Put the number of daughters as x and number of sons as y.
For the first half of the statement:
Number of sisters any daughter has = x-1 ( because you can't be your own sibling)
Number of brothers = y
y=x-1.
For the second half:
Number of sisters any son has = x
Number of brothers any son has = y-1
3(y-1)=x
Solve as system of equations.
3y-3=x
3(x-1)-3=x
3x-3-3=x
2x=6
x=3
y=x-1=3-1=2
3 daughters and 2 sons.
Hope this helps!
Your fortune is "You can't have everything...where would you put them all?"
If your solving for B then its B<43/3
Students on the minibus = 8
students that are boys = 5
fraction of the student that are boys = ?
we can write fraction as the numerator and denominator so here the fraction of boys students is = 5/8
this fraction means and shows that there are 5 students are boys from 8 students.
16/10=1.6 so the smallest number of combos would be 1.
I believe your answer is B, though I am not positive. I am doing the same question right now, so if it is correct I will let you know!
Hope this helps!