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Math Expert V
Joined: 02 Sep 2009
Posts: 60778
Re: Inequality and absolute value questions from my collection  [#permalink]

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ssshyam1995 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 cannot say for sure that 0<x<2 is true. Not sufficient.

If the combination range was 1<x<2, then will correct answer be option C??

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Absolutely. In that case we'd have an YES answer to the question.
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Joined: 27 Oct 2019
Posts: 54
Re: Inequality and absolute value questions from my collection  [#permalink]

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Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?

Notice that from $$y=|x|+x$$ it follows that y cannot be negative:
If $$x>0$$, then $$y=x+x=2x=2*positive=positive$$;
If $$x\leq{0}$$ (when x is negative or zero) then $$y=-x+x=0$$.

(1) $$x<0$$ --> $$y=|x|+x=-x+x=0$$. Sufficient.

(2) $$y<1$$. We found out above that y cannot be negative and we are given that y is an integer, hence $$y=0$$. Sufficient.

Why is it x<=0 and not x>=0??

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Joined: 27 Oct 2019
Posts: 54
Re: Inequality and absolute value questions from my collection  [#permalink]

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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.

Can we not solve the first statement showing positive and negative values when the mod is removed and hence prove the statement is not sufficient??

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Joined: 27 Oct 2019
Posts: 54
Re: Inequality and absolute value questions from my collection  [#permalink]

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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.

Shouldn't statement 2 be x>-2 and y<-2 to become true??

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Joined: 18 Nov 2018
Posts: 39
Re: Inequality and absolute value questions from my collection  [#permalink]

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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 tried using the number line to simulate A scenario and could get it something like below and this somehow looks fine.

--y/x-------- (-2)-----------

----------(-2)-------------x/y

I am not able to understand B scenario from the number line as we can have any distance between x and y. its just that the distance of x from -2 and y from -2 should be same and in opposite sides. How does the -4 come into the picture here ?

-------x<----------d------->(-2)<---------d-------->y---------

now this d here can be anything right ? Re: Inequality and absolute value questions from my collection   [#permalink] 27 Nov 2019, 01:37

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