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# Well Read

**Answer**: 7 girls

$b$ boys and $g$ girls, so $\frac{b+g}{3}$ teddy bears.

Books taken out: $$\begin{align}12b+17g+9\times\tfrac{b+g}{3}&=305\\

\Rightarrow 12b+17g+3(b+g)&=305\\ \Rightarrow15b+20g&=305\end{align}$$

Divide through by $5$: $$3b+4g=61$$ $b$ and $g$ must be whole numbers

Also need the number of teddy bears to be an integer so the total number of students is a multiple of $3$

$3+13=16$ no

$7 + 10 = 17$ no

$11 +7=18$ yes

$15+4=19$ no

$19+1=20$ no

So the number of girls is $7$.

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Age 14 to 16

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$b$ boys and $g$ girls, so $\frac{b+g}{3}$ teddy bears.

Books taken out: $$\begin{align}12b+17g+9\times\tfrac{b+g}{3}&=305\\

\Rightarrow 12b+17g+3(b+g)&=305\\ \Rightarrow15b+20g&=305\end{align}$$

Divide through by $5$: $$3b+4g=61$$ $b$ and $g$ must be whole numbers

$3b$ | $61-3b$ | Divisible by $4$? |

$3$ | $58$ | no |

$6$ | $55$ | no |

even multiples of $3$ give odd numbers so only try odd multiples of $3$ |
||

$9$ | $52$ | yes, $4\times13$; $3$ boys and $13$ girls |

$15$ | $46$ | no |

$21$ | $40$ | yes, $4\times10$; $7$ boys and $10$ girls |

$27$ | $34$ | no |

go up/down in $12$s (multiple of $4$) not $6$s to hit the multiples of $4$ |
||

$33$ | $28$ | yes, $4\times7$; $11$ boys and $7$ girls |

$45$ | $16$ | yes, $4\times4$; $15$ boys and $4$ girls |

$57$ | $4$ | yes, $4\times1$; $19$ boys and $1$ girl |

Also need the number of teddy bears to be an integer so the total number of students is a multiple of $3$

$3+13=16$ no

$7 + 10 = 17$ no

$11 +7=18$ yes

$15+4=19$ no

$19+1=20$ no

So the number of girls is $7$.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.