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Re: List A contains 7 consecutive multiples of 4 and nothing els [#permalink]
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Let the list be: 4n, 4(n+1), 4(n+2), 4(n+3), 4(n+4), 4(n+5), 4(n+6)

A great fact to remember is that given a list of numbers evenly spaced the mean is the same as the median of the list.
Since we have a list of consecutive multiples of 4 then our list is evenly spaced. Namely the gap between consecutive numbers is 4.

The mean of the last 3 numbers in a our list would be the median of this list and therefore it would be 4(n+5)
Therefore 4(n+5)=80
n+5=80/4=20
n=15

The mean of the first 5 numbers in the list is again the median of this list. Namely 4(n+2)
So 4(n+2)=4(15+2)=4(17)=68

Answer is choice C
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Re: List A contains 7 consecutive multiples of 4 and nothing els [#permalink]
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easier way would be --
80 x 3 = 240
so sum of last 3 consecutive nos is 240

4(x-1) + 4x + 4(x+1) = 240

so 12x = 240
x = 20

so the sequence of the largest 3 nos is - (76,80,84)

so the sequence of 5 smallest will be from 15x4 to 19x4
which is (60,64,68,72,76)

average is 68 therefore
ans is C.
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Re: List A contains 7 consecutive multiples of 4 and nothing els [#permalink]
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