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17 Oct 2017, 23:40
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
84% (00:57) correct
16%
(01:17)
wrong
based on 106
sessions
History
Date
Time
Result
Not Attempted Yet
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 [ 3.66 KiB | Viewed 3066 times ]
Bunuel
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 \(AC^2-BC^2<0 = > AC<BC\). Sufficient
Statement 2: Nothing mentioned about distances of any point, hence insufficient
Option A
18 Oct 2017, 07:13
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S1- OA>OB
OC is the same.
Therefore A is closer to C than B.
Sufficient
S2- No useful information.
Insufficient.
Answer A.
OC is the same.
Therefore A is closer to C than B.
Sufficient
S2- No useful information.
Insufficient.
Answer A.
