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# Inequality and absolute value questions from my collection

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Math Expert
Joined: 02 Sep 2009
Posts: 43894
Inequality and absolute value questions from my collection [#permalink]

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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) $$x^3< 0$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653690

2. If y is an integer and $$y = |x| + x$$, is $$y = 0$$?
(1) $$x < 0$$
(2) $$y < 1$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653695

3. Is $$x^2 + y^2 > 4a$$?
(1) $$(x + y)^2 = 9a$$
(2) $$(x – y)^2 = a$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653697

4. Are x and y both positive?
(1) $$2x-2y=1$$
(2) $$\frac{x}{y}>1$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653709

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653731

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653740

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653783 AND http://gmatclub.com/forum/inequality-an ... l#p1111747

8. $$a*b \neq 0$$. Is $$\frac{|a|}{|b|}=\frac{a}{b}$$?
(1) $$|a*b|=a*b$$
(2) $$\frac{|a|}{|b|}=|\frac{a}{b}|$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653789

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653792

10. If n is not equal to 0, is |n| < 4 ?
(1) $$n^2 > 16$$
(2) $$\frac{1}{|n|} > n$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653796

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653853

12. Is r=s?
(1) $$-s \leq r \leq s$$
(2) $$|r| \geq s$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653870

13. Is $$|x-1| < 1$$?
(1) $$(x-1)^2 \leq 1$$
(2) $$x^2 - 1 > 0$$

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.

PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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Re: Inequality and absolute value questions from my collection [#permalink]

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20 May 2014, 03:43
JusTLucK04 wrote:
Bunuel wrote:
JusTLucK04 wrote:

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..

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..?

I think mine is less error-prone. Notice that this question also has "zero trap"...
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Re: Inequality and absolute value questions from my collection [#permalink]

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22 May 2014, 17:55
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.

Hi Bunnel,

I don't quite get the part on "substitute x". Does it mean $$\frac{1}{y}(x-y)>0$$ and so (x-y) is zero?

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

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23 May 2014, 00:50
pretzel wrote:
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.

Hi Bunnel,

I don't quite get the part on "substitute x". Does it mean $$\frac{1}{y}(x-y)>0$$ and so (x-y) is zero?

No, we are substituting $$x=y+\frac{1}{2}$$ into $$\frac{x-y}{y}>0$$.

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

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 (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.

Graphic approach for this question: inequality-and-absolute-value-questions-from-my-collection-86939-260.html#p1269802

Hope this helps.
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Re: Inequality and absolute value questions from my collection [#permalink]

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23 May 2014, 01:00
pretzel wrote:
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.

Hi Bunnel,

I don't quite get the part on "substitute x". Does it mean $$\frac{1}{y}(x-y)>0$$ and so (x-y) is zero?

Hi Pretzel,
$$2x-2y=1$$ --> $$x=y+\frac{1}{2}$$ ------------- (i)

$$\frac{x}{y}>1$$ --> $$\frac{x}{y - 1}>0$$ --->$$\frac{(x-y)}{y}>0$$-----(ii)

substitute x from (i) into (ii):
-->$$(y+\frac{1}{2} -y)/y> 0$$ --> $$\frac{1}{2*y} > 0$$ --> $$\frac{1}{y} > 0$$ --> $$y > 0$$
i.e y is positive ---> from (i), x is positive.

So combining both the statements, sufficient to say whether x and y both are positive

Hope it helps

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

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04 Jun 2014, 07:12
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Hi Bunell,

Can you please explain the 2nd statement with some numbers. like let's a=-3,b=2 then |-3|/|2|=3/2=1.5 and similarly , |-3/2|=|-1.5|=1.5

Please let me know what i am doing wrong here
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Posts: 43894
Re: Inequality and absolute value questions from my collection [#permalink]

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04 Jun 2014, 07:49
Manik12345 wrote:
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Hi Bunell,

Can you please explain the 2nd statement with some numbers. like let's a=-3,b=2 then |-3|/|2|=3/2=1.5 and similarly , |-3/2|=|-1.5|=1.5

Please let me know what i am doing wrong here

(2) |a|/|b|=|a/b|.

If a=b=1, then answer is YES.
If a=1 and b=-1, then answer is NO.

Not sufficient.
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Re: Inequality and absolute value questions from my collection [#permalink]

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04 Jun 2014, 08:36
Bunuel wrote:
Manik12345 wrote:
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Hi Bunell,

Can you please explain the 2nd statement with some numbers. like let's a=-3,b=2 then |-3|/|2|=3/2=1.5 and similarly , |-3/2|=|-1.5|=1.5

Please let me know what i am doing wrong here

(2) |a|/|b|=|a/b|.

If a=b=1, then answer is YES.
If a=1 and b=-1, then answer is NO.

Not sufficient.

Hi Bunell,

But if a=1,b=-1 then |a|/|b|=|1|/|-1|=1 and |a/b|=|1/-1|=|-1|=1 as absolute value of any negative integer is positive

Please let me know what i am doing wrong here
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Posts: 43894
Re: Inequality and absolute value questions from my collection [#permalink]

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04 Jun 2014, 08:55
Manik12345 wrote:
Bunuel wrote:
Manik12345 wrote:

Hi Bunell,

Can you please explain the 2nd statement with some numbers. like let's a=-3,b=2 then |-3|/|2|=3/2=1.5 and similarly , |-3/2|=|-1.5|=1.5

Please let me know what i am doing wrong here

(2) |a|/|b|=|a/b|.

If a=b=1, then answer is YES.
If a=1 and b=-1, then answer is NO.

Not sufficient.

Hi Bunell,

But if a=1,b=-1 then |a|/|b|=|1|/|-1|=1 and |a/b|=|1/-1|=|-1|=1 as absolute value of any negative integer is positive

Please let me know what i am doing wrong here

So? Doesn't a=1 and b=-1 satisfy the second statement (|a|/|b|=|a/b|) and give a NO answer to the question whether |a|/|b|=a/b? While a=b=1, also satisfies the second statement (|a|/|b|=|a/b|) and gives an YES answer to the question whether |a|/|b|=a/b. Isn't it what we wanted, to establish that (2) is not sufficient?
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Re: Inequality and absolute value questions from my collection [#permalink]

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28 Jun 2014, 01:45
hi. what does the * sign denote here? doesnt look like multiplication.
Also can you please explain the solution once again in simpler way for someone who is not a math buff at all. I dont understand how the quadratic equation changed from positive to negative suddenly ((x-3)^2

thanks

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 $$x=3$$ or/and $$y=0$$. 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 $$xy=0$$ $$or$$ 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.

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

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28 Jun 2014, 03:29
PSMBAPS wrote:
hi. what does the * sign denote here? doesnt look like multiplication.
Also can you please explain the solution once again in simpler way for someone who is not a math buff at all. I dont understand how the quadratic equation changed from positive to negative suddenly ((x-3)^2

thanks

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 $$x=3$$ or/and $$y=0$$. 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 $$xy=0$$ $$or$$ 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.

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.

1. * denotes multiplication.

2. $$(a-b)^2 = a^2 - 2ab + b^2,$$ thus $$(x-3)^2=x^2-6x+9$$.

Hope it helps.

Those are hard questions. You need to brush up fundamentals before attempting them.

Theory on Algebra: algebra-101576.html

DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29
PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50

Special algebra set: new-algebra-set-149349.html

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

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13 Jul 2014, 03:13
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

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.

HI Bunuel,

Could you please explain me following statement. here how you are considering 1<x<=2. as I got doube here if some value such as 1 is between 0 and 2 it doesn't mean it is between 1 and 2.

(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.

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

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13 Jul 2014, 04:22
PathFinder007 wrote:
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

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.

HI Bunuel,

Could you please explain me following statement. here how you are considering 1<x<=2. as I got doube here if some value such as 1 is between 0 and 2 it doesn't mean it is between 1 and 2.

(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.

Thanks.

Sorry, I couldn't decode your question...
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Re: Inequality and absolute value questions from my collection [#permalink]

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24 Aug 2014, 15:34
Bunuel: for the question absvalue(x+2) = absvalue(y+2) with the first statement xy<o! what if x=-1 and y =2 won't that make both abs values positive and satisfy the constraints of xy<o?
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Re: Inequality and absolute value questions from my collection [#permalink]

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25 Aug 2014, 04:27
bankerboy30 wrote:
Bunuel: for the question absvalue(x+2) = absvalue(y+2) with the first statement xy<o! what if x=-1 and y =2 won't that make both abs values positive and satisfy the constraints of xy<o?

How is |-1 + 2| = 1 equal to |2 + 2| = 4 ?
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Re: Inequality and absolute value questions from my collection [#permalink]

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25 Aug 2014, 05:33
In that scenario isn't x=y?

Abs(x+2) = abs(y+2) if x>-2 and y>-2 then x =y wouldn't that be the scenario if x were -1 and y was so positive value. I think I wa just getting confused because in that case x would -1 and y would be -1 which wouldn't satisfy the condition correct?

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

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25 Aug 2014, 05:52
bankerboy30 wrote:
In that scenario isn't x=y?

Abs(x+2) = abs(y+2) if x>-2 and y>-2 then x =y wouldn't that be the scenario if x were -1 and y was so positive value. I think I wa just getting confused because in that case x would -1 and y would be -1 which wouldn't satisfy the condition correct?

Posted from my mobile device

Stem says that |x+2|=|y+2|. You CANNOT plug values which does not satisfy this.

Check another solution here: http://gmatclub.com/forum/inequality-an ... l#p1111747

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]

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27 Aug 2014, 16:30
Bunuel wrote:
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

2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

[

3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

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

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

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

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

8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

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

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

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

12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.

Wow, I am new for this forum. But it is easy to see Bunuel is absolutely one of gods of the forum
Thanks for your set of problems.
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Re: Inequality and absolute value questions from my collection [#permalink]

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30 Sep 2014, 22:42
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 $$x=3$$ or/and $$y=0$$. 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 $$xy=0$$ $$or$$ 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.

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.

Hello Bunuel

I tried a more generic approach by substituting y=x+3 in the equation from which I got a cubic expression=0 equation.Solving it we get x=3.
I did not divide the equality by y so there is no question of me running into the issue of y being 0.In such a case do we still have to think of the y=0 scenario,as while solving the equation by substitution only we don't run into the y=0 situation at all.
Not sure if my way of solving is correct.
Please do let me know your thoughts.
Thanks!
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Re: Inequality and absolute value questions from my collection [#permalink]

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20 Oct 2014, 05:52
I doubt the 4th q answer. PLugging in negetive values like x=-1.5 and y=-2 also satisfies .so i feel it should be E.
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Joined: 02 Sep 2009
Posts: 43894
Re: Inequality and absolute value questions from my collection [#permalink]

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20 Oct 2014, 05:56
bmetikal wrote:
I doubt the 4th q answer. PLugging in negetive values like x=-1.5 and y=-2 also satisfies .so i feel it should be E.

(-1.5)/(-2) = 0.75 < 1. So, these values don't satisfy the second statement. Check good discussion on this question here: are-x-and-y-both-positive-1-2x-2x-1-2-x-y-63377.html

Hope it helps.
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Re: Inequality and absolute value questions from my collection   [#permalink] 20 Oct 2014, 05:56

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