Hello, I have another extra credit problem: A number of girls form a singing group. All but 3 girls are blondes, all but 4 are brunettes, and all but 5 are redheads. How many girls in the group? I started with 5 girls are not redheads so that means they are blondes or brunettes. 3 girls are not blondes so the 5 that are not redheads, 2 are blondes. 4 are not brunettes so 1 of the five could be brunette. That leaves 5 not red, 3 of the 5 not red or blonde, 4 of the 5 not brunett or red.Now I am stuck despite various equations I have come up with.Any help would be great. Thank you
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Looks easy, but it is not
- Thread starter irene12
- Start date
Let b=# blondes, d=#brunettes, r=# redheads.
All BUT 3 are blonde:
b+d+r-3=b
All BUT 4 have dark hair:
b+d+r-4=d
All BUT 4 are redheads:
b+d+r-5=r
This creates a system of equations:
\(\displaystyle d+r=3\)
\(\displaystyle b+r=4\)
\(\displaystyle b+d=5\)
Solve the little system for b,d, and r.
Shouldn't be too bad.
All BUT 3 are blonde:
b+d+r-3=b
All BUT 4 have dark hair:
b+d+r-4=d
All BUT 4 are redheads:
b+d+r-5=r
This creates a system of equations:
\(\displaystyle d+r=3\)
\(\displaystyle b+r=4\)
\(\displaystyle b+d=5\)
Solve the little system for b,d, and r.
Shouldn't be too bad.
Hello, irene12!
\(\displaystyle \text{Let }x \,=\, \text{number of blondes.}\)
\(\displaystyle \text{Let }y \,=\,\text{number of brunettes.}\)
\(\displaystyle \text{Let }z \,=\,\text{number of redheads.}\)
\(\displaystyle \begin{array}{ccccccc}\text{All but 3 are blondes: }\; & y + z \:=\: 3 & [1] \\ \text{All but 4 are brunettes:} & x + z \:=\:4 & [2] \\ \text{All but 5 are redheads: } & x + y \:=\:5 & [3] \end{array}\)
\(\displaystyle \text{Add the equations: }\:2x + 2y + 2z \:=\:12 \quad\Rightarrow\quad x + y + z \:=\:6\)
\(\displaystyle \text{Therefore, thre are 6 girls in the group.}\)
A number of girls form a singing group.
All but 3 girls are blondes, all but 4 are brunettes, and all but 5 are redheads.
How many girls in the group?
\(\displaystyle \text{Let }x \,=\, \text{number of blondes.}\)
\(\displaystyle \text{Let }y \,=\,\text{number of brunettes.}\)
\(\displaystyle \text{Let }z \,=\,\text{number of redheads.}\)
\(\displaystyle \begin{array}{ccccccc}\text{All but 3 are blondes: }\; & y + z \:=\: 3 & [1] \\ \text{All but 4 are brunettes:} & x + z \:=\:4 & [2] \\ \text{All but 5 are redheads: } & x + y \:=\:5 & [3] \end{array}\)
\(\displaystyle \text{Add the equations: }\:2x + 2y + 2z \:=\:12 \quad\Rightarrow\quad x + y + z \:=\:6\)
\(\displaystyle \text{Therefore, thre are 6 girls in the group.}\)
A number of girls form a singing group. All but 3 girls are blondes, all but 4 are brunettes, and all but 5 are redheads. How many girls in the group?
Assuming there is at least 1 redhead, then the group is at least 6.
If all but 3 were blondes, then there were 6 - 3 = 3 blondes making the remaining 2 brunettes yielding
Red....Brun....Brun....Blon....Blon....Blon
Assuming there is at least 1 redhead, then the group is at least 6.
If all but 3 were blondes, then there were 6 - 3 = 3 blondes making the remaining 2 brunettes yielding
Red....Brun....Brun....Blon....Blon....Blon