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Founder  Joined: 18 Apr 2015
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If the average (arithmetic mean) of 16, 20, and n is between [#permalink]
Expert's post 00:00

Question Stats: 93% (00:57) correct 6% (00:00) wrong based on 15 sessions
If the average (arithmetic mean) of 16, 20, and n is between 18 and 21, inclusive, what is the greatest possible value of n ?

(A) 18

(B) 21

(C) 27

(D) 54

(E) 63
[Reveal] Spoiler: OA

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Director Joined: 20 Apr 2016
Posts: 908
WE: Engineering (Energy and Utilities)
Followers: 11

Kudos [?]: 673 , given: 148

Re: If the average (arithmetic mean) of 16, 20, and n is between [#permalink]
Carcass wrote:
If the average (arithmetic mean) of 16, 20, and n is between 18 and 21, inclusive, what is the greatest possible value of n ?

(A) 18

(B) 21

(C) 27

(D) 54

(E) 63

Since we need the greatest value of n so the Av arithmetic mean should be the greatest = 21

Now, we can write as = $$\frac {6 + 20 + n}3$$ = 21

or 36 + n = 63

or n = 27
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Director Joined: 09 Nov 2018
Posts: 508
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Kudos [?]: 27 , given: 1

Re: If the average (arithmetic mean) of 16, 20, and n is between [#permalink]
16+20=36
With upper limit of average,
total, 21*3=63
Now
63-36=27 Re: If the average (arithmetic mean) of 16, 20, and n is between   [#permalink] 11 Jan 2019, 08:19
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