# Finding a unified ratio from two separate ratios

I'm self-studying with Stroud & Booth's amazing Engineering Mathemathics, 7th edition. I'm stuck at a problem set that gives me two ratios of variables A and B, and B and C respectively, an expects me to "unify" them, so to speak, in order to produce an A:B:C ratio. To be more precise:

A, B and C are mixed according to the ratios $$A:B = 2:5$$ and $$B:C = 10:11$$. Find the common ratio $$A:B:C$$.

Just please keep in mind that I'm still on the first chapter, Arithmetic.

• Please don't use codeblocks for anything that is not actual code. It makes reading difficult or impossible for a variety of users, in particular low-vision users. – Nij Mar 9 at 7:31

$$A:B = 2:5$$ and $$B:C = 10:11$$

$$A:B = 4:10$$ and $$B:C = 10:11$$

$$A:B:C = 4:10:11$$

I work this kind of problem out by imagining totals that turn the ratios into amounts. I choose totals to make the arithmetic easy. Since I see $$2$$ and $$5$$ and $$10$$ and $$11$$ I'll suppose I have $$2 \times 10 \times 11 = 220$$ things. Then using $$x$$ to represent the number of things of type $$X$$ I know $$a + b + c = 220.$$ I also know that since $$a/b = 2/5$$, $$2b = 5a$$ and, similarly, $$11b = 10c.$$ Now solve those three linear equations ih three unknowns.

If you don't like the $$220$$ total I started with you can use $$a + b + c = 1$$ instead. Then the answers will be the appropriate fractions of the whole.

You have 2 of A per every 5 of B and you have 10 of B per every 11 of C. It's not hard to see that you need twice the amount of A per B to cover the amount of B per C:

$$2\cdot 2:2 \cdot 5:11\implies\\ 4:10:11$$