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# Inequality and absolute value questions from my collection

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Math Expert
Joined: 02 Sep 2009
Posts: 43789
Inequality and absolute value questions from my collection [#permalink]

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16 Nov 2009, 10:33
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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: http://gmatclub.com/forum/inequality-an ... ml#p653690

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

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

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

Solution: http://gmatclub.com/forum/inequality-an ... ml#p653709

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

Solution: http://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: http://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: http://gmatclub.com/forum/inequality-an ... ml#p653783 AND http://gmatclub.com/forum/inequality-an ... 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: http://gmatclub.com/forum/inequality-an ... ml#p653789

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

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

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

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

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

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

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

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

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

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Posts: 43789
Re: Inequality and absolute value questions from my collection [#permalink]

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26 Jul 2017, 20:11
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.

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]

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28 Jul 2017, 07:15
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.

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?
Math Expert
Joined: 02 Sep 2009
Posts: 43789
Re: Inequality and absolute value questions from my collection [#permalink]

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28 Jul 2017, 07:28
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.

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]

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29 Jul 2017, 05:40
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.

For (2), how do you get x<-1?
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Joined: 02 Sep 2009
Posts: 43789
Re: Inequality and absolute value questions from my collection [#permalink]

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29 Jul 2017, 07:20
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Expert's post
pclawong 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.

For (2), how do you get x<-1?

x^2 - 1 > 0

x^2 > 1

|x| > 1

x < -1 or x > 1.
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Re: Inequality and absolute value questions from my collection [#permalink]

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30 Jul 2017, 06:52
Bunuel wrote:
pclawong 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.

For (2), how do you get x<-1?

x^2 - 1 > 0

x^2 > 1

|x| > 1

x < -1 or x > 1.

That is clear to me. Thank you!
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Re: Inequality and absolute value questions from my collection [#permalink]

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02 Aug 2017, 08:19
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.

Hi Bunuel,

I did this question exactly the way you explained. And mathematically it makes perfect sense.

Now when I am trying to visualize the situation when we combine (1) and (2), I get a bit confused.

The question asks --> Is the distance of x from 1 smaller than 1 unit? Combining, (1) and (2), my answer is YES. Because 1<x<=2 dictates so. The maximum distance of x from 1 could be 0.999999....... but never 1!

I feel a bit stupid because I know I am missing something out. Could you please help me?
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Re: Inequality and absolute value questions from my collection [#permalink]

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29 Aug 2017, 08:25
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.

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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Posts: 43789
Re: Inequality and absolute value questions from my collection [#permalink]

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29 Aug 2017, 08:31
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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.

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]

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29 Aug 2017, 08:35
Bunuel wrote:
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.

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.

Thank you for the quick response.
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Re: Inequality and absolute value questions from my collection [#permalink]

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30 Aug 2017, 06:41
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.

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?

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Posts: 43789
Re: Inequality and absolute value questions from my collection [#permalink]

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30 Aug 2017, 09:34
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.

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?

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]

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13 Oct 2017, 19:21
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.

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

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14 Oct 2017, 00:42
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.

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 ?

|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]

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14 Oct 2017, 00:42
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.

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 ?

|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]

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14 Oct 2017, 05:13
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.

What if we deduct 2 from 1? We will get a value of xy and substitute in 1 and check whether x^2 + y^2 > 4a? then the answer would be C?
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Re: Inequality and absolute value questions from my collection [#permalink]

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14 Oct 2017, 05:22
tejas0999 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.

What if we deduct 2 from 1? We will get a value of xy and substitute in 1 and check whether x^2 + y^2 > 4a? then the answer would be C?

I cannot follow what are you trying to say but the answer is E, not matter how you approach it.
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Re: Inequality and absolute value questions from my collection [#permalink]

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16 Nov 2017, 22:23
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.

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]

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16 Nov 2017, 22:29
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.

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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21 Nov 2017, 00:27
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.

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

Now instead of adding, if i reduce (2) from (1),or, just equate the two, => 9a-2xy = a+2xy. => 4xy = 8a or 2xy = 4a.

Substituting 2xy = 4a in (1) or (2), => x^2 +y^2=5a
Hence (C) X^2+Y^2 > 4a

Kindly check my solution.
Re: Inequality and absolute value questions from my collection   [#permalink] 21 Nov 2017, 00:27

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