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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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05 Oct 2015, 22:22

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.

Answer: D.

Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.
_________________

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.

Answer: D.

Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.

Your question is not clear... Also, notice that we are given that y is an integer, so it cannot be 0.5.
_________________

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.

Answer: D.

Hi Bunuel,

I do agree with option 1 but donot agree with option 2, as the x value can be positive value such as 0.25 etc.so the Y value shall be 0.50, in this case how do we deduce.

Statement 2: \(y<1\)

Since we know that \(y = |x| + x\)

case 1: x>0.... In this case y = 2x and will be positive case 2: x<0.... In this case y = 0 i.e. Y can never be Negative

This statement tells us that Y is an Integer less than 1 therefore 0 is the only possible value of y

Hence, SUFFICIENT

I hope this helps!
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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

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08 Oct 2015, 13:34

Bunuel wrote:

jayaddula 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.

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.

Hence E. I know I am going wrong some where, please help.

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.

Answer: D.

Hope it's clear.

Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

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

Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

I think this question requires more of observation about given information

|x+2|=|y+2| will be true for two possible cases of x and y

Case 1: when x = y Case 2: For Values like (x=1 and y=-5) or (x=2 and y=-6) or (x=3 and y=-7) ...etc.

Case 1 gives inconsistent answers because for each different value of x and y, x+ywill be different but Case 2 always gives a consistent value of x+y=-4 (Check all set of values mentioned above in Case 2)

Statement 1 suggests that x and y are not equal (for x and y equal, their product must be Non-Negative) i.e. Case 2 prevails which always gives us x+y=-4 i.e. SUFFICIENT

Statement 2 also rules out the scenario in which x and y may be equal i.e. Case 2 prevails again leading to a consistent value of x+y=-4 i.e. SUFFICIENT

Answer: Option D
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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

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10 Oct 2015, 07:36

GMATinsight wrote:

petocities wrote:

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

Hi @Buenel, i'm having a really hard time understanding this question. First, I don't understand why x=y should imply a unique answer for x+y. Same for the second stantement, I don't fully understand why having the equation x+y=-4 ensures a unique answer. Maybe I am missing some steps. Would greatly appreciate your help, or any1 else's help (maybe different approaches will help me understand better).

I think this question requires more of observation about given information

|x+2|=|y+2| will be true for two possible cases of x and y

Case 1: when x = y Case 2: For Values like (x=1 and y=-5) or (x=2 and y=-6) or (x=3 and y=-7) ...etc.

Case 1 gives inconsistent answers because for each different value of x and y, x+ywill be different but Case 2 always gives a consistent value of x+y=-4 (Check all set of values mentioned above in Case 2)

Statement 1 suggests that x and y are not equal (for x and y equal, their product must be Non-Negative) i.e. Case 2 prevails which always gives us x+y=-4 i.e. SUFFICIENT

Statement 2 also rules out the scenario in which x and y may be equal i.e. Case 2 prevails again leading to a consistent value of x+y=-4 i.e. SUFFICIENT

Answer: Option D

Thanks a lot Bunuel, it's crystal clear now. Greatly appreciate your help. Would you say this question is 600-700?
_________________

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

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18 Oct 2015, 13:18

Bunuel wrote:

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Hi Bunuel, I've solved this one correctly, but have one question. A is ok - no questions. Can we manipulate Statement 2 and say |n|*n<1 as |n| is always positive we must be able to do this - but |n|*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.
_________________

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Share some Kudos, if my posts help you. Thank you !

10. If n is not equal to 0, is |n| < 4 ? (1) n^2 > 16 (2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Hi Bunuel, I've solved this one correctly, but have one question. A is ok - no questions. Can we manipulate Statement 2 and say |n|*n<1 as |n| is always positive we must be able to do this - but |n|*n can be also positive as it's not stated that n must be an integer, let's say 1/2*1/2<1 and it can be also any negative value as stated above.

\(\frac{1}{| n |}>n\) --> multiply by \(|n|\) (we can safely do that since |n|>0): \(n*|n| < 1\).

If \(n>0\), then we'll have \(n^2<1\) --> \(-1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\). If \(n<0\), then we'll have \(-n^2<1\) --> \(n^2>-1\). Which is true for any n from the range we consider. So, \(n*|n| < 1\) holds true for any negative value of n.

Thus \(\frac{1}{| n |}>n\) holds true if \(n<0\) and \(0<n<1\).
_________________

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

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20 Oct 2015, 08:45

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.

Can you please tell me where I am going wrong:

(1)+(2)

x-y = 0.5 AND x>y 1.0 - 0.5 = 0.5 --- x>y and x,y --- both positive -1.5 - (-2) = 0.5 --- x>y and x,y --- both negative.

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.

Can you please tell me where I am going wrong:

(1)+(2)

x-y = 0.5 AND x>y 1.0 - 0.5 = 0.5 --- x>y and x,y --- both positive -1.5 - (-2) = 0.5 --- x>y and x,y --- both negative.

So my answer was E.

The red part above is your mistake. When you are given, x/y > 1 --> this may or may not mean that x>y. This only means that both x and y have the same sign.

Consider the 2 cases:

x/y > 1 ---> y=1, x=3 ---> x>y but x/y>1 ---> x= -3, y = -1 ---> x < y

This is where you are going wrong. Make sure to not multiple or divide by variables in DS questions when you dont know explicitly the signs of the variables.

(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.

I have a doubt regarding this one. The equation in the question is of a circle whose radius is \(2\sqrt{a}\). We should find if the point (x,y) is outside this region or not. Which is also equivalent to finding the distance of the point (x,y) from origin as the center of the given circle is the origin.

stmt-1: it represents a straight line \(y=-x+3\sqrt{a}\). Now, the nearest point on this line from origin will be at a distance equal to the perpendicular distance from origin to this line. so that distance is \(3\sqrt{a}/ \sqrt{2}\)..... and this is greater than \(2\sqrt{a}\). so no point represented by this line lies inside the circle. hence stmt 1 is sufficient.

stmt-2: it represents a straight line \(y=x-\sqrt{a}\). The perpendicular distance from origin to this line is \(\sqrt{a}/\sqrt{2}\) and is less than the radius of circle. Hence this line intersects circle at two points and this is not sufficient to say that every point on this line is outside the circle.

(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.

I have a doubt regarding this one. The equation in the question is of a circle whose radius is \(2\sqrt{a}\). We should find if the point (x,y) is outside this region or not. Which is also equivalent to finding the distance of the point (x,y) from origin as the center of the given circle is the origin.

stmt-1: it represents a straight line \(y=-x+3\sqrt{a}\). Now, the nearest point on this line from origin will be at a distance equal to the perpendicular distance from origin to this line. so that distance is \(3\sqrt{a}/ \sqrt{2}\)..... and this is greater than \(2\sqrt{a}\). so no point represented by this line lies inside the circle. hence stmt 1 is sufficient.

stmt-2: it represents a straight line \(y=x-\sqrt{a}\). The perpendicular distance from origin to this line is \(\sqrt{a}/\sqrt{2}\) and is less than the radius of circle. Hence this line intersects circle at two points and this is not sufficient to say that every point on this line is outside the circle.

therefore Ans: A

Please correct me if i am wrong.

Please read the whole thread. Lot of useful posts there including solutions for this problem.

For example, consider a case when x=y=0 for a NO answer and x=1, y=2 and a=1 for an YES answer.
_________________

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

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04 Nov 2015, 19:46

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.

How did you figure out that \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.

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.

How did you figure out that \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages: \(\frac{x}{y}>1\)

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

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05 Nov 2015, 14:34

Bunuel wrote:

Johnbreeden85 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.

How did you figure out that \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages: \(\frac{x}{y}>1\)

\(\frac{x}{y}-1>0\)

\(\frac{x}{y}-\frac{y}{y}>0\)

\(\frac{x-y}{y}>0\).

Hope it's clear.

Thank you. The only part I'm still unclear on is transitioning from \(\frac{x}{y}-1>0\) to \(\frac{x}{y}-\frac{y}{y}>0\) . I'm not sure how you would know that -1 is equal to \(\frac{y}{y}\) or what law/concept allows for the conversion of -1 to \(\frac{y}{y}\) . Thank you again.

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

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08 Nov 2015, 07:34

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.

Hi Bunuel,

This was my thought process, is it correct?

Stmt 1: -s<=r<=s => |r| <= s => Not sufficient since r can also be less than s Stmt 2: |r|>=s => Not sufficient since r can be greater than s

Combining: |r| <= s and |r| >= s --> |r| = s --> r = s or r = -s --> Not sufficient

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

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08 Nov 2015, 07:39

Johnbreeden85 wrote:

Bunuel wrote:

Johnbreeden85 wrote:

How did you figure out that \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) ? Any help is appreciated. Thank you.

This is explained couple of times on the previous pages: \(\frac{x}{y}>1\)

\(\frac{x}{y}-1>0\)

\(\frac{x}{y}-\frac{y}{y}>0\)

\(\frac{x-y}{y}>0\).

Hope it's clear.

Thank you. The only part I'm still unclear on is transitioning from \(\frac{x}{y}-1>0\) to \(\frac{x}{y}-\frac{y}{y}>0\) . I'm not sure how you would know that -1 is equal to \(\frac{y}{y}\) or what law/concept allows for the conversion of -1 to \(\frac{y}{y}\) . Thank you again.

Hi,

Forget about thinking how -1 is equal to \(\frac{y}{y}\) You must be aware of cross multiplication, so here it goes:

\(\frac{x}{y}-1>0\) cross multiply now to get a common denominator: \(\frac{x - y}{y}>0\)

Visualise it with numbers: \(\frac{9}{5}-1>0\) cross multiply now to get a common denominator: \(\frac{9 - 5}{5}>0\)

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