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xxMikexx [17]
2 years ago
10

How many ounces of 50% alcohol solution must be mixed with 80 ounces of a 20% alcohol solution to make a 40% alcohol solution?

Mathematics
1 answer:
Paraphin [41]2 years ago
8 0
X = amount of 50% solution
y = total amount after mixing

Note that the amount of alcohol in the 20% solution is 80 (0.2) = 16 oz.

We have 2 unknowns, so we need 2 equations:
first, we can write the equation for the total volume:
y = x + 80

next, apply the percentages to get an equation for the amount of alcohol:

0.4 y = 0.5 x + 16

Finally, solve the pair of equations (in this case, by substitution of the 1st into the second:
0.4 (x + 80) = 0.5x + 16
0.4 x + 32 = 0.5X + 16
rearrange and combine like terms:
0.1x = 16
x = 160
from the 1st equation:
y = 160 + 80 = 240
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