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The average age of a group of n people is 15 yrs. One more person aged

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The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 19 Aug 2015, 01:01
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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 19 Aug 2015, 02:06
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Bunuel wrote:
The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13



Ans: C

Solution: (15n+39)/(n+1) = 17
this gives us n=11
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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 19 Aug 2015, 02:19
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Bunuel wrote:
The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

Kudos for a correct solution.


Total age of n people = 15n
(15n+39)/(n+1)=17, given
solving, we get n=11
The correct option is C
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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 23 Aug 2015, 11:41
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Bunuel wrote:
The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

First tell me, if the age of the additional person were 15 yrs, what would have happened to the average? The average would have remained the same since this new person’s age would have been the same as the age that represents the group. But his age is 39 – 15 = 24 more than the average. We know that we need to evenly split the extra among all the people to get the new average. When 24 is split evenly among all the people (including the new guy), everyone gets 2 extra (since average age increased from 15 to 17). There must be 24/2 = 12 people now (including the new guy) i.e. n must be 11 (without including the new guy).
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The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 08 Dec 2016, 23:27
1
Great Question.
Here is my solution to this one =>

This can be solved via two ways:-

Method 1-->
Using the Formula =>
\(Mean =\frac{Sum}{#}\)

\(Sum =15n\)
\(\frac{15n+39}{n+1} => 17\)
Hence \(2n=22\)
\(n=11\)


Method 2-->
Here Old mean =15
New mean =17
hence 2 is getting added to each n+1 terms.
Now difference between 15 and 39 is 24
hence 2(n+1)=24
So.n=11

Hence C

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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 12 Oct 2018, 12:41
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Bunuel wrote:
The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

Kudos for a correct solution.


The average age of a group of n people is 15 yrs.
So, (sum of all n ages)/n = 15
Multiply both sides by n to get: (sum of all n ages) = 15n

One more person aged 39 joins the group and the new average is 17 yrs.
IMPORTANT: Once this 39-year-old is added to the group, the (NEW sum of ages) = 15n + 39 [We just added 39 to the old sum]
Since we now have n+1 people, we can write: (15n + 39)/(n + 1) = 17
Multiply both sides by n to get: 15n + 39 = 17(n + 1)
Expand right side: 15n + 39 = 17n + 17
Subtract 15n from both sides to get: 39 = 2n + 17
Subtract 17 from both sides to get: 22 = 4n
Solve: n = 11

Answer: C

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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 13 Oct 2018, 00:37
let \(S\) be the sum of \(n\) peoples's ages.
\(\frac{S}{n}=15\)
\(S=15n\)

after adding one more person:
\(\frac{S+39}{n+1}=17\)
\(S+39=17n+17\)
\(15n+39=17n+17\)
\(2n=22\)
\(n=11\)
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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 29 Oct 2018, 08:17
Bunuel wrote:
The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

\(? = n\)

Perfect opportunity to use the homogeneity nature of the average:

\(\left. \begin{gathered}
\sum\nolimits_n { = 15 \cdot n} \hfill \\
39 + \sum\nolimits_n { = \sum\nolimits_{n + 1} { = \,\,\,} 17 \cdot \left( {n + 1} \right)\,\,\,} \, \hfill \\
\end{gathered} \right\}\,\,\,\,\,39 + 15n = 17n + 17\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,2n = 22\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = 11\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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New post 30 Oct 2018, 18:09
Bunuel wrote:
The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13



We can create the equation:

(15n + 39)/(n + 1) = 17

15n + 39 = 17n + 17

22 = 2n

11 = n

Answer: C
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Re: The average age of a group of n people is 15 yrs. One more person aged  [#permalink]

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