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!
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.