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(16 + 20 + n) / 3 = 21
36 + n = 21*3
n = 63 - 36
n =27

Hence, C.
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Bunuel
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

\(18 < \frac{16 + 20 + n}{3} < 21\)

Or, \(54 < 36 + n < 63\)

Or, \(18 < n < 27\), Among the given options only (B) fits it, Hence Answer must be (B)
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Theory

    ➡ Average = Sum of all the Values / Total Number of Values
    ➡ Sum of All the values = Average * Total Number of Values

The average (arithmetic mean) of 16, 20, and n is between 18 and 21, inclusive

Average = \(\frac{Sum}{3}\) = \(\frac{16 + 20 + n }{3}\) = \(\frac{36 + n}{3}\)

18 ≤ Average ≤ 21
=> 18 ≤ \(\frac{36 + n}{3}\) ≤ 21

Multiplying all the sides by 3 we get
18*3 ≤ 36 + n ≤ 21*3
54 ≤ 36+n ≤ 63

Subtracting 36 from all the sides we get
54 - 36 ≤ n ≤ 63 - 36
18 ≤ n ≤ 27

=> Greatest possible value of n = 27

So, Answer will be C.
Hope it helps!

Watch the following video to Learn the Basics of Statistics

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