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Re: Inequality and absolute value questions from my collection
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18 Nov 2009, 09:10



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18 Nov 2009, 09:12
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=11 > y=8 y>=3 > 3+y=11 > y=14
Two values for y. Not sufficient.
(1)+(2) y>=2, hence y=14. Sufficient.
Answer: C. just to chime in here your thanks for all this..it's really useful



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18 Nov 2009, 09:25
6. If x and y are integer, is y > 0? (1) x +1 > 0 (2) xy > 0 (1) x+1>0 > x>1. As x is an integer x can take the following values 0,1,2,... But we know nothing about y. Not sufficient. (2) xy>0. x and y have the same sign (both positive OR both negative) and neither x nor y is zero. Not sufficient. (1)+(2) x is positive, as from (1) it's 0,1,2.. and from (2) x is not zero. Hence xy to be positive y also must be positive. Sufficient. Answer: C.
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18 Nov 2009, 11:26
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=y2 > 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 viseversa 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.
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18 Nov 2009, 14:48
11. Is x+y>xy? (1) x > y (2) xy < 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 xy? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of xy. As x+y when they have the same sign will contribute to each other and xy will not. 5+3=8 and 53=2 OR 53=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) xy < 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 viseversa is not right, meaning that x and y can have the same sign but x can be less than xy, but if x>xy the only possibility is x and y to have the same sign.) Answer: B.
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18 Nov 2009, 15:42
12. Is r=s?(1) s<=r<=s (2) r>=s This one is tough. (1) s<=r<=s, we can conclude two things from this statement: A. s is either positive or zero, as s<=s; B. r is in the range (s,s) inclusive, meaning that r can be s as well as s. But we don't know whether r=s or not. Not sufficient. (2) r>=s, clearly insufficient. (1)+(2) s<=r<=s, s is not negative, r>=s > r>=s or r<=s. This doesn't imply that r=s, from this r can be s as well. Consider: s=5, r=5 > 5<=5<=5 5>=5 s=5, r=5 > 5<=5<=5 5>=5 Both statements are true with these values. Hence insufficient. Answer: E.
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18 Nov 2009, 16:43
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=11 > y=8 y>=3 > 3+y=11 > y=14
Two values for y. Not sufficient.
(1)+(2) y>=2, hence y=14. Sufficient.
Answer: C. Just curious if my thinking is correct. on the 2nd part I get y = 8 and y =14 Then I substituted the values into the first equation: 3x^24=10 the answer will never give 10/3 do the same for 14 3x^24=12 x = 0 using my methodology I also got C, but is my thinking correct?



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18 Nov 2009, 18:45
Awesome stuff Bunuel! Hats off to you dude. +5 from me.



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Re: Inequality and absolute value questions from my collection
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18 Nov 2009, 20:02
Awesome, not only have u put the question, but solution to all the problems. I am learning a lot. Thanks to Bunuel. Bunuel, more questions please.



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Re: Inequality and absolute value questions from my collection
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20 Nov 2009, 13:45
tihor wrote: Bunuel, two questions: shouldnt x1 < 1 be 0<x<2? and not 2<x<2? secondly, how does this happen: x(x2)<=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: x1 < 1 means 0<x<2. Already edited the post. As for the second one: x(x2)<=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(x2) is "smiling" parabola, and the intersections with Xaxis are at x=0 and x=2, the range between will be below Xaxis. Hope it helps.
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Re: Inequality and absolute value questions from my collection
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01 Dec 2009, 10:58
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=11 > y=8 y>=3 > 3+y=11 > y=14
Two values for y. Not sufficient.
(1)+(2) y>=2, hence y=14. Sufficient.
Answer: C. Bunuel, I tried to solve this in another way. 1) 3x^2 4 = y  2 if (x^2 4) is positive, we can rewrite above as 3(x^2 4) = y  2 => 3x^2y = 10 > Eqn. 1 if (x^2 4) is negative, we can rewrite above as 3(4x^2) = y  2 => 3x^2y = 14 > Eqn. 2 Adding equation 1 and 2, we get 2y = 4 or y = 2. So (A) as the answer is tempting. I know this is not correct and carries the assumption that y is an integer which is not the case here. If y indeed were an integer in the question, do you think the above approach had any problems ? I am a little confused because every inequality problem appears to have a different method for solving it! Thanks



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Re: Inequality and absolute value questions from my collection
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01 Dec 2009, 11:57
kaptain wrote: Bunuel, I tried to solve this in another way.
1) 3x^2 4 = y  2 if (x^2 4) is positive, we can rewrite above as 3(x^2 4) = y  2 => 3x^2y = 10 > Eqn. 1 if (x^2 4) is negative, we can rewrite above as 3(4x^2) = y  2 => 3x^2y = 14 > Eqn. 2 Adding equation 1 and 2, we get 2y = 4 or y = 2. So (A) as the answer is tempting.
I know this is not correct and carries the assumption that y is an integer which is not the case here.
If y indeed were an integer in the question, do you think the above approach had any problems ? I am a little confused because every inequality problem appears to have a different method for solving it!
Thanks This approach is not correct not because we are not told that y is an integer, but because you can not add inequalities like you did. 3(x^2 4) = y  2 OR 3(4x^2) = y  2, in fact these equation are derived from one and from them only one is right. It's not that we have 3(x^2 4) = y  2 AND 3(4x^2) = y  2 and we are asked to solve fro unknowns. If it were then your solution would be right. Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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13 Dec 2009, 18:52
lagomez wrote: Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
I'm getting B for this one 1. (x1)^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 hi how would mod(1x)<1 would resolve, i mean the interval of x



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22 Dec 2009, 12:55
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=2. B. x+2=y2 > 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 positive only. Hence if xy is not positive we have scenario B and x+y=4. Also note that viseversa 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. hey Bunuel!! first i would like to thank you for posting such wonderful questions.. regarding a question that you posted above, i got a small doubt.. x+2=y+2 so lets say x+2=y+2=k (some 'k') now x+2=k =====> x+2=+/ k and x+2= +k, iff x>2 x+2= k, iff x<2 also we have y+2=k =====> y+2=+/ k and y+2= +k, iff y>2 y+2= k, iff y<2 so x+2=y+2 ===> x=y , iff (x>2 and y>2) or (x<2 and y<2)eq1 and x+y=4, iff (x<2 and y>2) or (x>2 and y<2)eq2 now coming to the options, 1) xy<0 i.e., (x=ve and y=+ve) or (x=+ve and y=ve) (x=ve and y=+ve): this also means that x and y can have values, x=1 and y= some +ve value. so eq2 cannot be applied, x+y#4. if x=3 and y=some +ve value, x+y=4. two cases. data insuff. (x=+ve and y=ve): this also means that x and y can have values, x=+ve value and y=1.so eq2 cannot be applied, x+y#4. if x=some +ve value and y=3, x+y=4. two cases. data insuff. 2)x>2,y>2 for this option too we cannot judge the value of x+y, with the limits of x and y being different in the question and the answer stem. so data insuff. so i have a doubt that, why the answer cannot be E?? plz point out if i made any mistake..
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