First, I calculated the prices:

$£24 - 25 \%*£24 = £18$
$£18 - 1/3*£18 = £12$
$£18 - 50\%*£18 = £9$

I used the following notation:

• $x$ - the number of CDs sold by the first store for $£24$
• $y$ - the number of CDs sold by the first store for $£18$
• $z$ - the number of CDs sold by the first store for $£12$
• $u$ - the number of CDs sold by the second store for $£9$

From the first store, with what it sold, I can write the following equation:

$24x + 18y + 12z = 2010$

which can be written as:

$4x + 3y + 2z = 335$ (eqn. $1$)

If the store managed to sell all the CDs, I would have had:

$24x + 18y + 30*18 = 2370$

or

$4x + 3y = 305$ (eqn. $2$)

From the second store, I have the following information:

$18x + 24y + 9u = 2010$ (eqn. $3$)

However, I know that:

$x + y + 30 = x + y + u$

So, $u = 30$

Now, I substitute $u$ in equation ($3$):

$18x + 24y + 9*30 = 2010$

or

$3x + 4y = 290$ (eqn. 4)

Now, I have a system of $3$ equations (($1$), ($2$), ($4$)) with $3$ unknowns ($x$, $y$ and $z$), so that I can calculate them all. First, I use equations ($2$) and ($4$).

From equation ($2$), I write $x$ in function of $y$:

$x = (305 - 3y)/4$ (eqn. $5$)

Now, I add the piece of information from equation ($5$) in equation ($4$), obtaining:

$3 (305 - 3y)/4 + 4y = 290$ (eqn. $6$)

And from equation ($6$), I calculate $y$:

$(16y - 9y + 915) = 1160$
$7y = 245$
$y = 35$ (eqn. $7$)

Now, I calculate $x$ from equation ($5$):

$x = 50$ (eqn. $8$)

And I calculate $z$ from equation ($1$):

$z = 15$ (eqn. $9$)

From $7, 8$ and $9$ we have:

$50$ CDs sold for $£24$
$35$ CDs sold for $£35$
$15$ CDs sold for $£12$.