Question

#3
every week i buy four 2-litre bottles of lemonade whilst shopping at my local supermarket. last week the bottles on the shelf were limited edition “2.5 litres for the price of 2 litres” bottles, and the supermarket also had a “buy 3 bottles, get another one free” offer.

in terms of cost per litre, by what percentage was the cost of my four bottles of lemonade lower than usual last week?

Explanation:

Step1: Define Usual Cost and Volume

Let the usual price for a 2 - litre bottle be \( P \). Usual number of bottles bought per week: 4, each 2 - litre. So usual total volume \( V_{usual}=4\times2 = 8 \) litres. Usual total cost \( C_{usual}=4P \) (since 4 bottles at price \( P \) each).

Step2: Calculate Last Week's Cost and Volume

Last week: "buy 3, get 1 free" for 4 bottles, and each bottle is 2.5 - litre (price of 2 - litre). So for 4 bottles, we pay for 3 bottles. Each paid bottle has volume 2.5 L, but costs \( P \) (price of 2 - litre). Total volume last week \( V_{last}=4\times2.5=10 \) litres. Total cost last week \( C_{last}=3P \) (paid for 3 bottles).

Step3: Calculate Cost per Litre for Both Cases

Usual cost per litre: \( \frac{C_{usual}}{V_{usual}}=\frac{4P}{8}=\frac{P}{2} \) per litre.

Last week's cost per litre: \( \frac{C_{last}}{V_{last}}=\frac{3P}{10} \) per litre.

Step4: Calculate Percentage Decrease

The formula for percentage decrease is \( \text{Percentage Decrease}=\frac{\text{Usual Cost per Litre}-\text{Last Week Cost per Litre}}{\text{Usual Cost per Litre}}\times100\% \)

Substitute the values:

\( \text{Usual Cost per Litre}-\text{Last Week Cost per Litre}=\frac{P}{2}-\frac{3P}{10}=\frac{5P - 3P}{10}=\frac{2P}{10}=\frac{P}{5} \)

Then, \( \text{Percentage Decrease}=\frac{\frac{P}{5}}{\frac{P}{2}}\times100\%=\frac{P}{5}\times\frac{2}{P}\times100\%=\frac{2}{5}\times100\% = 40\% \)

Answer:

40%