Copyright © University of Cambridge. All rights reserved.

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

Show menu


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
Tim Beth
8 2
6 4
4 6
2 8

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!