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Inequality and absolute value questions from my collection [#permalink]
 16 Nov 2009, 11:33
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Re: Inequality and absolute value questions from my collection [#permalink]
25 Jun 2012, 07:33
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.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486data-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535Hope 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.htmlSo 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]
09 Aug 2012, 17:06
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]
09 Aug 2012, 17:21
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.
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Re: Inequality and absolute value questions from my collection [#permalink]
10 Aug 2012, 08:26
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 jay
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Re: Inequality and absolute value questions from my collection [#permalink]
26 Aug 2012, 01:23
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]
05 Sep 2012, 03:04
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]
05 Sep 2012, 04:53
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]
03 Oct 2012, 05:39
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]
03 Oct 2012, 07:13
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.htmlHope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
03 Oct 2012, 12:26
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]
04 Oct 2012, 03:39
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]
04 Oct 2012, 03:53
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:  
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Last edited by carcass on 04 Oct 2012, 04:03, edited 2 times in total.
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Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 04:13
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]
04 Oct 2012, 04:36
-1<n^2 is the same n^2 > -1 (the reverse) 1>n^2 is the same n^2 < 1 (the reverse) I got it  Many Thanksss
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Re: Inequality and absolute value questions from my collection [#permalink]
11 Oct 2012, 03:11
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]
11 Oct 2012, 03:22
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]
23 Dec 2012, 04:40
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, 05:29
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.
Answer: C. Hello Bunuel, I got A as the answer to the Q. From St1, we have -n=|-n|---> -n=n (As Mod value is +ve)---> we have 2n=0 or -2n=0. In both case we can say that n=0 and hence Ans should be A. From your explanation, it is very clear that either n<0 or n=0. Could you tell me what was your approach to this Question. I mean did you assume values of 1. n as less than zero, 2. ngreater than zero and 3. n equal to zero and check under which condition the St1 holds true. If so, would this be a standard way of doing a modulus Question because clearly I just considered only 1 of the above conditions here. Your inputs please Thanks Mridul
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Re: Inequality and absolute value questions from my collection [#permalink]
23 Dec 2012, 05:37
mridulparashar1 wrote: 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 y=|x|+x = -x+x=0, 1. we know any value in modulus is positive then ideally the above should be interpreted as y=|x|+x--> y=x-x=0.
2.Also if from St 1 if we x<0 then y=|x|+x= -x-x=-2x
3. Where as we also know that |x|= -x for X<0 and |x|= x for X>/ 0So can you please tell me where am I going wrong with the concept. Thanks Mridul When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|\leq{-(some \ expression)}. For example: |-5|=5=-(-5); When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|\leq{some \ expression}. For example: |5|=5; So, if x<0, then |x|=-x and y=|x|+x=-x+x=0. For more check here: math-absolute-value-modulus-86462.html
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Re: Inequality and absolute value questions from my collection [#permalink]
23 Dec 2012, 05:40
mridulparashar1 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.
Answer: C. Hello Bunuel, I got A as the answer to the Q. From St1, we have -n=|-n|---> -n=n (As Mod value is +ve)---> we have 2n=0 or -2n=0. In both case we can say that n=0 and hence Ans should be A. First of all: |-n|=|n|, so -n=|-n| is the same as -n=|n|, which means that n\leq{0}.
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Re: Inequality and absolute value questions from my collection
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