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# If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what

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If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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18 Jan 2018, 02:23
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If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what is the average of m + 4 and n – 2 ?

(A) 9
(B) 13
(C) 14
(D) 18
(E) 26
[Reveal] Spoiler: OA

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Joined: 16 Jan 2018
Posts: 7
Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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18 Jan 2018, 02:29
(3+6+10+m+n)/5 = 9
Solving we get m+n = 26

(m+4+n-2)/2 = (m+n+2)/2 = 28/2 = 14.
Option C.

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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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18 Jan 2018, 06:45
Bunuel wrote:
If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what is the average of m + 4 and n – 2 ?

(A) 9
(B) 13
(C) 14
(D) 18
(E) 26

3 + 6 + 10 + m + n = 45

Or, ( m + n ) + 19 = 45

Or, m + n = 26

Now, the average of m + 4 and n – 2 is

= $$\frac{(m + n ) + 4 - 2}{2}$$

= $$\frac{28}{2}$$

= $$14$$, Answer will be (C)
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If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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18 Jan 2018, 07:41
Bunuel wrote:
If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what is the average of m + 4 and n – 2 ?

(A) 9
(B) 13
(C) 14
(D) 18
(E) 26

Average of m+4 and n-2 is $$\frac{m+n+4-2}{2} = \frac{m+n}{2} + 1$$ = Average of m,n + 1

It has been given that the average of 5 numbers is 9, making their sum 45.
Of the 5 numbers, we know that three of the numbers total 19. So, m+n=26
Average of m and n is 13.

Therefore, the average of m+4 and n-2 is 13+1 = 14(Option C)
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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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20 Jan 2018, 06:39
Bunuel wrote:
If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what is the average of m + 4 and n – 2 ?

(A) 9
(B) 13
(C) 14
(D) 18
(E) 26

We can use the formula for the arithmetic mean to create the equation:

(3 + 6 + 10 + m + n)/5 = 9

19 + m + n = 45

m + n = 26

We need to determine the average of m + 4 and n – 2; thus:

(m + 4 + n - 2)/2 = (m + n + 4 - 2)/2 = (26 + 4 - 2)/2 = 28/2 = 14

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Joined: 28 Dec 2017
Posts: 32
Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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20 Jan 2018, 11:54
Find average of 3, 6, 10, m and n first.
(3+6+10+m+n)/5 = 9
Simplify the equation:
3+6+10+m+n = 45
m+n = 26

Problem is asking for average of (m+4) and (n-2):
[(m+4)+(n-2)]/2 = ?
(m+n+2)/2 = ?
Substituting m+n into the equation, (26+2)/2 = 14 (Option C)
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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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20 Feb 2018, 18:15
Bunuel wrote:
If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what is the average of m + 4 and n – 2 ?

(A) 9
(B) 13
(C) 14
(D) 18
(E) 26

This question would be absolutely punishing if 7 were an answer choice.
VP
Joined: 26 Mar 2013
Posts: 1440
Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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20 Feb 2018, 19:21
Bunuel wrote:
If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what is the average of m + 4 and n – 2 ?

(A) 9
(B) 13
(C) 14
(D) 18
(E) 26

$$\frac{3+6+10+m+n}{5}$$ = 9

19 + m +n = 45...........m+n = 26 & m++4+n-2 = m +n +2 = 28

Average = $$\frac{28}{2}$$ =14

Intern
Joined: 20 Feb 2018
Posts: 2
Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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20 Feb 2018, 19:41
[{(9-3)+(9-6)+(9-10)+4-2}/2]+9

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Intern
Joined: 20 Feb 2018
Posts: 2
Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink]

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20 Feb 2018, 19:47
3 is 6 less than 9(average) & 6 is 3 less and 10 is 1 more so we got 8 less that has to be covered by m and n so average of m and n is 9+(8/2)= 13 and then m+4 and n-2 we have 2 as extra other than m &n and by dividing it with 2 we get 1 which is added to average of m & n so final answer is 14

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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what   [#permalink] 20 Feb 2018, 19:47
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# If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what

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