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Inequality and absolute value questions from my collection

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Inequality and absolute value questions from my collection [#permalink] New post 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: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747

8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

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

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 21 Jun 2012, 01:21
Got it. Thanks Bunuel.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 24 Jun 2012, 12:45
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.

Answer: C.



I solved it this way. I hope i am assuming it correct

is x=y= +ve = ?

(A) 2x-2y=1

y=x-1/2 equation of line,

So, y=0 then x=1/2
x=0 then y=-1/2

thus, the line passes through points (1/2,0) & (0,-1/2) i.e - I, III & IV quadrant

(+,+) or (-,-) or (+,-) values of x & y

Not sufficient infromation

(B)

x/y > 1
if y>0 x>y
if y<0 x<y
not sufficient infomration

(C)


combine (A) and (B)

if y>0 i.e line passes only through Ist quadrant & x>y because y= x - 1/2

x>y ==> therefore x=y=+ve satisfies only first quadrant


condition y<0 from (B) is not true for x<y because

when line pass through IV the quadrant x is +ve and y is -ve, so we can ignore this condition as it doesnot satisfy (A) and (B) combined

Therefore left with only y>0 and x>y where line passes thr' Ist quadrant: C
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Re: Inequality and absolute value questions from my collection [#permalink] New post 24 Jun 2012, 17:41
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.

Answer: D.



Vote for D [Need help in understanding part (B) below]

I solved it this way. I really find it difficult to open mod sign, I hope i am doing it correctly

(A) xy < 0

So we have (x = +ve and y = -ve) or (x = -ve and y = +ve)

|x + 2| = |y + 2|

critical point (x > -2 & y < -2) or (x < -2 & y > -2)

if (x > -2 & y < -2)
then

(x + 2) = -(y + 2)
x + 2 = -y -2
x + y = -4

sufficient

(B) x > 2 and y < 2

when y < -2
(x + 2) = - (y + 2)
x + y = -4

when 2 > y > -2 we get x - y = 0 - Cannot be true statement [Not sure if this is correct. Need help here]

and considered this as sufficient
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Re: Inequality and absolute value questions from my collection [#permalink] New post 24 Jun 2012, 18:48
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.



Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???

Here is what i did

Question: |x-1| < 1
critical point x>1 or x<1

when x>1 then (x-1)<1 x<2
when x<1 then -(x-1)<1 -x<0 therefore x>0
to prove 0<x<2 --? [understood this]

(A)(x-1)^2 <= 1
x^2 - 2x + 1 <= 1
x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]

my thoughts [what m i doing wrong??]
when (x = 0) then x-2<=0 therefoe x<=2
when (x-2 = 0) then x<=0 ????? i am confused here ?????


(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]

Thank you
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Re: Inequality and absolute value questions from my collection [#permalink] New post 25 Jun 2012, 06:33
Bunuel wrote:
kuttingchai 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.



Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???

Here is what i did

Question: |x-1| < 1
critical point x>1 or x<1

when x>1 then (x-1)<1 x<2
when x<1 then -(x-1)<1 -x<0 therefore x>0
to prove 0<x<2 --? [understood this]

(A)(x-1)^2 <= 1
x^2 - 2x + 1 <= 1
x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]

my thoughts [what m i doing wrong??]
when (x = 0) then x-2<=0 therefoe x<=2
when (x-2 = 0) then x<=0 ????? i am confused here ?????


(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]

Thank you


Check the following links:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535

Hope it helps.


Thank you Bunuel for the link below that helped : if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html

So for x (x-2) <= 0
first check for point of intersection on x-axis

x=0 and x=2
then as equation is <= 0 points between (0,2) are below x axis and so we have 0<=x<=2
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Re: Inequality and absolute value questions from my collection [#permalink] New post 09 Aug 2012, 16:06
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.

Answer: D.


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
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Re: Inequality and absolute value questions from my collection [#permalink] New post 10 Aug 2012, 07:26
Bunuel wrote:
jayaddula 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.

Answer: D.


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.

Answer: D.

Hope it's clear.



Thanks Bunuel.
Its a shame that i did all the work and completely misread the question.

thanks
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Re: Inequality and absolute value questions from my collection [#permalink] New post 26 Aug 2012, 00:23
dvinoth86 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.



The answer to this one is C right? B alone is not sufficient.


Don't forget that y*(x-3)^2=0.
B alone is definitely sufficient.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 05 Sep 2012, 02:04
Expert's post
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.

Answer: C.


I do not wanna to be wrong but when we substitute x in x-y/y>0 we should not have 1/2y>0 ??? and not 1/y>0.......then the meaning that y is positive eitherway not change.

Let me know. Thanks
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Re: Inequality and absolute value questions from my collection [#permalink] New post 05 Sep 2012, 03:53
Expert's post
carcass 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.

Answer: C.


I do not wanna to be wrong but when we substitute x in x-y/y>0 we should not have 1/2y>0 ??? and not 1/y>0.......then the meaning that y is positive eitherway not change.

Let me know. Thanks


Check this post: inequality-and-absolute-value-questions-from-my-collection-86939-60.html#p666191

"I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them.

Hope it's clear. "
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Re: Inequality and absolute value questions from my collection [#permalink] New post 03 Oct 2012, 04:39
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!
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Re: Inequality and absolute value questions from my collection [#permalink] New post 03 Oct 2012, 06:13
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liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 03 Oct 2012, 11:26
Quote:
liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.


Great.. I get it now.. Thanks Bunuel.

I am also stuck at Q10 :

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

We need to see if |n|<4 (this means -4<n<4)

1) n^2>16 => n<-4 and n>4
So from n<-4, |n|<|-4| = |n|<4 (works)
But n>4 does not work.
Doesn't that make 1) Insufficient?

Could you please tell me what I am doing wrong here ??
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Re: Inequality and absolute value questions from my collection [#permalink] New post 04 Oct 2012, 02:39
Expert's post
liarish wrote:
Quote:
liarish wrote:
Hi Bunuel,
I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

1) Insufficient . First reduced equation to x-y=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 3-2.5=0.5 works.. then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5. works again. So x and y can be both +ve and -ve. So 1) is Insufficient.
2) x/y>1. Just tells us that both x and y have same sign. both are -ve or both are +ve. So Insufficient.

Now Combining,
Picking the same values used in 1) x=3, y=2.5. both signs positive and x-y=0.5. works
then chose x= -1, y=-1.5 => -1 - (-1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or -ve. So my answer is E.

Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!


If x= -1 and y=-1.5, then x/y=2/3<1, so these values don't satisfy the second statement.

This question is also discussed here: are-x-and-y-both-positive-1-2x-2y-1-2-x-y-93964.html

Hope it helps.


Great.. I get it now.. Thanks Bunuel.

I am also stuck at Q10 :

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

We need to see if |n|<4 (this means -4<n<4)

1) n^2>16 => n<-4 and n>4
So from n<-4, |n|<|-4| = |n|<4 (works)
But n>4 does not work.
Doesn't that make 1) Insufficient?

Could you please tell me what I am doing wrong here ??


If n<-4, then n, for example can be -4.5 --> |-4.5|=4.5>4, so |n|<4 doesn't hold true.

If n is not equal to 0, is |n| < 4 ?

Question basically asks whether -4<n<4, so whether n is some number from this range.

(1) n^2>16. This implies that either n>4 or n<-4. No number from these ranges is between -4 and 4, thus the answer to the question whether -4<n<4 is NO. Since we have a definite answer then this statement is sufficient.

Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 04 Oct 2012, 02:53
Expert's post

Last edited by carcass on 04 Oct 2012, 03:03, edited 2 times in total.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 04 Oct 2012, 03:13
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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 those information and we are not sure of course of - 4 < n < 4.

Correct ???

Thanks
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Re: Inequality and absolute value questions from my collection [#permalink] New post 04 Oct 2012, 03:36
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Re: Inequality and absolute value questions from my collection [#permalink] New post 11 Oct 2012, 02:11
Bunuel wrote:
11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2
OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.


Hi Bunuel - Can this solved in the below way?

Is |x+y|>|x-y|?

Since both sides are +ve we can square both side of the inequality....
On squaring we get xy>0?

statement 1

(1) |x| > |y|

This is NS as xy can be opp sign as well as same sign

(2) |x-y| < |x|

Squaring on both sides we get y^2 < 2xy
Y cannot be zero otherwise the inequality cannot hold so Y^2 is +ve hence xy is +ve
So we can answer the question xy>0

Is this fine Bunuel?
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Re: Inequality and absolute value questions from my collection [#permalink] New post 11 Oct 2012, 02:22
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Jp27 wrote:
Bunuel wrote:
11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2
OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

Answer: B.


Hi Bunuel - Can this solved in the below way?

Is |x+y|>|x-y|?

Since both sides are +ve we can square both side of the inequality....
On squaring we get xy>0?

statement 1

(1) |x| > |y|

This is NS as xy can be opp sign as well as same sign

(2) |x-y| < |x|

Squaring on both sides we get y^2 < 2xy
Y cannot be zero otherwise the inequality cannot hold so Y^2 is +ve hence xy is +ve
So we can answer the question xy>0

Is this fine Bunuel?


Yes, this approach is correct.
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Re: Inequality and absolute value questions from my collection [#permalink] New post 23 Dec 2012, 03:40
Bunuel wrote:
2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as y=|x|+x then y is never negative. For x>{0} then y=x+x=2x and for x\leq{0} then (when x is negative or zero) then y=-x+x=0.

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

(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence y=0. Sufficient.

Answer: D.



Hi Bunuel,

Thanks for the explanation to the above Q.

Regarding st 1 i.e X less than zero then [m]y=|x|+x = -x+x=0,

1. we know any value in modulus is positive then ideally the above should be interpreted as [m]y=|x|+x--> [m]y=x-x=0.
2.Also if from St 1 if we x<0 then [m]y=|x|+x= -x-x=-2x

3. Where as we also know that |x|= -x for X<0 and |x|= x for X>/ 0

So can you please tell me where am I going wrong with the concept.

Thanks
Mridul
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Re: Inequality and absolute value questions from my collection   [#permalink] 23 Dec 2012, 03:40
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