Finding percents 6% of 300
2 answers:
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So hmm the food C = 6p + 3q and the food D = 4.5p + 1.5q
now, if we check how much is the ratios of C/D for each component, then, mixture M = 144p + 60q, must contain the same ratio for each "p" and "q" component
so, if we divide the 144p by 4+3, or 7 even pieces, 4 must belong to food C and 3 to food D, retaining the ratio of 4/3
and we do the same for 60q, we divide it in 2+1 or 3 even pieces, but that one is very clear, 2 must belong to food C and 1 to food D, 60/3 is clear ends up with a ratio of 40 and 20
now the "p" part... ends up as
that's what I see it, as the ratio of 4/3 and 2/1 being retained in the mixture M
now, if we check how much is the ratios of C/D for each component, then, mixture M = 144p + 60q, must contain the same ratio for each "p" and "q" component
so, if we divide the 144p by 4+3, or 7 even pieces, 4 must belong to food C and 3 to food D, retaining the ratio of 4/3
and we do the same for 60q, we divide it in 2+1 or 3 even pieces, but that one is very clear, 2 must belong to food C and 1 to food D, 60/3 is clear ends up with a ratio of 40 and 20
now the "p" part... ends up as
that's what I see it, as the ratio of 4/3 and 2/1 being retained in the mixture M