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

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New post 19 Aug 2015, 02:03
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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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New post 23 Aug 2015, 12:58
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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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New post 19 Aug 2015, 11:29
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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

Answer:


(A) 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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New post 19 Aug 2015, 02:34
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Answer according to me: A

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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New post 19 Aug 2015, 02:55
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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New post 19 Aug 2015, 12:56
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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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New post 08 Dec 2016, 21:40
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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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Re: When a person aged 39 is added to a group of n people, the average age  [#permalink]

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New post 11 Feb 2018, 12:44
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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New post Updated on: 09 Jun 2019, 18:30
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=total/average→t/n=a
equation 1: t+39/(n+1)=a+2
equation 2: t+15/(n+1)=a-1
subtracting 2 from 1,
24/(n+1)=3→
n=7
A

Originally posted by gracie on 11 Feb 2018, 20:32.
Last edited by gracie on 09 Jun 2019, 18:30, edited 2 times in total.
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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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New post 12 Feb 2018, 00:03
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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New post 13 Feb 2018, 08:05
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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

Answer: 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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New post 14 Feb 2018, 10:58
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

Answer: 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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New post Updated on: 07 Apr 2019, 13:29
Hi All,

We're told that the average age of a group of N people INCREASES BY 2 if one more person aged 39 joins the group and instead DECREASES BY 1 if one more person aged 15 joins the group. 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 TESTing THE ANSWERS.

To start, it's interesting the average increases or decreases by an exact INTEGER (often, we'd see a change that creates a decimal), so the numbers 39 and 15 impact the average in a specific way - and the DIFFERENCE in those values (39 and 15) is something that we have to consider. Including an extra 39 instead of an extra 15 would increase the SUM by 24... and again, would lead to an increase in average of 3. This implies that the NEW number of people (the original N people + 1) is a factor of 24. Based on the 5 answer choices, the likely answer is either 7 (which would be 8 people when the extra person is added) or 11 (which would be 12 people when the extra person is added).

Notice how 24 = (8)(3)... which would be the new number of people and the increase in the average (between when an extra 15 is included and when an extra 39 is included), so this is far more likely to be the situation that we're dealing with. Here's how to visualize that difference:

IF....
N=7 and the sum of those ages is X....
adding a 39 would give us a sum of (X+39)
adding a 15 instead would gives us a sum of (X+15)

The difference in those 2 situations is (X+39) - (X+15) = 24.
With 8 total people, the average would INCREASE by 24/8 = 3
This is exactly what happens in the situation described about, so this must be the answer.

Final Answer:

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Originally posted by EMPOWERgmatRichC on 28 Oct 2018, 20:47.
Last edited by EMPOWERgmatRichC on 07 Apr 2019, 13:29, edited 1 time in total.
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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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New post 29 Oct 2018, 09:26
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.

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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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New post 03 Nov 2018, 15:03
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?

I'm very confused here. Thank you for your help in advance!
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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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New post 14 Mar 2019, 06:07
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.


Bunuel
I am not getting the logic here. Since we are taking difference of 15 and 39. So in a way we are considering two extra person instead of one. so the answer by this process should be 8-2=6. Can you please help me understand it?
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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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New post 07 Apr 2019, 13:11
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shuvodip04 rencsee

Imagine a number line (if you are familiar with VeritasKarishma posts on weighted average)... think about it like we are dealing with absolute value i.e. the 'distance' from the Average.

___1x<------->2x___
15 ------ Avg ------- 39

Adding 15 gives Avg - 1, the Avg is 'pulled' to the left because 15 is less than the Avg
Adding 39 gives Avg + 2, Avg is 'pulled' to the right because 39 is more than the Avg

Now, subtract the 15,

___1x<------->2x___
0 ------ Avg ------- 24

The 'pull' from 0 to the Avg is still 1, and the 'pull' from the Avg to 24 is still 2. This is because the distance covered stays the same in a ratio of 1x:2x.
Thus, the total from 0 to 24 would be 3x = 24 --> x = 8, hence n+1 = 8 and originally n = 7, A.


Btw, this is the same thing that happens when we combine the 2 equations algebraically:

A+2 = S+39/n+1
A-1 = S+15/n+1
- ____________
3 = 24/n+1

3n+3 = 24
3n = 21
n = 7

Notice that the total change in Average is 3 when 24 is divided by the number of people + 1. Another way to think about it is that the 24 has to be distributed equally in 3's among the number of people + 1, some of which are below and some of which are above the average. The main idea is that we're not adding people, but rather using the 'distance' between the 2 given values to find the relationship between the number of people and the change in the Sum, i.e. how many people we need to distribute the total 'distance' to. We still don't know the actual Average or the Sum.

Hope that helps and that experts can chime in if there's an error in my explanation.
P.S. EMPOWERgmatRichC I think you misread the problem
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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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New post 08 Apr 2019, 03:44
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. You know that n = 7.

When that '+1' is at 39 yrs, the average of 8 people goes up by 2. So 39 has 2*8 = 16 more than the average.
39 - 16 = 23 (original average)
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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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New post 08 Apr 2019, 03:51
shuvodip04 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.


Bunuel
I am not getting the logic here. Since we are taking difference of 15 and 39. So in a way we are considering two extra person instead of one. so the answer by this process should be 8-2=6. Can you please help me understand it?


Note the word "instead" in the question.
"When a person aged 15 is added instead, the average age ..."

Instead of the person aged 39, when we add a person aged 15...
So we are talking about adding only one person. If he is 39 yrs old, avg increases by 2. But if he is 15 yrs old, avg decreases by 1.
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Re: When a person aged 39 is added to a group of n people, the average age   [#permalink] 08 Apr 2019, 03:51

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

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