# Average of age in a family

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.

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.

• part of me feels the easiest way to solve this is to set up a spreadsheet, but that wouldn't help for the general case. the real difficulty is of course the unknown age of the kids, i'd suggest writing down exactly what you know and what you dont know Feb 26 '17 at 19:26
• Are we allowed to set an age limit for the children based on the age of the father? Feb 26 '17 at 19:29
• @mrnovice i suspect so Feb 26 '17 at 19:30
• @mrnovice Yes, if the father is 38, his children obviously can't be older than him. Feb 26 '17 at 19:30

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.

• for your first equation, why have you divided by k + 2, surely it should be k + 1, because the average is total age (s) / total people (his kids plus him, or K + 1), making the second one all over k? Feb 26 '17 at 19:38
• @Cursed1701 I think you forgot to take the mother into the equation. Total people should be father + mother + kids or k + 2. Feb 26 '17 at 19:45
• @Janekmuric ah that does make sense Feb 26 '17 at 19:49

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.

• The problem asks for the number of children in that family, not the total number of members Feb 26 '17 at 19:44
• but from there you just take away 2 for the parents, gaining the correct answer of 4 children Feb 26 '17 at 19:49

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.