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# If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ

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Manager
Joined: 31 Oct 2018
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If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ  [#permalink]

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18 Apr 2019, 09:44
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45% (medium)

Question Stats:

68% (02:51) correct 32% (03:01) wrong based on 63 sessions

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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

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Re: If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ  [#permalink]

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18 Apr 2019, 17:35
|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 both are odd, or even if ether 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.
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Re: If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ  [#permalink]

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18 Apr 2019, 21:32
btrg wrote:
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

|a-5/2| = 1/2

Using the distance concept of absolute values discussed here: https://www.veritasprep.com/blog/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

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If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ  [#permalink]

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18 Apr 2019, 23:00
btrg wrote:
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

Firstly, solve |a - 5/2| = 1/2

1st case :- a - 5/2 = 1/2
a= 6/2 = 3

or 2nd case :- -(a - 5/2) = 1/2
-a + 5/2 = 1/2
a= 2;

a can be 2 or 3 that means both odd and even.

n is odd.

b can be odd or even.

even * even = even
even * odd = even

So look for the option which has one term even becuase then the product will also be even no matter what other terms are.

Option 1 :- abn is odd --> can be false also...when either a or b is even...

Option 2 :- a^2 * b^2 *n^2 = even....can be odd also...when both a and b are odd...n is already odd...So the product will be odd...Not always true...

Option 3 :- a*b*(n^2 +n) is even... Since (n^2 + n) is even a*b*(n^2 +n) will be even no matter what a and b are...

C is correct...Option 3 must be true..

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Re: If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ  [#permalink]

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19 Apr 2019, 01:47
btrg wrote:
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

solve for |a-5/2| = 1/2
we get a= 3 & 2
also b is meadin of n odd integers so value of n is odd , which means value of b can be odd or even
1. $$abn$$ is odd ; all values of a,b,n are odd , but a can be even also ; insufficient
2. $$a^2b^2n^2$$ is even ; a can be even or odd , b & n are odd so ; insufficient
3. $$ab(n^2+n)$$ ;we know n is odd so n^+n will be even always ; sufficient
IMO C
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Re: If |a-5/2| = 1/2 and b is the median of a set of n consecutive integ   [#permalink] 19 Apr 2019, 01:47
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