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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]  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) 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#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|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) 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<=r<=s
(2) |r|>=s

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

13. Is |x-1| < 1?
(1) (x-1)^2 <= 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]  16 Dec 2011, 20:24
Awesome questions Bunuel...thanks a ton for sharing...
Wonderful explanations as well..
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Re: Inequality and absolute value questions from my collection [#permalink]  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]  24 Jan 2012, 12:47
Expert's post
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]  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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Re: Inequality and absolute value questions from my collection [#permalink]  10 Feb 2012, 06:14
Expert's post
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]  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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Re: Inequality and absolute value questions from my collection [#permalink]  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]  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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Re: Inequality and absolute value questions from my collection [#permalink]  25 Feb 2012, 08:06
Expert's post
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]  28 Mar 2012, 22:53
Nice set of questions....... superlike.......
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Re: Inequality and absolute value questions from my collection [#permalink]  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

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Re: Inequality and absolute value questions from my collection [#permalink]  29 Mar 2012, 00:03
Expert's post
vdbhamare wrote:
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

Actually the answer is E, not C: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

The links to the OA's with solutions are given in the initial post: inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]  29 Mar 2012, 02:56
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

1. r is between -s & s
2. r>s or r <-s

1 & 2 together to be true r=s 0r r=-s hence E
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Re: Inequality and absolute value questions from my collection [#permalink]  31 Mar 2012, 08:34
I got

===========================
1 - A

6*x*y = x^2*y + 9*y
can be simplified to x = 3
==================================
2 - A

y = |x| + x

for y to be = 0 ; x should be either 0 or -ve
==================================
3 - E
==================================
4 - C

(A)

(x-y) = 1/2

x can be either (0 or -1/2)
y can be either (1 or 1/2)

(B)

x/y > 1

both x & y can be either -ve
or
both x & y can be either +ve

(c)

works for both

==================================

5- E

==================================

6 - C

==================================

7 - A

|x+2|=|y+2|
we have
if -(x+2) = (y+2)
then x+y = -4

if (x+2) = -(y+2)
then x+y = -4

(A) xy<0 - sufficient data

==================================

8 - A

(A) |a*b|=a*b

if (a & b ) = +ve then |a*b|=a*b = true
|a|/|b|=a/b data sufficient

if (a & b ) = -ve then |a*b|=a*b = true
|a|/|b|=a/b data sufficient

(B)

|a|/|b|=|a/b|
if(a = -ve and b = + ve) then |a|/|b|=|a/b| = true
|a|/|b|=a/b data NOT sufficient

if(a = -ve and b = - ve) then |a|/|b|=|a/b| = true
|a|/|b|=a/b data sufficient

==================================

9 - A

(A) -n = |-n|

if ( n = -ve) then -n = |-n| is true and n < 0 true - data sufficient

if ( n = +ve) then -n = |-n| is false

(B) n^2=16
n = + or - 4 not sufficient

==================================
10 - A

(A) n^2 > 16

n > 4 therefore: |n| < 4 = no (Data sufficient)

(B) 1/|n| > n

if n = -2
then 1/|-2|> -2 true --> |-2| < 4 = true

if n = -8
then 1/|-8| > -8 true --> |-8| < 4 = false

==================================

11- E

(A) |x| > |y|

if (X = 3 && Y = 2)

then |x| > |y| = true
|x+y|>|x-y| = true

if (X = -3 && Y = 2)

then |x| > |y| = true
|x+y|>|x-y| = false

if (X = -3 && Y = -2)

then |x| > |y| = true
|x+y|>|x-y| = true

(B) |x-y| < |x|

if (X = 3 && Y = 2)

then |x-y| < |x| = true
|x+y|>|x-y| = true

if (X = -3 && Y = -2)

then |x-y| < |x| = true
|x+y|>|x-y| = true

(C)

if (X = 3 && Y = 2)

then |x| > |y| = true
then |x-y| < |x| = true
|x+y|>|x-y| = true

if (X = -3 && Y = -2)

then |x| > |y| = true
then |x-y| < |x| = true
|x+y|>|x-y| = true

=================================

12 - C

(A) -s<=r<=s

if s = 3

then -3<= r <= 3

r = [-3,-2,-1,0,1,2,3]

[not sufficient]

(B) |r|>=s
if s=3

r = {3 or-3}
[not sufficient]

(C)

if s = 3 then r should be 3

================================

13 - E

(A)

(x-1)^2 <= 1

x^2 - 2x + 1 <= 1
x^2 - 2x < = 0

x(x-2)<= 0
x<=0 or x <= 2

not sufficient

(B)

x^2 - 1 > 0

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

x>-1

not sufficient

(C)

2>= x > -1

not sufficient
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Re: Inequality and absolute value questions from my collection [#permalink]  02 Apr 2012, 19:09
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as $$y=|x|+x$$ then $$y$$ is never negative. For $$x>{0}$$ then $$y=x+x=2x$$ and for $$x\leq{0}$$ then (when x is negative or zero) then $$y=-x+x=0$$.

(1) $$x<0$$ --> $$y=|x|+x=-x+x=0$$. Sufficient.

(2) $$y<1$$, as we concluded y is never negative, and we are given that $$y$$ is an integer, hence $$y=0$$. Sufficient.

for evaluating 2nd Option- y<1, try with value x= 0.25, then Y would be |0.25|+0.25 = 0.5, which is less than 1. It is not mentioned that x is also an integer.

And also as you mentioned for negative value of x, it will be equal to zero. So it doesnt give definite ans whether y=0. So option (2) not sufficient.
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Re: Inequality and absolute value questions from my collection [#permalink]  02 Apr 2012, 23:37
Expert's post
roshan1985 wrote:
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as $$y=|x|+x$$ then $$y$$ is never negative. For $$x>{0}$$ then $$y=x+x=2x$$ and for $$x\leq{0}$$ then (when x is negative or zero) then $$y=-x+x=0$$.

(1) $$x<0$$ --> $$y=|x|+x=-x+x=0$$. Sufficient.

(2) $$y<1$$, as we concluded y is never negative, and we are given that $$y$$ is an integer, hence $$y=0$$. Sufficient.

for evaluating 2nd Option- y<1, try with value x= 0.25, then Y would be |0.25|+0.25 = 0.5, which is less than 1. It is not mentioned that x is also an integer.

And also as you mentioned for negative value of x, it will be equal to zero. So it doesnt give definite ans whether y=0. So option (2) not sufficient.

Welcome to GMAT Club.

The red part in your reasoning is not correct: y cannot be 0.25 since given that y is an integer. So, answer D is correct.

The links to the OA's with solutions are given in the initial post: inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]  03 Apr 2012, 06:47
h2polo wrote:
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 I reduced the given equation (divided out the y) and solved for x:
6*x*y = x^2*y + 9*y
6*x = x^2 + 9
0 = x^2 - 6*x + 9
0 = (x-3)^2
x = 3

Statement 1:

y-x=3
y-3=3
y=6
xy=3*6=18

SUFFICIENT

Statement 2:

x^3<0

We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?

INSUFFICIENT

solved the same. a doubt!
when this is a DS , cant we take statement 1 with 2 variables 'x' and 'y' will give the answer for xy?
but the statement 2 wont.
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Re: Inequality and absolute value questions from my collection [#permalink]  03 Apr 2012, 09:08
Expert's post
kashishh wrote:
h2polo wrote:
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 I reduced the given equation (divided out the y) and solved for x:
6*x*y = x^2*y + 9*y
6*x = x^2 + 9
0 = x^2 - 6*x + 9
0 = (x-3)^2
x = 3

Statement 1:

y-x=3
y-3=3
y=6
xy=3*6=18

SUFFICIENT

Statement 2:

x^3<0

We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?

INSUFFICIENT

solved the same. a doubt!
when this is a DS , cant we take statement 1 with 2 variables 'x' and 'y' will give the answer for xy?
but the statement 2 wont.

Answer to this question is B, not A.

The links to the OA's with solutions are given in the initial post: inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806
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Re: Inequality and absolute value questions from my collection [#permalink]  20 Jun 2012, 17:43
sriharimurthy wrote:
Quote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

hi sriharimurthi,

If we have the statements -s <= r < = s and -s >= r >= s, there could be 2 possible answers; r=s or r=-s. Both the possible answers satisfy the 2 equations/inequalities. So isn't the answer E and not C?

Thanks v much.
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Re: Inequality and absolute value questions from my collection [#permalink]  21 Jun 2012, 00:13
Expert's post
shivamayam wrote:
sriharimurthy wrote:
Quote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

hi sriharimurthi,

If we have the statements -s <= r < = s and -s >= r >= s, there could be 2 possible answers; r=s or r=-s. Both the possible answers satisfy the 2 equations/inequalities. So isn't the answer E and not C?

Thanks v much.

Sriharimurthi solution is not correct. OA for this question is E, not C. The links to the OA's and solutions are given in the initial post (if the links does not work, then switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p652806 will lead you to the solutions).

As for this question:
12. Is r=s?

(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;
B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.
But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well.
Consider: s=5, r=5 --> -5<=5<=5 |5|>=5
s=5, r=-5 --> -5<=-5<=5 |-5|>=5
Both statements are true with these values. Hence insufficient.

Hope it helps.
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Re: Inequality and absolute value questions from my collection   [#permalink] 21 Jun 2012, 00:13

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