Can you give us the solution to the OA

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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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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.

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.