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# Inequality and absolute value questions from my collection

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Math Expert
Joined: 02 Sep 2009
Posts: 44298
Inequality and absolute value questions from my collection [#permalink]

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16 Nov 2009, 11:33
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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.

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Posts: 44298
Re: Inequality and absolute value questions from my collection [#permalink]

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21 Nov 2017, 01:31
talismaaniac wrote:
Bunuel wrote:
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(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.

Hi Bunuel
(1) => x^2+y^2 = 9a - 2xy
(2) => x^2+y^2 = a + 2xy

Now instead of adding, if i reduce (2) from (1),or, just equate the two, => 9a-2xy = a+2xy. => 4xy = 8a or 2xy = 4a.

Substituting 2xy = 4a in (1) or (2), => x^2 +y^2=5a
Hence (C) X^2+Y^2 > 4a

Kindly check my solution.

This is explained many, many times on previous pages.
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Re: Inequality and absolute value questions from my collection [#permalink]

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21 Nov 2017, 01:58
Bunuel wrote:

Kindly check my solution.

This is explained many, many times on previous pages.[/quote]

Oh yes! you are right. ! How could i miss reading that! Thanks!
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Concentration: Finance, Technology
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Re: Inequality and absolute value questions from my collection [#permalink]

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27 Dec 2017, 10:12
Bunuel 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.

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.

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.

Hope it's clear.

Could you explain what is the criteria for squaring when modulus involved? Can we do it with any modulus equations or it has to be done when certain things are in place.
Id like to get to that level on thinking, when in 2 mins i find out okay instead of plugging in values i should instead of square both sides. or vice versa.
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Re: Inequality and absolute value questions from my collection [#permalink]

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27 Dec 2017, 15:10
Bunuel wrote:
tihor wrote:
Bunuel, two questions:
shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2?
secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2?
does this not translate into x<=0 or x<=2?

thank you very much for all the questions and solutions.

Thank you very much for this catch. +1. There was a typo. So you are right with the first one:
|x-1| < 1 means 0<x<2. Already edited the post.

As for the second one:
x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.

x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.

Hope it helps.

Hi

I have the same question, could explain this again.

When we do this algebraically why does the range come out to be $$x<= 0$$ or $$x <=2$$

$$(x-1)^2 <= 1$$
$$(x^2 - 2x + 1) <=1$$
$$x^2 -2x <= 1-1$$
$$x^2 -2x<= 0$$
$$x(x-2) <= 0$$
$$x<=0$$ or $$x<=2$$

Where am i going wrong in this. It would be helpful if you could explain in detail.
Math Expert
Joined: 02 Sep 2009
Posts: 44298
Inequality and absolute value questions from my collection [#permalink]

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27 Dec 2017, 21:25
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mtk10 wrote:
Bunuel wrote:
Could you explain what is the criteria for squaring when modulus involved? Can we do it with any modulus equations or it has to be done when certain things are in place.
Id like to get to that level on thinking, when in 2 mins i find out okay instead of plugging in values i should instead of square both sides. or vice versa.

You can always square if you have absolute values on both sides, so if you have |something| = |something|, then you can square. Squaring allows you to get rid of the modulus. Often times you are left with quadratics and it could be easier to solve.

If you have |something| = something, then squaring might give wrong solution(s). For example, |x - 1| = 2x - 1 --> x = 2/3 but if you square you get (x - 1)^2 = (2x - 1)^2, which gives x = 2/3 or x = 0.
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Re: Inequality and absolute value questions from my collection [#permalink]

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27 Dec 2017, 21:35
mtk10 wrote:
Bunuel wrote:
tihor wrote:
Bunuel, two questions:
shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2?
secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2?
does this not translate into x<=0 or x<=2?

thank you very much for all the questions and solutions.

Thank you very much for this catch. +1. There was a typo. So you are right with the first one:
|x-1| < 1 means 0<x<2. Already edited the post.

As for the second one:
x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.

x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.

Hope it helps.

Hi

I have the same question, could explain this again.

When we do this algebraically why does the range come out to be $$x<= 0$$ or $$x <=2$$

$$(x-1)^2 <= 1$$
$$(x^2 - 2x + 1) <=1$$
$$x^2 -2x <= 1-1$$
$$x^2 -2x<= 0$$
$$x(x-2) <= 0$$
$$x<=0$$ or $$x<=2$$

Where am i going wrong in this. It would be helpful if you could explain in detail.

First of all, $$x \leq 0$$ or $$x\leq 2$$ does not make any sense. What are the values in this case? Could x be 1 because you say $$x\leq 2$$ or it should be $$x \leq 0$$.

Next, if you solve the way you do then:

$$x(x-2) \leq 0$$ --> x and x - 2 have the opposite signs:

$$x \geq 0$$ and $$x -2 \leq 0$$ --> $$x \geq 0$$ and $$x \leq 2$$ --> $$0 \leq x \leq 2$$
$$x \leq 0$$ and $$x -2 \geq 0$$ --> $$x \leq 0$$ and $$x \geq 2$$ --> no intersection, so no solution in this case (x cannot be simultaneously less than 0 and more than 2).

Finally, the questions in this set are very tough and tricky. You should be very familiar at least with standard ways of solving inequality questions, modulus, quadraticsbefore attempting. Hope the links below help:

9. Inequalities

For more check Ultimate GMAT Quantitative Megathread

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

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28 Dec 2017, 02:09
Bunuel wrote:
mtk10 wrote:
Bunuel wrote:
Could you explain what is the criteria for squaring when modulus involved? Can we do it with any modulus equations or it has to be done when certain things are in place.
Id like to get to that level on thinking, when in 2 mins i find out okay instead of plugging in values i should instead of square both sides. or vice versa.

You can always square if you have absolute values on both sides, so if you have |something| = |something|, then you can square. Squaring allows you to get rid of the modulus. Often times you are left with quadratics and it could be easier to solve.

If you have |something| = something, then squaring might give wrong solution(s). For example, |x - 1| = 2x - 1 --> x = 2/3 but if you square you get (x - 1)^2 = (2x - 1)^2, which gives x = 2/3 or x = 0.

Thanks Bunuel this cleared my confusion
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Re: Inequality and absolute value questions from my collection [#permalink]

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14 Feb 2018, 04:18
[quote="Bunuel"]11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

Hi Bunuel,
Can we deduce from 2 that after squaring both sides, y^2-2xy<0,
or, y(y-2x)<0
case 1, if y<0, y-2x must be >0,
=>both x,y should have same sign i.e -ve in this case.
Case 2,
y>0, y-2x<0,
=> both x,y should have same sign i.e +ve in this case.
Hence sufficient. B
Re: Inequality and absolute value questions from my collection   [#permalink] 14 Feb 2018, 04:18

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