From question stem it is asking for \(\frac{(a-b)^2}{9}\)

Statement 1) \(\frac{(a-b)}{3} = q + \frac{2}{3}\) this can be converted to \(a-b = 3q + 2\)

if q = 0 then a-b = 2 and \((a-b)^2\) = 4 as a result it is 4/9

if q = 1 then a-b = 5 and \((a-b)^2\) = 25 as a result it is \(\frac{25}{9} = 2 \frac{7}{9}\)

Insufficient.

Statement 2) \(\frac{(a-b)}{9} = q + \frac{2}{9}\) this can be converted to \(a-b = 9q + 2\)

if q = 0 then a-b = 2 and \((a-b)^2\) = 4 as a result it is 4/9

if q = 1 then a-b = 11 and \((a-b)^2\) = 121 as a result it is \(\frac{121}{9} = 13 \frac{4}{9}\)

Sufficient.

answer choice B

Dear Salsanousi , chetan2u

I read both your explanations but did not really understand why statement 2 is sufficient, could you please explain in more simplistic manner?

(May be I cannot visualize the statement 2 - remainder is 2 always)
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(1) The remainder when \(a-b\) is divided by \(3\) is \(2\): insufic.
\(remainder:\frac{(a-b)^2}{9}…\frac{(3m+2)^2}{9}…\frac{9m+4+12m}{9}…\frac{4+12m}{9}\)
\(m=0:\frac{4}{9}…remainder=4\)
\(m=1:\frac{4+12}{9}=\frac{16}{9}…remainder=7\)

(2) The remainder when \(a-b\) is divided by \(9\) is \(2\): sufic.
\(remainder:\frac{(a-b)^2}{9}…\frac{(9n+2)^2}{9}…\frac{9n+4+36n}{9}…\frac{4}{9}\)
\(remainder=4\)

Answer (B)