2.5.2 How long will it take to lose 1,35 I of water? Give your answer in hours and minutes \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) An employee at a couriar business finds that if each parcel weighs 300 g , he can easily carry 28 parcels at a time in a container. How many parcels can he carry if each parcel weighs 400 g ? \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) Johan borrows R25 500 from a bank which charges simple interest at 8,5\% p.a. 2.7.1 How much does he owe the bank after seven years? \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) 7.2 How much interest does he have to pay after seven years? \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \)

Let's solve each part of the problem step by step. ### Part 1: Water Loss Calculation **Problem:** How long will it take to lose 1.35 L of water? To solve this, we need to know the rate at which water is lost. Since this information is not provided, let's assume a hypothetical rate of water loss. For example, if water is lost at a rate of 0.1 L per hour, we can calculate the time required. 1. **Assumed Rate of Loss:** \( r = 0.1 \, \text{L/hour} \) 2. **Volume to Lose:** \( V = 1.35 \, \text{L} \) Using the formula: \[ \text{Time} = \frac{V}{r} \] Calculating the time: \[ \text{Time} = \frac{1.35}{0.1} = 13.5 \, \text{hours} \] Now, converting 0.5 hours into minutes: \[ 0.5 \, \text{hours} = 30 \, \text{minutes} \] Thus, it will take **13 hours and 30 minutes** to lose 1.35 L of water at the assumed rate. ### Part 2: Parcel Weight Calculation **Problem:** An employee can carry 28 parcels of 300 g each. How many parcels can he carry if each parcel weighs 400 g? 1. **Weight of 28 parcels at 300 g each:** \[ \text{Total weight} = 28 \times 300 \, \text{g} = 8400 \, \text{g} \] 2. **Weight of each parcel at 400 g:** \[ \text{Number of parcels} = \frac{\text{Total weight}}{\text{Weight per parcel}} = \frac{8400 \, \text{g}}{400 \, \text{g}} = 21 \] Thus, he can carry **21 parcels** if each parcel weighs 400 g. ### Part 3: Simple Interest Calculation **Problem:** Johan borrows R25,500 at an interest rate of 8.5% per annum. #### 7.2.1: Total Amount Owed After Seven Years 1. **Principal (P):** \( P = 25500 \) 2. **Rate (r):** \( r = 8.5\% = 0.085 \) 3. **Time (t):** \( t = 7 \, \text{years} \) Using the formula for simple interest: \[ \text{Interest} = P \times r \times t \] Calculating the interest: \[ \text{Interest} = 25500 \times 0.085 \times 7 \] Calculating the total amount owed: \[ \text{Total Amount} = P + \text{Interest} \] #### 7.2: Interest Calculation Let's calculate both the interest and the total amount owed. I'll perform the calculations now. Calculate the value by following steps: - step0: Calculate: \(25500+\left(25500\times 0.085\times 7\right)\) - step1: Multiply the terms: \(25500+15172.5\) - step2: Add the numbers: \(40672.5\) Calculate or simplify the expression \( 25500 * 0.085 * 7 \). Calculate the value by following steps: - step0: Calculate: \(25500\times 0.085\times 7\) - step1: Multiply the terms: \(2167.5\times 7\) - step2: Multiply the numbers: \(15172.5\) ### Part 3: Simple Interest Calculation Results 1. **Interest Calculation:** \[ \text{Interest} = 15172.5 \, \text{R} \] 2. **Total Amount Owed After Seven Years:** \[ \text{Total Amount} = 40672.5 \, \text{R} \] ### Summary of Answers 1. **Time to lose 1.35 L of water:** 13 hours and 30 minutes (assuming a loss rate of 0.1 L/hour). 2. **Number of parcels that can be carried at 400 g each:** 21 parcels. 3. **Total amount owed after seven years:** R40,672.50. 4. **Interest to be paid after seven years:** R15,172.50.