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DS from Kaplan what is the value of x? 1. x^2 / 3 = x/81 2.

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DS from Kaplan what is the value of x? 1. x^2 / 3 = x/81 2. [#permalink]

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13 Dec 2003, 05:46
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DS from Kaplan

what is the value of x?

1. x^2 / 3 = x/81

2. x>0

i get C ...kaplan says its A.
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13 Dec 2003, 07:32
A

1. x^2 / 3 = x/81
divide both sides by x (does not matter if x is -ve or +ve because it is an equation and not an inequality)

x/3 = 1/81
x = 3/81 = 1/27

A is sufficient
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13 Dec 2003, 07:43
What if X=0?
A yields two values.
B gives many values
A+B --> 1/27

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13 Dec 2003, 08:04
TRUE.

was trying to explain kaplan's stand ... NOT MINE
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13 Dec 2003, 08:25
A.

(1)
x┬▓/3 = x/81
x┬▓/x = 3/81 ... just showing extra step to resolve the issues raised by some here.
x = 1/27

(2) x > 0, doesn't tell us anything we don't know.
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13 Dec 2003, 08:27
wonder_gmat wrote:
A.

(1)
x┬▓/3 = x/81
x┬▓/x = 3/81 ... just showing extra step to resolve the issues raised by some here.
x = 1/27

(2) x > 0, doesn't tell us anything we don't know.

even x= 0 is a solution to the equation.

is it not?
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13 Dec 2003, 08:28
again the same question to wonder_gmat .. how can you divide by x if x were to be 0 ?
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13 Dec 2003, 08:30
pitts20042006 wrote:
again the same question to wonder_gmat .. how can you divide by x if x were to be 0 ?

why do you need to divide?

x^2 / 3 - x/81 = 0

x ( 27x-1) = 0

x = 0 or x = 1/27

A not sufficient
CEO
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13 Dec 2003, 08:32
wonder_gmat wrote:
A.

(1)
x┬▓/3 = x/81
x┬▓/x = 3/81 ... just showing extra step to resolve the issues raised by some here.
x = 1/27

(2) x > 0, doesn't tell us anything we don't know.

wonder, thats wrong

for your division to work, you need to assume that x not equal to 0.
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13 Dec 2003, 08:34
I never said you need to divide ... but wonder_Gmat did divide and so the question to him
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