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The average of 8 numbers is A, and one of the numbers is 14. If 14 is

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The average of 8 numbers is A, and one of the numbers is 14. If 14 is [#permalink]

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New post 20 May 2017, 05:19
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The average of 8 numbers is A, and one of the numbers is 14. If 14 is replaced with 28, then what is the new average in terms of A ?

(A) A + 7/4

(B) A + 1/2

(C) A + 2

(D) 2A + 1

(E) A + 4
[Reveal] Spoiler: OA

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Re: The average of 8 numbers is A, and one of the numbers is 14. If 14 is [#permalink]

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New post 20 May 2017, 06:07
Average = Sum/n, where n= number of terms

Here A= Sum/8 or Sum = 8*A

When a number is changed from 14 to 28, overall there is an increase of (28-14) = 14 in the sum
So, new Sum = 8A + 14
New Average = (8A + 14)/8 = 8A/8 + 14/8 = A + 7/4

Hence answer is A
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Re: The average of 8 numbers is A, and one of the numbers is 14. If 14 is [#permalink]

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New post 20 May 2017, 06:35
Lets assume value of A to be 10
So the total of the 8 numbers is 80.
Since, one of the numbers that was 14 increases to 28, we will have a net addition of 14.
The total of these 8 numbers will become 94. The average : 94/8(47/4) or 10 + 7/4 (Option A)
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The average of 8 numbers is A, and one of the numbers is 14. If 14 is [#permalink]

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New post 20 May 2017, 12:15
The average of 8 numbers is A, and one of the numbers is 14. If 14 is replaced with 28, then what is the new average in terms of A ?
(A) A + 7/4
(B) A + 1/2
(C) A + 2
(D) 2A + 1
(E) A + 4

Let numbers be 1,1,1,1,1,1,1,14. (Hence One number is 14, we can put all other numbers as 1)

Average (A) = \(\frac{1+1+1+1+1+1+1+14}{8}\) = \(\frac{7+ 14}{8}\) = \(\frac{21}{8}\)

14 is replaced with 28; Therefore the numbers would be => 1,1,1,1,1,1,1,28

New Average = \(\frac{1+1+1+1+1+1+1+28}{8}\) = \(\frac{7+ 28}{8}\) = \(\frac{35}{8}\)

New Average is greater than old Average by = \(\frac{35}{8}\) - \(\frac{21}{8}\) = \(\frac{14}{8}\) = \(\frac{7}{4}\)

Therefore New Average in terms of A = A + \(\frac{7}{4}\)
Answer A...

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The average of 8 numbers is A, and one of the numbers is 14. If 14 is   [#permalink] 20 May 2017, 12:15
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