When the prices are both increased by £20, the ratio becomes 5 : 2.When the prices are both reduced by £5, the ratio becomes 5 : 1.

To solve this problem, we can set up the equations based on the information given about the prices and their ratios. We will denote the original prices of the two items as $x$ and $y$.

1. Break Down the Problem

We have two conditions that give us two equations based on the changes in price and the resulting ratios:

• Condition 1: When both prices are increased by £20, the ratio becomes $\frac{5}{2}$.
• Condition 2: When both prices are reduced by £5, the ratio becomes $\frac{5}{1}$.

2. Relevant Concepts

We can express the conditions in the form of equations:

1. For the first condition: $\frac{x + 20}{y + 20} = \frac{5}{2}$ This can be rearranged as: $2(x + 20) = 5(y + 20)$

2. For the second condition: $\frac{x - 5}{y - 5} = \frac{5}{1}$ This can be rearranged as: $x - 5 = 5(y - 5)$

3. Analysis and Detail

Now we will expand and simplify both equations:

1. From the first equation: $2x + 40 = 5y + 100 \\ \Rightarrow 2x - 5y = 60 \quad (i)$

2. From the second equation: $x - 5 = 5y - 25 \\ \Rightarrow x - 5y = -20 \quad (ii)$

We now have a system of equations:

• Equation (i): $2x - 5y = 60$
• Equation (ii): $x - 5y = -20$

4. Verify and Summarize

We can solve this system using substitution or elimination. Let's use substitution. From equation (ii), we can isolate $x$:

$x = 5y - 20$

We substitute $x$ into equation (i):

$2(5y - 20) - 5y = 60 \\ 10y - 40 - 5y = 60 \\ 5y - 40 = 60 \\ 5y = 100 \\ y = 20$

Now substituting $y$ back into $x = 5y - 20$:

$x = 5(20) - 20 \\ x = 100 - 20 \\ x = 80$

Final Answer

The original prices are:

• $x = £80$
• $y = £20$