The water behind a dam is normally released at a rate of 250000 litres per second - Leaving Cert Mathematics - Question 10 - 2021

With the increased flow rate, we have:

$$\text{Increased Flow Rate} = 250,000 \text{ litres/second} \times 1.1 = 275,000 \text{ litres/second}.$$

Converting this to cubic metres:

$$275,000 \text{ litres/second} = \frac{275,000}{1000} = 275 \text{ m³/second}.$$

Now, calculating the volume of water released in 24 hours:

$$\text{Volume} = 275 \text{ m³/second} \times 60 \text{ seconds/min} \times 60 \text{ min/hour} \times 24 \text{ hours} = 2376000 \text{ m³}.$$

This can be written in the required form as:

$$2.376 \times 10^6 \text{ m³ (to 3 significant figures)}.$$

First, calculate the circumference of the circular trail:

$$D = 2\pi r = 2\pi(0.5) \approx 3.14 \text{ km}.$$

Now, for 3 circuits:

$$\text{Total Distance} = 3 \times 3.14 = 9.42 \text{ km}.$$

Time taken at a speed of 6 km/h:

$$T = \frac{D}{S} = \frac{9.42 \text{ km}}{6 \text{ km/h}} \approx 1.57 \text{ hours} \approx 94 \text{ minutes (to the nearest minute)}.$$

Mary's walking distances over the 5 day period are:

• Day 1: 3 km
• Day 2: 3 km × 1.15 = 3.45 km
• Day 3: 3.45 km × 1.15 = 3.97 km
• Day 4: 3.97 km × 1.15 ≈ 4.56 km
• Day 5: 4 km/h is her speed.

Now calculate the time taken on Day 5:

$$T = \frac{D}{S} = \frac{4.56 \text{ km}}{4 \text{ km/h}} = 1.14 \text{ hours} = 68.4 \text{ minutes} \approx 69 \text{ minutes (to the nearest minute)}.$$

When John meets Mary, they continue walking at their respective speeds. Let’s denote the time taken for them to meet again as 't'.

In that time:

• John travels at 6 km/h.
• Mary travels at 4 km/h.

They are walking in opposite directions, so their relative speed is:

$$\text{Relative Speed} = 6 + 4 = 10 \text{ km/h}.$$

The total distance before they meet is equal to the circle's circumference (3.14 km):

$$\text{Time to meet again} = \frac{3.14 \text{ km}}{10 \text{ km/h}} \approx 0.314 \text{ hours} \approx 18.84 \text{ minutes}.$$

In this time, John travels:

$$\text{Distance} = D = 6 \text{ km/h} \times t \approx 6 \text{ km/h} \times 0.314 \text{ h} \\ = 1.884 \text{ km} \approx 1885 ext{ m (to the nearest metre)}.$$