross posting, but this may be of use to someone confused like me!)

+1 from me.

cheers

Sure when I said that you can not add this way I meant: in this case. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
09 Aug 2012, 16:21

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jayaddula wrote:

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.

Hi Bunuel,

I am getting E and just cannot understand D. Please see my solution below - I used number picking.

A. xy<0, x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value or x=- and y=+ - This condition doesn't satisfy the modulus condiotion

B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.

Hence E. I know I am going wrong some where, please help.

thanks jay

In your example, both pairs give the same value for x+y: 2-6=-4 and 3-7=-4.

We can solve this question in another way:

7. |x+2|=|y+2| what is the value of x+y?

Square both sides: x^2+4x+4=y^2+4y+4 --> x^2-y^2+4x-4y=0 --> (x+y)(x-y)+4(x-y)=0 --> (x-y)(x+y+4)=0 --> either x=y or x+y=-4.

(1) xy<0 --> the first case is not possible, since if x=y, then xy=x^2\geq{0}, not <0 as given in this statement, hence we have the second case: x+y=-4. Sufficient.

(2) x>2 and y<2. This statement implies that x\neq{y}, therefore x+y=-4. Sufficient.

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 02:15

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gmat620 wrote:

Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.

These questions are from various sources. Couple of questions might be from MGMAT CAT or Gmat Prep, but not more than that.

I'll provide OA in a day or two, after discussions. Tell me if you want the answers for the specific questions earlier than that and I'll mail you. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 06:38

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3) I) (x+y)^2=9a x^2+y^2=9a-2xy NS II) (x-y)^2=a x^2+y^2=a+2xy NS Together 2(x^2+y^2)=10a x^2+y^2=5a If either x or y are larger than 0, the stem would be true, but if they’re both zero the stem is false, hence E

4) I don’t get the two clues; they seem to be mutually exclusive

5) I) 3|x^2-4|=y-2 either y=3x^2-10 or y=14-3x^2 NS II) |3-y|=11 either y=-8 or y=14 NS Together -8=3x^2-10 so 3x^2=2 ok 14=3x^2-10 so 3x^2=28 ok, hence E

6) I) x+1>0 so x={0, 1, 2, …} NS II) xy>0 so x and y have the same sign and none of them is zero NS Together, x={1, 2, 3, ..} and y has the same sign, hence C

7) |x+2|=|y+2| either x+2=y+2 or x+2=-y-2 (the other two combinations can be transformed into these by multiplying by -1) Reordering: x-y=0 or x+y=-4 I)xy<0, hence x and y have different signs and none of them is zero. The only possibility is x+y=-4 S II) x>2, y<2 hence x#y. The only possibility is x+y=-4 S, therefore D

8)a*b#0, hence a and b are both non-zero I) |a*b|=a*b a and b have the same sign and the stem is always true S II) |a|/|b|=|a/b| this is true regardless of the values of a and b, and nothing can be said about the stem NS, therefore A

9) I) –n=|-n| n<=0 NS II) n^2=16 n=+/-4 NS Together n=-4 therefore C

10)n#0 I) n^2>16, so |n|>4 S II) 1/|n|>n true for n<-1 NS, therefore A

11) Plugging in numbers I get B, but there’s no rime or reason to my solution

12) I) –s<=r<=s obviously NS. Since s>=-s, s is either positive or zero II)|r|>=s obviously NS Together: I) tells us that s>=0; II) tells us that r>=s or r<=-s. The only case in which I and II are simultaneously satisfied is r=s, therefore C

13) x=(0:2) with 0 and 2 excluded I) (x-1)^2<=1, hence x=[0:2] with 0 and 2 included, hence NS II) x^2-1>0 x<-1 or x>1. For x=1.5 the stem is true, for x=3 it is false, hence NS Together, for x=1.5 the stem is true, for x=2 it is false, hence E

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:07

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Marco83 wrote:

4) I don’t get the two clues; they seem to be mutually exclusive

Yes there was a typo in 4. Edited. Great job Marco83. Even though not every answer is correct, you definitely know how to deal with this kind of problems. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:34

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Bunuel wrote:

5. What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11

Statement 1:

Two equations, two unknowns... INSUFFICIENT

Statement 2:

|3 - y| = 11 (3-y)=11 or (3-y)=-11 y=-8, 14

INSUFFICIENT

Statements 1 and 2:

y must be 14 because 3|x^2 -4| can never be a negative value (no matter what you plug in for x, you will get a positve value because of the absolute value signs).

SUFFICIENT

ANSWER: C.

Last edited by h2polo on 17 Nov 2009, 09:54, edited 1 time in total.

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 13:43

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Quote:

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

Question stem : Note: Since the equations is symmetrical, there will only be two distinct cases. However, for the sake of explanation, I have illustrated all 4. (a) When both x and y are greater than - 2 ; x + 2 = y + 2 ; x = y (b) When both x and y are less than - 2 ; - x - 2 = - y - 2 ; x = y (c) When x is less than -2 and y is greater than -2 ; - x - 2 = y + 2 ; x + y = - 4 (d) When x is greater than -2 and y is less than -2 ; x + 2 = - y - 2 ; x + y = - 4

St. (1) : xy < 0 This implies that one is negative and the other is positive. Therefore, in order for xy to be less than 0, x cannot be equal to y. Thus in order to satisfy the question stem, it can only be cases (c) and (d). Thus Sufficient.

St. (2) : x > 2 ; y < 2 Again, this implies that x and y cannot be equal. Thus, in order to satisfy the question stem it can only be cases (c) and (d). Thus Sufficient.

Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 16:19

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Bunuel wrote:

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

First let's simplify given expression 6*x*y = x^2*y + 9*y:

y*(x^2-6x+9)=0 --> y*(x-3)^2=0. Note here that we CAN NOT reduce this expression by y, as some of you did. Remember we are asked to determine the value of xy, and when reducing by y you are assuming that y doesn't equal to 0. We don't know that.

Next: we can conclude that either x=3or/andy=0. Which means that xy equals to 0, when y=0 and x any value (including 3), OR xy=3*y when y is not equal to zero, and x=3.

(1) y-x=3. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case xy=18. But if y=0 then x=-3 and xy=0. Two possible scenarios. Not sufficient.

(2) x^3<0. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 13:09

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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.

1 and 2 are insuff for the above mentioned (by Bunuel) reasons, now taking 1 and 2 together, here's another approach: 2x-2y=1 2(x-y)=1 (x-y)=1/2

y(x/y -1)=1/2 now we know that from option 2 x/y>1

so, y(x/y -1)=1/2====> y[(a value >1) - 1]=1/2 so, [y*(+ve value)]=1/2 hence y= +ve

now since y=+ve and (x/y)>1, we have x=+ve therefore, option c. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
21 May 2010, 02:06

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ManishS wrote:

I am a little confused on this one . Can the answer be E??

From A: 2x-2y=1 => x-y= 0.5 INSF

From B x/y > 1 => x > y INSF

From A & B x-y =0.5 and x > y

If x = -0.5 and y = -1 then x > y and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5 Hence both x and y can be negative

If x= 1 and y = 0.5 then x > y and x- y = 1 -0.5 = 0.5 Hence both x and y can be positive

Ans = E ??

Problem with your solution is that the red part is not correct.

\frac{x}{y}>1 does not mean that x>y. If both x and y are positive, then x>y, BUT if both are negative, then x<y.

From (2) \frac{x}{y}>1, we can only deduce that x and y have the same sigh (either both positive or both negative).

When we consider two statement together:

From (1): 2x-2y=1 --> x=y+\frac{1}{2}

From (2): \frac{x}{y}>1 --> \frac{x}{y}-1>0 --> \frac{x-y}{y}>0 --> substitute x from (1) --> \frac{y+\frac{1}{2}-y}{y}>0--> \frac{1}{2y}>0 (we can drop 2 as it won't affect anything here and write as I wrote \frac{1}{y}>0, but basically it's the same) --> \frac{1}{2y}>0 means y is positive, and from (2) we know that if y is positive x must also be positive.

OR: as y is positive and as from (1) x=y+\frac{1}{2}, x=positive+\frac{1}{2}=positive, hence x is positive too.

Re: Inequality and absolute value questions from my collection [#permalink]
23 Feb 2012, 11:00

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shankar245 wrote:

Quote:

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.

In the first statement from 2x-2y=1 --> we can sat x-y=1/2

So it cud be 8.5-8 or 0.25 - (-0.25) HOw can we say both x and y are positive?

similarily statement 2

x/y>1 =>x>y

how can we be sure x and y have the same sign we can have 8>7 or 8>-8

Bunel can you pls xplain..or am i missing sumtin fundamental?

For (1) we are not saying that x and y are positive, we saying that if they were both positive then it would mean that the line y=x-1/2 is only in I quadrant, which is not possible. So from (1) we cannot say that x and y are both positive.

Or consider following numbers: x=2, y=1.5 and x=-2, y=-2.5.

For (2). First of all you can not multiply x/y>1 by y and write x>y, since you don't know the sign of y: if it's positive then x>y but if it's negative then when multiplying by negative number you should flip the sing of the inequality and write x<y. Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Next, since x/y>1 then x/y>0, which means that x and y are either both positive or both negative, hence they have the same sign.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???

Here is what i did

Question: |x-1| < 1 critical point x>1 or x<1

when x>1 then (x-1)<1 x<2 when x<1 then -(x-1)<1 -x<0 therefore x>0 to prove 0<x<2 --? [understood this]

(A)(x-1)^2 <= 1 x^2 - 2x + 1 <= 1 x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]

my thoughts [what m i doing wrong??] when (x = 0) then x-2<=0 therefoe x<=2 when (x-2 = 0) then x<=0 ????? i am confused here ?????

(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]

Re: Inequality and absolute value questions from my collection [#permalink]
25 Aug 2012, 06:26

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Bunuel wrote:

SOLUTIONS:

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

First let's simplify given expression 6*x*y = x^2*y + 9*y:

y*(x^2-6x+9)=0 --> y*(x-3)^2=0. Note here that we CAN NOT reduce this expression by y, as some of you did. Remember we are asked to determine the value of xy, and when reducing by y you are assuming that y doesn't equal to 0. We don't know that.

Next: we can conclude that either x=3or/andy=0. Which means that xy equals to 0, when y=0 and x any value (including 3), OR xy=3*y when y is not equal to zero, and x=3.

(1) y-x=3. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case xy=18. But if y=0 then x=-3 and xy=0. Two possible scenarios. Not sufficient.

OR:

y-x=3 --> x=y-3 --> y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0 --> either y=0 or y=6 --> if y=0, then x=-3 and xy=0or if y=6, then x=3 and xy=18. Two different answers. Not sufficient.

(2) x^3<0. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B.

The answer to this one is C right? B alone is not sufficient. _________________

Encourage me by pressing the KUDOS if you find my post to be helpful.

Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 03:00

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carcass wrote:

Bunuel I have very clear the 1 statement but not so much the secon one : 1/|n|< n this is true only for negative value. So we could have i. e. : -1 or -6 so insuff but how we you arrive to this conclusion:

Given: 1/|n| > n. Now, 1/|n| is always positive (because of the absolute value in the denominator), so if n is negative then this inequality will always hold true: 1/|n|=positive>n=negative.

Hope it's clear.

P.S. Complete solution for 1/|n| > n is n<0 or 0<n<1. _________________

Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 03:20

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carcass wrote:

Correct me if I'm wrong because the concepts are always the same but the gmat blonds them and as consequence blow your mind.

in other words, you are saying

1/|n| > n 2 cases

1/-n > n ----> n^2 > -1 this implies that any squared number is positive and therefore greater than -1, all negative n values work as solutions. your n < 0

1/n > n -----> 1 > n^2 ----> n^ 2 < 1 --------> -1 < n < 1 . your second range. so in the end we have all this information and we are not sure of course of - 4 < n < 4.

Correct ???

Thanks

1/|n| > n --> 2 cases:

If n<0, then |n|=-n, so we'll have that 1/-n>n --> multiply by n and flip the sign (since we consider negative n): -1<n^2 --> which holds true for any n from this range, so for any negative n.

If n>0, then |n|=n, so we'll have that 1/n>n --> multiply by positive n, this time: 1>n^2 --> -1<n<1, since we consider n>0, then finally we'll get 0<n<1.

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