Math worksheet featuring story problems on greatest common factor and least common multiple.

You’ve uploaded a worksheet titled “Greatest Common Factor and Least Common Multiple Story Problems”. Since I can’t see the actual image (as per system constraints), I’ll solve the problems based on the text you provided, which appears to be a transcription of the worksheet.

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## 📄 Problem 1:
> A florist has 36 roses, 27 tulips, and 18 carnations she must use to create bouquets. She wants the largest number of bouquets she can make without having any flowers left over.

✔ Goal:

Find the greatest number of identical bouquets she can make using all flowers, with no leftovers.

This is a Greatest Common Factor (GCF) problem — we need to find the largest number that divides 36, 27, and 18 evenly.

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🔢 Step-by-step Solution:

Prime factorization:

- 36 = 2² × 3²
- 27 = 3³
- 18 = 2 × 3²

GCF = product of lowest powers of common primes

→ Only common prime is 3, lowest power is 3² = 9

✔ Answer: 9 bouquets

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🌹 How many of each flower per bouquet?

- Roses: 36 ÷ 9 = 4 roses per bouquet
- Tulips: 27 ÷ 9 = 3 tulips per bouquet
- Carnations: 18 ÷ 9 = 2 carnations per bouquet

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## 📄 Problem 2:
> Say you have 60 pencils, 90 pens and 120 notebooks you need to make packages of pencils, pens and notebooks to donate to your school for students who cannot afford these items. You want to make the same number of packages with the same number of each item in them, and how many pencils, pens and notebooks will be in each package?

✔ Goal:

Find the maximum number of identical packages using all items, with no leftovers → again, this is a GCF problem.

Numbers: 60, 90, 120

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🔢 Step-by-step Solution:

Prime factorization:

- 60 = 2² × 3 × 5
- 90 = 2 × 3² × 5
- 120 = 2³ × 3 × 5

GCF = product of lowest powers of common primes

→ Common primes: 2, 3, 5
→ Lowest powers: 2¹, 3¹, 5¹
→ GCF = 2 × 3 × 5 = 30

✔ Answer: 30 packages

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🖊️ How many of each item per package?

- Pencils: 60 ÷ 30 = 2 pencils per package
- Pens: 90 ÷ 30 = 3 pens per package
- Notebooks: 120 ÷ 30 = 4 notebooks per package

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## 📄 Problem 3:
> One day around a running track is 440 yards. One jogger can complete one lap in 4 minutes; the other can complete it in 6 minutes. They begin at the same time and start jogging in the same direction together. If they start at the same time, when will they meet again at the starting point?

✔ Goal:

Find the first time both joggers are back at the starting point together → this is a Least Common Multiple (LCM) problem.

Times: 4 minutes and 6 minutes

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🔢 Step-by-step Solution:

Prime factorization:

- 4 = 2²
- 6 = 2 × 3

LCM = product of highest powers of all primes

→ LCM = 2² × 3 = 4 × 3 = 12

✔ Answer: They will meet again at the starting point after 12 minutes.

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## 🧾 Final Answers Summary:

Problem 1:

- Number of bouquets: 9
- Per bouquet: 4 roses, 3 tulips, 2 carnations

Problem 2:

- Number of packages: 30
- Per package: 2 pencils, 3 pens, 4 notebooks

Problem 3:

- Time until they meet again: 12 minutes

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## 💡 Key Concept Recap:

- GCF (Greatest Common Factor): Used when you want to divide things into the largest possible equal groups with nothing left over.
- LCM (Least Common Multiple): Used when you want to find the first time two or more events coincide (like runners meeting again).

Let me know if you’d like visual diagrams or step-by-step breakdowns for younger learners! 🎯