To keep ratios unchanged, How much white colour do we have to add?

Question

The ratio of the colours given below.

$\dfrac{R}{B} = \dfrac {3}{4} \space , \dfrac{R}{W} = \dfrac{12}{5}$

Where $B$ is blue colour, $R$ is red colour and $W$ is white colour. $48$ gr of the blue colour is added into the mixture that is made by using the ratio of the colours. To keep the ratios unchanged, How much white colour do we have to add?

Let's recall $R = 12k$, $B = 16k$ and $W = 15k$ then we have

$$12k + 16k + 15k = 43k $$

However, there will be no solution from here. Could you take a look at it?

Regards

Answer 1

To maintain the ratio, add colors in the same ratio. If you add $48$ blue, then add $36$ red, because $$\frac{36}{48}=\frac{3}{4}=\frac{R}{B}$$

Next, since you're adding $36$ red, you need to add $15$ white, because $$\frac{36}{15}=\frac{12}{5}=\frac{R}{W}$$

Answer 2

This $$12k + 16k + 15k = 43k$$ should have been $$12k + 16k + 5k = 33k$$

Now if you add $48g$ to your black then you need to add $36g$ to you red and $15g $to your white to keep the same ratio.

Answer 3

==== new answer: easier ======

Maybe easier:

If you are adding $48$ g of blue, you must add $r$ g of red in proportion to $\frac 34$. So we need $\frac r{48} = \frac 34$. So what is $r$?

And if we are adding $r$ g of red, you must add $w$ g of white in proportion to $\frac {12}{5}$. So we need $\frac rw = \frac {12}{5}$. So what is $w$?

It doesn't actually matter what we started with. To keep proportions, we must add in proportion.

===== old answer: slightly harder, more detail but more thorough =====

$12k + 16k + 15k = 43k$

That is not correct. It should be $12k + 16k + 5k = 33k$.

$\frac RB = \frac 34$ so if you have $3n$ grams of red for some unknown quantity of $n$, you must have $4n$ grams of blue. So that the proportions are $\frac 34$. That is $\frac {3n}{4n} = \frac 34$.

$\frac RW = \frac {12}{5}$ so if you have $12k$ grams of red for some unknown quantity of $k$, you must have $5k$ grams of blue. $\frac {12k}{5k} = \frac {12}{5}$.

Now you have $3n$ grams of red and that is the same as $12k$ grams of red. So $n = 4k$. So you have $4n = 4*4k = 16k$ of blue.

So you have $12k$ of red; $16k$ of blue, and $5k$ of white.

So you have $12k + 16k + 5k = 33k $ grams total. But that's completely irrelevant.

Now you add $48$ grams of blue so you know have $16k + 48$ grams of blue.

But $\frac {12k}{16k + 48} \ne \frac 34$. We must add some red to keep the proportions correct.

So we must add $r$ grams of red to get:

$$\frac {12k + r}{16k + 48} = \frac 34$$

But now $\frac {12k + r}{5k} \ne \frac {12}{5}$! We must add some white to keep the proportions.

So we must add $w$ grams of white to get:

$$\frac {12k + r}{5k + w} = \frac {12}{5}$$

Now, we do have the equation $(12k + r) + (16k+48)+(5k + w) = 33k + r + 48 + w$ but... that is completely irrelevant.

So if $\frac {12k + r}{16k + 48} = \frac 34$ then

$4(12k + r) = 3(16k + 48)$.

and if $\frac {12k + r}{5k + w} = \frac {12}{5}$ then

$5(12k + r) = 12(5k + w)$.

Solve for $w$.

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Note: You will never know what $k$ is. That is not the point.