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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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25 Sep 2016, 15:20

[quote="Bunuel"]9. Is n<0? (1) -n=|-n| (2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.[/quote

How does statement 1 imply that n is negative? Doesn't it mean that |-n| = negative (i.e. -n) and hence when n has a positive value, say 4, |-4|=-4 rather than |-(-4)| which would be 4 (and would not be = -4)?

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

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25 Sep 2016, 19:23

prvy wrote:

Bunuel wrote:

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

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

Answer: C.[/quote

How does statement 1 imply that n is negative? Doesn't it mean that |-n| = negative (i.e. -n) and hence when n has a positive value, say 4, |-4|=-4 rather than |-(-4)| which would be 4 (and would not be = -4)?

Please guide.

Please note that any number n denoted as |n| means only its magnitude from 0 so |-n| is also the magnitude from 0 is always equal to n and |-n| or |n| can only be thus >=0 so |-n|=n thus if we write |n-2| then it denotes for any number n having its magnitude in range 2 to the left side of n as well right side of n

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

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26 Sep 2016, 04:10

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.

I am sorry am asking a very basic question:

The 1st statement as you said is (1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

But cant it be:

As per the definition of modulus, |x| is -x when x<0. Then dosent the 1st statement translate to:

|x|+x= -x+(-)x = -2x [ Here as x is negative, therefore a negative sign (-)x ]

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

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27 Dec 2016, 05:11

[quote="Bunuel"]SOLUTIONS:

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 let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

OR:

\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B. This one was quite tricky and was solved incorrectly by all of you.

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here. From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3? That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

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 let's simplify given expression \(6*x*y = x^2*y + 9*y\):

\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.

Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.

(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.

OR:

\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.

(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

Answer: B. This one was quite tricky and was solved incorrectly by all of you.

Hi Bunuel,

I am a big fan of your posts and all your easy explanations, just want to point out a possible correction in your explanation of question 1 here. From the statement (ii) x^3<0, it is clear that x<0. But how did you arrive at the value of x=3 without combining statement (i) which gives two values of x=3,-3? That is the reason I think the answer is (C) when we get x=-3 using both statement (i) and statement (ii) and hence the value of xy.

Let me know if I am correct.

Thanks!

No, you are not correct.

From \(y*(x-3)^2=0\) it follows that either \(x=3\) or/and \(y=0\). (2) says that \(x^3<0\), thus x is not 3, therefore y must be 0 --> xy = 0.
_________________

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, thank you so much for such an amazing post, so so so helpful.

A quick query, regarding statement 2 here: (2) \(\frac{1}{|n|} > n\), Shouldn't it be true for all values of n such that n<1 (n#0) ?

Eg: n =1/2: \(\frac{1}{|(1/2)|} > \frac{1}{2}\) : \(2 > \frac{1}{2}\)

PS: It doesn't alter the final answer though.

Yes, but the fact that it's true for all negative n's was enough to discard this statement. So, we did not need to find the actual range. Still if interested here it is:

\(\frac{1}{| n |}>n\) --> 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 \(0<n<1\).
_________________

(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.

Option E is correct only because of Zero? If we put any integer value then LHS > RHS every time. But if x,y and z are Zero then the Q doesn't hold true. So, Zero is the only option that warrants E?

(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.

Option E is correct only because of Zero? If we put any integer value then LHS > RHS every time. But if x,y and z are Zero then the Q doesn't hold true. So, Zero is the only option that warrants E?

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.

While solving Statement-1, I did following: x(x-2)<=0 x<=0 and x-2<=0. so, x<=0 and x<=2. I always make this kind of solution based on understanding of solving linear equations. Ok! When I solved while typing this I recognised my error. If I put X as -ve in x(x-2) it will not be less than Zero. So now-onwards if I see x(x-4)<=0 I must consider x is between 0 and 4, both including. Right?

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

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12 Apr 2017, 17:52

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.

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

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12 Apr 2017, 19:05

rachitasetiya 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.

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

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25 Jul 2017, 13:48

Bunuel wrote:

8. a*b#0. Is |a|/|b|=a/b? (1) |a*b|=a*b (2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Answer: A.

Hi Bunuel, From 2, can't we say, |a|/|b|>=0, in this case, both a & B will be either +ve or -ve. Hence, sufficient? Pls pardon my ignorance.

8. a*b#0. Is |a|/|b|=a/b? (1) |a*b|=a*b (2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Answer: A.

Hi Bunuel, From 2, can't we say, |a|/|b|>=0, in this case, both a & B will be either +ve or -ve. Hence, sufficient? Pls pardon my ignorance.

Absolute value of a number is non-negative, so both |a| and |b| are >=0. So, naturally |a|/|b|>=0. But we cannot know whether a and b are positive or negative? Consider a = 1 and b = 2 OR a = -1 and b = 2.
_________________

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

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26 Jul 2017, 02:15

Bunuel wrote:

ManishKM1 wrote:

Bunuel wrote:

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

Absolute value of a number is non-negative, so both |a| and |b| are >=0. So, naturally |a|/|b|>=0. But we cannot know whether a and b are positive or negative? Consider a = 1 and b = 2 OR a = -1 and b = 2.

Thanks a lot Bunuel. Struggling with Absolute & Inequalities DS a lot, any tip will be a great help.

(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.

Bunuel,

Combined 1+2 statement together, I got a=0 if a=0, 1) (x+y)^2=9a (x+y)^2= 0

question is x^2+y^2>4a? the answer is no, because both side is 0

so I think the answer for the question is c, can you point out how wrong I am lol? Thanks

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