Here’s an example. Suppose the problem is: “How much 85% alcohol do you have to add to 1 kilogram of 40% alcohol to get 55% alcohol?” Ethanol has a cute property where “Mixing equal volumes of ethanol and water results in only 1.92 volumes of mixture”, which would be a distraction here, so let’s specify that we’re working with mass fractions, aka alcohol by weight here.
We’re starting with 1 kg of 40% alcohol, which contains .4 kg alcohol in 1 kg of total fluid. We’re adding N kg of 85% alcohol, which contains (.85 N) kg alcohol in N kg of total fluid. So we’ll end up with (.4 + .85 N) kg of alcohol in (1 + N) kg of total fluid. We want 55% alcohol, so this gives us an equation:
(.4 + .85 * N) kg alcohol / (1 + N) kg total fluid = 55% alcohol
We specified that “55% alcohol” means dividing kilograms of alcohol by kilograms of total fluid, so the division is correct here. (Dimensional analysis means more than just making sure that you don’t try to add kilograms to liters. Dividing kg total fluid by kg alcohol would give us a dimensionless number, but not one that could be referred to as “percent alcohol”. Same for dividing kg alcohol by kg water.)
Now we can nuke the units and grind out the math:
(.4 + .85 * N) / (1 + N) = .55
(.4 + .85 * N) = .55 * (1 + N)
.4 + .85 * N = .55 + .55 * N
.4 + .85 * N - .55 * N = .55
.85 * N - .55 * N = .55 - .4
.3 * N = .15
N = .15 / .3
N = .5
Verify:
1 kg of 40% alcohol contains 1 * .4 = .4 kg alcohol
.5 kg of 85% alcohol contains .5 * .85 = .425 kg alcohol
.4 kg alcohol + .425 kg alcohol = .825 kg alcohol
1 kg total fluid + .5 kg total fluid = 1.5 kg total fluid
.825 kg alcohol / 1.5 kg total fluid = .55 YAY!
The way I’d try to do this problem mentally would be:
Relative to the desired concentration of 55%, each unit of 40% is missing .15 units of alcohol, and each unit of 85% has .3 extra units of alcohol. .15:.3=1:2, so to balance these out we need (amount of 40%):(amount of 85%)=2:1, i.e. we need twice as much 40% as 85%. Since we’re using 1kg of 40%, this means 0.5kg of 85%.
That’s clever! Changing your frame of reference is a useful tool—there are a lot of problems which become simpler if you use measurements from a ‘zero’ that you pick.
To put it more generally, for the specified class of problem, you have 2 mixtures of the same 2 substances. Classify one substance (e.g. grams of salt) as the numerator, and either the other substance (e.g. grams of water) or their sum (e.g. grams of saltwater) as the denominator. But be sure to pick one representation and stick with it!
Add the numerators and denominators separately to express the problem algebraically, then solve.
Dimensional analysis is an important tool.
Here’s an example. Suppose the problem is: “How much 85% alcohol do you have to add to 1 kilogram of 40% alcohol to get 55% alcohol?” Ethanol has a cute property where “Mixing equal volumes of ethanol and water results in only 1.92 volumes of mixture”, which would be a distraction here, so let’s specify that we’re working with mass fractions, aka alcohol by weight here.
We’re starting with 1 kg of 40% alcohol, which contains .4 kg alcohol in 1 kg of total fluid. We’re adding N kg of 85% alcohol, which contains (.85 N) kg alcohol in N kg of total fluid. So we’ll end up with (.4 + .85 N) kg of alcohol in (1 + N) kg of total fluid. We want 55% alcohol, so this gives us an equation:
(.4 + .85 * N) kg alcohol / (1 + N) kg total fluid = 55% alcohol
We specified that “55% alcohol” means dividing kilograms of alcohol by kilograms of total fluid, so the division is correct here. (Dimensional analysis means more than just making sure that you don’t try to add kilograms to liters. Dividing kg total fluid by kg alcohol would give us a dimensionless number, but not one that could be referred to as “percent alcohol”. Same for dividing kg alcohol by kg water.)
Now we can nuke the units and grind out the math:
Verify:
The way I’d try to do this problem mentally would be:
Relative to the desired concentration of 55%, each unit of 40% is missing .15 units of alcohol, and each unit of 85% has .3 extra units of alcohol. .15:.3=1:2, so to balance these out we need (amount of 40%):(amount of 85%)=2:1, i.e. we need twice as much 40% as 85%. Since we’re using 1kg of 40%, this means 0.5kg of 85%.
That’s clever! Changing your frame of reference is a useful tool—there are a lot of problems which become simpler if you use measurements from a ‘zero’ that you pick.
To put it more generally, for the specified class of problem, you have 2 mixtures of the same 2 substances. Classify one substance (e.g. grams of salt) as the numerator, and either the other substance (e.g. grams of water) or their sum (e.g. grams of saltwater) as the denominator. But be sure to pick one representation and stick with it!
Add the numerators and denominators separately to express the problem algebraically, then solve.