# Homework help: Ratios in Mathematics [closed]

I'm trying to solve this ratio problem. Can't seem to get around in solving it.

$750g$ of salt is to be divided between three storage cans $A, B$ and $C$ in the ratios $A:B = 5:2$ and $B:C = 3:2$. What mass of salt will be stored in each can?

A quick solution would be much appreciated.

## closed as off-topic by Dan Rust, user63181, M Turgeon, user7530, Davide GiraudoJan 6 '14 at 22:05

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• maybe its easier to just build a single ratio. multiply $A:B$ by $3$ and $B:C$ by 2, then put them together. $A:B = 15:6$, $B:C=6:4$, so $A:B:C=15:6:4$. – John Joy Sep 17 '14 at 15:22

This is, in my opinion, the most intuitive and easiest solution. Set up the 3 equations

$$\frac{A}{B} = \frac{5}{2} \implies B=\frac{2}{5}A\\ \frac{B}{C} = \frac{3}{2} \implies C = \frac{2}{3}B\\ A+B+C = 750$$

Solving these 3 equations with 3 unknowns yield $A=450,\hspace{5px} B= 180, \hspace{5px} \text{and } C= 120.$ I'm assuming that you know how to solve 3 equations with 3 unknowns, but I'm perfectly willing to help you out if not.

If you use A, B and C as the quantities of salt that end up in each can, then you already know the sum of those quantities. Since both ratios include B in terms of one other quantity, you can do a little algebra to find expressions for A and C in terms of B. Substituting those into your sum should then only require some simple algebra to get your answer.

Hint $\$ Find simple integers $\,a,b,c\,$ in given ratio, then scale them to have the desired sum.

$\rm \color{#c00}2a = 5b\,\Rightarrow\, \color{#c00}2\mid b,\,\ 2b = \color{#0a0}3c\ \Rightarrow \color{#0a0}3\mid b,\,$ so choose $\rm \, b = \color{#c00}2\cdot\color{#0a0}3,\$ so $\rm\ a = 5b/2 = 15,\ c = 2b/3 = 4$.

Then $\rm\, a+b+c = 15+6+4 = 25$ but we want $\,750 = 25\cdot 30,\,$ so scale $\rm\,a,b,c\,$ by $\,30\,$ (which preserves their ratios, i.e. $\rm\, 30a/30b = a/b,\,\ 30b/30c = b/c).$

Remark $\$ By exploiting innate scaling symmetry, i.e. that the ratios are invariant under scalings $\rm\,(a,b,c)\mapsto (an,bn,cn),$ we have greatly simplified the arithmetic, so that only arithmetic of small integers (vs. fractions) is required. The problem simplifies so much that it can be solved in a minute of mental arithmetic (with a little practice). This is but one small example of the general principle that one should always look for special innate structure (here scaling symmetry) before diving head-first into brute force calculations.