# Ratios/Proportions of numbers: problem solving

In a bag, there are red and blue cubes in the ratio $4 : 7$. $\quad$ (red : blue) $\;\;4 : 7$.

I add $10$ more red cubes to the bag.

Now there are red and blue cubes in the ratio of $\quad$ (red : blue) $\;\;6 : 7$.

How many blue cubes are in the bag?

I'm not sure, how would you work it out if you haven't got the amount of cubes altogether?

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Thanks I everyone. –  user61406 May 8 '13 at 17:32
Don't forget to accept $\;\large \checkmark\;$ an answer! (But you can only accept one answer, even when lots are helpful. (But you can upvote $\;(\,\uparrow\,)\;$ as many helpful answers as you'd like.) –  amWhy May 8 '13 at 17:38

Let $\color{red}{r}$ be the number of red cubes (at the start), and $\color{blue}{b}$ be the number of blue cubes. From the first statement, we know that: $$\frac{\color{red}{r}}{\color{blue}{b}} = \frac 4 7$$

From the second statement, we know that $$\frac{\color{red}{r + 10}}{\color{blue}{b}} = \frac 6 7$$

Cross multiplying both of these, we obtain: $$7\color{red}r = 4\color{blue}b\tag{1}$$ $$7(\color{red}{r+10}) = 6\color{blue}b\tag{2}$$ Distribute the $7$ through in equation $(2)$: $$7\color{red}{r} + 70 = 6\color{blue}b\tag{3}$$ Now plug the $7\color{red}r$ from equation $(1)$ into equation $(3)$: $$4\color{blue}b + 70 = 6\color{blue}b\tag{4}$$

Collect like terms: $$70 = 2\color{blue}b$$ Dividing the two over: $$\color{blue}b = 35$$

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how would you pug 7r from equations into 3. I got the rest just not sure how you got 4b. –  user61406 May 8 '13 at 17:32
@HonkyHanka In equation $(1)$, we have the result that $7\color{red}r = 4\color{blue}b$. So, wherever we see a $7\color{red}r$, we can replace it with $4\color{blue}b$. We have a $7\color{red}r$ in equation $(3)$, so we replace it with $4\color{blue}b$. –  anorton May 8 '13 at 17:37

The following is the most mechanical approach I can think of.

The current ratio is $4:7$. So there are $4k$ red and $7k$ blue for some unknown number $k$.

When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.

So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.

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Let $x$ be the number of red cubes and $y$ be the number of blue cubes.

To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:

$7x = 4y$.

When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:

$7(x+10) = 6y$.

Expanding, we get a system:

$7x = 4y$

$7x + 70 = 6y$

Can you solve the system of equations from here?

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Let $4x$ be the number of red cubes. Since the ratio is 4:7, this means we have $7x$ blue cubes. If you add 10 more red cubes to the bag, then we have $4x + 10$ red cubes, and still $7x$ blue cubes.
Now the new ratio is 6:7, so let the number of red cubes be $6y$. Then the number of blue cubes is $7y$. This tells us that
$$4x + 10 = 6y$$ $$7x = 7y.$$ Hence we have a system of equations we can solve for. We see that $x = y$ and so
$$4y + 10 = 6y \rightarrow 10 = 2y \rightarrow y = 5,$$ hence $x = 5$ as well. So the number of blue cubes in the bag was $\boxed{7 \cdot 5 = 35}$.