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The average (arithmetic mean) of 9 scores is N. If a 10th score is the

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New post Updated on: 30 Oct 2018, 01:01
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The average (arithmetic mean) of 9 scores is N. If a 10th score is then included with the original 9, the average of the 10 scores is T. Which of the following expressions represents the value of the 10th score?


(A) \(10(T - N)\)

(B) \(10T - 9N\)

(C) \(\frac{10T - 9N}{10}\)

(D) \(\frac{10T - 9N}{9}\)

(E) \(\frac{10T - 9N}{2}\)

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Originally posted by pushpitkc on 30 Oct 2018, 00:56.
Last edited by Bunuel on 30 Oct 2018, 01:01, edited 1 time in total.
Edited the question.
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New post Updated on: 30 Oct 2018, 01:17
The average of the nine numbers is given by \(\frac{a1+a2+a3+...+a9}{9}=N\).

Therefore, \(a1+a2+a3...+a9= 9N\). (Let this be equation 1)

Now the average of the nine numbers and an added 10th number is given=T.
9
Therefore \(\frac{a1+a2+a3+...+a9+a10}{10}=T\) (Let this be equation 2)

From equation 1, we can substitute a1+a2+...a9 with 9N in equation 2; giving us:

\(\frac{9N+a10}{10}=T.\)

Therefore, \(a10=10T-9N\) (Choice B)

Originally posted by manavivarma on 30 Oct 2018, 01:13.
Last edited by manavivarma on 30 Oct 2018, 01:17, edited 1 time in total.
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Re: The average (arithmetic mean) of 9 scores is N. If a 10th score is the  [#permalink]

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New post 30 Oct 2018, 01:16
Sum initially=9*N
Sum finally=9N+x where x is the 10th term
Given,
(9N+x)/10=T
So, x=10T-9N
B should be the correct choice

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Re: The average (arithmetic mean) of 9 scores is N. If a 10th score is the   [#permalink] 30 Oct 2018, 01:16
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