L1, L2 and L3 are three solutions of alcohol and water, with the alcohol percentages (by volume) in them being 20%, c% and 2c%, respectively. If the alcohol percentage (by volume) in the mixture obtained by mixing 1 litre, 3 litres and 2 litres of L1, L2 and L3, respectively, is 22%, find the value of c
Question
L1, L2 and L3 are three solutions of alcohol and water, with the alcohol percentages (by volume) in them being 20%, c% and 2c%, respectively. If the alcohol percentage (by volume) in the mixture obtained by mixing 1 litre, 3 litres and 2 litres of L1, L2 and L3, respectively, is 22%, find the value of c
Solution
To solve this problem, we can use the concept of weighted averages.
Step 1: Understand the problem We have three solutions L1, L2, and L3 with alcohol percentages of 20%, c%, and 2c%, respectively. We mix 1 litre of L1, 3 litres of L2, and 2 litres of L3 to get a mixture with an alcohol percentage of 22%.
Step 2: Set up the equation The total alcohol in the final mixture is the sum of the alcohol in each of the individual solutions. We can express this as:
(1 litre * 20%) + (3 litres * c%) + (2 litres * 2c%) = (1 litre + 3 litres + 2 litres) * 22%
Step 3: Simplify the equation 0.2 litres + 0.03c litres + 0.04c litres = 1.32 litres
Step 4: Solve for c Combine like terms on the left side of the equation to get:
0.2 litres + 0.07c litres = 1.32 litres
Subtract 0.2 litres from both sides to isolate the term with c:
0.07c litres = 1.12 litres
Finally, divide both sides by 0.07 to solve for c:
c = 1.12 litres / 0.07 = 16
So, the value of c is 16. This means that the alcohol percentage in solution L2 is 16%, and in solution L3 it is 2*16% = 32%.
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