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

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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, 04:12
rencsee wrote:
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!


Think of it in another way:

There are n people.
When you add one person aged 15, you get n+1 people.
Say the average of these n+1 people is p. (which is 1 less than original)
If instead of 15 yrs old, this (n+1)th person were 39 yrs old, average would be 2 more than original. Can we say that this would be (p + 3) ?

So if you increase the age of a person from 15 to 39 (i.e. add 24), the average increases by 3. So you must be giving 3 extra to 8 people to account for the extra 24.

Hence (n+1) = 8
So n = 7
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Veritas Prep GMAT Instructor
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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, 04:14
energetics wrote:
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.




Sound logic!
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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, 04:14

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