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08 Nov 2011, 19:49
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:43) correct
28%
(01:46)
wrong
based on 709
sessions
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If \(a+b=200\) and \(a<b\), is \(a+b>c+d\)?
(1) \(c+d<200\)
(2) \(b+c+d=300\)
(1) \(c+d<200\)
(2) \(b+c+d=300\)
08 Nov 2011, 23:39
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Statement 1: If a+b=200 and c+d<200, then a+b>c+d. Sufficient.
Statement 2: If a+b = 200 and a<b => b>100. Therefore if b+c+d = 300 => c+d <200 which translates into the same case as in statement 1. Sufficient.
(D) is the answer.
Statement 2: If a+b = 200 and a<b => b>100. Therefore if b+c+d = 300 => c+d <200 which translates into the same case as in statement 1. Sufficient.
(D) is the answer.
General Discussion
27 Nov 2016, 14:10
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kostyan5
If \(a+b=200\) and \(a<b\), is \(a+b>c+d\)?
(1) \(c+d<200\)
(2) \(b+c+d=300\)
(1) \(c+d<200\)
(2) \(b+c+d=300\)
\(a+b=200\) and \(a<b\)
Thus, \(a + b < 2b\)
Or, \(200 < 2b\)
Or, \(b > 100\)
FROM STATEMENT - I ( SUFFICIENT )
Given , \(a+b=200\) & \(c+d<200\)
Thus, we can safely conclude - \(a+b>c+d\)
FROM STATEMENT - II ( SUFFICIENT )
I will try to plug in some value and check here, \(a+b=200\) & \(b > 100\)
Say \(b = 110\) & \(a = 90\)
Given, \(b+c+d=300\) , if \(b = 110\) , \(c + d = 190\)
It is given, \(a+b=200\) & we have \(c + d = 190\)
So, we can safely conclude here as well that \(a+b>c+d\)
Thus, EACH statement ALONE is sufficient to answer the question asked, answer will be (D)
