GCF and LCM Word Problems worksheet with six math scenarios requiring students to calculate greatest common factors and least common multiples.

Problem Analysis and Solutions

The provided worksheet involves problems related to the Greatest Common Factor (GCF) and the Least Common Multiple (LCM). Let's solve each problem step by step.

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Section 1: Greatest Common Factor (GCF)

#### Problem 1:
Sally had 10 purple beads, 25 yellow beads, and 30 red beads. She is going to make identical bracelets for her friends. What is the maximum number of bracelets she can make if she wanted to use all of her beads?

Solution:
To determine the maximum number of identical bracelets Sally can make, we need to find the GCF of the numbers of beads:
- Purple beads: 10
- Yellow beads: 25
- Red beads: 30

Step 1: Find the prime factorization of each number:
- \( 10 = 2 \times 5 \)
- \( 25 = 5 \times 5 \)
- \( 30 = 2 \times 3 \times 5 \)

Step 2: Identify the common prime factors:
- The only common prime factor is \( 5 \).

Step 3: The GCF is the product of the common prime factors:
- GCF = \( 5 \)

Step 4: Interpret the result:
- Sally can make a maximum of 5 identical bracelets, using all her beads.

Answer:
\[
\boxed{5}
\]

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#### Problem 2:
The Student Council collected 60 pens, 90 pencils, and 45 packs of paper to donate to students who cannot afford the supplies. They made packages with the same number of supplies in each. What is the maximum number of packages the group could make if they used all of the supplies?

Solution:
To determine the maximum number of identical packages, we need to find the GCF of the numbers of supplies:
- Pens: 60
- Pencils: 90
- Packs of paper: 45

Step 1: Find the prime factorization of each number:
- \( 60 = 2^2 \times 3 \times 5 \)
- \( 90 = 2 \times 3^2 \times 5 \)
- \( 45 = 3^2 \times 5 \)

Step 2: Identify the common prime factors:
- The common prime factors are \( 3 \) and \( 5 \).

Step 3: The GCF is the product of the lowest powers of the common prime factors:
- GCF = \( 3 \times 5 = 15 \)

Step 4: Interpret the result:
- The Student Council can make a maximum of 15 identical packages, using all the supplies.

Answer:
\[
\boxed{15}
\]

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#### Problem 3:
The PTA put together baskets for a silent auction. The students donated the items for baskets. Altogether the PTA collected 100 movie tickets, 50 boxes of microwave popcorn, and 75 bottles of soda. What is the maximum number of identical baskets they can make if they used all of the items collected?

Solution:
To determine the maximum number of identical baskets, we need to find the GCF of the numbers of items:
- Movie tickets: 100
- Boxes of popcorn: 50
- Bottles of soda: 75

Step 1: Find the prime factorization of each number:
- \( 100 = 2^2 \times 5^2 \)
- \( 50 = 2 \times 5^2 \)
- \( 75 = 3 \times 5^2 \)

Step 2: Identify the common prime factors:
- The common prime factor is \( 5^2 = 25 \).

Step 3: The GCF is the product of the lowest powers of the common prime factors:
- GCF = \( 25 \)

Step 4: Interpret the result:
- The PTA can make a maximum of 25 identical baskets, using all the items.

Answer:
\[
\boxed{25}
\]

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Section 2: Least Common Multiple (LCM)

#### Problem 4:
Robbie and Cynthia are cousins. Robbie visits his grandmother every 6 days. Cynthia visits her grandmother every 5 days. Over the next 40 days, how many will the two cousins visit their grandmother on the same day?

Solution:
To determine how many times Robbie and Cynthia visit their grandmother on the same day, we need to find the LCM of their visiting cycles:
- Robbie: every 6 days
- Cynthia: every 5 days

Step 1: Find the prime factorization of each number:
- \( 6 = 2 \times 3 \)
- \( 5 = 5 \)

Step 2: Identify the highest power of each prime factor:
- For \( 2 \): \( 2^1 \)
- For \( 3 \): \( 3^1 \)
- For \( 5 \): \( 5^1 \)

Step 3: The LCM is the product of these highest powers:
- LCM = \( 2 \times 3 \times 5 = 30 \)

Step 4: Interpret the result:
- Robbie and Cynthia will visit their grandmother on the same day every 30 days.

Step 5: Determine how many times this happens in 40 days:
- In 40 days, they will visit on the same day once (on day 30).

Answer:
\[
\boxed{1}
\]

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#### Problem 5:
Peggy has three plants. She waters the cactus every 5 days, the spider plant every 3 days, and the violet every 2 days. She watered all three plants on October 1st. What is the next day that she will water all three plants on the same day?

Solution:
To determine the next day Peggy will water all three plants on the same day, we need to find the LCM of the watering cycles:
- Cactus: every 5 days
- Spider plant: every 3 days
- Violet: every 2 days

Step 1: Find the prime factorization of each number:
- \( 5 = 5 \)
- \( 3 = 3 \)
- \( 2 = 2 \)

Step 2: Identify the highest power of each prime factor:
- For \( 2 \): \( 2^1 \)
- For \( 3 \): \( 3^1 \)
- For \( 5 \): \( 5^1 \)

Step 3: The LCM is the product of these highest powers:
- LCM = \( 2 \times 3 \times 5 = 30 \)

Step 4: Interpret the result:
- Peggy will water all three plants on the same day every 30 days.

Step 5: Determine the next day after October 1st:
- Since she watered all three plants on October 1st, the next day will be 30 days later, which is October 31st.

Answer:
\[
\boxed{\text{October 31st}}
\]

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#### Problem 6:
Gary made 20 treat bags for his Halloween party. In every 3rd bag he put lollipops, in every 4th bag he put Hershey Kisses, and in every 5th bag he put fake eyeballs. How many of the 20 bags have all three items in them?

Solution:
To determine how many bags have all three items, we need to find the LCM of the intervals at which Gary puts the items:
- Lollipops: every 3rd bag
- Hershey Kisses: every 4th bag
- Fake eyeballs: every 5th bag

Step 1: Find the prime factorization of each number:
- \( 3 = 3 \)
- \( 4 = 2^2 \)
- \( 5 = 5 \)

Step 2: Identify the highest power of each prime factor:
- For \( 2 \): \( 2^2 \)
- For \( 3 \): \( 3^1 \)
- For \( 5 \): \( 5^1 \)

Step 3: The LCM is the product of these highest powers:
- LCM = \( 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60 \)

Step 4: Interpret the result:
- A bag will have all three items every 60 bags.

Step 5: Determine how many such bags exist in 20 bags:
- Since 60 is greater than 20, there are no bags that contain all three items within the first 20 bags.

Answer:
\[
\boxed{0}
\]

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Final Answers:

1. \(\boxed{5}\)
2. \(\boxed{15}\)
3. \(\boxed{25}\)
4. \(\boxed{1}\)
5. \(\boxed{\text{October 31st}}\)
6. \(\boxed{0}\)