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Inequality and absolute value questions from my collection

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Bunuel
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Inequality and absolute value questions from my collection [#permalink] 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: https://gmatclub.com/forum/inequality-a ... ml#p653690

2. If y is an integer and \(y = |x| + x\), is \(y = 0\)?
(1) \(x < 0\)
(2) \(y < 1\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653695

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

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653697

4. Are x and y both positive?
(1) \(2x-2y=1\)
(2) \(\frac{x}{y}>1\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653709
Graphic approach: https://gmatclub.com/forum/inequality-a ... l#p1269802

5. What is the value of y?
(1) \(3|x^2 -4| = y - 2\)
(2) \(|3 - y| = 11\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653731

6. If x and y are integer, is y > 0?
(1) \(x +1 > 0\)
(2) \(xy > 0\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653740

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

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653783 AND https://gmatclub.com/forum/inequality-a ... l#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: https://gmatclub.com/forum/inequality-an ... ml#p653789

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

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653792

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

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653796

11. Is \(|x+y|>|x-y|\)?
(1) \(|x| > |y|\)
(2) \(|x-y| < |x|\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653853

12. Is r=s?
(1) \(-s \leq r \leq s\)
(2) \(|r| \geq s\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653870

13. Is \(|x-1| < 1\)?
(1) \((x-1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)

Solution: https://gmatclub.com/forum/inequality-an ... ml#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.


PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   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.

Let me know if I am correct.

Thanks!
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Re: Inequality and absolute value questions from my collection [#permalink] New post   27 Dec 2016, 06:52
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naren01 wrote:
Bunuel wrote:
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.

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.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 Dec 2016, 07:50
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, 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.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 Dec 2016, 08:09
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RR88 wrote:
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, 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 \(0<n<1\).
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Re: Inequality and absolute value questions from my collection [#permalink] New post   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.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   25 Jul 2017, 20:55
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ManishKM1 wrote:
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.


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.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   26 Jul 2017, 21:01
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.


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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Re: Inequality and absolute value questions from my collection [#permalink] New post   26 Jul 2017, 21:11
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pclawong 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.


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


The question asks whether x^2 + y^2 > 4a. If x = y = a = 0, then the answer is NO but if this is not so then the answer is YES. Please read the whole thread. This was discussed many, many, many times before.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   28 Jul 2017, 08:15
lagomez wrote:
Bunuel wrote:
5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8
y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.



Answer: C.



just to chime in here your thanks for all this..it's really useful


hi, could you just explain how you came to y>= 2?
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Re: Inequality and absolute value questions from my collection [#permalink] New post   28 Jul 2017, 08:28
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achira wrote:
lagomez wrote:
Bunuel wrote:
5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8
y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.



Answer: C.



just to chime in here your thanks for all this..it's really useful


hi, could you just explain how you came to y>= 2?


\(3|x^2 -4| = y - 2\)

The left hand side of the equation (3|x^2 -4|) is an absolute value, which cannot be negative, so the right hand side also cannot be negative. Thus, \(y - 2 \geq 0\), which gives \(y \geq 2\).

Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   29 Jul 2017, 06:40
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 (2), how do you get x<-1?
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Re: Inequality and absolute value questions from my collection [#permalink] New post   29 Aug 2017, 09:25
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.


for (2) can't n be equal to 1/2 and the statement still holds true? Even though it doesn't change the answer.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   29 Aug 2017, 09:31
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goenkashreya wrote:
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.


for (2) can't n be equal to 1/2 and the statement still holds true? Even though it doesn't change the answer.


As you can see we don't really want the complete range for (2) to see that this statement is not sufficient, but still if interested:

1/|n| > n --> n*|n| < 1.

If n<0, then we'll have -n^2<1 --> n^2>-1. Which is true. So, n*|n| < 1 holds true for any negative value of n.
If n>0, then we'll have n^2<1 --> -1<n<1. So, n*|n| < 1 also holds true for 0<n<1.

Thus 1/|n| > n holds true if n<0 and 0<n<1.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 Aug 2017, 07:41
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.


Hi Bunuel,

I have a query for this question.
We are given that |x-1|<=1. Since modulus cannot be negative, can we write x-1>=0? So x>=1. This is prerequisite condition from question stem.
Now from statement 1, only x=2 will satisfy (x-1)^2<=1. because from question stem, x>=1.

So why can't be answer A?

Thanks in advance.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   30 Aug 2017, 10:34
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goalMBA1990 wrote:
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.


Hi Bunuel,

I have a query for this question.
We are given that |x-1|<=1. Since modulus cannot be negative, can we write x-1>=0? So x>=1. This is prerequisite condition from question stem.
Now from statement 1, only x=2 will satisfy (x-1)^2<=1. because from question stem, x>=1.

So why can't be answer A?

Thanks in advance.


No, this is not correct. Yes, |x| cannot be negative but x itself can be. If we are given say that |x| < 2, then it means that -2 < x < 2. For any x from this range |x| < 2.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   13 Oct 2017, 20:21
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.


Bunuel, Thank you for these awesome questions and solutions.
I have a question here : Don't we take x < 0 when |x| = -x , but in statement 1 , you have taken two possibilities : 0 and negative ? Is there a gap in my understanding ?
Please explain.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   14 Oct 2017, 01:42
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Manku 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.


Bunuel, Thank you for these awesome questions and solutions.
I have a question here : Don't we take x < 0 when |x| = -x , but in statement 1 , you have taken two possibilities : 0 and negative ? Is there a gap in my understanding ?
Please explain.


|0| = -0 = 0.

|x| = -x, when x <= 0.
|x| = x, when x >= 0.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   14 Oct 2017, 01:42
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Bunuel wrote:
Manku 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.


Bunuel, Thank you for these awesome questions and solutions.
I have a question here : Don't we take x < 0 when |x| = -x , but in statement 1 , you have taken two possibilities : 0 and negative ? Is there a gap in my understanding ?
Please explain.


|0| = -0 = 0.

|x| = -x, when x <= 0.
|x| = x, when x >= 0.


9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink] New post   16 Nov 2017, 23:23
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.


Hi Bunuel. When checking both A and B together, the intersection of the ranges is the value 2. For the value 2, we are sure that 1 is not less than 1. Hence shouldnt the answer be C?
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Re: Inequality and absolute value questions from my collection [#permalink] New post   16 Nov 2017, 23:29
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harshab wrote:
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.


Hi Bunuel. When checking both A and B together, the intersection of the ranges is the value 2. For the value 2, we are sure that 1 is not less than 1. Hence shouldnt the answer be C?


The question asks is 0<x<2 true?

When combining: if x = 1.5, then the answer is YES but if x = 2, then the answer is NO.
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Re: Inequality and absolute value questions from my collection [#permalink]
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