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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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20 May 2017, 06:19
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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

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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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20 May 2017, 07:07
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

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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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20 May 2017, 07:35
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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20 May 2017, 13:15
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 + $$\frac{7}{4}$$

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The average of 8 numbers is A, and one of the numbers is 14. If 14 is &nbs [#permalink] 20 May 2017, 13:15
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