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# Inequality and absolute value questions from my collection

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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: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#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: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
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Re: Inequality and absolute value questions from my collection [#permalink]

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08 Aug 2011, 10:39
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.

duh! missed out on the 0 and thought C was the ans.

Is this what you call the ZIP trap.. if so, then i sure did get zipped.. dang
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08 Aug 2011, 18:53
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, devide the whole equation by y, and
6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3
(1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient
(2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient
B

There is only one answer from the equation that you formed above from the question stem.
x^2 - 6xy +9 = 0

X =3

There for statement 1 is sufficient

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08 Aug 2011, 20:54
manishgeorge 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, devide the whole equation by y, and
6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3
(1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient
(2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient
B

There is only one answer from the equation that you formed above from the question stem.
x^2 - 6xy +9 = 0

X =3

There for statement 1 is sufficient

read Bunuel's and ImJun's post for this problem.

He writes that we can't divide both sides by 'y' since that would mean that we are assuming that y is not equal to 0.

The question prompt asks us the value of xy

Sol. would be:

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

6*x*y = x^2*y + 9*y
x^2*y - 6*x*y + 9*y = 0
y (x^2 - 6*x + 9) = 0
y (x - 3)^2 = 0

Thus y could be = 0 or some other no.
X = -3 and +3

(1) y – x = 3

If x = 3, then y = 6
x = -3, then y = 0

thus two solutions, xy = 18 and xy = 0

Insuff

(2) x^3< 0

if x^3 < 0, we know that x = -3. Thus y = 0

Ans: xy = 0

Stmt 2 suff

Ans B

Hope this helps.
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08 Aug 2011, 21:12
I understand now. Thanks for explaining
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15 Aug 2011, 05:09
great effort..thanks
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25 Aug 2011, 20:53
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

Ans:

From the initial equation we get: y*(x^2-6x+9)=0 i.e y=0 or (x-3)^2=0
means either y=0 or (x-3)^2=0 or both
A does not give any idea about
B states that x^3<0 which means that x is not equal to 3.hence in that case y needs to be zero. so statement 2 itself is sufficient to answer the question.
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07 Sep 2011, 20:47
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

Hi Bunuel,

Different approach and I got the answer C. Please correct if I am wrong!

6*x*y= x^2*y + 9*y

(1) y - x = 3 => y = x+3
Substitue y into stem, we have:
6*x*(x+3) = x^2*(x+3) + 9*(x+3)
6*x^2 + 18*x = x^3 + 3*x^2 + 9*x +27
x^3 - 3*x^2 - 9*x + 27 = 0
(x-3)^2*(x+3) = 0
x = 3 or x = -3 (Insufficient)

(2) x^3 <0

(1) + (2), we have x = -3, y =0, then x*y = 0

Then the answer must be C
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16 Dec 2011, 20:24
Awesome questions Bunuel...thanks a ton for sharing...
Wonderful explanations as well..
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20 Jan 2012, 13:30
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I accumulated bunnuel's answers together with absolute and inequality questions
Attachment:

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24 Jan 2012, 08:01
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=-2.
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.
(...)

Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!
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Re: Inequality and absolute value questions from my collection [#permalink]

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24 Jan 2012, 12:47
metallicafan 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=-2.
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.
(...)

Bunuel, how did you figure out that |x+2|=|y+2| can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0
Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!

We have $$|x+2|=|y+2|$$.
If both absolute values expand with + or - sign we'll get: $$x+2=y+2$$ (notice that $$-(x+2)=-(y+2)$$ is exactly the same);
If they will expand with different signs we'll get: $$-(x+2)=y+2$$ (notice that $$x+2=-(y+2)$$ is exactly the same).

So only two forms.

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

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10 Feb 2012, 02:40
For question 3:

I think I found a different answer. My answer is C (DATA 1 AND 2 are SUFFICIENT together).

x^2+y^2 >4a ?

From (1), (x+y)^2=9a means that x^2+y^2 = 9a-2xy

To satisfy our equation, must 5a > 2xy

From (2), x^2+y^2 = a +2xy

To satisfy our equation, must a + 2xy > 4a, therefore we must have 2xy > 3a

Hence, 3a < 2xy < 5a (aggregating both inequalities 1 and 2)

Then we know that 9a-2xy = a +2xy (from 1 and 2)

and this gives us 2a = xy, or 2xy = 4a

Conclusion: 2xy=4a satisfies our constraint, hence X^2+Y^2 > 4a

Let me know if you see a mistake here, thanks.
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10 Feb 2012, 06:14
grotily90 wrote:
For question 3:

I think I found a different answer. My answer is C (DATA 1 AND 2 are SUFFICIENT together).

x^2+y^2 >4a ?

From (1), (x+y)^2=9a means that x^2+y^2 = 9a-2xy

To satisfy our equation, must 5a > 2xy

From (2), x^2+y^2 = a +2xy

To satisfy our equation, must a + 2xy > 4a, therefore we must have 2xy > 3a

Hence, 3a < 2xy < 5a (aggregating both inequalities 1 and 2)

Then we know that 9a-2xy = a +2xy (from 1 and 2)

and this gives us 2a = xy, or 2xy = 4a

Conclusion: 2xy=4a satisfies our constraint, hence X^2+Y^2 > 4a

Let me know if you see a mistake here, thanks.

Welcome to GMAT Club. Let me assist you with this set of problems: first post on each page contains links to the detailed solutions of the questions with OA's.

OA for the 3rd question is E, not C. Please refer to this post for explanation: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

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

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23 Feb 2012, 10:41
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?
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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.

Hope it's clear.
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24 Feb 2012, 19:01
Thanks a lot buneuel..You are just too good +1.

This is a great take away!
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Re: Inequality and absolute value questions from my collection [#permalink]

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25 Feb 2012, 06:51
Quote:
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.

when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

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25 Feb 2012, 08:06
shankar245 wrote:
Quote:
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.

when we have x+y=-4
Acc to the explanation , we have
This will occur when either x or y is less then -2 and the other is more than -2

so x can be be -3,-4,-5..(X<-2)
and y can be -1,0,1...(y>-2)

a solution could that -2-2=-4

in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption

When we consider ranges to expand an absolute value we should put equal sign (=) in either of the range. For our question we put equal sign for the first range (case A) when we are analyzing the case when x and y are both $$\geq{-2}$$ than -2 OR both $$\leq{-2}$$. So the scenario when $$x=y=-2$$ ($$x+y=-2-2=-4$$) is included in case A.

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

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28 Mar 2012, 22:53
Nice set of questions....... superlike.......
Manager
Joined: 28 Apr 2011
Posts: 195
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Kudos [?]: 8 [0], given: 6

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

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28 Mar 2012, 23:25
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

Tricky one......
now i know its C for sure
nice explanation bunnel

Re: Inequality and absolute value questions from my collection   [#permalink] 28 Mar 2012, 23:25

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