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Mixing More Paints

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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

Combining paints C ($1:3$) and D ($1:7$):

Required
Ratio
Amount of
paint C
Amount of
paint D
$2:9$ $5$ $3$
$3:14$ $7$ $5$
$10:43$ $27$ $13$

They are based on the misconception that you can add the ratios to work out the necessary combinations. The solutions given have assumed that the 'parts' in the ratios are of equal size so that a can in the ratio $1:3$ contains half the amount of the one in the ratio $1:7$.

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 red in can C has the same volume as the one part red 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 red and $1/8$ of can D is red.

To compare equal quantities we will need to express the ratio of the colours in can C as $2:6$, so we have:
in can C: $2/8$ red and $6/8$ white
in can D: $1/8$ red and $7/8$ white

Combining one can of each paint will now give us
$3/8$ red and $13/8$ white,
that is, paint in the ratio $3:13$

Andy from Clitheroe Royal Grammar School sent us his work on this question.

Going with the initial example, note that if we call a single tin of paint '$1$ unit', $5$ tins of Paint A gives $1$ unit of red paint and $4$ of white. Similarly, $6$ tins of Paint B gives $1$ unit of red paint and $5$ of white.

Mixing $5$ tins of A with $6$ tins of B gives $1$ red and $4$ white + $1$ red and $5$ white $= 2$ red and $9$ white $= 2:9$. Similarly $5$ A with $12$ B gives $(1R4W) + 2(1R5W) = 3R14W = 3:14$, and $35$ A to $18$ B is $7(1R4W) + 3(1R5W) = 10R43W = 10:43$. Since $5$ and $6$, $5$ and $12$, and $35$ and $18$ are respectively coprime, these are the 'minimums'.

Similarly, $4$ tins of Paint C gives $1$ unit of red and $3$ units of white, and $8$ tins of D give $1$ unit of red and $7$ of white. To get $2:9$ this time is slightly harder.

We set up the identity $c(R + 3W) + d(R + 7W) \equiv z(2R + 9W)$,where $c$ and $d$ represent the number of tins of Paint C and Paint D divided by $4$ and $8$ respectively.

We get $c + d = 2z$ and $3c + 7d = 9z$, which resolves to $5d = 3c$, to which the minimum solution is $d=3, c=5$ ($20$ tins of C and $24$ tins of D). This gives $5(1R 3W) + 3(1R 7W) = 8R:36W = 2:9$. Since $20$ and $24$ are not coprime, this isn't the minimum solution, instead we can divide both sides by $4$ to give $5$ tins of C and $6$ of D, which is the minimum solution.

Similarly, to get $3:14$, we use $c=7, d=5$, giving $28$ and $40$ tins of C and D, or $7$ of C and $10$ C (minimum), and $27$ tins of C and $26$ of D to get 10:43.

To get a general solution, "combining two paints made up in the ratios $1:x$ and $1:y$ and turn them into paint made up in the ratio $a:b$", we set up the following identity.

Let $e$ be the number of tins of $1:x$ paint and f be the number of tins of $1:y$ paint, divided by $(x+1)$ and $(y+1)$ respectively.
$$e(R + xW) + f(R + yW) = z(aR + bW)$$
$$e + f = az$$
$$ex + fy = bz$$
$$be + bf = abz = aex + afy$$
$$e(b - ax) = f(ay - b)$$ (if $b-ax$ and $ay-b$ are negative, take the modulus)
This yields a solution of $f = b-ax$, $e = ay-b$, giving $(ay-b)(x+1)$ as the number tins of E, and $(b-ax)(y+1)$ as the number of tins of F, which may or may not be a minimum. Going back to the final example, where we used paints of $1:3$ and $1:7$ to get $10:43$:
$$x = 3$$ $$y = 7$$ $$a = 10$$$$b = 43$$
We use $(ay-b)(x+1) = (70-43)(4) = 27 \times 4 = 108$ tins of E, and $(b-ax)(y+1) = (43-30)(8) = 13 \times 8 = 104$ tins of F. This simplfies to $27$ E to $26$ F, as found previously.
Therefore, using the general formula, it is always possible to combine paints to give desired ratios, unless the desired ratio is outside the range of the original paints (eg $1:2$ from paints of $1:3$ and $1:4$), in which case only one of $(b-ax)$ or $(ay-b)$ is negative, rather than both or none.