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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N

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Bunuel
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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   14 Oct 2015, 21:22
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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   11 Feb 2021, 20:26
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   14 Oct 2015, 21:35
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3+15+32+N+1= 18 x 4 = 72
=>N +51 = 72
=>N=21
Answer C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   14 Oct 2015, 21:45
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Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


Kudos for a correct solution.


The average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18
So, (sum of all 4 numbers)/4 = 18
Multiply both sides by 4 to get: sum of all 4 numbers = 72
This means that 3 + 15 + 32 + (N+1) = 72
Simplify: N + 51 = 72
Solve to get: N = 21

Answer: C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   08 May 2016, 06:35
(3+15+32+N+1)/4=18
51+N=18*4
N=72-51
N=21

Is this approach correct?
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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   08 May 2016, 06:38
paidlukkha wrote:
(3+15+32+N+1)/4=18
51+N=18*4
N=72-51
N=21

Is this approach correct?


Total of 4 numbers is 14*8 = 72

Total of 3 numbers is 50 (3+15+32)

So, N+1 = 22

Or N = 21
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   08 May 2016, 08:45
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3+15+32+N+1/4= 18

N= 18*4 - 51= 21
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   09 Dec 2016, 02:09
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Here is my solution to this one-->

Suing the concept of mean =>

\(Mean =\frac{Sum}{#}\)



\(\frac{51+n}{4}=18\)
Hene n=72-51 = 21
Hence C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   10 Dec 2016, 01:39
Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


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3 + 15 + 32 + ( N + 1 ) = 72

Or, N + 51 = 72

So, N = 21

Hence, answer will be (C) 21
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   18 Aug 2018, 22:01
To answer this, we need to find the sum of the 4 numbers which equals 51+N

we know that average x no. of numbers = sum hence 18 * 4= 51+ N, solving which gives us N=21
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   14 Jun 2019, 15:54
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Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


Kudos for a correct solution.


(3 + 15 + 32 + (N + 1))/4 = 18

51 + N = 72

N = 21

Answer: C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   05 Aug 2019, 11:08
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Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29
Kudos for a correct solution.


A different way for practice, consider how far each number is from the average of 18:

-15 + -3 + 14 + (n+1) = 18
-4 + (n+1) = 18
n - 3 = 18
n = 21
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   16 Sep 2019, 08:22
Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


Kudos for a correct solution.


If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

3 + 15 + 32 + (N+1) = 18*4
51 + N = 72
N = 72 -51 = 21

IMO C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   16 Sep 2019, 11:23
Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


Kudos for a correct solution.

\(3 + 15 + 32 + (N + 1) = 72\)

Or, \(51 + N = 72\)

Or, \(N = 21\) , Answer must be (C)
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   17 Jan 2020, 21:50
3, 15,32 , N+1 /4 = 18

So 50+N+1 = 18*4=>72

So N= 72-50-1= 21
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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   Updated on: 06 May 2021, 11:13
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Bunuel wrote:
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N + 1) is 18, then N =

(A) 19
(B) 20
(C) 21
(D) 22
(E) 29


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Answer: Option C

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Originally posted by GMATinsight on 02 May 2021, 22:16.
Last edited by GMATinsight on 06 May 2021, 11:13, edited 1 time in total.
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   03 May 2021, 01:26
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Let all numbers are 18 so that average is 18.

3 : 3 - 18 (-15 less)
15 : 15 - 18 (-3 less)
32 : 32 - 18 (14 more)

Therefore, we have -15 - 3 + 14 = -4

We are 4 short and therefore N + 1 should be 4 more than 18 which is 22.

=> N + 1 = 22 and therefore N = 21.

Answer C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   12 Aug 2021, 16:00
Hey! Another way of solving this question: our list is [3, 15, 32, N+1], the average of the two terms [3, 32] +1 is 18 (3+32+1 /2 =18). So, 15+N should be 36 (because 15+N /2 must also be 18), then N=21.
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N [#permalink] New post   19 Feb 2023, 02:58
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