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If x>0, y<0 and z<0, (x+y+z)^2=?
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Updated on: 01 Mar 2016, 15:48
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If x>0, y<0 and z<0,\((x+y+z)^2\)=? A. \(x^2+y^2+z^2+2xy+2yz+2zx\) B.\(x^2+y^2+z^2+2xy2yz+2zx\) C.\(x^2+y^2+z^22xy+2yz2zx\) D. \(x^2+y^2+z^22xy2yz2zx\) E.\(x^2y^2z^2+2xy+2yz+2zx\) * A solution will be posted in two days.
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Originally posted by MathRevolution on 18 Jan 2016, 21:50.
Last edited by Bunuel on 01 Mar 2016, 15:48, edited 4 times in total.
Edited the question.



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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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19 Jan 2016, 01:41
MathRevolution wrote: If x>0, y<0 and z<0,\((x+y+z)^2\)=?
A. \(x^2+y^2+z^2+2xy+2yz+2zx\) B.\(x^2+y^2+z^2+2xy2yz+2zx\) C.\(x^2+y^2+z^22xy+2yz2zx\) D. \(x^2+y^2+z^22xy2yz2zx\) E.\(x^2y^2z^2+2xy+2yz+2zx\)
* A solution will be posted in two days. Hi, We do not require to know the formula for this but realize that there are two ive qty,z and y.. the mod of x does not have any effect as it is positive.. lets see the eq.. \((x+y+z)^2\)... all terms in the Eq when expanded should be +ive..When we see the choices all terms are same except change in sign at few places.. What will be any term with single z and single y, 2yz, it will be +ive as y is ive and z is ive..What will be any term with only one of single z or single y, 2yx or 2xz, it will be ive as x is +ive and other y or z is ive..so for 2xz to be positive, it should be 2xz and 2xy should be 2xy.. So, we have to look for a choice where all terms are positive apart from terms containing single power of z or single power of y but not both.. lets see the choices A. \(x^2+y^2+z^2+2xy+2yz+2zx\)  all terms are +ive, making 2xz and 2xy negative.. eliminate B.\(x^2+y^2+z^2+2xy2yz+2zx\)2xy, 2yz, 2zx all are ive .. eliminate C.\(x^2+y^2+z^22xy+2yz2zx\) 2xy, 2yz, 2zx all are +ive .. CORRECT D. \(x^2+y^2+z^22xy2yz2zx\) 2yz is ive .. eliminate E.\(x^2y^2z^2+2xy+2yz+2zx\)2xy,y^2,z^2, 2zx all are ive .. eliminate ans C
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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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20 Jan 2016, 18:24
MathRevolution wrote: If x>0, y<0 and z<0,\((x+y+z)^2\)=?
A. \(x^2+y^2+z^2+2xy+2yz+2zx\) B.\(x^2+y^2+z^2+2xy2yz+2zx\) C.\(x^2+y^2+z^22xy+2yz2zx\) D. \(x^2+y^2+z^22xy2yz2zx\) E.\(x^2y^2z^2+2xy+2yz+2zx\)
* A solution will be posted in two days. Because all its the square of the sum of absolute values, it MUST be >=0. Only C is the option which has outcome of each combination as positive. So i went with C as answer.



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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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20 Jan 2016, 21:35
If x>0, y<0 and z<0, (x+y+z)^2=? A. x^2+y^2+z^2+2xy+2yz+2zx B. x^2+y^2+z^2+2xy2yz+2zx C. x^2+y^2+z^22xy+2yz2zx D. x^2+y^2+z^22xy2yz2zx E. x^2y^2z^2+2xy+2yz+2zx ==> A=A when A>0 ,and A=A when A<0. So, (x+y+z)^2=(xyz)^2=x^2+(y)^2+(z)^2+2x(y)+2(y)(z)+2(z)x =x^2+y^2+z^22xy+2yz2zx. Therefore, the answer is C.
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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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01 Mar 2016, 09:38
Are you sure C is the answer? The question says that y and z are negative, agree. But here we need to find the square of the sm of three absolute values, not three integers. If the question asked for (x+y+z)^2 I would have agreed. But here we have three absolute values which are always positive or equal to zero by definition. Probably I'm missing something...could you please explain? Thanks



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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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01 Mar 2016, 10:20
elenap818 wrote: Are you sure C is the answer? The question says that y and z are negative, agree. But here we need to find the square of the sm of three absolute values, not three integers. If the question asked for (x+y+z)^2 I would have agreed. But here we have three absolute values which are always positive or equal to zero by definition. Probably I'm missing something...could you please explain? Thanks You are correct in your definition of the absolute value but you can not have (x+y+z)^2 = 0 as x, y ,z are NON zero numbers. What should according to you be the answer if not for C?



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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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01 Mar 2016, 10:23
elenap818 wrote: Are you sure C is the answer? The question says that y and z are negative, agree. But here we need to find the square of the sm of three absolute values, not three integers. If the question asked for (x+y+z)^2 I would have agreed. But here we have three absolute values which are always positive or equal to zero by definition. Probably I'm missing something...could you please explain? Thanks Hi, you are correct it will not effect the mod values .. But the choices are not in mod values but integers, and that i swhy we have to see if the choices match the given conditions..
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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22 May 2016, 10:30
I am confused too! How will I know whether or not to take mods seriously when the correct answer is otherwise! The question says tis asking us to find the square of three absolute values, not three integers. Where am i wrong here??



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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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22 May 2016, 10:40
smritidabas wrote: chetan2u wrote: elenap818 wrote: Are you sure C is the answer? The question says that y and z are negative, agree. But here we need to find the square of the sm of three absolute values, not three integers. If the question asked for (x+y+z)^2 I would have agreed. But here we have three absolute values which are always positive or equal to zero by definition. Probably I'm missing something...could you please explain? Thanks Hi, you are correct it will not effect the mod values .. But the choices are not in mod values but integers, and that i swhy we have to see if the choices match the given conditions..
Answer can be A Hi, substitute x as 2 and y and z as 1 to check the answer.. If x>0, y<0 and z<0,\((x+y+z)^2= (2+1+1)^2 = 16\).. But lets substitute values in A. \(x^2+y^2+z^2+2xy+2yz+2zx = 2^2+(1)^2+(1)^2+2*2*(1)+2*(1)*(1)+2*2*(1) = 4+1+14+24=0\)...NOT equal to 16.. lets see C \(x^2+y^2+z^22xy+2yz2zx = 2^2+(1)^2+(1)^22*2*(1)+2*(1)*(1)2*2*(1) = 4+1+1+4+2+4=16\)... equal to 16.. So C is CORRECT
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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25 Jul 2016, 00:32
MathRevolution wrote: If x>0, y<0 and z<0,\((x+y+z)^2\)=?
A. \(x^2+y^2+z^2+2xy+2yz+2zx\) B.\(x^2+y^2+z^2+2xy2yz+2zx\) C.\(x^2+y^2+z^22xy+2yz2zx\) D. \(x^2+y^2+z^22xy2yz2zx\) E.\(x^2y^2z^2+2xy+2yz+2zx\)
* A solution will be posted in two days. s Remember that these kind of question can be tricky if you don't apply the last trick after sorting your reasoning. The reasoning is anything that comes out of mod is always positive . It can not be negative. But the last part of the trick that many of us forget to apply is to check whether your answers depends on the absolute value (value that comes out of the mod) or on the "raw value" that is the original original value of the integer without the mod (values with original polarity ve or +ve) IN THIS QUESTION: value of y and z are negative SO the expression \((x+y+z)^2\) will always be positive You don't have to worry about terms with square. because squaring kills negative polarity so x^2, y^2 and z^2 are not our worries. Now to get a positive value from the options that contains "Raw values" and not the "absolute values" you must remember that x is +ve and y and z are ve , so if they are multiplied with x their product will be negative. You have to remove that negative polarity . What is the easiest way of removing a negative polarity ? multiply it with 1 or simply  Therefore all terms that contains either y or z should be multiplied with 1 to give us terms that are positive, so our answer should contains 2xy and 2xz Also because y and z will multiply to give +yz we don't have to multiply it with 1 finally our expression should look like \(x^2+y^2+z^2  2xy +2yz2xz\) THAT IS OPTION C C is the answer
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Re: If x>0, y<0 and z<0, (x+y+z)^2=?
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