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## 'Put Out the Flags' printed from http://nrich.maths.org/

Aditya sent us a very well explained solution
to this problem:

Tim has $50$% blue, $35$% red, $10$% white and $5$%
union jacks.

Beth has $40$% blue, $32$% red, $20$% white and $8$% union jacks.
In fractions, this is:

T = $\frac{1}{2}$ blue, $\frac{7}{20}$red, $\frac{1}{10}$ white
& $\frac{1}{20}$ union jack.

= $\frac{10}{20}$ blue, $\frac{7}{20}$ red, $\frac{2}{20}$ white
and $\frac{1}{20}$ union jacks.

Therefore Tim has $20$
flags .

B = $\frac{2}{5}$ blue, $\frac{8}{25}$ red, $\frac{1}{5}$ white
& 2/25 $\frac{2}{25}$ union jacks.

= $\frac{10}{25}$ blue, $\frac{8}{25}$ red, $\frac{5}{25}$ white
and $\frac{2}{25}$ union jacks.

Therefore Beth has $25$
flags .
Now, we know that Beth has more flags than Tim. Beth has one
more red flag, and both have the same number of blue flags. Between
them, they have $3$ union jacks.

The second part of the problem:

Out of every $20$ flags Tim would have $1$ union jack.

Out of every $25$ flags Beth would have $2$ union jacks.

So first we thought about how many different ways you could
make $10$ union jacks

For each of these we then need toknow the total number of flags
:

Tim |
Beth |
Tim has this number of flags |
Beth has this number of flags |
Total number of flags |

8 |
2 |
160 |
25 |
185 |

6 |
4 |
120 |
50 |
170 |

4 |
6 |
80 |
75 |
155 |

2 |
8 |
40 |
100 |
140 |

That is assuming that they both have some flags!