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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N
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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 Kudos for a correct solution.
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N
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14 Oct 2015, 21:35
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
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14 Oct 2015, 21:45
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 18So, (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
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08 May 2016, 06:35
(3+15+32+N+1)/4=18 51+N=18*4 N=7251 N=21
Is this approach correct?



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If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N
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08 May 2016, 06:38
paidlukkha wrote: (3+15+32+N+1)/4=18 51+N=18*4 N=7251 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
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08 May 2016, 08:45
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
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09 Dec 2016, 02:09
Here is my solution to this one>
Suing the concept of mean =>
\(Mean =\frac{Sum}{#}\)
\(\frac{51+n}{4}=18\) Hene n=7251 = 21 Hence C
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N
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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
Kudos for a correct solution. 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
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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
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14 Jun 2019, 15:54
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
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05 Aug 2019, 11:08
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
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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
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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
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