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Math Expert V
Joined: 02 Sep 2009
Posts: 61412
If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N  [#permalink]

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Difficulty:   5% (low)

Question Stats: 89% (00:54) correct 11% (01:05) wrong based on 1517 sessions

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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  [#permalink]

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1
3+15+32+N+1= 18 x 4 = 72
=>N +51 = 72
=>N=21
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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N  [#permalink]

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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

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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N  [#permalink]

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(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]

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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]

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

N= 18*4 - 51= 21
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GRE 1: Q169 V154 Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N  [#permalink]

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1
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]

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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 ) = 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]

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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]

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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

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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N  [#permalink]

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1
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]

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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.

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]

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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.

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

Or, $$51 + N = 72$$

Or, $$N = 21$$ , Answer must be (C)
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Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

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Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N  [#permalink]

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3, 15,32 , N+1 /4 = 18

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

So N= 72-50-1= 21 Re: If the average (arithmetic mean) of the four numbers 3, 15, 32, and (N   [#permalink] 17 Jan 2020, 20:50
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