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What is the value of x, where x=36,500(1.05)^n?

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What is the value of x, where x=36,500(1.05)^n? [#permalink] New post 02 Mar 2005, 22:57
15. What is the value of x, where x=36,500(1.05)^n?
(1)n^2-5n+6=0
(2) n - 2 ≠ 0
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 [#permalink] New post 02 Mar 2005, 23:44
The answer would be C ; Using I we know that n is either 2 or 3. Using II we know that n is not 2 .. so n could be anything . Combining both n=3

So C has to be the answer ..
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 [#permalink] New post 03 Mar 2005, 02:24
to define the two values of n we obtain from (1) we need to know that n is different from either of the two
So C


In case (2) said n is different from 4 or 7 (not roots of the equations) we would get (E) ??
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 [#permalink] New post 03 Mar 2005, 03:04
C

i. we can rewrite i. as (n-2)(n-3)=0
so not suff as n=2 or n=3

ii. not suff as no clue about n but n<>2

both i & ii tells us that only n=3
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 [#permalink] New post 03 Mar 2005, 18:59
C...

first time I can be 100% sure.

:lol:
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 [#permalink] New post 04 Mar 2005, 07:07
C

simply look at the statements,

ask yourself do we get a unique value for n?

statement 1

we get n=3 or n=2....therefore Insuff

statement 2
n <>2

taking together, we know that n=3, therefore we will know value of X!

this kind of a problem should take less than 25sec to solve!
  [#permalink] 04 Mar 2005, 07:07
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