A building has a leaking roof and, while it is raining, water drips into a 12 litre bucket - AQA - A-Level Maths Pure - Question 7 - 2021 - Paper 3

The sequence described forms a geometric progression where each term after the first is obtained by multiplying the previous term by a constant factor (in this case, 0.98, representing the 2% reduction).

The first term in this sequence ($W_1$) is 30 ml. Thus:

• $A = 30$
• This means the general term can be expressed as:

$W_n = A \times 0.98^{n-1}$

To determine the total increase in water in the bucket after 15 minutes, we need to calculate the sum of the series for the first 15 terms:

$S_{15} = \sum_{n=1}^{15} W_n = \sum_{n=1}^{15} (30 \times 0.98^{n-1})$ Using the formula for the sum of a geometric series:

$S_N = A \times \frac{1 - r^N}{1 - r}$ where $A = 30$, $r = 0.98$, and $N = 15$:

$S_{15} = 30 \times \frac{1 - 0.98^{15}}{1 - 0.98}$ Calculating this gives:

$S_{15} \approx 30 \times \frac{1 - 0.7433}{0.02} \approx 30 \times 12.8352 \approx 385.06$

So, rounding to the nearest millilitre, the increase in the water in the bucket is approximately 385 ml.

To find the maximum amount of water, we need to determine the sum of the series as $n \to \infty$:

$S_{\infty} = \frac{A}{1 - r}$ Where $A = 30$ and $r = 0.98$:

$S_{\infty} = \frac{30}{1 - 0.98} = \frac{30}{0.02} = 1500$

Hence, the maximum amount of water in the bucket is 1500 ml.

1. The model assumes that water will continue to drip from the roof indefinitely, but in reality, there may be limitations on the total water available from the puddle and the drip.

2. Environmental factors such as evaporation may lead to a reduction in the amount of water in the bucket over time, preventing it from reaching the theoretical maximum calculated.