what you have to do is track the amount of solute. If you add up all the smaller amounts, they must total the total amount.
Now, 5 gal of a 12% solution contains 5*.12 = .6 gal of solute. Thinking along those lines,
70L of 40% antifreeze contains 28L of antifreeze
70L of 20% antifreeze contains 14L of antifreeze
nL of 100% antifreeze contains nL of antifreeze
Now, if we drain nL of 20% solution, we end up with (70-n)L
(70-n)*.2 + n = 70*.4
14 - .2n + n = 28
.8n = 14
n=17.5
so, if we add 17.5L of 100% antifreeze to 52.5L of 20%, we end up with
52.5*.2 + 17.5 = 10.5+17.5 = 28L of antifreeze in 70L, or 40% antifreeze
Help!! I do not understand how to work these types of problems. My instructor gives us really long formulas and charts for these and i don't get it!! Im good at algebra, I have an A in the class, so this is irritating to me!
The radiator in a certain make of car needs to contain 70 liters of 40% antifreeze. the radiator now contains 70 liters of 20% antifreeze. How many liters of this solvent must be drained and replaced with 00% antifreeze to get the desired strength?
1 answer