Solve Riddle With Algebra

Question

There is a riddle and I believe it can be solved by algebra - please assist

A boy has as many sisters as brothers, but each sister has only half as many sisters as brothers. How many brothers and sisters are there in the family

Here is algebra, but I am stuck

$b=brother$

$s=sister$

$t=total$

boy has as many brothers as sisters

$b + b + s = t$

each sister has only half as many sisters as brothers

$s + s + b = t$

$s + \frac{1}{b} + b = t$

Hence

$b + b + s = s + \frac{1}{b} + b$

$2b + s = s + \frac{3b}{2}$

$2b = \frac{3b}{2}$

$4b = 3b$

Please assist.

Answer 1

$b$ = number of boys = number of brothers each girl has

$g$ = number of girls = number of sisters each boy has

$b-1$ = number of brothers each boy has

$g-1$ = number of sisters each girl has

$$b-1 = g, \qquad (\text{Equation 1})$$

$$g-1 = \dfrac{b}{2}, \qquad (\text{Equation 2})$$

Plugging in for $g=b-1$ into the second equation:

$$b-1-1 = \dfrac{b}{2} \Longrightarrow b=4, g=3$$

Answer 2

The problem is you're using $b$ for the number of the boy's brothers and also for the number of the girl's brothers.

Instead say there are $b$ boys and $g$ girls. Then a boy has $b-1$ brothers, so $$g=b-1.$$Similarly $$g-1=b/2.$$

Answer 3

Although this is probably not intentional, this problem has two different solutions. If you assume there are $s>0$ sisters, the boy also has $s$ brothers making $s+1$ boys in all. Each boy being a brother of any of the girls, each having $s-1$ sisters, the equation $s+1=2(s-1)$ easily solves to $s=3$ (four boys and three girls in all).

However, the question does not say that $s\neq0$; the boy need not have any sisters. In that case each of the sisters has (whatever) is true; there aren't any sisters so that is vacuously true. So $s=0$ is another solution: a family with one boy and no girls. Technically, in this case there are no brothers and sisters at all in the family (which is what the question was asking), as the one child that is there is neither a sister nor a brother. Actually, nowadays this is the more likely solution, I would think.

Answer 4

I think it'd be easiest to do

$b = $ number of boys

$g = $ number of girls and the essential insight is to notice:

A boy has $b-1$ brothers and $g$ sisters, and a girl has $b$ brother and $g-1$ sisters.

So "A boy has as many sisters as brothers" means $b-1 = g$.

And "each sister has only half as many sisters as brothers" means $ g-1 = \frac 12 b$

So solve the two equations two unknowns:

$b-1 = g$

$g-1 =\frac 12 b$

......

If you want to use variables for brothers and sisters you can do

$b_b = $ number of brothers a boy has

$b_g = $ number of brothers a girl has

$s_b = $ number of sisters a boy has and

$s_g = $ number of sisters a girl has.

Then our riddle is $b_b = s_b$ and $s_g = \frac 12 b_s$.

But to solve it we need to not the unstated but essential a girl has $1$ fewer sister than a boy, and a boy has one fewer brother than a girl. So

$b_b = b_g -1$ and $s_g = s_b - 1$.

So solve the four equations four unknowns

$b_b = s_b$

$s_g = \frac 12b_s$

$b_b = b_s -1$

$s_g = s_b - 1$.