CD Heaven
All CD Heaven stores were given the same number of a popular CD to sell for £24. In their two week sale each store reduces the price of the CD by 25% ... How many CDs did the store sell at each price?
Problem
All CD Heaven stores were given the same number of a popular CD to sell for £24.
In their two week sale each store reduced the price of the CD by 25% and any further reductions were then left to the discretion of the store managers.
At the end of the two week sale one store still had 30 CDs left and the manager decided to try to clear the remainder by reducing the price by a further $\frac{1}{3}$ but still did not manage to sell them all. The total sales for the CDs was £2010.
If the store had managed to sell all its CDs by the end of the two week sale period it would have received £2370.
Another store sold the same number of CDs before and during the first two weeks of the sale as the first store but in reverse order (the number they sold for £24 was the same as the number sold in the two weeks of the sale by the first store and vice versa).
At the end of the two weeks the shop manager decided to reduce the price of the CDs by a further 50% and managed to sell all the remaining 30 CDs making her total sales £2010.
How many CDs did the first store sell at each price?
Getting Started
There are four unknown numbers of CDs for each of the different price brackets (including nothing) so you will need four equations.
Student Solutions
Here is a well laid out solution from Andrei, School 205, Bucharest. Well done Andrei.
First, I calculated the prices:
£$24 - 25 \% \times $ £$24 = $ £$18$
£$18 - \frac13 \times $ £$18 = $ £$12$
£$18 - 50\% \times $ £$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:
$24x + 18y + 30\times18 = 2370$
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 ($3$):
$18x + 24y + 9\times30 = 2010$
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 $x$ as a function of $y$:
$x = (305 - 3y)/4$ (eqn. $5$)
Now, I substitute 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 £$18$
$15$ CDs sold for £$12$.
Teachers' Resources
Why do this problem :
This problem creates a context where students are challenged to pick out detail from a large block of text, represent that data and the relationships between the data as a set of equations, which need solving and interpreting back into the practical context.
Possible approach :
This kind of work can be hard and not always rewarding for individuals not motivated just because it's a puzzle. It may be that the problem achieves most to develop a problem solving culture in the classroom if it is approached with plenty of discussion between students and in the spirit of a group effort. Less motivated students can still verify and critique the proposals of others.
Pair students to extract the relevant data and relationships from the text. Check orally that all the group has all the needed data and debate items that are not universally agreed upon.
Ask pairs to express these relationships as equations, again pooling all and discussing elements of contention.
Set students to work looking for a solution, which is then written up poster-size for display around the room.
Students peruse each others results and presentation of method, before a whole group discussion of the routes followed and the comparative efficiency of the methods deployed. Finally, ask different students to explain how the algebraic results work as a meaningful solution to the practical problem context.
Key questions :
- Pick out all the data that you are going to need from this text.
- Try to get clear about exactly how these values relate to each other and express those relationships as algebra (equation)
- What does it mean to solve those equations and can you see how to manage that ?
Possible support :
The style of approach suggested above should allow a wide range of ability within a group to function effectively as a problem-solving culture - where abler students explain more, and less able students providing verification of results and feedback on effectiveness of communication.
Possible extension :
Matchless is an interesting challenge which pushes students into a deeper understanding of simultaneous equations.