calculate proportion of 15%, 8%, & 3% alcohol required to make 6% alcohol
Answers
The required proportion will be 1:3:5
Explanation:
Let the proportion be x:y:z
Hence total alcohol present in the solution
= 15x + 8y + 3z
Total volume of the solution = 100x + 100y + 100z = 100(x+y+z)
since the final solution will be 6% alcohol
=> volume of alcohol / total volume = 6/100
=> (15x + 8y + 3z)/100(x+y+z) = 6/100
=> (15x + 8y + 3z) = 6(x+y+z)
=> 9x + 2y = 3z
keeping x = 1 as it is the largest concentration and the final concentration is less than it, we get;
9 + 2y = 3z
The nearest possible values that satisfy this relation are, y= 3 , z = 5
so final volume of alcohol = 15x + 8y + 3z
= 15 + 24 + 15 = 54
Final volume of the solution = 100(x+y+z) = 100( 1+3+5) = 900
Hence concentration = 54/900 = 6/100 = 6%
Hence the proportion of 15%, 8%, & 3% alcohol required to make 6% alcohol will be 1:3:5