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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]
24 Dec 2012, 14:16
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Re: Inequality and absolute value questions from my collection [#permalink]
25 Dec 2012, 05:12
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Re: Inequality and absolute value questions from my collection [#permalink]
25 Dec 2012, 05:19
Bunuel wrote: debayan222 wrote: Great collection Bunuel...Kudos..  Are these Qs. included in your signature or they exist as separate entity? merry Xmas...Happy Holidays. Yes, they are in Inequalities set. Thanks a lot Bunuel.. Well I guess, whatever Qs come from you directly to the forum, are included in you Sig. ? Hope I got you right..
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Re: Inequality and absolute value questions from my collection [#permalink]
21 Feb 2013, 11:57
Thanks Bunuel. These questions are of great value indeed!
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Re: Inequality and absolute value questions from my collection [#permalink]
21 Feb 2013, 20:55
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.
Answer: E. I got C. Can you please explain why this is incorrect? Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?
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Re: Inequality and absolute value questions from my collection [#permalink]
21 Feb 2013, 21:23
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.
Answer: C. 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.
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Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 01:18
JJ2014 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.
Answer: E. I got C. Can you please explain why this is incorrect? Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong? The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a. Below posts might help with this question: inequality-and-absolute-value-questions-from-my-collection-86939-80.html#p687991inequality-and-absolute-value-questions-from-my-collection-86939-100.html#p746278Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 07:14
Bunuel wrote: JJ2014 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.
Answer: E. I got C. Can you please explain why this is incorrect? Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2 Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong? The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a. Below posts might help with this question: inequality-and-absolute-value-questions-from-my-collection-86939-80.html#p687991inequality-and-absolute-value-questions-from-my-collection-86939-100.html#p746278Hope it helps. So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again.
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Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 07:26
Bunuel wrote: 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.
Answer: C. 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. This is clear. Thank you!!
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Re: Inequality and absolute value questions from my collection [#permalink]
22 Feb 2013, 08:09
JJ2014 wrote: Bunuel wrote: JJ2014 wrote:
I got C. Can you please explain why this is incorrect?
Statement 1 gives x^2+2xy+y^2 = 9a, and I rewrote it to 9a-2xy = x^2+y^2
Statement 2 gives x^2-2xy+y^2=a, and i rewrote this to x^2+y^2= a +2xy
Together, I have 9a-2xy = a+2xy, which leads to 8a = 4xy, but 8a is also equal to x^2+y^2? ... this last part must be wrong?
The red part is not right. If you sum the two equations you'll get 2(x^2+y^2)=10a --> x^2+y^2=5a. Below posts might help with this question: inequality-and-absolute-value-questions-from-my-collection-86939-80.html#p687991inequality-and-absolute-value-questions-from-my-collection-86939-100.html#p746278Hope it helps. So I understand that you summed the equations and got that answer. But why is setting them equal to each other wrong in this case? I'm trying to figure out what concept I'm missing so that I don't end up doing it again. What are you trying to get when setting "them" equal? Anyway, you won't be able to solve two equations with three unknowns.
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Re: Inequality and absolute value questions from my collection [#permalink]
28 Feb 2013, 06:35
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2
The solution seem confusing to me as I see four cases:
a] x<-2, y<-2
b]x>-2, y>-2
c] x<-2, y>-2
d]x>-2, y<-2
case [a] and [b] support x=y while case [c] and [d] support x+y=-4
when xy<0, the case [c]or[d] always do not apply, for example: x=-3 and y=3 would come under case[c] and x=-1 and y=3 would come under case [b] , so it is insufficient.
when x>2 , y<2, we have a case [b] with x=3, y=-1 and a case [d] with x=3,y=-3. So insufficient
when we combine(1)+(2) , we have a case as shown above , it is also insufficient.
So my answer choice would be E.
Can somebody help if I am wrong.
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Apr 2013, 04:18
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.
Answer: E. Hi Bunnel Solved the 1st statement like this - (x + y)^2 = 9aSince x^2 + y^2 >= 2xyx^2 + y^2 + x^2 + y^2 >= 9a2(x^2 + y^2) >= 9ax^2 + y^2 >= 4.5aNow this would have been sufficient if a is not = 0 had been given in the stem Is this approach to the problem alright?? Is this st sufficient if it is given that a is not equal to 0 Thanks
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Re: Inequality and absolute value questions from my collection [#permalink]
17 Apr 2013, 06:34
Dipankar6435 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.
Answer: E. Hi Bunnel Solved the 1st statement like this - (x + y)^2 = 9aSince x^2 + y^2 >= 2xyx^2 + y^2 + x^2 + y^2 >= 9a2(x^2 + y^2) >= 9ax^2 + y^2 >= 4.5aNow this would have been sufficient if a is not = 0 had been given in the stem Is this approach to the problem alright?? Is this st sufficient if it is given that a is not equal to 0 Thanks Yes, but we don't know whether x is 0.
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Re: Inequality and absolute value questions from my collection [#permalink]
24 Apr 2013, 05:04
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.
Answer: E. hi Bunuel, Thank you very much for all the explanations. I have a query on this one Combining both we get x^2+y^2=5a or x,y,a = 0 aren't those sufficient to answer the question is x^2+y^2>4a Is the first case where x^2+y^2=5a, the answer is yes Second case where x,y,a=0, the answer is no Kindly do elaborate. Thanks
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Re: Inequality and absolute value questions from my collection [#permalink]
24 Apr 2013, 05:26
Transcendentalist 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.
Answer: E. hi Bunuel, Thank you very much for all the explanations. I have a query on this one Combining both we get x^2+y^2=5a or x,y,a = 0 aren't those sufficient to answer the question is x^2+y^2>4a Is the first case where x^2+y^2=5a, the answer is yes Second case where x,y,a=0, the answer is no Kindly do elaborate. Thanks First of all when we combine we get that x^2+y^2=5a. If xya\neq{0}, then the answer is YES but if xya={0}, then the answer is NO. Next, it's a YES/NO DS question. In a Yes/No Data Sufficiency question, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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