First we must work out the number of tickets of each colour. Let the numbers of red, white and blue tickets be $$r$$, $$w$$ and b respectively; we have $$r : w = 5 : 4$$ and $$w : b = 8 : 7$$. Writing $$r : w$$ as $$10 : 8$$, we can combine the ratios to find $$r : w : b = 10 : 8 : 7$$. Hence for a total of $$100$$ tickets, there must be $$40$$ red, $$32$$ white and $$28$$ blue tickets. Now the largest number of red tickets one can select without having half of them is $$19$$, and likewise $$15$$ and $$13$$ for white and blue tickets. So it would be possible to select $$19 + 15 + 13 = 47$$ tickets and still not have half or more of any particular colour of ticket. However, the very next ticket, the $$48\text{th}$$, would have to be a red, a white or a blue one, and so one of the colours would now have at least half of its number selected. 
A bag contains $$100$$ raffle tickets in red, white and blue. The ratio of red to white tickets is $$5:4$$ and the ratio of white to blue tickets is $$8:7$$. What is the smallest number of tickets that would have to be picked to make sure having selected at least half of \textbf{one} colour? 
Options:
a: $$48$$ 
b: $$49$$ 
c: $$50$$ 
d: $$51$$ 
e: $$52$$ 