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18 Apr 2019, 09:44
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C
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:
73% (02:42) correct
27%
(02:43)
wrong
based on 534
sessions
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Not Attempted Yet
If |a-5/2| = 1/2 and b is the median of a set of n consecutive integers, where n is odd, which of the following must be true?
I. \(abn\) is odd
II . \(a^2b^2n^2\) is even
III. \(ab(n^2+n)\) is even
A. I only
B. II only
C. III only
D. I and II
E. I,II and III
I. \(abn\) is odd
II . \(a^2b^2n^2\) is even
III. \(ab(n^2+n)\) is even
A. I only
B. II only
C. III only
D. I and II
E. I,II and III
18 Apr 2019, 17:35
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|a-2.5|=0.5
a-2.5=+-0.5
a=2 or 3[ a can be even or odd]
b can also have either odd or even value
n is definitely odd
1. abn can be odd if a and b both are odd, or even if either of them is even
2. (a^2)(b^2)(n^2) can be both odd or even same as case 1
3. ab{(n^2)+n} is always even because n^2+n=odd+odd= even.
a-2.5=+-0.5
a=2 or 3[ a can be even or odd]
b can also have either odd or even value
n is definitely odd
1. abn can be odd if a and b both are odd, or even if either of them is even
2. (a^2)(b^2)(n^2) can be both odd or even same as case 1
3. ab{(n^2)+n} is always even because n^2+n=odd+odd= even.
18 Apr 2019, 21:32
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btrg
|a-5/2| = 1/2
Using the distance concept of absolute values discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... edore-did/
a is a point 1/2 away from 5/2. It is either at 6/2 = 3 or at 4/2 = 2. This means a is 2 or 3. So a can be even or odd.
b is the median of odd number of consecutive integers so b must be an integer too. It can be even or odd.
n is odd
I. \(abn\) is odd
a and b can be even or odd. So not necessary.
II . \(a^2b^2n^2\) is even
a and b can be even or odd. If both are odd, this will be odd.
III. \(ab(n^2+n)\) is even
n is certainly odd so n^2 is odd too.
(n^2 + n) = Odd + Odd = Even
Hence \(ab(n^2+n)\) is even. True
Answer (C)
