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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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New post 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: 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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New post 28 Aug 2015, 12:55
Bunuel wrote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(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.

Answer: E.


Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 28 Aug 2015, 13:29
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SauravPathak27 wrote:
Bunuel wrote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(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.

Answer: E.


Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.


Hi SauravPathak27

I'm no expert, but hope I might be able to help.

your understanding that |x| <1 means -1<x<1 is correct.

(1) is telling us that \(r\) falls between \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)...... INSUFFICIENT
(2) is telling us that \(r\) falls outside \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)........ INSUFFICIENT
(1) & (2) together tells us that \(r\) must be equal to either \(-s\) or \(s\) but cannot determine which one........ INSUFFICIENT

Hopefully my freehand below makes it a little clearer.

If it helps.... Throw me a Kudos :)
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 28 Aug 2015, 13:31
DropBear wrote:
SauravPathak27 wrote:
Bunuel wrote:
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(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.

Answer: E.


Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.


Hi SauravPathak27

I'm no expert, but hope I might be able to help.

your understanding that |x| <1 means -1<x<1 is correct.

(1) is telling us that \(r\) falls between \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)...... INSUFFICIENT
(2) is telling us that \(r\) falls outside \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)........ INSUFFICIENT
(1) & (2) together tells us that \(r\) must be equal to either \(-s\) or \(s\) but cannot determine which one........ INSUFFICIENT

Hopefully my freehand below makes it a little clearer.

If it helps.... Throw me a Kudos :)


Indeed it helps. Thanks for the extra effort though!
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 08 Sep 2015, 11:08
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.

Answer: E.


Are 1 and to clearly insufficient because you can see at the first sight that x and y could be 0, leading to a clear No and that some positive values would immeadetly lead to a Yes. Can the case "won't be true" classify as a No to the question "is x^2+y^2>4a" or is it neutral?
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 08 Sep 2015, 11:13
reto 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.

Answer: E.


Are 1 and to clearly insufficient because you can see at the first sight that x and y could be 0, leading to a clear No and that some positive values would immeadetly lead to a Yes. Can the case "won't be true" classify as a No to the question "is x^2+y^2>4a" or is it neutral?


Yes, you are correct. When you combine both the statements, you get \(x^2+y^2 = 5a\).

Consider the following 2 cases:

Case 1: When a = -1, 4a=-4 and 5a = -5. In this case you get a "NO"

Case 2: When a = 1, 4a=4 and 5a = 5. In this case you get a "YES"

Takeaway from this question: make sure to check the ranges of the variables!
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 08 Sep 2015, 11:23
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.


\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\)

If you substitute x, the new fraction should look as follows: \(\frac{0.5}{y}>0\) since \(x=y+\frac{1}{2}\) ?
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 08 Sep 2015, 11:30
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reto 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.

Answer: C.


\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\)

If you substitute x, the new fraction should look as follows: \(\frac{0.5}{y}>0\) since \(x=y+\frac{1}{2}\) ?


Yes, you are correct.

After substituting for x, you get, \(\frac{0.5}{y} >0\) ---> \(\frac{1}{2y} >0\) ---> \(\frac{1}{y} >0\) ---> y >0
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Re: Inequality and absolute value questions from my collection [#permalink]

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New post 08 Sep 2015, 18:21
Simplifying the question stem gives me (x-3)^2 * y = 0 , which means either y= 0 or x-3 = 0

S1) it states that y-x = 3 , so if i take y= 0 from question stem then x= -3 and xy = 0 but if x-3 = 0 then x= 3 and y will be 6 so xy = 18 hence insufficient

S2) it tells me that x is negative so y = 0 hence xy = 0 always.
so B

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New post 08 Sep 2015, 21:34
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reto 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.

Answer: C.


\(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\)

If you substitute x, the new fraction should look as follows: \(\frac{0.5}{y}>0\) since \(x=y+\frac{1}{2}\) ?


Check this post: inequality-and-absolute-value-questions-from-my-collection-86939-60.html#p666191

"I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them.

Hope it's clear. "
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 09 Sep 2015, 10:31
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.


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. >> Both x and y can be -2 leading to -4. So why this statement that either x or y is less than -2 and the other is more?
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New post 09 Sep 2015, 10:40
reto 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.


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. >> Both x and y can be -2 leading to -4. So why this statement that either x or y is less than -2 and the other is more?


Because if x and y are both = -2, then you would end up with case A (x=y) and NOT case B.
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New post 24 Sep 2015, 05:50
Hi Bunuel, Are these from GMATclub tests?

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New post 24 Sep 2015, 07:20

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New post 25 Sep 2015, 03:15
Guys,

Is it ok plugging in in data sufficiency?

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New post 25 Sep 2015, 20:18
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.

Answer: E.


For the option (1) (x-1)^2 <= 1 can I do something like this?

(x-1)^2 <=1
|x-1| <= 1 Insufficient because |x-| can be equal to 1 also.

Thanks!

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New post 26 Sep 2015, 12:34
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.


I have a doubt about this problema.
I´m clear that x+y=-4, but at the end you can have as you said that x and y have differnts signs but also to have the same for example:

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

So you can have x and y with differents signs or equals sings ans still be able to commit with the equation x+y = -4

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

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New post 26 Sep 2015, 13:43
alfonsecar 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.


I have a doubt about this problema.
I´m clear that x+y=-4, but at the end you can have as you said that x and y have differnts signs but also to have the same for example:

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

So you can have x and y with differents signs or equals sings ans still be able to commit with the equation x+y = -4



Your question is not clear. Where is it written that x and y can have same and different signs? Can you repost your question as it is difficult to understand what exactly are you trying to say.
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New post 04 Oct 2015, 12:37
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a


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

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Answer is C (Together, they can answer the initial question).

Am I wrong, or right?

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New post 04 Oct 2015, 13:34
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RafaelPina 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 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Answer is C (Together, they can answer the initial question).

Am I wrong, or right?


You are correct till \(x^2+y^2=5a\) but what if x=y=0 giving you a=0. In this case, \(x^2+y^2\) will be = 4a and NOT > 4a. This is the reason why E is the correct answer.

If you were given "is \(x^2+y^2 \geq 4a\) instead of just >4a", then yes, you would have marked C as the correct answer but you are asked >4a which may or may not be true.

Hope this helps.
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New post 04 Oct 2015, 13:56
mrslee wrote:
Guys,

Is it ok plugging in in data sufficiency?


Yes, but you need to recognize where plugging in will be required and where algebraic equations will work. This realization can only come from practice. Try to solve some of the DS questions that you have solve using algebra by plugging in certain values and see which one gives you correct answer in average time frame.
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Re: Inequality and absolute value questions from my collection   [#permalink] 04 Oct 2015, 13:56

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