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

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 old sum is 8A. The new sum is 8A - 14 + 28 = 8A + 14. Therefore, the new average is (8A + 14)/8 = A + 7/4.

Answer: A