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7. Nancy buys 15 m of ribbon. Her sister buys 6 m less ribbon but pays twice as much per metre. Together they pay R66. What is the price per metre of each type of ribbon? 8. Sam bought some apples at 60 c each and sold all but 5 of them at R1,00 each. If he made a profit of R5,00, how many apples did he buy? 9. If you are allowed to choose 20 coins from a collection of 50 cent coins and R1 coins, how many coins must be R1 coins if the total value is R15?

Ask by Elliott Wagner. in South Africa
Mar 15,2025

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- **Problem 7:** Nancy's ribbon costs R2 per metre, her sister's costs R4 per metre. - **Problem 8:** Sam bought 25 apples. - **Problem 9:** There must be 10 R1 coins.

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To solve Nancy's ribbon problem, let's denote the price per metre of Nancy’s ribbon as R. Nancy buys 15 m of ribbon, equating to 15R. Her sister buys 6 m less, which is 9 m, and pays 2R per metre, amounting to 9(2R) or 18R. Together, they spend R66, leading to the equation 15R + 18R = 66, giving us 33R = 66, so R = R2. This means Nancy's ribbon costs R2 per metre, while her sister's ribbon costs R4 per metre. For Sam's apple conundrum, knowing he got 5 apples left hints at how many he sold: if he bought 'x' apples and sold 'x - 5', we can write the cost equation. The total cost of apples is 60c each, translating to 0.6x. By selling at R1 each for 'x - 5', he earned R1(x - 5). His profit of R5 gives us the formula R1(x - 5) - 0.6x = 5, leading to x = 15. Therefore, Sam bought 15 apples in total. To tackle the coin challenge, let's label the number of R1 coins as 'y' and the 50 cent coins as 'x'. We have x + y = 20 for the total coins and 0.5x + 1y = 15 for the total value equation. Solving these simultaneously, we can substitute x = 20 - y into the value equation, leading to half of that equation and reworking until we find y = 10. Hence, Sam needs 10 R1 coins in that mix for the total of R15!

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