# Mixing Paints

*Mixing Paints printable worksheet *

A decorator can buy pink paint from two manufacturers.

- Paint A is made up from red and white paint in the ratio
**$1:3$** - Paint B is made up from red and white paint in the ratio
**$1:7$**

The decorator can mix the paints to produce different shades of pink.

If Paint A and Paint B come in the same size cans, what is the least number the decorator would need of each type in order to produce pink paint containing red and white in the following ratios?

$1:5$

$1:6$

*You may wish to use the interactivity below to check your answers.*

Another decorator buys pink paint from two different manufacturers:

- Paint C is made up from red and white paint in the ratio
**$1:4$** - Paint D is made up from red and white paint in the ratio
**$1:9$**

What is the least number the decorator would need of each type in order to produce pink paint containing red and white in the following ratios?

$1:6$

$1:7$

$1:8$

*Once again, you might like to use the interactivity to check your answers.*

For each of the types of paint, did you have a strategy for mixing the new paint colour?

**Is it always possible to combine two paints made up in the ratios**

**$1:x$**and**$1:y$**and turn them into paint made up in the ratio**$1:z$**? (where $x < z < y$)*You might like to try Mixing More Paints next.*

Experiment with different combinations of paints A and B (and paints C and D).

The ratio $1:4 = 2:8 = 3:12 = 4:16$ ...

The ratio $1:5 = 2:10 = 3:15 = 4:20$ ...

B is whiter than A (and D is whiter than C), so what will be special about the combinations that achieve the paler shades of pink?

TK and Stewart from Bangkok Patana School in Thailand warned of some calculations which would give incorrect results:

Some may try to just add together 1 + 1 red : 3 + 7 white

This would be incorrect as the sum of 1 and 3 is 4; the sum of 1 and 7 is 8. 4$\neq$8 (so the cans of paint would not have been the same size).

This method is incorrect. You would not add $\frac{1}{3}$ and $\frac{1}{7}$ by adding the numerators and denominators together; it makes no sense to add together the numbers left of the colon and those right of it to create a new ratio.

Here is another way to go about solving this problem.

We know that one can contains 1 litre of paint.

In the instance of Can A, the ratio is 1:3

This means that there is 250ml of red paint in Can A and 750ml of white paint.

In Can B the ratio is 1:7

From this, we know that there is 125ml of red paint in Can B and 875ml of white paint.

If we were to mix a Can of A and a Can of B, we would have 375ml of red paint and 1625ml of white paint.

Deana, TK and Stewart from Bangkok Patana School, Moon from British International School Phuket in Thailand, Gabriel and Wooyong from BSKL in Malaysia and Abraham from Harrow International School Hong Kong worked out the quantites of Paint A and Paint B needed to get a ratio of 1 : 4. This is Moon's work:

Take 80 litres to be in the cans therefore:

Paint Can A has

Red: White

$20 : 60$

Paint Can B

Red: White

$10 : 70$

And a combination of cans is given by:

$x$(Can A) + $y$(Can B)

So the total red amount is given by $20x + 10y$

And the total white amount is given by $60x + 70y$

i.e the ratio is

Red White

$20x + 10y : 60x + 70y$

So if we want a ratio of $1:4$ this means we have the equation:

$4(20x + 10y) = 60x + 70y$

$\Rightarrow 2x = 3y$

the LCM of $2$ and $3$ is $6,$ so we solve this by $x = 3$ and $y = 2.$

i.e we need $3$ Can As and $2$ Can Bs.

Gabriel used a slightly different method. This is Gabriel's work to find quantities of Paint A and Paint B to get a red:white ratio of 1:5.

Paint A has a red to white ratio of 1:3 and paint B has a red to white ratio of 1:7. Paint A is $\frac14$ red and $\frac34$ white and paint B is $\frac18$ red and $\frac78$ white. By mixing $x$ cans of paint A and $y$ cans of paint B, we get paint with a red to white ratio of $\left(\frac x4+\frac y8\right)$ to $\left(\frac{3x}4+\frac{7y}8\right)$ or $(2x+y)$ to $(6x+7y).$

We need to find integers $x$ and $y$ such that $\dfrac{\frac x4+\frac y8}{\frac{3x}4+\frac{7y}8}=\frac15.$

To do that, plot $\dfrac{\frac x4+\frac y8}{\frac{3x}4+\frac{7y}8}=\frac15$ and find the first point $(x,y)$ at which $x$ and $y$ are integers.

Gabriel plotted this using www.desmos.com. Click here to see Gabriel's graph.

For 1:5, the point is $(1,2) \Rightarrow$ solution is 1 can of paint A and 2 cans of paint B.

Wooyong moved straight on to the ratio $1:k$ for red paint : white paint.

For any value of k, we use the added ratio, $\frac14x + \frac18y : \frac34x + \frac78y$, to solve it (by forming the equation $k\left(\frac14x + \frac18y\right) = \frac34x + \frac78y$).

Abraham used very similar mathematics to answer the questions about Paint C and Paint D, and to generalise:

**Why do this problem?**

This problem requires students to think very carefully about what to use as their basic unit and how to combine ratios. Many students (and teachers!) initially try combining $1:3$ and $1:7$ and assume that the $1$ in each case represents the same quantity of paint, rather than considering a volume of paint and the proportion that is made up of each shade. This offers a great opportunity for
discussion about the importance of checking assumptions carefully!

As students work towards generalisation, they will need to show resilience and perseverance to succeed.**Possible approach***This printable worksheet may be useful: Mixing Paints*

Introduce the two paints from the problem:

- Paint A is made up from red and white paint in the ratio 1:3
- Paint B is made up from red and white paint in the ratio 1:7

"What would the ratio of red to white paint be if the painter mixed one can of Paint A and one can of Paint B?" Give students some thinking time, then invite them to check that they agree with their partner, and finally share answers with the whole class.

Some students will initially give the answer $1:5$, making the argument that if A is $1:3$ and B is $1:7$, A+B is $2:10$ which simplifies to $1:5$. One way to help them to spot their error is to choose a particular size of paint can (1 litre, perhaps):

1 litre of Paint A: 250ml red, 750ml white.

1 litre of Paint B: 125ml red, 875ml white.

Mix them together: 375ml red, 1625ml white, which is in a ratio $3:13$.

Once everyone is happy, invite students to choose some combinations of paints A and B and work out the ratios. Then, once they have had a chance to explore, challenge them to find ways of combining A and B to give the ratios $1:4$, $1:5$, and $1:6$.

If computers/tablets are available, students could use the interactivity from the problem to experiment or to check their answers.

Some students may wish to practise any strategies they have developed by working on paint sets C and D and using the second interactivity. Others might want to get started on the final challenge in the problem: can they come up with a method to combine two paints made up in the ratios 1:x and 1:y and turn them into paint made up in the ratio 1:z?

Allow some time at the end of the lesson for students to share any strategies they came up with for finding particular targets, and if appropriate, to look at a general strategy.

### Key questions

Why might it be sensible to consider cans that hold 8 litres?

$1:3 = 2:6 = 3:9 = ...$ Which of these ratios might be the most useful when mixing paint A with paint B?

What would the paint colours look like? Which would be darkest? Which would be lightest?

Is it possible to make paint D ($1:9$) from paint A and paint B?

### Possible support

Students could use different coloured counters or multilink cubes to represent the cans of paint, being sure to use the same total for each can of paint, or use diagrams by shading squares on squared paper.

### Possible extension

Mixing More Paints invites students to find a strategy for making any ratio.