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Bunuel
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If the average (arithmetic mean) of 16, 20, and n is between 18 and 21 07 Aug 2018, 04:05
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C
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92% (01:06) correct 8% (01:24) wrong based on 109 sessions

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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
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C
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If the average (arithmetic mean) of 16, 20, and n is between 18 and 21 07 Aug 2018, 04:17
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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
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what is the greatest possible value of n ?

Greatest possible value of n is possible when mean is 21

So 21*3 = 63

63-(16+20) = 27

Hence C

Another Approach,

By logic D & E are not possible bcoz mean = between 18 and 21

and if you take n= 18 it will be not greatest possible value of n
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If the average (arithmetic mean) of 16, 20, and n is between 18 and 21 07 Aug 2018, 04:35
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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
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Note : we must take such a value of n for which the average must be in between 18 to 21, inclusive. It means 21 can be the average . We are looking fro greatest value of n. So, we must take 21 as average of this 3 numbers.


16 + 20 + n = 3*21

36 + n = 63

n = 27.

The best answer is C.
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If the average (arithmetic mean) of 16, 20, and n is between 18 and 21 07 Aug 2018, 07:29
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(16 + 20 + n) / 3 = 21
36 + n = 21*3
n = 63 - 36
n =27

Hence, C.
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If the average (arithmetic mean) of 16, 20, and n is between 18 and 21 07 Aug 2018, 07:35
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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
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\(18 < \frac{16 + 20 + n}{3} < 21\)

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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If the average (arithmetic mean) of 16, 20, and n is between 18 and 21 04 Aug 2022, 00:52
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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!

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