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

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Bunuel
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If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   18 Jan 2018, 03:23
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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
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C
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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   09 Mar 2018, 08:33
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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


The average (arithmetic mean) of 3, 6, 10, m and n is 9
So, (3 + 6 + 10 + m + n)/5 = 9
Multiply both sides by 5 to get: 3 + 6 + 10 + m + n = 45
Simplify: 19 + m + n = 45
Subtract 19 from both sides to get: m + n = 26

What is the average of m + 4 and n – 2 ?
We get: average = [(m + 4) + (n - 2)]/2
Simplify: average = (m + n + 2)/2
Replace m + n to get: average = (26 + 2)/2
Simplify: average = (28)/2 = 14

Answer: C

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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   18 Jan 2018, 03:29
(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] New post   18 Jan 2018, 07:45
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] New post   18 Jan 2018, 08:41
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] New post   20 Jan 2018, 07:39
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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

Answer: C
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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   20 Jan 2018, 12:54
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] New post   20 Feb 2018, 19:15
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.
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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   20 Feb 2018, 20:21
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

Answer: C
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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   20 Feb 2018, 20:41
[{(9-3)+(9-6)+(9-10)+4-2}/2]+9

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Re: If the average (arithmetic mean) of 3, 6, 10, m and n is 9, then what [#permalink] New post   20 Feb 2018, 20:47
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] New post   30 Jun 2022, 10:31
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Theory: Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\)

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

=> Sum of all the values= 3 + 6 + 10 + m + n = 19 + m + n

Using, Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\), We get
9 = \(\frac{19 + m + n}{5}\) (diving by 5 as there are 5 values)
=> 19 + m + n = 9*5 = 45
=> m + n = 45-19 = 26

We need to find the average of m+4 and n-2

Average = \(\frac{Sum}{2}\) = \(\frac{m+4 + n-2}{2}\) = \(\frac{m + n + 2}{2}\) = \(\frac{26+2}{2}\) = \(\frac{28}{2}\) = 14

So, Answer will be C
Hope it helps!

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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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