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Some days ago my friend sent me this problem, and I couldn't solve it. It's a pretty simple problem, but I'm struggling with it.
It reads:
The average age in a family (mother, father and their children) is $18$. If we don't take the father, who is $38$, into the average, it drops to $14$. How many kids are in that family?
So how many kids are in that family? Any help is very much appreciated.
If you say $s$ is the sum of the ages, and $k$ is the number of children, then we have the following two equations:
$\frac s {k+2}=18$
$\frac {s-38}{k+1}=14$
Solving them gives $k=4$, so there are 4 children.
Let $X$ be the sum of the ages (of the family members) and let $n$ be the number of members. We have:
$18=\frac{X}{n}$ and $14=\frac{X-38}{n-1}$
We conclude $18n=X$ and $14(n-1)=X-38$. From this, we conclude $4n=24$ and so $n=6$.
Since $n$ is the number of family members, we can subtract 2 (parents) from $n$ and get the number of children, 4.
We know the average of everyone besides the father is $14$. So if the father were fourteen, the average of every one would still be $14$. Now let's increase the hypothetical $14$ year old father's age by $24$ years up to $38$. The family's average age increases by $4$ years to $18$. Therefore the father's age has a $4/24$ (that is, $1/6$) weight in determining the average, so there are $6$ people total in the family, and therefore $4$ children.
We have one person of age $38$ and an unspecified number of people with average age $14.$ The average age of the entire group is $18.$
When averaging any set of numbers, the sum of all deviations from the average (taking deviations above average as positive, deviations below as negative) will be zero. The father has a deviation of $20$ years above the average, so the total net deviation of all other members of the family from the average age is $-20.$ But the average deviation of the other $n$ members of the family from the whole-family average is $14 - 18 = -4.$ In order for $n$ people with an average deviation of $-4$ to add up to a total net deviation of $-20,$ we must have $n = (-20)/(-4) = 5.$
Therefore there are $5$ other family members, consisting of the mother and $4$ children.
Let's say that $a$ is the average and $g$ is the total number of family members. Since $$a = \frac{\sum_{i = 1}^{g}x_{i}}{g} = 18$$ where $x_{i}$ denotes the age of each individual member of the family, with $x_{g} = 38$, we can see that $$\frac{ag - 38}{g-1}= \frac{18g - 38}{g-1}$$ is the average of all the ages without the father's age, and therefore $$\frac{18g - 38}{g-1} = 14$$
Solving for $g$ we find that $g = 6$ which is, again, the total number of family members, and therefore $g - 2 = 4$ is the number of children in the family. Therefore, there are 4 children in the family.
