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# When a person aged 39 is added to a group of n people, the average age

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Math Expert
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When a person aged 39 is added to a group of n people, the average age  [#permalink]

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19 Aug 2015, 01:03
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When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Kudos for a correct solution.

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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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23 Aug 2015, 11:58
4
5
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.
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When a person aged 39 is added to a group of n people, the average age  [#permalink]

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19 Aug 2015, 10:29
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When a person aged 39 is added to a group of n people, the average age increases by 2.
(Sum+39)/(n+1) = avg+2

When a person aged 15 is added instead, the average age decreases by 1.
(Sum+15)/(n+1) = avg-1

We can take the 1st equation and subtract from the 2nd equation to isolate n.
24=3n+3
n=7

(A) 7
##### General Discussion
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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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19 Aug 2015, 01:34
2
2

Let's assume that the total no. of people is n and the initial average is x.

So we can assume that the total sum of ages would be nx initially.

When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:

(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2

we get,

2n+x=37

When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:

(nx+15)/(n+1) = x-1

we get

x-n=16

solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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19 Aug 2015, 01:55
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Ans: A

Solution: lets say avg is x so total is (xn):
now with the first condition
(nx+39)/n+1 = x+2
2n+x = 37 Eq.1

with second condition
(nx+15)/n+1 = x-1
x-n = 16 Eq.2

by solving Eq1 and Eq2 we can find out n=7
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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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19 Aug 2015, 11:56
3
5
A simple and elegant solution.

As addition of 39, shifts mean by 2, and addition of 15, shifts mean by 1 to the other side, we have the mean lying between 39 & 15, and in a ratio of 2:1

39-15 = 24
24 divide by 3 is 8.

Meaning mean of the n terms is 15+8 = 39-16 = 23

Now, from first statement, When a person aged 39 is added to a group of n people, the average age increases by 2.

n*23 +39 = 25*(n+1)

n = 7

Ans. (A)
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When a person aged 39 is added to a group of n people, the average age  [#permalink]

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08 Dec 2016, 20:40
1
1
Here is my solution to this one -->
Let the original Mean = $$p$$
Thus original sum = $$p*n$$
Where $$n$$ is the number of people in the original set.
we need to get

As per question =>
$$\frac{p*n+39}{n+1} = p+2$$
$$p+2n=37$$-> Equation 1

Also
$$\frac{p*n+15}{n+1} = p-1$$

Hence $$p-n=16$$--> Equation 2

Hence using the two equations (1 and 2)
$$n=7$$

Hence A
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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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09 Dec 2016, 03:14
One more method -->

Let Mean=p
APQ=> 2(n+1)=39-p
Hence p+2n=37

Also -1(n+1)=15-p
Hence p=16+n
Clearly Both of these Equations are as the above method.

To know Everything about Mean or Statistics in general, refer to this Post -->

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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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11 Feb 2018, 11:44
Bunuel wrote:
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

Would it be possible to figure out the initial average in a problem like this?
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When a person aged 39 is added to a group of n people, the average age  [#permalink]

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Updated on: 31 Oct 2018, 17:32
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

let t=total age of n people
(t+39)/(n+1)-(t+15)/(n+1)=3→
3(n+1)=24→
n=7
A

Originally posted by gracie on 11 Feb 2018, 19:32.
Last edited by gracie on 31 Oct 2018, 17:32, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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11 Feb 2018, 23:03
mjbobak wrote:
Bunuel wrote:
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

Would it be possible to figure out the initial average in a problem like this?

Yes. Initial average is 23. From: 7x = 8*(x-1)-15 --> x = 23.
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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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13 Feb 2018, 07:05
1
Top Contributor
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Kudos for a correct solution.

Let m = mean of ORIGINAL n people
This means the SUM of the ages of the ORIGINAL n people = nm

When a person aged 39 is added to a group of n people, the average age increases by 2
In other words: new mean (with extra person) of n+1 people = original mean + 2
Rewrite as: (nm + 39)/(n+1) = m + 2
Cross multiply to get: nm + 39 = (n + 1)(m + 2)
Simplify: nm + 39 = nm + 2n + m + 2
Simplify: 39 = 2n + m + 2
Rearrange to get: 2n+ m = 37

When a person aged 15 is added instead, the average age decreases by 1
In other words: new mean (with extra person) of n+1 people = original mean - 1
Rewrite as: (nm + 15)/(n+1) = m - 1
Cross multiply to get: nm + 15 = (n + 1)(m - 1)
Simplify: nm + 15 = nm - n + m - 1
Simplify: 15 = -n + m - 1
Rearrange to get: -n + m = 16

We now have the following system:
2n+ m = 37
-n + m = 16

Subtract the bottom equation from the top equation to get: 3n = 21
Solve: n = 7

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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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14 Feb 2018, 09:58
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

We can let the original average = x, and thus, the original sum is nx; thus:

x + 2 = (39 + nx)/(n + 1)

(n + 1)(x + 2) = 39 + nx

nx + x + 2n + 2 = 39 + nx

x + 2n = 37 (Eq 1)

and

x - 1 = (15 + nx)/(n + 1)

(n + 1)(x - 1) = 15 + nx

nx + x - n - 1 = 15 + nx

x - n = 16

x = 16 + n (Eq 2)

Substituting Eq 2 into Eq 1, we have:

16 + n + 2n = 37

3n = 21

n = 7

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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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28 Oct 2018, 19:47
Hi All,

We're told that the average age of a group of N people is 15 yrs and when one more person aged 39 joins the group, the new average is 17 yrs. We're asked for the value of N. This question can be solved in a number of different ways, including with a bit of 'number logic' and a little bit or arithmetic.

Since the 39-year-old raises the average of the group from 15 to 17, the 39 essentially raises each age '2' years AND accounts for another '17' in the group.

39 - 17 = 22 and those 22 extra years would mean that there are 22/2 = 11 people who were ORIGINALLY in the group.

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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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29 Oct 2018, 08:26
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

$$? = n$$

Perfect opportunity to use the homogeneity nature of the average:

$$\sum\nolimits_n { = \mu \cdot n\,\,\,\,\,\,\,\left( {\mu = {\text{original}}\,\,{\text{average}}} \right)\,\,\,}$$

$$\left. \begin{gathered} 39 + \sum\nolimits_n { = \sum\nolimits_{n + 1} { = \,\,\,} \left( {\mu + 2} \right) \cdot \left( {n + 1} \right)\,\,\,} \, \hfill \\ 15 + \sum\nolimits_n { = \sum\nolimits_{n + 1} { = \,\,\,} \left( {\mu - 1} \right) \cdot \left( {n + 1} \right)\,\,\,} \hfill \\ \end{gathered} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,39 - 15 = \left( {n + 1} \right)\left[ {\left( {\mu + 2} \right) - \left( {\mu - 1} \right)} \right] = 3\left( {n + 1} \right)$$

$$n + 1 = \frac{39 - 15}{3} = 8\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = n = 7$$

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

Regards,
Fabio.
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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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03 Nov 2018, 14:03
Bunuel wrote:
What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

Hi Bunuel,

I'm trying to understand the distribution of the numbers deviating from the average, but I'm far away from confident with this approach yet.

In case of this post, it is clear that the mean is between [15,39]

If (n*Av+39)/(n+1)=Av+2 -> adding 39, creates a difference of 2 in the Av (each number increased by 2)
If (n*Av+15)/(n+1)=Av-1 -> adding 15, creates a difference of 1 in the Av (each number decreased by 1)

using both of the given numbers -> 39-15=24 -> the Av is around 24 with a difference of 3 in the Av? How do we get this 3? Because we used both 39 and 15, the absolute difference from Av is 2+1?

Re: When a person aged 39 is added to a group of n people, the average age &nbs [#permalink] 03 Nov 2018, 14:03
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# When a person aged 39 is added to a group of n people, the average age

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