Challenge Level

Start from the brother - he must have at least one brother and one sister.

What about the sister? Add another sister and brother so that she has twice as many brothers as sisters.

Need to add another brother so that the brother has the same number of brothers and sisters

Need to add more so that the sister has twice as many brothers as sisters

This works for the brother too.

Let $b$ represent the number of brothers in the family and $s$ represent the number of sisters in the family.

Each brother has $b-1$ brothers and $s$ sisters.

Each sister has $b$ brothers and $s-1$ sisters.

The boy has the same number of brothers as sisters, so $b-1=s$.

Each sister has half as many sisters as brothers, so $s-1=\dfrac{1}{2}b$.

Solving by substitution or elimination (see below), $b=4$ and $s=3$, so there are $7$ siblings in total.

$s-1=\dfrac{1}{2}b$, so $2s-2=b$ (by multiplying by 2).

Substituting $b=2s-2$ into $b-1=s$ gives $2s-2-1=s$, so $2s-3=s$, so $s=3$.

Substituting $s=3$ into $b-1=s$ gives $b=4$.

Therefore there are $7$ siblings in total.

Subtracting $s-1=\dfrac{1}{2}b$ from $s=b-1$ gives $s-(s-1)=b-1-\left(\dfrac{1}{2}b\right)$.

This simplifies to $s-s+1=b-\dfrac{1}{2}b-1$, so $1=\dfrac{1}{2}b-1$.

Adding 1 to both sides, $2=\dfrac{1}{2}b$, and multiplying by 2 gives $4=b$.

Substituting $b=4$ into $b-1=s$ gives $s=3$.

Therefore there are $7$ siblings in total.

You can find more short problems, arranged by curriculum topic, in our short problems collection.