Brothers and Sisters

Answer: 4 brothers and 3 sisters


Working it out starting from small numbers
 

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.





Using algebra
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.

Solving by substitution
$s-1=\dfrac{1}{2}b$, 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.

Solving by elimination
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.