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29 Apr 2016, 02:47
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
63% (02:24) correct
37%
(01:55)
wrong
based on 59
sessions
History
Date
Time
Result
Not Attempted Yet
Balajikarthick1990
30 Apr 2016, 04:29
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Bunuel
Is a > b?
(1) −6 + a/3 = b − 8
(2) 9b^2 ≥ 16
(1) −6 + a/3 = b − 8
(2) 9b^2 ≥ 16
I will go with C
i took b=2 and -2 and substituted in Eq1 ---- for both cases it resulted in NO
30 Apr 2016, 06:12
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Balajikarthick1990
Bunuel
Is a > b?
(1) −6 + a/3 = b − 8
(2) 9b^2 ≥ 16
(1) −6 + a/3 = b − 8
(2) 9b^2 ≥ 16
I will go with C
i took b=2 and -2 and substituted in Eq1 ---- for both cases it resulted in NO
Hi balaji don't take two values and if they give you answer YES or NO, believe it to be suff, there may be other pair which may give you opposite answer..
Always solve equation when required..
lets see the statements-
(1) \(−6 + \frac{a}{3} = b − 8\)
-18+a=3b-24..
a-3b= -6...
(i) a>b if b>3
(ii) a<b if b<3
(iii) a=b if b=3
Insuff
(2) \(9b^2 ≥ 16\)
\(b^2≥ \frac{16}{9}\)........\(|b|≥\frac{4}{3}\)
So, \(b≥ \frac{4}{3}\) or \(b<= - \frac{4}{3}\)..
Again nothing about 'a'..
Insuff
Combined..
\(b<= - \frac{4}{3}\), then a<b
\(b>3≥ \frac{4}{3}\), then a>b
Insuff
E
