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# The average of a set of seven consecutive integers is (X+1)

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Senior Manager
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The average of a set of seven consecutive integers is (X+1) [#permalink]

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10 Feb 2007, 12:31
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The average of a set of seven consecutive integers is (X+1) and that of a different set of seven integers is (X-1). Find the average of all the integers in both the sets considering all the common integers only once.

Sorry, I posted the wrong answer choice. . I will just delete my choices then.

Last edited by devilmirror on 10 Feb 2007, 13:37, edited 1 time in total.
SVP
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10 Feb 2007, 13:33
I find x ... So, which answer choice is it?

Set 1 : { x-2, x-1, x, x+1, x+2, x+3, x+4 }
Set 2 : { x-4, x-3, x-2, x-1, x, x+1, x+2 }

So the final set is, without repeats of common integers:

Set final : { x-4, x-3, x-2, x-1, x, x+1, x+2, x+3, x+4 }

X is the average of the set
VP
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12 Feb 2007, 10:18
Let S1 = Sum of 7 Integers

Let S2 = Sum of OTHER 7 integers

S1 / 7 = x + 1

S1 = 7X + 7

S2 / 7 = x - 1

S2 = 7X - 7

ok to find the average we add up the sums of both sets and divide by the total number 14 ( 7 + 7 )

S1 + S2 / 14

((7X + 7) + ( 7X - 7)) / 14

14X/14

Average = X
Senior Manager
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12 Feb 2007, 10:55
terp26 wrote:
Let S1 = Sum of 7 Integers

Let S2 = Sum of OTHER 7 integers

S1 / 7 = x + 1

S1 = 7X + 7

S2 / 7 = x - 1

S2 = 7X - 7

ok to find the average we add up the sums of both sets and divide by the total number 14 ( 7 + 7 )

S1 + S2 / 14

((7X + 7) + ( 7X - 7)) / 14

14X/14

Average = X

i'd say that this is a wrong solution, because it doesn't take into account numbers that may appear in both groups only once (and not twice).

also, Fig's solution might be wrong. the second set is not said to be of 7 CONSECUTIVE integers, but just 7 integers...
if this is not just a sloppy wording, and the author really meant that the second set may consists of non-consecutive integers - then i'd say that the answer cannot be determined by the data given.
if the second set is indeed of consecutive integers... then Fig's solution is correct, and by chance give the same solution as terp's.
12 Feb 2007, 10:55
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