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8. A company offers a nut mixture with 6 peanuts for every 4 almonds. The company changes the mixture to have 10 peanuts for every 2 almonds, but the number of nuts per container does not change. How many nuts are in the smallest possible container?

8. A company offers a nut mixture with 6 peanuts for every 4 almonds. The company changes the mixture to have 10 peanuts for every 2 almonds, but the number of nuts per container does not change. How many nuts are in the smallest possible container?

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Problem 8: Nut Mixture** A company offers a nut mixture with a ratio of 6 peanuts for every 4 almonds. The company changes the mixture to have a ratio of 10 peanuts for every 2 almonds, but the total number of nuts per container does not change. How many nuts are in the smallest possible container? **Solution Explanation:** To solve this problem, we first need to understand the original and new ratios: - **Original Ratio:** 6 peanuts : 4 almonds - **Simplified Original Ratio:** 3 peanuts : 2 almonds (dividing by 2) - **New Ratio:** 10 peanuts : 2 almonds - **Simplified New Ratio:** 5 peanuts : 1 almond (dividing by 2) Next, we determine the total nuts in one iteration of each ratio: - **Original Total:** 3 + 2 = 5 nuts - **New Total:** 5 + 1 = 6 nuts Since the total number of nuts per container remains constant, find the smallest common total by equating multiples of the original and new totals. The smallest number that 5 and 6 can both divide into is 30. - Original mix to reach 30 nuts: - Multiply components by 6: \(3 \times 6 = 18\) peanuts and \(2 \times 6 = 12\) almonds - New mix to reach 30 nuts: - Multiply components by 5: \(5 \times 5 = 25\) peanuts and \(1 \times 5 = 5\) almonds Therefore, the smallest possible container has 30 nuts in total.
Transcribed Image Text:**Problem 8: Nut Mixture** A company offers a nut mixture with a ratio of 6 peanuts for every 4 almonds. The company changes the mixture to have a ratio of 10 peanuts for every 2 almonds, but the total number of nuts per container does not change. How many nuts are in the smallest possible container? **Solution Explanation:** To solve this problem, we first need to understand the original and new ratios: - **Original Ratio:** 6 peanuts : 4 almonds - **Simplified Original Ratio:** 3 peanuts : 2 almonds (dividing by 2) - **New Ratio:** 10 peanuts : 2 almonds - **Simplified New Ratio:** 5 peanuts : 1 almond (dividing by 2) Next, we determine the total nuts in one iteration of each ratio: - **Original Total:** 3 + 2 = 5 nuts - **New Total:** 5 + 1 = 6 nuts Since the total number of nuts per container remains constant, find the smallest common total by equating multiples of the original and new totals. The smallest number that 5 and 6 can both divide into is 30. - Original mix to reach 30 nuts: - Multiply components by 6: \(3 \times 6 = 18\) peanuts and \(2 \times 6 = 12\) almonds - New mix to reach 30 nuts: - Multiply components by 5: \(5 \times 5 = 25\) peanuts and \(1 \times 5 = 5\) almonds Therefore, the smallest possible container has 30 nuts in total.
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