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 35% of the total.

Question

Answer is 5.

Answer 1

Amount of quinine is 60% of 35=60/100*35=3/5*35=21lb. The amount of "adulteration" is 40% (100-60%). If we add x lb of quinine we get 21+x lb of quinine and the mass has increased by x to 35+x lb, so the percentage of quinine is now (21+x)/(35+x)=65/100, because the "adulteration" has reduced from 40/100 to 35/100, and the amount of quinine must be 1-35/100=65/100.

Multiply equation by (35+x): 21+x=0.65(35+x)=22.75+0.65x; this gives us: 0.35x=22.75-21=1.75. So x=1.75/0.35=5lb. 

Answer 2

Quantity in the mixture =35 lb
Quantity of quinine in the mixture =35*60/100=21lb
Quantity of adulteration in the mixture = 35-21=14lb
now, supose x lb of pure quinine is added in the mixture so that the ratio of pure quinine and the aduterant becomes 65:35
=>21+x=65*14/35=26lb
x=26-21=5lb

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