A ratio of red beads to black is $r:b=4:3$. Why doesn't this translate to $4r=3b$ instead of $3r=4b$?

Question

Suppose that for every 4 red beads ($r$), there are 3 black beads ($b$)

The ratio of red beads to black beads is $r:b=4:3$

But then why is this ratio not equal to $4r=3b$ and instead $3r=4b$?

I have that 4 red beads ($4r$) is the same as 3 black beads ($3b$) after all....

Answer 1

I can think of two ways to think about this:

1. First think about writing the relation as a fraction:

For every $4$ red beads, there are $3$ blue beads. Then we have $$\frac{r}{b} = \frac{4}{3}$$ You can then cross multiply to get $3r = 4b$.

2. Lowest Common Multiple

"I have that $4$ red beads ($4r$) is the same as $3$ blue beads ($3b$) after all..."

This suggests you don't really get the relationship intuitively, so let me try to explain by thinking about lowest common multiples

Consider the problem where red beads are sold in packs of $4$ and blue beads are sold in packs of $3$.

How many packs of each colour do you need to buy to have the same number of red and blue beads?

This involves finding the lowest common multiple of $4$ and $3$ which is $12$.

You can therefore buy $3$ packs of red beads and $4$ packs of blue beads to give you the same amount.

The ratio is exactly the same. You have $4$ red beads for every $3$ blue beads.

So if you have $4$ red beads, there are $3$ blue beads. If you have $8$ red beads, there are $6$ blue beads. If you have $12$ red beads, there are $9$ blue beads. If you have $16$ red beads, there are $12$ blue beads.

So you can immediately see that $3 \times 4$ red beads = $4 \times 3$ blue beads.

Generalising, it's just $3 \times$ number of red beads = $4 \times$ number of blue beads, so $3r = 4b$

Hope that helps.

Answer 2

$$r:b=4:3$$ $$\frac rb=\frac43$$ $$3r=4b$$

Answer 3

If you take 4 red beads at a time, you'll have 3 black beads . The ratio is then simply: $$r:b=4:3$$ So: \begin{align} \frac rb&=\frac 43 \\[2ex] \implies3r&=4b \; \;\;\;\text{(by cross multiplication)} \end{align}