Here is a well laid out solution from Andrei Lazanu, School 205, Bucharest. Well done Andrei.

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.