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In a 35 lbs. mixture, the percentage of pure quinine is 60.how much pure quinine should be added to the mixture so that the adulteration is 36%of the total?

In 35 lbs, 60% is quinine, therefore 40% is adulteration.

i.e. amount of adulteration is 0.4 * 35 = 14 lbs

and amount of quinine is 35 – 14 = 21 lbs

Now add x lbs of pure quinine to the mix.

This gives us 35 + x lbs of mix

Percentage of adulteration is 14/(35+x) * 100 = 36 (i.e. 36%)

14/(35+x) = 0.36

14 = 0.36(35+x)

14 = 12.6 + 0.36x

1.4 = 0.36x

x = 1.4/0.36 = 3.8888... lbs

Answer: Amount of quinine to be added, to give 36% adulteration, would be 3.89 lbs

by Level 11 User (81.5k points)

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