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# (The average of five consecutive integers starting from m) –

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(The average of five consecutive integers starting from m) – [#permalink]  16 Apr 2018, 11:44
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Question Stats:

66% (01:49) correct 33% (01:15) wrong based on 6 sessions
(The average of five consecutive integers starting from m) – (the average of six consecutive integers starting from m) =

(A) $$\frac{-1}{4}$$

(B) $$\frac{-1}{2}$$

(C) $$0$$

(D) $$\frac{1}{2}$$

(E) $$\frac{1}{4}$$
[Reveal] Spoiler: OA

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Re: (The average of five consecutive integers starting from m) – [#permalink]  18 Apr 2018, 01:54
Let,
1st integer = $$m$$
2nd integer = $$m+1$$ and so on
5th integer = $$m+4$$
6th integer = $$m+5$$

average of the consecutive five integers = $$\frac{m+m+1+m+2+m+3+m+4}{5} = \frac{5(m+2)}{5} = m+2$$
average of the consecutive six integers = $$\frac{5m +10 +m +5}{6} = \frac{6m + 15}{6} = m +2.5$$
therefore the difference = $$m+2 - m - 2.5 = -0.5$$
option B
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Re: (The average of five consecutive integers starting from m) – [#permalink]  23 Apr 2018, 00:11
Let m = 0

Then, we have set 1 {0,1,2,3,4) and set 2 (0,1,2,3,4,5)
Now, mean1 - mean2 = (0+1+2+3+4)/5 - (0+1+2+3+4+5)/6 = 2 - 15/6 = 2 - 2.5 = -1/2

--> B
Re: (The average of five consecutive integers starting from m) –   [#permalink] 23 Apr 2018, 00:11
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