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Joined: 07 Jun 2017
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Re: Inequality and absolute value questions from my collection  [#permalink]

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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
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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.

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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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?
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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.

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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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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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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GMAT 1: 710 Q46 V41 GMAT 2: 700 Q48 V37 Re: Inequality and absolute value questions from my collection  [#permalink]

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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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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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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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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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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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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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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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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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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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Bunuel 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.

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.

Hope it's clear.

Could you explain what is the criteria for squaring when modulus involved? Can we do it with any modulus equations or it has to be done when certain things are in place.
Id like to get to that level on thinking, when in 2 mins i find out okay instead of plugging in values i should instead of square both sides. or vice versa.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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Bunuel wrote:
tihor wrote:
Bunuel, two questions:
shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2?
secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2?
does this not translate into x<=0 or x<=2?

thank you very much for all the questions and solutions.

Thank you very much for this catch. +1. There was a typo. So you are right with the first one:
|x-1| < 1 means 0<x<2. Already edited the post.

As for the second one:
x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.

x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.

Hope it helps.

Hi

I have the same question, could explain this again.

When we do this algebraically why does the range come out to be $$x<= 0$$ or $$x <=2$$

$$(x-1)^2 <= 1$$
$$(x^2 - 2x + 1) <=1$$
$$x^2 -2x <= 1-1$$
$$x^2 -2x<= 0$$
$$x(x-2) <= 0$$
$$x<=0$$ or $$x<=2$$

Where am i going wrong in this. It would be helpful if you could explain in detail.
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1
mtk10 wrote:
Bunuel wrote:
Could you explain what is the criteria for squaring when modulus involved? Can we do it with any modulus equations or it has to be done when certain things are in place.
Id like to get to that level on thinking, when in 2 mins i find out okay instead of plugging in values i should instead of square both sides. or vice versa.

You can always square if you have absolute values on both sides, so if you have |something| = |something|, then you can square. Squaring allows you to get rid of the modulus. Often times you are left with quadratics and it could be easier to solve.

If you have |something| = something, then squaring might give wrong solution(s). For example, |x - 1| = 2x - 1 --> x = 2/3 but if you square you get (x - 1)^2 = (2x - 1)^2, which gives x = 2/3 or x = 0.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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mtk10 wrote:
Bunuel wrote:
tihor wrote:
Bunuel, two questions:
shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2?
secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2?
does this not translate into x<=0 or x<=2?

thank you very much for all the questions and solutions.

Thank you very much for this catch. +1. There was a typo. So you are right with the first one:
|x-1| < 1 means 0<x<2. Already edited the post.

As for the second one:
x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.

x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.

Hope it helps.

Hi

I have the same question, could explain this again.

When we do this algebraically why does the range come out to be $$x<= 0$$ or $$x <=2$$

$$(x-1)^2 <= 1$$
$$(x^2 - 2x + 1) <=1$$
$$x^2 -2x <= 1-1$$
$$x^2 -2x<= 0$$
$$x(x-2) <= 0$$
$$x<=0$$ or $$x<=2$$

Where am i going wrong in this. It would be helpful if you could explain in detail.

First of all, $$x \leq 0$$ or $$x\leq 2$$ does not make any sense. What are the values in this case? Could x be 1 because you say $$x\leq 2$$ or it should be $$x \leq 0$$.

Next, if you solve the way you do then:

$$x(x-2) \leq 0$$ --> x and x - 2 have the opposite signs:

$$x \geq 0$$ and $$x -2 \leq 0$$ --> $$x \geq 0$$ and $$x \leq 2$$ --> $$0 \leq x \leq 2$$
$$x \leq 0$$ and $$x -2 \geq 0$$ --> $$x \leq 0$$ and $$x \geq 2$$ --> no intersection, so no solution in this case (x cannot be simultaneously less than 0 and more than 2).

Finally, the questions in this set are very tough and tricky. You should be very familiar at least with standard ways of solving inequality questions, modulus, quadraticsbefore attempting. Hope the links below help:

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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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.

Hi Bunuel, I solved it as below
(1) 2x-2y=1 -> x-y= 1/2. This means x>y by 1/2 but x can be 1/2 and y=0 so this is insufficient
(2) x/y=1 -> x>y which is no different from (1) so insufficient again

But I didn't follow your approach on combining the 2 statements, how did you get x to be substituted by 1/y? The fact that x is positive is already proved by (1) isn't it? Re: Inequality and absolute value questions from my collection   [#permalink] 05 Apr 2018, 08:37

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