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Which of the following sets includes ALL of the solutions of [#permalink]
 16 Jul 2003, 02:02
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Which of the following sets includes ALL of the solutions of x that will satisfy the equation:
|x тАУ 2| - |x тАУ 3| = |x тАУ 5|?
(A) {-6, -5, 0, 1, 7, 8}
(B) {-4, -2, 0, 10/3, 4, 5}
(C) {-4, 0, 1, 4, 5, 6}
(D) {-1, 10/3, 3, 5, 6, 8}
(E) {-2, -1, 1, 3, 4, 5}
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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Manager
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D.
Simple, but effective way - I found that 6 is the solution. so, I eliminated wrong answers. then I tryed 0 in the remaining ones and ended up this D.
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GMAT Instructor
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Any more tries?
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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SVP
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I have just solved the equation and got only 4 and 6.
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Manager
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Ans is C.
Take 2 < x < 3.
So the equation will be,
x -2 +x -3 = -x +5.
So, 3x = 10.
So, x = 10/3 not possible.
Take 3 < x < 5
x =4,
Take x >5
So x = 6.
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GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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evensflow wrote: Ans is C.
Take 2 < x < 3.
So the equation will be,
x -2 +x -3 = -x +5.
So, 3x = 10.
So, x = 10/3 not possible.
Take 3 < x < 5
x =4,
Take x >5
So x = 6.
Your solution has the proper approach. You forgot to check x<2, but there is no solution in that interval so your answer is correct.
BTW, I composed the problem this way so that it would be somewhat timeconsuming to backsolve it (i.e., check all the answers).
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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Manager
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Well i did for x <2 also on paper..
but felt lazy to type the answer as yes the soultion is not possible, since x = 6.
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Manager
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Sorry, I don┬┤t get it
In C, you have number 0, and that does not satisfy the equation, how can be right?
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GMAT Instructor
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MBA04 wrote: Sorry, I don┬┤t get it
In C, you have number 0, and that does not satisfy the equation, how can e right?
I asked for the set that contained all of the solutions, not the set where egvery member was a solution. The only solutions are 4 and 6. C has both of them.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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Senior Manager
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evensflow wrote: Ans is C.
Take 2 < x < 3.
So the equation will be,
x -2 +x -3 = -x +5.
So, 3x = 10.
So, x = 10/3 not possible.
Take 3 < x < 5
x =4,
Take x >5
So x = 6.
Hi Akamai,
In the above approach, while selecting various range, should we consider one of the extreme as part of the range? For exampl, instead of range x > 5, should we consider x >= 5. So that just in case if we arrive at x = 5 then that can be considered valid solution? I know it is not applicable in this problem but what if the quation is designed that way?
Thanks
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GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
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If the solution is right on the value, then it will be a valid solution in both intervals (before and after) so >= is fine unless the variable is in the denominator and will equal zero.
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Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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