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(The average of five consecutive integers starting from m) –
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16 Apr 2018, 11:44
16 Apr 2018, 11:44
Expert Reply
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Question Stats:
80% (01:22) correct
19% (01:05) wrong based on 36 sessions
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(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) \(\frac{1}{4}\)
(A) \(\frac{-1}{4}\)
(B) \(\frac{-1}{2}\)
(C) \(0\)
(D) \(\frac{1}{2}\)
(E) \(\frac{1}{4}\)
Re: (The average of five consecutive integers starting from m) –
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18 Apr 2018, 01:54
18 Apr 2018, 01:54
1
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
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
Re: (The average of five consecutive integers starting from m) –
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23 Apr 2018, 00:11
23 Apr 2018, 00:11
1
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
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
