Homework Help: Solving Ratios in Mathematics?

The first step is to combine the two ratios A:B and B:C into one ratio A:B:C.

A:B = 5:2 = 15:6
B:C = 3:2 = 6:4
A:B:C = 15:6:4

Now, this means that if you had 25 grams of salt, then 15g would be in can A, 6g in B, and 4g in C.

You have 750 = 25*30 grams of salt, so these numbers must also be multiplied by 30.

450g in A
180g in B
120g in C

@PhiNotPi‘s solution is quicker, but this is probably what schools teach. If you took Pre-Algebra, you might have run into a problem-solving method called substitution. Important parts in bold.

Describe the problem: 750 = A + B + C (it comes out and tells you this)

You got two ratios, so cross-multiple them (this looks nicer written out). B is in both ratios, so let’s solve each ratio for B. All you need to do is cross-multiply:

First ratio:
A/B=5/2
2A=5B
A = 2.5B
basically, we’re saying A is 2.5 times bigger than B

Second Ratio:
B/C = 3/2
2B=3C
C = ⅔B
that’s a 2 divided by 3, in case your font is tiny – we’re saying C is two-thirds the size of B

Using this information in bold, we can substitute A and C in a way that is described in terms of B and use basic algebra to find the value of B:
750 = A + B +C
750 = 2.5B + B + ⅔B
750=(25/6)B
B = 180
this is how much salt is in container B

Then you can pretty easily solve those two cross multiplied problems in bold from before:
A = 2.5B = 2.5(180) = 450

C = ⅔B = ⅔(180) = 120

And we already figured out B = 180

Without seeing @PhiNotPi ‘s neat trick, this is how I would do it.

B = 3/2 C
A = 5/2 B = 5/2×3/2 C = 15/4 C
A + B + C = 750
C + 3/2 C + 15/4 C = 750
25/4 C = 750
C = 4/25×750 = 120 g
B = 3/2 C = 180 g
A = 15/4 C = 450 g