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Which of the following sets includes ALL of the solutions of

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Which of the following sets includes ALL of the solutions of [#permalink] New post 16 Jul 2003, 01:02
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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
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 [#permalink] New post 16 Jul 2003, 02:10
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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 [#permalink] New post 16 Jul 2003, 02:35
Any more tries?

:twisted:
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 [#permalink] New post 16 Jul 2003, 03:17
I have just solved the equation and got only 4 and 6.
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 [#permalink] New post 16 Jul 2003, 03:19
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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 [#permalink] New post 16 Jul 2003, 03:41
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
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 [#permalink] New post 16 Jul 2003, 03:51
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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 [#permalink] New post 16 Jul 2003, 11:14
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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 [#permalink] New post 16 Jul 2003, 12:34
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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 [#permalink] New post 12 Feb 2004, 15:04
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?

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 [#permalink] New post 14 Feb 2004, 01:34
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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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

  [#permalink] 14 Feb 2004, 01:34
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