There was definitely a method of using Linear Programming.
We clearly have 2 equations:
6x+3y+0.1z=100
x+y+z=100
But, to solve the 3 unknowns, we need 3 equations. Assumed a value of any of the options, and check for equality.
I used a logic that the Blue should be in multiples of ten.
And, considering the total needed to be 100, a majority had to be from the blue.
I started off by assuming Blue = 60, then 70, then 80.
60 seemed too little, because the remaining 40 needed to be filled by the Yellow and Red, and it was not possible.
70 Seemed right, but so did 80. I had a feeling it should be 70, and tried to find a combination for it.
Putting B=70, I had 30 places left.. and $93. Dividing 93 by 3, I got 31. This meant 29 Yellow, and 1 Red.
Crude, but effective. I managed to solve it quicker than it seems, so not a bad effort