Bunuel wrote:

AB is perpendicular to CO. Is A or B closer to C?

(1) OA is less than OB.

(2) ACBD is not a parallelogram.

Attachment:

2017-10-18_1039.png

We need to find which is less \(AC\) or \(BC\)

\(AC^2= OC^2+OA^2\) and \(BC^2=OC^2+OB^2\). subtract the two to get

\(AC^2-BC^2=OA^2-OB^2\)-------------------(1)

Statement 1: Statement itself is sufficient to answer the question, because position of C is fixed.

Mathematically it can be deduced as - given \(OA<OB\), square both sides to get \(OA^2<OB^2\), or \(OA^2-OB^2<0\)

Hence equation 1 becomes \(AC^2-BC^2<0 = > AC<BC\).

SufficientStatement 2: Nothing mentioned about distances of any point, hence

insufficientOption

A