Crazy hard math?

Question

Antifreeze A is 18% alcohol. Antifreeze B is 10% alcohol. How many liters of each should be mixed to get 20 liters of a mixture that is 15% alcohol?


Pleeeease explain how in the world you got this answer!

Please explain all of it so I learn it!

Answer 1

This is a "system of equations" problem, where you have to solve for two different equations. Denoting A as the number of liters of antifreeze A and B for antifreeze B, the first equation is...

--- A+B=20

which should be obvious. The second equation (concerning only the alcohol) would then be...

--- 0.18A+0.1B=0.15(A+B)

because for every liter of A, there's 0.18 liters of alcohol and for every liter of B, there's 0.1 liters of alcohol and you're looking to set it up such that 0.15 of the total liters is alcohol. Note that since we already know A+B to be 20 we can simplify the second equation to...

--- 0.18A+0.1B=0.15(20)
--- 0.18A+0.1B=3

So all you need to do is to find A and B so that they fit that equation and that they add up to 20. If A+B=20, then B=20-A, which we can substitute in the latest equation to get...

--- 0.18A + 0.1(20-A)=3 (then just do some algebra)
--- 0.18A + 2 - 0.1A=3
--- 2 + 0.08A=3
--- 0.08A=1
--- A= 1/0.08 = 12.5

The mixture should have 12.5 liters of A and 20-12.5, or 7.5 liters of B. You can then check the answer if you wish...

--- 0.18(12.5) + 0.1(7.5) = 2.25 + 0.75 = 3

Answer 2

let vol of A=x
and of B=20-x,so
(18*x/100)+[10*(20-x)]/100
=15*20/100
solve this for x