*C*is a set of positive integers where 4 is least. If the average (arithmetic mean) of

*C*equals its range, then how many positive integers does set

*C*have?

(1) The elements in

*C*are consecutive integers.

(2) The average (arithmetic mean) of C is 8.

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Weird question

ReplyDelete(1) avg of consecutive int = range = int

So, the set size must be odd. Hence, the avg is odd if it starts with 4 (or any even). However, the range is even, given 4 is the least. This is a contradiction.

(2) least 4, max 12, avg 8. From this, we can say the avg of remaining set, whose elements can be from 5 to 11 is 8. Insuff

Check ur question pls

Hey Blaoism, I think you are wrong, see that the

ReplyDeleteset can be 4 5 6 7 8 9 10 11 12, and you are right, the number of terms is odd, because there

is a central element, but the average does not

have to be odd, as in this set (notice that both

average and range are eight.

The formula to arrive at shi set is, if you call n the number of elements:

(4 + 4 + n - 1)/2 = 4 + n - 1 -4, and n = 8.

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ReplyDeleteI think the answer is A.

ReplyDelete(2) Avg. is 8, highest element = 4 + 8 = 12

Set Possible = {4, 12}

or, { 4, 8, 12 } or, {4,5,6,7,8,9,10,11,12}

-> insufficient

(1) Numbers are consecutive.

Set can be represented as {4, 4+n, 4+2n.... 4+kn}

(n=1, consecutive)

no. of elements = k+1

Range = 4+kn-4 = kn

Avg. = (4 + 4+kn)/2 = (8+kn)/2

Range = avg.

kn = 8 => k = 8

No. of terms = 9

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ReplyDeleteExplanation of answer choice A

ReplyDeleteF = First number

L = Last number

1:

AVG: (F+L)/2

Range = L-F

L=3F or L=12

4,5,6,7,8,9,10,11,12 : Avg =8, Range=8

2: Average is 8 . Not sufficient (Eventhough it is the right average) Per this the soln could be

4,8,12, or 4,5,6,7,8,9,10,11,12

@vikram: Pl note that avg= (F+L)/2 only if they are said to be consecutive...

ReplyDeleteeg: 2,6,8 avg is not 5 but 5.333

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