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# Inequality and absolute value questions from my collection

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

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04 Oct 2012, 04:00
1
carcass wrote:
Bunuel I have very clear the 1 statement but not so much the secon one : 1/|n|< n this is true only for negative value. So we could have i. e. : -1 or -6 so insuff but how we you arrive to this conclusion:

Given: 1/|n| > n. Now, 1/|n| is always positive (because of the absolute value in the denominator), so if n is negative then this inequality will always hold true: 1/|n|=positive>n=negative.

Hope it's clear.

P.S. Complete solution for 1/|n| > n is n<0 or 0<n<1.
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04 Oct 2012, 04:20
1
carcass wrote:
Correct me if I'm wrong because the concepts are always the same but the gmat blonds them and as consequence blow your mind.

in other words, you are saying

1/|n| > n 2 cases

1/-n > n ----> n^2 > -1 this implies that any squared number is positive and therefore greater than -1, all negative n values work as solutions. your n < 0

1/n > n -----> 1 > n^2 ----> n^ 2 < 1 --------> -1 < n < 1 . your second range. so in the end we have all this information and we are not sure of course of - 4 < n < 4.

Correct ???

Thanks

1/|n| > n --> 2 cases:

If n<0, then |n|=-n, so we'll have that 1/-n>n --> multiply by n and flip the sign (since we consider negative n): -1<n^2 --> which holds true for any n from this range, so for any negative n.

If n>0, then |n|=n, so we'll have that 1/n>n --> multiply by positive n, this time: 1>n^2 --> -1<n<1, since we consider n>0, then finally we'll get 0<n<1.

So, 1/|n| > n holds true for n<0 and 0<n<1.

Hope it's clear.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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22 Feb 2013, 01:27
1
JJ2014 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.

Bunuel, I think I need some conceptual help. Why should we not solve statement 1 by rewriting the two statements and then adding them together? (Besides the fact that it's time consuming....) I rewrote them and found 3x^2 -10 = y for the positive absolute vlaue, and -3x^2+14=y for the negative abs value. From this, I added them together and got y=4..

Can you please explain what I'm getting wrong conceptually? Thanks so much!!!! I appreciate your kindness.

|x^2-4|=x^2-4 when x^2-4>0;
|x^2-4|=-(x^2-4) when x^2-4<=0.

So, the two equations you'll get from the original are relevant for different ranges of x. Hence, you cannot consider them as two separate equations and solve.

To put it simply: we cannot get the single value of y from 3|x^2 -4| = y - 2. Consider y=2 and x=2 OR y=11 and x=1.

Hope it's clear.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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22 Sep 2013, 06:00
1
StormedBrain 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.

Hi Bunuel ,

I know saying (1/|n|) < n will be true for all n<0 is quite clear logically. Still I want to reach this conclusion mathematically.

I got swayed solving for n|n| < 1 .

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.

Does this make sense?
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Re: Inequality and absolute value questions from my collection  [#permalink]

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02 Jun 2015, 03:30
1
EgmatQuantExpert wrote:
camlan1990 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.

Hi Bruel,

For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient?

Could you help me find out whether there is any mistake in my solution?
Thanks Bruel,

Dear camlan1990

The highlighted part in your solution above is wrong.

The correct expansion for $$(x+y)^2 = x^2 + y^2 + 2xy$$.

You on the other hand have wrongly written: $$(x+y)^2 = 2(x^2 + y^2)$$.

Hope this helped.

Best Regards

Japinder

Dear Japinder,

As I highlighted, (x+y)^2 is smaller or equal 2(x^2 + y^2)
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Re: Inequality and absolute value questions from my collection  [#permalink]

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03 Jun 2015, 07:14
1
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 tried substituting values for x and y and seem to be getting a different ans.

If we take x=-1 and y=-1.5 the ans to the question is NO, whereas if we take x=1.5 & y=1 the ans is YES. Since we are unable to ans the question,

shouldn't the ans be E? if my reasoning is flawed can you please point out the flaw.. Thanks in advance
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Re: Inequality and absolute value questions from my collection  [#permalink]

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04 Oct 2015, 13:34
1
RafaelPina 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 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

(2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a +2xy

So we can conclude that:
9a - 2xy = a + 2xy
8a = 4xy
4a = 2xy

Hence, in the first conclusion is:
x^2 + y^2 = 9a - 4a
x^2 + y^2 = 5a

Am I wrong, or right?

You are correct till $$x^2+y^2=5a$$ but what if x=y=0 giving you a=0. In this case, $$x^2+y^2$$ will be = 4a and NOT > 4a. This is the reason why E is the correct answer.

If you were given "is $$x^2+y^2 \geq 4a$$ instead of just >4a", then yes, you would have marked C as the correct answer but you are asked >4a which may or may not be true.

Hope this helps.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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05 Nov 2015, 01:28
1
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.

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

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23 Feb 2016, 00:55
1
nishantdoshi wrote:
Bunuel wrote:
nishantdoshi wrote:

hey bunuel
can you please clear my doubt?
in statement 1 you've written either n is negative OR n equals to zero but as per my knowledge shouldn't n be negative only because I've read it in many post that are on absolute value, here's a link: math-absolute-value-modulus-86462.html

correct me if i'm wrong!

An absolute value cannot be negative but it CAN be 0. For this particular case 0 fits:
-n=|-n| --> -0 = |-0| --> 0 = 0.

if x>=0 then |x|=x
and if x<0 then |x|=-x

am i wrong?

Yes, you are wrong.

We can say that when x<=0, then |x| is also equal to -x:

|0| = -0.
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29 Jul 2017, 08:20
1
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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27 Dec 2017, 21:25
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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25 Oct 2019, 00:37
1
Ajeet97 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.

If I were to substitute values instead, to check if C is correct

The first equation can be written as x-y = 0.5

Using second equation, I know x is greater than y.

If I substitute x = 2.5 and y = 2, both my values are positive and x-y = 0.5

However, if I use x = -2 and y = -2.5, then x - y = -2 - (-2.5) = 0.5 is also true.

In this case, both x and y are negative. So I thought the answer should be E.

$$\frac{x}{y}>1$$ does not mean that $$x>y$$. If both x and y are positive, then $$x>y$$, BUT if both are negative, then $$x<y$$.

From (2) $$\frac{x}{y}>1$$, we can only deduce that x and y have the same sigh (either both positive or both negative).[/quote]

If you check, you'll see that x = -2 and y = -2.5 do not satisfy x/y > 1.

Check more solutions of this question here: https://gmatclub.com/forum/are-x-and-y- ... 63377.html
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Re: Inequality and absolute value questions from my collection  [#permalink]

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16 Nov 2009, 13:08
Bunuel wrote:

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

I'm getting B for this one

1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient

2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient
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16 Nov 2009, 13:19
Bunuel wrote:

12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

I'm getting c

1. s can be 3 and r can be 3 which makes question yes
s can be 3 and r can be 2 which makes question no
insufficient

2. r can be 3 and s can be 3 makes question yes
r can be 3 s can be 2 makes question no
insufficient

combining:
|r|>=s means
r>=s or r<=-s

and -s<=r<=s means
-s<=r and r<=s

now we have -s<=r and -s>=r so -s = r or s = r
r>=s and r<=s so s = r
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16 Nov 2009, 15:33
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

because in number 2 n can be negative or a fraction
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Re: Inequality and absolute value questions from my collection  [#permalink]

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16 Nov 2009, 20:07
1
Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.
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Re: Inequality and absolute value questions from my collection  [#permalink]

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16 Nov 2009, 21:39
Quality questions as always... Thanks Bunuel! +1
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Re: Inequality and absolute value questions from my collection  [#permalink]

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16 Nov 2009, 22:46
lagomez wrote:
Bunuel wrote:

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

I'm getting B for this one

1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient

2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient

(1) (x-1)^2 <= 1
x is 0 to 2.
If x = 2, yes.
If x < 2, No.

(2) x^2 - 1 > 0
x cannot be -1 to 1 i.e. x<-1 or x>1. NSF.

From 1 and 2: x is >1 but <=2. NSF..

E.
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17 Nov 2009, 05:18
Bunuel wrote:
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 I reduced the given equation (divided out the y) and solved for x:
6*x*y = x^2*y + 9*y
6*x = x^2 + 9
0 = x^2 - 6*x + 9
0 = (x-3)^2
x = 3

Statement 1:

y-x=3
y-3=3
y=6
xy=3*6=18

SUFFICIENT

Statement 2:

x^3<0

We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?

INSUFFICIENT

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17 Nov 2009, 05:34
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Another way of looking at the problem is to ask, is x<0? Because if it is, then we know that y is zero. The only case in which y will not be zero is if x is positive.

Statement 1:

SUFFICIENT

Statement 2:

y<1

Because y is an integer, it must be one of the following values: 0, -1, -2, -3...

BUT |x| + x can never be a negative value. The lowest value that it can be is 0.

Hence, y can never be negative and the only possible value it can be then is 0.

SUFFICIENT

Re: Inequality and absolute value questions from my collection   [#permalink] 17 Nov 2009, 05:34

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