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Re: 6-2a / a-3 = 2/a [#permalink]
Can you give us the solution to the OA

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Re: 6-2a / a-3 = 2/a [#permalink]
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Carcass wrote:
The OA is B :(


Yes the OA is B. Its a trap :(

\(\frac{(6-2a)}{(a-3)}\) = \(\frac{2}{a}\)

Cross Multiply:
6a-2\(a^2 \)= 2a-6

By simplification we get,
2\(a^2\)-4a-6=0
2(\(a^2\)-2a-3)=0

Factorize:
\(a^2\)-3a+a-3=0
a=3 and a=-1

Now if we put positive value in original equation we get,

\(\frac{(6-2a)}{(a-3)}\) = \(\frac{2}{a}\)
\(\frac{6-2(3)}{(3-3)}\) = \(\frac{2}{3}\)

Both numerator and denominator are zero, hence we will get 0 which is smaller than 3, hence Quantity B is the answer
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Re: 6-2a / a-3 = 2/a [#permalink]
Farina wrote:
Carcass wrote:
The OA is B :(


Yes the OA is B. Its a trap :(

\(\frac{(6-2a)}{(a-3)}\) = \(\frac{2}{a}\)

Cross Multiply:
6a-2\(a^2 \)= 2a-6

By simplification we get,
2\(a^2\)-4a-6=0
2(\(a^2\)-2a-3)=0

Factorize:
\(a^2\)-3a+a-3=0
a=3 and a=-1

Now if we put positive value in original equation we get,

\(\frac{(6-2a)}{(a-3)}\) = \(\frac{2}{a}\)
\(\frac{6-2(3)}{(3-3)}\) = \(\frac{2}{3}\)

Both numerator and denominator are zero, hence we will get 0 which is smaller than 3, hence Quantity B is the answer


Sorry are we comparing the value of a or the value of the whole expression

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Re: 6-2a / a-3 = 2/a [#permalink]
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Olasunbo wrote:
Sorry are we comparing the value of a or the value of the whole expression


We are comparing "a" with 3 and "a" has two values so it should be D but we have to check the validity of equation as well by the values of "a" that's why its a trap question and that's what the explanation is given there. This question was in research section of Kaplan Test.
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Re: 6-2a / a-3 = 2/a [#permalink]
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Best way to handle variables in the denominator like these is to write them out before beginning the algebra.

So \(a\) cannot equal 3 or 0.

Once you get to the solutions of \(a=3\) and \(a=-1\), you already know it can't equal 3.
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Re: 6-2a / a-3 = 2/a [#permalink]
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Re: 6-2a / a-3 = 2/a [#permalink]
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