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If the average (arithmetic mean) of n and 11 is equal to 2n, what is

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Bunuel
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If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] New post   26 Aug 2018, 21:20
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If the average (arithmetic mean) of n and 11 is equal to 2n, what is the average of n and \(\frac{13}{3}\)?

(A) 4
(B) 8
(C) 11
(D) 14
(E) 19
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A
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Re: If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] New post   26 Aug 2018, 21:32
Bunuel wrote:
If the average (arithmetic mean) of n and 11 is equal to 2n, what is the average of n and \(\frac{13}{3}\)?

(A) 4
(B) 8
(C) 11
(D) 14
(E) 19


First we have to find out the value of n.
Given, average (arithmetic mean) of n and 11 is equal to 2n
Or, \(\frac{n+11}{2}=2n\)
Or, n+11=4n
Or, 3n=11
Or, \(n=\frac{11}{3}\)

Average of n and \(\frac{13}{3}\)=Average of \(\frac{11}{3}\) and \(\frac{13}{3}\)=\(\frac{\frac{11}{3}+\frac{13}{3}}{2}\)=\(\frac{8}{2}=4\)

Ans. (A)
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Re: If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] New post   26 Aug 2018, 21:51
From what is given :
(n+11)/2 = 2n
n = 11/3

Average of n and 13/3:

(11/3 + 13/3)/2 = 8/2 = 4

IMO A.
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Re: If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] New post   27 Aug 2018, 04:58
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Bunuel wrote:
If the average (arithmetic mean) of n and 11 is equal to 2n, what is the average of n and \(\frac{13}{3}\)?

(A) 4
(B) 8
(C) 11
(D) 14
(E) 19



n +11/2=2n
n=11/3

n+13/3/2=3n+ 13/6


-substituting value of n
we get average of n and 13/2 =4

Correct ans:A
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Re: If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] New post   27 Aug 2018, 05:07
n+11/2=2n
n=11/3

Now substituting it
n+13/3/2=?
11/3+13/3/2 = 24/3/2
Ans= 4(option A)
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Re: If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink]
27 Aug 2018, 05:07
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