Mixing More Paints
Mixing More Paints printable worksheet
This problem follows on from Mixing Paints.
A decorator can buy blue paint from two manufacturers.
 Paint A is made up from dark blue and white paint in the ratio $1:4$
 Paint B is made up from dark blue and white paint in the ratio $1:5$
She can mix the paints to produce different shades of blue.
What is the least number she would need of each type in order to produce blue paint containing dark blue and white in the following ratios:
$3:14$
$10:43$
You may wish to experiment with the interactivity below.
Another decorator buys blue paint from two different manufacturers:
 Paint C is made up from dark blue and white paint in the ratio $1:3$
 Paint D is made up from dark blue and white paint in the ratio $1:7$
$3:14$
$10:43$
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 $a:b$ ? Experiment with a few more examples.
Can you describe an efficient way of doing this?
Take a look at Mixing Paints
Experiment with different combinations of paints A and B (and paints C and D).
Different mixes produce paints that are somewhere in between the shades of paints A and B (or paints C and D).
We received several incorrect solutions like the ones below:
Combining paints A ($1:4$) and B ($1:5$):
Required
Ratio

Amount of
paint A

Amount of
paint B

$2:9$  $1$  $1$ 
$3:14$  $1$  $2$ 
$10:43$  $7$  $3$ 
Required
Ratio

Amount of
paint C

Amount of
paint D

$2:9$  $5$  $3$ 
$3:14$  $7$  $5$ 
$10:43$  $27$  $13$ 
However, one can of paint C and one can of paint D does not produce paint in the ratio $2:10$ (or $1:5$), since that would assume that the one part blue in can C has the same volume as the one part blue in can D.
This can't be the case since there are $4$ parts in can C and $8$ parts in can D,
so $1/4$ of can C is blue and $1/8$ of can D is blue.
Ian and Amos from South Island School in Hong Kong sent an picutre of their work for making paint in the ratio $2 : 9.$ There is also an explanation about what they have done.
$\frac{11}{30}$ refers to the case where $a$ and $b$ are both $1,$ so $1$ can of each paint. This is the amount of blue paint in the mixture.
The desired proportion of blue paint is $\frac 2 {11}.$ This is equivalent to $\frac{60}{330}.$ It is multiplied by $\frac{30}{30}$ because the amount of blue paint will be $\frac{6a+5b}{30}.$
The proportion of blue paint will be $\dfrac{\frac{6a+5b}{30}}{a+b}$ which is equal to $\dfrac{6a+5b}{(a+b)\times30},$ so they have equated this to $\frac{60}{330}.$ This gives $a+b=11$ and $6a+5b=60.$
Navjot from Sherborne Qatar used algebra to find a general solution. Navjot's solution begins by using the simpler case in Mixing Paints.
It is possible to find a ratio of paint $1:z$ with 2 colours of paint with ratios $1:x$ and $1:y$ where $x\lt z \lt y$.
To start off, we first need to know the fact that the mixing of $2$ paints essentially gives us the average of the two ratios. Eg; mixing $1:2$ and $1:5$ gives us $1:3,$ which is $\dfrac{\frac13+\frac16}2.$
So, let:The number of pots of paint A $= a,$
The number of pots of paint B $= b,$
The ratio of paint A $= 1:x,$
The ratio of paint B $= 1:y,$
And, the target ratio $= 1:z.$
Since we know that the mixing of paints A and B will give us the mean of the two ratios:
$\dfrac{a\frac1{x+1}+b\frac1{y+1}}{a+b} = \frac1{z+1}$
(I am using $x+1, y+1,$ and $z+1$ rather than $x, y,$ and $z$ because both terms of a ratio add up to make a whole. Eg; $1:3$ means that something is divided up into $\frac14$ and $\frac34,$ not $\frac13$ and $\frac33$)
Now, we split the equation above into 2 parts, one being the numerator of both sides of the equation and the other being the denominator:
$$\begin{align} & a\tfrac1{x+1}+b\tfrac1{y+1}=1\\
\Rightarrow &\tfrac{a(y+1)+b(x+1)}{(x+1)(y+1)}=1\\
\Rightarrow &a(y+1)+b(x+1)=(x+1)(y+1)\end{align}$$ and $$a+b = z+1\Rightarrow a=(z+1)b$$
Substituting 2) into 1),
$$\begin{align}& ((z+1)  b)(y+1) + b(x+1) = (x+1)(y+1)\\
\Rightarrow &(z+1)(y+1)  b(y+1) + b(x+1) = (x+1)(y+1)\\
\Rightarrow &b((x+1)(y+1)) = (x+1)(y+1)(z+1)(y+1)\\
\Rightarrow &b(xy) = (y+1)((x+1)(z+1))\\
\Rightarrow& b = \frac{(y+1)(xz)}{xy}\end{align}$$
Therefore, $a = z+1  \dfrac{(y+1)(xz)}{xy}$
However, this general solution does give answers in fractions or in unsimplified form.
The only difference here is that the target ratios are in the form $a:b$ rather than $1:x.$ However, this $m:n,$ a ratio simplified to its simplest terms and where $a$ and $b$ are natural
numbers, can be converted into a unitary ratio (a ratio in the form $1:x$), dividing both terms by $m$ gives us the ratio $1:\frac nm,$ which is in the form of $1:x.$
Using paints of ratios $1:4$ and $1:5$
Target ratio:
$2:9 = 1:\frac92 = 1:4.5$
$x =4, y = 5, z = 4.5$
$a = 5.5  \frac{6\times0.5}{1}=\frac52$
$b = \frac{6(0.5)}{1} = 3$
So the decorator needs $\frac52$ pots of paint A and $3$ pots of paint B.
However, the answer gives us a fraction rather than a natural number, and we can't buy parts of pots of paint, so we find a common denominator for both the numbers of pots and then remove the common factor, which would be the common denominator.
$\frac52, \frac31 $ are equivalent to $\frac52$ and $\frac 62$
Answer = $5$ pots of A and $6$ pots of B.
Keeping in mind how to handle the fractions we get in this solution, let's move on to the next answers.
$3:14 = 1:\frac{14}3$
$x = 4, y = 5, z = \frac{14}3$
$a = \frac{14}3+1  \dfrac{6\times\frac23}{1}=\frac53$
$b = \dfrac{6\times\frac23}{1}=4$
$\frac53$ of A and $4$ of B
$= 5$ of A and $12$ of B
$10:43 = 1: \frac{43}{10} = 1:4.3$
$x = 4, y = 5, z = 4.3$
$a = 5.3  \frac{6(0.3)}{1} = \frac72$
$b = \frac{6(0.3)}{1} = \frac95$
$\frac72$ pots of A and $\frac95$ pots of B
Or, $\frac{35}{10}$ pots of A and $\frac{18}{10}$ pots of B
Or, $35$ pots of A and $18$ pots of B
Using paints with ratios $1:3$ and $1:7$
Target ratio:
$2:9 = 1:4.5$
$x =3, y = 7, z = 4.5$
$a = 5.5  \frac{8(1.5)}{4} = \frac52$
$b = 8\frac{1.5}{4} = 3$
$\frac52$ of A and $3$ of B
Or, $5$ of A and $6$ of B
$3:14 = 1:\frac{14}3$
$x = 3, y = 7, z = \frac{14}3$
$a = \frac{14}3+1  \dfrac{8(\frac{5}3}{4} = \frac73$
$b = \dfrac{8(\frac{5}3}{4} = \frac{10}3$
$\frac73$ of A and $\frac{10}3$ of B
Or, $7$ of A and $10$ of B
$10:43 = 1:4.3$
$x = 3, y = 7, z = 4.3$
$a = 5.3  \frac{8(1.3)}{4} = \frac{27}{10}$
$b = \frac{8(1.3)}{4} = \frac{13}5$
$\frac{27}{10}$ pots of A and $\frac{13}5$ pots of B
Or, $27$ pots of A and $26$ pots of B
It is possible to achieve a ratio of $a:b$ by using paints of ratios $1:x$ and $1:y$ only if $x\lt\frac ba\lt y.$
In an algebraic sense too, let's consider the general solution for $b,$ which is $\frac{(y+1)(xz)}{xy}.$ Since $x \lt y,$ the denominator is negative, but if $z\lt x,$ the numerator remains positive, thus giving a negative answer, but we can't have negative number of pots.
If $z\gt y,$ the value for $b$ will be positive since the numerator will be negative like the denominator, but, the value for $b$ is greater than $z+1,$ and the general solution for $a = z+1  \frac{(y+1)(xz)}{xy} = (z+1)  b,$ since $b \gt z+1,$ the value for $a$ is negative, so it is not a valid solution.
Therefore, to get a ratio $a:b$ using paints with ratio $1:x$ and $1: y$ where $x\lt y, $ $x\lt \frac ab \lt y.$
Why do this problem?
This problem follows on from Mixing Paints and encourages students to think about ratio in new ways, explore using an interactive environment, and come up with some generalisations and proofs. The most general case will require perseverance to discover and prove.
Possible approach
This printable worksheet may be useful: Mixing More Paints
This problem could be used in a followup lesson after working on Mixing Paints, or as an extension activity for some students.
Students could start by using the interactivity to experiment, and then gradually move to penandpaper methods, only using the interactivity to check.
Key questions
If I am mixing $1:4$ paint with $1:5$ paint, why might it be useful to start with 30 litres of paint?
If I want to mix $1:x$ paint with $1:y$ paint, what would I start with instead of 30 litres?
Possible support
Make sure students are secure in any general strategies they came up with for Mixing Paints before embarking on this task.
Possible extension
Coming up with a general strategy for making any ratio from any paint is a challenging extension task.