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15 Oct 2021, 03:45
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E
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:
64% (01:34) correct
36%
(01:49)
wrong
based on 69
sessions
History
Date
Time
Result
Not Attempted Yet
For all integers a and b, where b ≠ 0, subtracting b from a must result in a positive integer if
(A) \(|a - b|\) is a positive integer
(B) \((a + b)\) is a positive integer
(C) \((\frac{a}{b})\) is a positive integer
(D) \((ab)\) is a positive integer
(E) \((b - a)\) is a negative integer
(A) \(|a - b|\) is a positive integer
(B) \((a + b)\) is a positive integer
(C) \((\frac{a}{b})\) is a positive integer
(D) \((ab)\) is a positive integer
(E) \((b - a)\) is a negative integer
24 Oct 2021, 21:36
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Bunuel
This is only possible when \(a > b \)
(A) \(|a - b|\) is a positive integer - Not necessarily true that \(a >b\) , \(a \) could less than \(b\) i.e. \(a=-4\) and \(b = 2\) in which case \(| -4 - 2 | = 6\) but \(a-b = -4 -2 = -6\) hence this condition does not always give \(a -b\) as positive integer.
(B) \((a + b)\) is a positive integer if \(a=-2\) and \(b= 4 ,\) ans to the stem is NO.
(C) \((\frac{a}{b})\) is a positive integer - > \(a\) could be \(< b\) i.e. \(\frac{-4}{-2}= 2\), in this case \(-4 + 2= -2\) will NOT be a positive integer.
(D) \((ab)\) is a positive integer -> \(a=-4 \) and \(b=-2\) then \(-4 +2 = -2\) , hence ans to the stem NO
(e)\((b - a)\) is a negative integer= This implies that \(a > b \) and hence \(a-b \) will always be \(a +ve\) integer. we can check with a couple of cases.
Verifying option e :
let \(a= -2\) and \(b=-4 , b-a = -2\) and \(a-b = 2\)
let \(a=4\) and \(b=2 , b-a = -2\) and \( a-b = 2\)
let \(a= 4\) and \(b=-2\), \(b-a = -6\) and \(a-b= 6\)
Hence option E is correct .
25 Oct 2021, 06:24
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Bunuel
Rephrase the question:
\(a - b > 0\) if ....
Breaking Down the Info:
We can rewrite \(a - b > 0\) as \(a > b\), or \(0 > b - a\) which is choice E.
Answer: E
