A runner runs a distance of 3 km in 20 min , what is the speed \( \mathrm{in} \mathrm{m} / \mathrm{s} \) A business stocks 300 packets of beef -flavoured chips and 200 packets of tomatoe-flavourd Chips. The tomatoe -flavoured chips are not selling well, so the business decides to to decrease the number, they stock by \( 20 \% \). They also decide to increase the number of beef-flavoured chips, which are selling well, by \( 10 \% \). 1.6.1 How many packets of tomatoe chips are there after the decrease? 1.6 .2 How many packets of beef chips do they stock after the increase? A clothing store sells old stock at a \( 20 \% \) discount of its original price. If a pair of jeans now costs R160, what was its original price?

To calculate the speed of the runner, we first convert the distance from kilometers to meters (3 km = 3000 m) and the time from minutes to seconds (20 min = 1200 s). Using the formula Speed = Distance / Time, we find that the speed is \( \frac{3000 \text{ m}}{1200 \text{ s}} = 2.5 \text{ m/s} \). After decreasing the number of tomato-flavoured chips by \( 20\% \), we calculate it as follows: \( 200 \times 0.20 = 40 \) packets; therefore, \( 200 - 40 = 160 \) packets remain. For the beef-flavoured chips after a \( 10\% \) increase: \( 300 \times 0.10 = 30 \) packets more, resulting in \( 300 + 30 = 330 \) packets stocked. To find the original price of the jeans that are currently sold at a \( 20\% \) discount, we set up the equation. The selling price is \( 80\% \) of the original price. If the jeans now cost R160, then \( 0.8 \text{ (original price)} = R160 \). Thus, the original price is \( \frac{R160}{0.8} = R200 \).