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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) \(x^3< 0\)

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

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27 Jun 2013, 20:01

Is x^2 + y^2 > 4a?

I ran into trouble when I instinctively got the absolute value of x^2 + y^2 > 4a ===> |x| + |y| > 4a and did the same for number one and two. We need to find the values of x, y and a to solve for this problem. if x = 10 and y = 20 and a = 2 then the inequality would hold true but if x = 0 and y = 0 and a = 0 then it wouldn't. I'm still having trouble figuring out why this problem is solved the way it is, as opposed to the way I outline below:

|x| + |y| > √4a x + y > 4a OR x + -y > 4a

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

|x + y| = √9a x + y = √9a y - x = √9a

However, this method is taking too long so I will solve the way Bunuel did (though I still don't know how I would know to solve it this way on the test!

(1) (x + y)^2 = 9a x^2 + 2xy + y^2 = 9a This tells us nothing about x, y and a INSUFFICIENT

(2) (x – y)^2 = a x^2 - 2xy + y^2 = a Again, this tells us nothing about x, y and a INSUFFICIENT

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

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28 Jun 2013, 07:08

6. If x and y are integer, is y > 0?

(1) x +1 > 0

Tells us nothing about the value of y. INSUFFICIENT

(2) xy > 0

x and y must both be positive or negative. We don't know if y is positive or negative. INSUFFICIENT

x + 1 > 0 xy > 0

x + 1 must be > 0. x cannot be negative as -x + 1 wouldn't be greater than zero. Therefore, x is an integer ≥ 0. xy must be > 0 which means xy = (+)*(+) or (-)*(-) but neither x or y can equal zero. So, if x and y cannot = zero then x must be an integer > 0. If x is positive then y also must be positive for xy > 0.

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

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28 Jun 2013, 13:58

7. |x+2|=|y+2| what is the value of x+y?

I. (x+2) = (y+2) ===> x-y = 0 (x = y) OR II. (x+2) = -y-2 ===> x + y = -4

(1) xy<0 if xy < 0 then x,y ≠ 0 and either x OR y is negative.

This is a VERY tricky problem!

To start, let's look at just |x+2|=|y+2|. #1) tells us that either x OR y is negative and that neither = 0. Of course, |x+2| must = |y+2| so we can try a few different values for x and y.

If we were told that xy>0, we wouldn't be able to solve. Now, back to the stem:

(x+2) = (y+2) ===> x-y = 0 (x = y) OR (x+2) = -y-2 ===> x + y = -4

x and y don't = 0. In scenario 1 we have x-y=0 or (x=y) #1) says that xy<0 meaning either x or y is negative and therefore x ≠ y.

(2) x>2 y<2

|x+2|=|y+2| tells us that x+2 must equal y+2. #2.) rules out the possibility of scenario #1) which means the only other possibility (|x+2|=|y+2| is stated as fact so one of the scenarios must hold true) is that x+y=-4. This can be confirmed by plugging in x values greater than 2 and y values less than -2 to get 0x+y=-4. SUFFICIENT

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

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28 Jun 2013, 16:18

8. a*b # 0. Is |a|/|b| = a/b?

a,b ≠ 0 |a|/|b| will always be positive are a,b BOTH positive or BOTH negative?

(1) |a*b|=a*b

|a*b|=a*b

a*b MUST be positive as it is equal to an absolute value. The only way for a*b to be positive is if a and b are BOTH positive or BOTH negative. SUFFICIENT.

(2) |a|/|b|=|a/b| We could have positive or negative values for a and b and get correct results:

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

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28 Jun 2013, 16:23

9. Is n<0?

(1) -n=|-n| if -n = |-n| then -n must be positive as it is equal to an absolute value. Therefore, -n = |-n| ===> -(-n) = |-n| N CAN BE ≥0 AS -(-0) = |-0| ===> 0=0! INSUFFICIENT

(2) n^2=16 |n| = 4 n=4 OR n=-4 N could be less than zero or greater than zero. INSUFFICIENT

1+2) #1 tells us that N is ≥0. #2 tells us that n could be 4 or -4. The only value of n that is ≥0 is 4. SUFFICIENT

There are many different ways |r|>=s holds true with different combinations of +/- r,s. INSUFFICIENT

-s≤r≤s ===> this means that s≥0. For example, s couldn't be negative for the following reason:

-(-s)≤r≤(-s) s≤r≤-s (not valid)

~ Obviously a positive number cannot be less than a negative number. ~ r can be equal to -s or s or it can be in between the range of -s and s.

|r|≥s

The absolute value of r may be greater than or equal to s, but r could be negative. If s was 5 r = -5 in which case: |r|≥s |-5|≥5 5≥5 In this case, r = -5 and s = 5, on the other hand |5|≥5 5≥5 In this case r = 5 and s = -5

r is somewhere between - s and s or it could = s or negative s. The absolute value of r is greater than or equal to s (positive s) so let's say r = -5 and s = 5. The absolute value or r would equal s and would satisfy the inequality however, r wouldn't = s as -5 ≠ 5. On the other hand, r could =5 and s could = 5. |5|≥5 is valid and r=s ans 5=5.

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

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29 Jun 2013, 11:35

13. Is |x-1| < 1?

Is 0 < x < 2

(1) (x-1)^2 <= 1

|x-1| ≤ 1 x-1 ≤ 1 ===> x≤2 OR -x+1 ≤ 1 x ≥ 0

0≤x≤2

The stem asks if |x-1| is less than one. #1 says it could be less than OR equal to 1. Furthermore, #1 says that 0≤x≤2 but the question asks if 0<x<2. INSUFFICIENT

(2) x^2 - 1 > 0 x^2 > 1 |x| > 1 x>1 OR x<-1

If x = 1.5 then |x-1| < 1 = |.5| < 1 Valid OR if x =1 4 then |x-1| < 1 = |4| < 1 Invalid INSUFFICIENT

1+2) let's say x =2, then: |2-1| ≤ 1 ==> |1|<1 but we're looking to see if |x-1| < 1 and in that case it is not, or x = 1.1 in which case |1.1-1| < 1 ===> .1 < 1

Another way to look at it is to examine the intersection of the ranges. From #1 the range is 0≤x≤2 and from #2 the range is x>1 or x<-1 so the combined range is 1<x≤2. In the stem x must be between 0 and 2 but 1+2 combined allows for the possibility of x = 2. INSUFFICIENT

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

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03 Jul 2013, 21:02

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.

Option 2 would mean that |x+2 | is == x + 2 but for -2<=y<2 |y+2| is == y + 2

==> x + 2 = y + 2 ==> x = y Only for y < -2 would |y+2| be == -y - 2 which would give x + y = -4 i.e. INSUFFICIENT Please clarify.

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.

Option 2 would mean that |x+2 | is == x + 2 but for -2<=y<2 |y+2| is == y + 2

==> x + 2 = y + 2 ==> x = y Only for y < -2 would |y+2| be == -y - 2 which would give x + y = -4 i.e. INSUFFICIENT Please clarify.

Notice that x>2 and y<2 means that \(x\neq{y}\). Now, in case of \(-2\leq{x}<2\) you get x=y which is contradicts \(x\neq{y}\), thus \(-2\leq{x}<2\) case is not possible (valid).

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

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06 Jul 2013, 07:09

Hi Bunuel,

Thanks for the excellent collection of questions and good to see this thread is active for more than 3 years

Please can you help me out with the below:

For (1), I started like this:

6*x*y = x^2*y + 9*y as the question is asking for xy, I converted the above eqn to ==> x*y = (x^2*y + 9*y)/6 and then tried finding the value of xy from the two statements, and ended up selecting (e) as my answer choice. If I had converted the equation to (x-3)^2*Y = 0, I am sure I would have done it right.

For (5) as the question is asking for y, I converted the equation in question to y=3x^2-10 or y=14-3x^2. then tried finding the value of Y and chose (e) as my answer as I was getting two diff answers. Again, if I had thought that Y>=2 as 3|x^2-4| will always be >=0 from the question stem, I would have got it right.

I am doing this mistake most often in inequalities and also in other topics . Please can you tell me what I am missing here. what parameters should I consider before converting the equation in the question, so that it helps to find the answer easily.

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

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09 Jul 2013, 20:54

Can someone pls help me understand the flaw in my reasoning for qn 3 ? :

(1) It is given that \((x+y)^2 =9a\) or \(x^2+y^2+2xy=9a\). Let us assume that \(x^2+y^2 < 4a\), then from (1) it follows that \(2xy >5a\) which means that \(2xy > x^2+y^2\) which reduces to \((x-y)^2 < 0\) which is not possible , thus \(x^2+y^2 > 4a\). Sufficient.

(2) Let us assume that \(x^2+y^2 < 4a\) , unlike above , this gives us no further info on the relation betn \(x^2+y^2\) and \(2xy\). Insufficient. Answer is (A).

Can someone pls help me understand the flaw in my reasoning for qn 3 ? :

(1) It is given that \((x+y)^2 =9a\) or \(x^2+y^2+2xy=9a\). Let us assume that \(x^2+y^2 < 4a\), then from (1) it follows that \(2xy >5a\) which means that \(2xy > x^2+y^2\) which reduces to \((x-y)^2 < 0\) which is not possible , thus \(x^2+y^2 > 4a\). Sufficient.

(2) Let us assume that \(x^2+y^2 < 4a\) , unlike above , this gives us no further info on the relation betn \(x^2+y^2\) and \(2xy\). Insufficient. Answer is (A).

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

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10 Jul 2013, 17:38

Hi Bunuel,

Thanks for the excellent collection of questions and good to see this thread is active for more than 3 years

Please can you help me out with the below:

For (1), I started like this:

6*x*y = x^2*y + 9*y as the question is asking for xy, I converted the above eqn to ==> x*y = (x^2*y + 9*y)/6 and then tried finding the value of xy from the two statements, and ended up selecting (e) as my answer choice. If I had converted the equation to (x-3)^2*Y = 0, I am sure I would have done it right.

For (5) as the question is asking for y, I converted the equation in question to y=3x^2-10 or y=14-3x^2. then tried finding the value of Y and chose (e) as my answer as I was getting two diff answers. Again, if I had thought that Y>=2 as 3|x^2-4| will always be >=0 from the question stem, I would have got it right.

I am doing this mistake most often in inequalities and also in other topics . Please can you tell me what I am missing here. what parameters should I consider before converting the equation in the question, so that it helps to find the answer easily.

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

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11 Jul 2013, 20:42

1

This post received KUDOS

Hi johncoffey , My two cents- for (1) - it is always useful to start out by factoring an expression if possible, especially when there is a variable in common ("y" in this example). Even though it does make sense to isolate the expression "xy" that we are being asked for- note that in this case that would give us more unknowns on the RHS. Hope tht helps.

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

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18 Jul 2013, 21:25

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

Hi Bunuel, Why wouldn't the answer be 'C'. I inferred from the question that we just need to prove X^2 + y^2 > 4a, which we deduced by adding the equations. Please clarify why we are considering the values x,y or a can hold? Thanks, Amit

Hi Bunuel, Why wouldn't the answer be 'C'. I inferred from the question that we just need to prove X^2 + y^2 > 4a, which we deduced by adding the equations. Please clarify why we are considering the values x,y or a can hold? Thanks, Amit

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