, 2013

We denote the number of tickets of each of the five colours by the first letter of the colour.

We are given that $b:g:r=1:2:4$ and that $g:y:o=1:3:6$.

Through multiplication by 2, the ratio $1:3:6$ is equivalent to the ratio $2:6:12$.

Thus, $g:y:o=2:6:12$.

We chose to scale this ratio by a factor of 2 so that the only colour common to the two given ratios, green, now has the same number in both of these ratios.

That is, $b:{\bf g}:r=1:{\bf2}:4$ and ${\bf g}:y:o={\bf2}:6:12$ and since the term $g$ is 2 in each ratio, then we can combine these to form a single ratio, $b:g:r:y:o=1:2:4:6:12$.

This ratios tells us that for every blue ticket, there are 2 green, 4 red, 6 yellow, and 12 orange tickets.

Thus, if there was only 1 blue ticket, then there would be $1+2+4+6+12=25$ tickets in total.

However, we are given that the box contains 400 tickets in total.

Therefore, the number of blue tickets in the box is $\frac{400}{25}=16$.

Through multiplication by 16, the ratio $b:g:r:y:o=1:2:4:6:12$ becomes $b:g:r:y:o=16:32:64:96:192$.

(Note that there are $16+32+64+96+192=400$ tickets in total.)

Next, we must determine the smallest number of tickets that must be drawn to ensure that at least 50 tickets of one colour have been selected.

It is important to consider that up to 49 tickets of any one colour could be selected without being able to ensure that 50 tickets of one colour have been selected.

That is, it is possible that the first 195 tickets selected could include exactly 49 orange, 49 yellow, 49 red, all 32 green, and all 16 blue tickets ($49+49+49+32+16=195$).

Since all green and blue tickets would have been drawn from the box, the next ticket selected would have to be the 50$^{th}$ orange, yellow or red ticket.
Thus, the smallest number of tickets that must be drawn to ensure that at least 50 tickets of one colour have been selected is 196.