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 $bax$ and
$ayb$ are negative, take the modulus)
This yields a solution of $f = bax$, $e = ayb$, giving
$(ayb)(x+1)$ as the number tins of E, and $(bax)(y+1)$ 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 $(ayb)(x+1) = (7043)(4) = 27 \times 4 = 108$
tins of E, and $(bax)(y+1) = (4330)(8) = 13 \times 8 = 104$ 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 $(bax)$ or $(ayb)$ is
negative, rather than both or none.