Solve Riddle With AlgebraCan someone please explain the catch behind this question?How do I solve two equations in two unknowns?Generalization of a certain riddle and ultrafilters (?)Determining how many sibling a person hasCounting Siblings Word Problem: Linear Algebra SystemsWhich of these two methods provides the correct answer for this probability riddle?Riddle to solve. Grouping of integers.Translate riddle to math“A Queer Coincidence,” riddle from Dudeney's bookCounting siblings problem

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Solve Riddle With Algebra


Can someone please explain the catch behind this question?How do I solve two equations in two unknowns?Generalization of a certain riddle and ultrafilters (?)Determining how many sibling a person hasCounting Siblings Word Problem: Linear Algebra SystemsWhich of these two methods provides the correct answer for this probability riddle?Riddle to solve. Grouping of integers.Translate riddle to math“A Queer Coincidence,” riddle from Dudeney's bookCounting siblings problem













5












$begingroup$


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




$s + s + b = t$



$s + frac1b + b = t$



Hence



$b + b + s = s + frac1b + b$



$2b + s = s + frac3b2$



$2b = frac3b2$



$4b = 3b$



Please assist.






algebra-precalculus puzzle word-problem







share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    Better to choose variables for the number of boys and girls. Remember that a boy has one fewer brother than a girl has.
    $endgroup$
    – lulu
    Jun 7 at 16:17










  • $begingroup$
    A boy (who is one of the brothers) has as many brothers as sisters. So we know $total= boys + girls$ and we know $boys = girls + 1$. Each sister has half as many sisters as brothers. So one sister, could have 1 sister and 2 brothers, or 2 sisters and 4 brothers, etc. but only 2 sister and 4 brothers fits with the previous. So there are 3 girls and 4 boys
    $endgroup$
    – Benedict W. J. Irwin
    Jun 7 at 16:20











  • $begingroup$
    better to write better description of variables. "b=brothers" and "t=total" mean nothing. "Let $b$ be the number of brothers X has" (where you need to specify X too) is a much better description of what you are trying to do so you and anyone reading your work is less likely to be confused.
    $endgroup$
    – user10354138
    Jun 7 at 16:21















5












$begingroup$


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




$s + s + b = t$



$s + frac1b + b = t$



Hence



$b + b + s = s + frac1b + b$



$2b + s = s + frac3b2$



$2b = frac3b2$



$4b = 3b$



Please assist.






algebra-precalculus puzzle word-problem







share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    Better to choose variables for the number of boys and girls. Remember that a boy has one fewer brother than a girl has.
    $endgroup$
    – lulu
    Jun 7 at 16:17










  • $begingroup$
    A boy (who is one of the brothers) has as many brothers as sisters. So we know $total= boys + girls$ and we know $boys = girls + 1$. Each sister has half as many sisters as brothers. So one sister, could have 1 sister and 2 brothers, or 2 sisters and 4 brothers, etc. but only 2 sister and 4 brothers fits with the previous. So there are 3 girls and 4 boys
    $endgroup$
    – Benedict W. J. Irwin
    Jun 7 at 16:20











  • $begingroup$
    better to write better description of variables. "b=brothers" and "t=total" mean nothing. "Let $b$ be the number of brothers X has" (where you need to specify X too) is a much better description of what you are trying to do so you and anyone reading your work is less likely to be confused.
    $endgroup$
    – user10354138
    Jun 7 at 16:21













5












5








5





$begingroup$


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




$s + s + b = t$



$s + frac1b + b = t$



Hence



$b + b + s = s + frac1b + b$



$2b + s = s + frac3b2$



$2b = frac3b2$



$4b = 3b$



Please assist.






algebra-precalculus puzzle word-problem







share|cite|improve this question











$endgroup$




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




$s + s + b = t$



$s + frac1b + b = t$



Hence



$b + b + s = s + frac1b + b$



$2b + s = s + frac3b2$



$2b = frac3b2$



$4b = 3b$



Please assist.






algebra-precalculus puzzle word-problem




algebra-precalculus puzzle word-problem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 7 at 16:22









J. W. Tanner

7,9591723




7,9591723










asked Jun 7 at 16:15









MariumMarium

1573




1573







  • 3




    $begingroup$
    Better to choose variables for the number of boys and girls. Remember that a boy has one fewer brother than a girl has.
    $endgroup$
    – lulu
    Jun 7 at 16:17










  • $begingroup$
    A boy (who is one of the brothers) has as many brothers as sisters. So we know $total= boys + girls$ and we know $boys = girls + 1$. Each sister has half as many sisters as brothers. So one sister, could have 1 sister and 2 brothers, or 2 sisters and 4 brothers, etc. but only 2 sister and 4 brothers fits with the previous. So there are 3 girls and 4 boys
    $endgroup$
    – Benedict W. J. Irwin
    Jun 7 at 16:20











  • $begingroup$
    better to write better description of variables. "b=brothers" and "t=total" mean nothing. "Let $b$ be the number of brothers X has" (where you need to specify X too) is a much better description of what you are trying to do so you and anyone reading your work is less likely to be confused.
    $endgroup$
    – user10354138
    Jun 7 at 16:21












  • 3




    $begingroup$
    Better to choose variables for the number of boys and girls. Remember that a boy has one fewer brother than a girl has.
    $endgroup$
    – lulu
    Jun 7 at 16:17










  • $begingroup$
    A boy (who is one of the brothers) has as many brothers as sisters. So we know $total= boys + girls$ and we know $boys = girls + 1$. Each sister has half as many sisters as brothers. So one sister, could have 1 sister and 2 brothers, or 2 sisters and 4 brothers, etc. but only 2 sister and 4 brothers fits with the previous. So there are 3 girls and 4 boys
    $endgroup$
    – Benedict W. J. Irwin
    Jun 7 at 16:20











  • $begingroup$
    better to write better description of variables. "b=brothers" and "t=total" mean nothing. "Let $b$ be the number of brothers X has" (where you need to specify X too) is a much better description of what you are trying to do so you and anyone reading your work is less likely to be confused.
    $endgroup$
    – user10354138
    Jun 7 at 16:21







3




3




$begingroup$
Better to choose variables for the number of boys and girls. Remember that a boy has one fewer brother than a girl has.
$endgroup$
– lulu
Jun 7 at 16:17




$begingroup$
Better to choose variables for the number of boys and girls. Remember that a boy has one fewer brother than a girl has.
$endgroup$
– lulu
Jun 7 at 16:17












$begingroup$
A boy (who is one of the brothers) has as many brothers as sisters. So we know $total= boys + girls$ and we know $boys = girls + 1$. Each sister has half as many sisters as brothers. So one sister, could have 1 sister and 2 brothers, or 2 sisters and 4 brothers, etc. but only 2 sister and 4 brothers fits with the previous. So there are 3 girls and 4 boys
$endgroup$
– Benedict W. J. Irwin
Jun 7 at 16:20





$begingroup$
A boy (who is one of the brothers) has as many brothers as sisters. So we know $total= boys + girls$ and we know $boys = girls + 1$. Each sister has half as many sisters as brothers. So one sister, could have 1 sister and 2 brothers, or 2 sisters and 4 brothers, etc. but only 2 sister and 4 brothers fits with the previous. So there are 3 girls and 4 boys
$endgroup$
– Benedict W. J. Irwin
Jun 7 at 16:20













$begingroup$
better to write better description of variables. "b=brothers" and "t=total" mean nothing. "Let $b$ be the number of brothers X has" (where you need to specify X too) is a much better description of what you are trying to do so you and anyone reading your work is less likely to be confused.
$endgroup$
– user10354138
Jun 7 at 16:21




$begingroup$
better to write better description of variables. "b=brothers" and "t=total" mean nothing. "Let $b$ be the number of brothers X has" (where you need to specify X too) is a much better description of what you are trying to do so you and anyone reading your work is less likely to be confused.
$endgroup$
– user10354138
Jun 7 at 16:21










4 Answers
4






active

oldest

votes


















5












$begingroup$

$b$ = 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



$$b-1 = g, qquad (textEquation 1)$$



$$g-1 = dfracb2, qquad (textEquation 2)$$



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



$$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Yet you are missing a solution.
    $endgroup$
    – Marc van Leeuwen
    Jun 7 at 16:30










  • $begingroup$
    Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
    $endgroup$
    – InterstellarProbe
    Jun 7 at 17:27










  • $begingroup$
    A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
    $endgroup$
    – Marc van Leeuwen
    Jun 7 at 19:34










  • $begingroup$
    @MarcvanLeeuwen Agree to disagree?
    $endgroup$
    – InterstellarProbe
    Jun 7 at 19:36











  • $begingroup$
    $$ Agree
    $endgroup$
    – Marc van Leeuwen
    Jun 8 at 6:44


















5












$begingroup$

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 $sneq0$; 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.






share|cite|improve this answer











$endgroup$




















    3












    $begingroup$

    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 $$g-1=b/2.$$






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      There is another solution.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 16:30


















    2












    $begingroup$

    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



    $b_b = s_b$



    $s_g = frac 12b_s$



    $b_b = b_s -1$



    $s_g = s_b - 1$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
      $endgroup$
      – InterstellarProbe
      Jun 7 at 17:32












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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    $b$ = 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



    $$b-1 = g, qquad (textEquation 1)$$



    $$g-1 = dfracb2, qquad (textEquation 2)$$



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



    $$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Yet you are missing a solution.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 16:30










    • $begingroup$
      Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
      $endgroup$
      – InterstellarProbe
      Jun 7 at 17:27










    • $begingroup$
      A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 19:34










    • $begingroup$
      @MarcvanLeeuwen Agree to disagree?
      $endgroup$
      – InterstellarProbe
      Jun 7 at 19:36











    • $begingroup$
      $$ Agree
      $endgroup$
      – Marc van Leeuwen
      Jun 8 at 6:44















    5












    $begingroup$

    $b$ = 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



    $$b-1 = g, qquad (textEquation 1)$$



    $$g-1 = dfracb2, qquad (textEquation 2)$$



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



    $$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Yet you are missing a solution.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 16:30










    • $begingroup$
      Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
      $endgroup$
      – InterstellarProbe
      Jun 7 at 17:27










    • $begingroup$
      A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 19:34










    • $begingroup$
      @MarcvanLeeuwen Agree to disagree?
      $endgroup$
      – InterstellarProbe
      Jun 7 at 19:36











    • $begingroup$
      $$ Agree
      $endgroup$
      – Marc van Leeuwen
      Jun 8 at 6:44













    5












    5








    5





    $begingroup$

    $b$ = 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



    $$b-1 = g, qquad (textEquation 1)$$



    $$g-1 = dfracb2, qquad (textEquation 2)$$



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



    $$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$






    share|cite|improve this answer









    $endgroup$



    $b$ = 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



    $$b-1 = g, qquad (textEquation 1)$$



    $$g-1 = dfracb2, qquad (textEquation 2)$$



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



    $$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 7 at 16:20









    InterstellarProbeInterstellarProbe

    4,545931




    4,545931











    • $begingroup$
      Yet you are missing a solution.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 16:30










    • $begingroup$
      Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
      $endgroup$
      – InterstellarProbe
      Jun 7 at 17:27










    • $begingroup$
      A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 19:34










    • $begingroup$
      @MarcvanLeeuwen Agree to disagree?
      $endgroup$
      – InterstellarProbe
      Jun 7 at 19:36











    • $begingroup$
      $$ Agree
      $endgroup$
      – Marc van Leeuwen
      Jun 8 at 6:44
















    • $begingroup$
      Yet you are missing a solution.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 16:30










    • $begingroup$
      Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
      $endgroup$
      – InterstellarProbe
      Jun 7 at 17:27










    • $begingroup$
      A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
      $endgroup$
      – Marc van Leeuwen
      Jun 7 at 19:34










    • $begingroup$
      @MarcvanLeeuwen Agree to disagree?
      $endgroup$
      – InterstellarProbe
      Jun 7 at 19:36











    • $begingroup$
      $$ Agree
      $endgroup$
      – Marc van Leeuwen
      Jun 8 at 6:44















    $begingroup$
    Yet you are missing a solution.
    $endgroup$
    – Marc van Leeuwen
    Jun 7 at 16:30




    $begingroup$
    Yet you are missing a solution.
    $endgroup$
    – Marc van Leeuwen
    Jun 7 at 16:30












    $begingroup$
    Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
    $endgroup$
    – InterstellarProbe
    Jun 7 at 17:27




    $begingroup$
    Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
    $endgroup$
    – InterstellarProbe
    Jun 7 at 17:27












    $begingroup$
    A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
    $endgroup$
    – Marc van Leeuwen
    Jun 7 at 19:34




    $begingroup$
    A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
    $endgroup$
    – Marc van Leeuwen
    Jun 7 at 19:34












    $begingroup$
    @MarcvanLeeuwen Agree to disagree?
    $endgroup$
    – InterstellarProbe
    Jun 7 at 19:36





    $begingroup$
    @MarcvanLeeuwen Agree to disagree?
    $endgroup$
    – InterstellarProbe
    Jun 7 at 19:36













    $begingroup$
    $$ Agree
    $endgroup$
    – Marc van Leeuwen
    Jun 8 at 6:44




    $begingroup$
    $$ Agree
    $endgroup$
    – Marc van Leeuwen
    Jun 8 at 6:44











    5












    $begingroup$

    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 $sneq0$; 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.






    share|cite|improve this answer











    $endgroup$

















      5












      $begingroup$

      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 $sneq0$; 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.






      share|cite|improve this answer











      $endgroup$















        5












        5








        5





        $begingroup$

        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 $sneq0$; 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.






        share|cite|improve this answer











        $endgroup$



        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 $sneq0$; 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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jun 7 at 16:39

























        answered Jun 7 at 16:26









        Marc van LeeuwenMarc van Leeuwen

        90.2k6112235




        90.2k6112235





















            3












            $begingroup$

            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 $$g-1=b/2.$$






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              There is another solution.
              $endgroup$
              – Marc van Leeuwen
              Jun 7 at 16:30















            3












            $begingroup$

            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 $$g-1=b/2.$$






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              There is another solution.
              $endgroup$
              – Marc van Leeuwen
              Jun 7 at 16:30













            3












            3








            3





            $begingroup$

            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 $$g-1=b/2.$$






            share|cite|improve this answer









            $endgroup$



            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 $$g-1=b/2.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 7 at 16:20









            David C. UllrichDavid C. Ullrich

            63.2k44298




            63.2k44298







            • 1




              $begingroup$
              There is another solution.
              $endgroup$
              – Marc van Leeuwen
              Jun 7 at 16:30












            • 1




              $begingroup$
              There is another solution.
              $endgroup$
              – Marc van Leeuwen
              Jun 7 at 16:30







            1




            1




            $begingroup$
            There is another solution.
            $endgroup$
            – Marc van Leeuwen
            Jun 7 at 16:30




            $begingroup$
            There is another solution.
            $endgroup$
            – Marc van Leeuwen
            Jun 7 at 16:30











            2












            $begingroup$

            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



            $b_b = s_b$



            $s_g = frac 12b_s$



            $b_b = b_s -1$



            $s_g = s_b - 1$.






            share|cite|improve this answer











            $endgroup$








            • 1




              $begingroup$
              Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
              $endgroup$
              – InterstellarProbe
              Jun 7 at 17:32
















            2












            $begingroup$

            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



            $b_b = s_b$



            $s_g = frac 12b_s$



            $b_b = b_s -1$



            $s_g = s_b - 1$.






            share|cite|improve this answer











            $endgroup$








            • 1




              $begingroup$
              Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
              $endgroup$
              – InterstellarProbe
              Jun 7 at 17:32














            2












            2








            2





            $begingroup$

            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



            $b_b = s_b$



            $s_g = frac 12b_s$



            $b_b = b_s -1$



            $s_g = s_b - 1$.






            share|cite|improve this answer











            $endgroup$



            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



            $b_b = s_b$



            $s_g = frac 12b_s$



            $b_b = b_s -1$



            $s_g = s_b - 1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 7 at 19:59

























            answered Jun 7 at 16:56









            fleabloodfleablood

            74.4k22892




            74.4k22892







            • 1




              $begingroup$
              Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
              $endgroup$
              – InterstellarProbe
              Jun 7 at 17:32













            • 1




              $begingroup$
              Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
              $endgroup$
              – InterstellarProbe
              Jun 7 at 17:32








            1




            1




            $begingroup$
            Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
            $endgroup$
            – InterstellarProbe
            Jun 7 at 17:32





            $begingroup$
            Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
            $endgroup$
            – InterstellarProbe
            Jun 7 at 17:32


















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