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
$begingroup$
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
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
asked Jun 7 at 16:15
MariumMarium
1573
$endgroup$
add a comment |
$begingroup$
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
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
asked Jun 7 at 16:15
MariumMarium
1573
$endgroup$
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
add a comment |
$begingroup$
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
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
asked Jun 7 at 16:15
MariumMarium
1573
$endgroup$
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
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
asked Jun 7 at 16:15
MariumMarium
1573
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
asked Jun 7 at 16:15
MariumMarium
1573
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
edited Jun 7 at 16:22
J. W. Tanner
7,9591723
7,9591723
asked Jun 7 at 16:15
MariumMarium
1573
asked Jun 7 at 16:15
MariumMarium
1573
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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$b$ = 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:
$$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
$endgroup$
$begingroup$
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
$begingroup$
$$ Agree
$endgroup$
– Marc van Leeuwen
Jun 8 at 6:44
add a comment |
$begingroup$
Although this is probably not intentional, this problem has two different solutions. If you assume there are $s>0$ sisters, the boy also has $s$ brothers making $s+1$ boys in all. Each boy being a brother of any of the girls, each having $s-1$ sisters, the equation $s+1=2(s-1)$ easily solves to $s=3$ (four boys and three girls in all).
However, the question does not say that $sneq0$; the boy need not have any sisters. In that case each of the sisters has (whatever) is true; there aren't any sisters so that is vacuously true. So $s=0$ is another solution: a family with one boy and no girls. Technically, in this case there are no brothers and sisters at all in the family (which is what the question was asking), as the one child that is there is neither a sister nor a brother. Actually, nowadays this is the more likely solution, I would think.
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
$endgroup$
add a comment |
$begingroup$
The problem is you're using $b$ for the number of the boy's brothers and also for the number of the girl's brothers.
Instead say there are $b$ boys and $g$ girls. Then a boy has $b-1$ brothers, so $$g=b-1.$$Similarly $$g-1=b/2.$$
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
$endgroup$
1
$begingroup$
There is another solution.
$endgroup$
– Marc van Leeuwen
Jun 7 at 16:30
add a comment |
$begingroup$
I think it'd be easiest to do
$b = $ number of boys
$g = $ number of girls and the essential insight is to notice:
A boy has $b-1$ brothers and $g$ sisters, and a girl has $b$ brother and $g-1$ sisters.
So "A boy has as many sisters as brothers" means $b-1 = g$.
And "each sister has only half as many sisters as brothers" means $ g-1 = frac 12 b$
So solve the two equations two unknowns:
$b-1 = g$
$g-1 =frac 12 b$
......
If you want to use variables for brothers and sisters you can do
$b_b = $ number of brothers a boy has
$b_g = $ number of brothers a girl has
$s_b = $ number of sisters a boy has and
$s_g = $ number of sisters a girl has.
Then our riddle is $b_b = s_b$ and $s_g = frac 12 b_s$.
But to solve it we need to not the unstated but essential a girl has $1$ fewer sister than a boy, and a boy has one fewer brother than a girl. So
$b_b = b_g -1$ and $s_g = s_b - 1$.
So solve the four equations four unknowns
$b_b = s_b$
$s_g = frac 12b_s$
$b_b = b_s -1$
$s_g = s_b - 1$.
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
$endgroup$
1
$begingroup$
Your notation is not clear. You state $b_b=b_g-1$ and then below that, you have $b_b=b_g$. This implies $0=-1$, a contradiction.
$endgroup$
– InterstellarProbe
Jun 7 at 17:32
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$b$ = number of boys = number of brothers each girl has
$g$ = number of girls = number of sisters each boy has
$b-1$ = number of brothers each boy has
$g-1$ = number of sisters each girl has
$$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$$
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
$endgroup$
$begingroup$
Yet you are missing a solution.
$endgroup$
– Marc van Leeuwen
Jun 7 at 16:30
$begingroup$
Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
$endgroup$
– InterstellarProbe
Jun 7 at 17:27
$begingroup$
A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
$endgroup$
– Marc van Leeuwen
Jun 7 at 19:34
$begingroup$
@MarcvanLeeuwen Agree to disagree?
$endgroup$
– InterstellarProbe
Jun 7 at 19:36
$begingroup$
$$ Agree
$endgroup$
– Marc van Leeuwen
Jun 8 at 6:44
add a comment |
$begingroup$
$b$ = 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:
$$b-1-1 = dfracb2 Longrightarrow b=4, g=3$$
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
$endgroup$
$begingroup$
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
$begingroup$
$$ Agree
$endgroup$
– Marc van Leeuwen
Jun 8 at 6:44
add a comment |
$begingroup$
$b$ = number of boys = number of brothers each girl has
$g$ = number of girls = number of sisters each boy has
$b-1$ = number of brothers each boy has
$g-1$ = number of sisters each girl has
$$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$$
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
$endgroup$
$b$ = number of boys = number of brothers each girl has
$g$ = number of girls = number of sisters each boy has
$b-1$ = number of brothers each boy has
$g-1$ = number of sisters each girl has
$$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$$
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
answered Jun 7 at 16:20
InterstellarProbeInterstellarProbe
4,545931
4,545931
$begingroup$
Yet you are missing a solution.
$endgroup$
– Marc van Leeuwen
Jun 7 at 16:30
$begingroup$
Am I though? It depends on how you read the OP's question. The "missing" solution is not an algebraic one. It requires an understanding of logic and set theory. The OP did not explicitly ask for all solutions. The OP asked for a solution derived algebraically.
$endgroup$
– InterstellarProbe
Jun 7 at 17:27
$begingroup$
A solution is a solution, it exists independently of any derivation. And besides, one is always using logic to derive the algebraic conditions from the stated facts before solving them. In doing so, one cannot just assume that something that has to be true for every girl is actually instantiated at least once; that is just bad deduction.
$endgroup$
– Marc van Leeuwen
Jun 7 at 19:34
$begingroup$
@MarcvanLeeuwen Agree to disagree?
$endgroup$
– InterstellarProbe
Jun 7 at 19:36
$begingroup$
$$ Agree
$endgroup$
– Marc van Leeuwen
Jun 8 at 6:44
add a comment |
$begingroup$
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
add a comment |
$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.
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
$endgroup$
add a comment |
$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.
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
$endgroup$
add a comment |
$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.
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
$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.
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
edited Jun 7 at 16:39
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
answered Jun 7 at 16:26
Marc van LeeuwenMarc van Leeuwen
90.2k6112235
90.2k6112235
add a comment |
add a comment |
$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.$$
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
$endgroup$
1
$begingroup$
There is another solution.
$endgroup$
– Marc van Leeuwen
Jun 7 at 16:30
add a comment |
$begingroup$
The problem is you're using $b$ for the number of the boy's brothers and also for the number of the girl's brothers.
Instead say there are $b$ boys and $g$ girls. Then a boy has $b-1$ brothers, so $$g=b-1.$$Similarly $$g-1=b/2.$$
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
$endgroup$
1
$begingroup$
There is another solution.
$endgroup$
– Marc van Leeuwen
Jun 7 at 16:30
add a comment |
$begingroup$
The problem is you're using $b$ for the number of the boy's brothers and also for the number of the girl's brothers.
Instead say there are $b$ boys and $g$ girls. Then a boy has $b-1$ brothers, so $$g=b-1.$$Similarly $$g-1=b/2.$$
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
$endgroup$
The problem is you're using $b$ for the number of the boy's brothers and also for the number of the girl's brothers.
Instead say there are $b$ boys and $g$ girls. Then a boy has $b-1$ brothers, so $$g=b-1.$$Similarly $$g-1=b/2.$$
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
answered Jun 7 at 16:20
David C. UllrichDavid C. Ullrich
63.2k44298
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
add a comment |
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
add a comment |
$begingroup$
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
$b_b = s_b$
$s_g = frac 12b_s$
$b_b = b_s -1$
$s_g = s_b - 1$.
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
$endgroup$
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
add a comment |
$begingroup$
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
$b_b = s_b$
$s_g = frac 12b_s$
$b_b = b_s -1$
$s_g = s_b - 1$.
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
$endgroup$
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
add a comment |
$begingroup$
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
$b_b = s_b$
$s_g = frac 12b_s$
$b_b = b_s -1$
$s_g = s_b - 1$.
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
$endgroup$
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
$b_b = s_b$
$s_g = frac 12b_s$
$b_b = b_s -1$
$s_g = s_b - 1$.
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
edited Jun 7 at 19:59
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
answered Jun 7 at 16:56
fleabloodfleablood
74.4k22892
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
add a comment |
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
add a comment |
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$begingroup$
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