# Brothers and sisters

Can you work out how many brothers and sisters are in the family?

A boy has the same number of sisters as brothers.

Each of his sisters has only half as many sisters as brothers.

How many brothers and sisters are in the family altogether?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

*This problem is adapted from the World Mathematics Championships*

**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.