Brothers and sisters

Answer: 4 brothers and 3 sisters

Working it out starting from small numbers

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

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

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

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

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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.