Worksheet titled "GCF and LCM | Word Problems" featuring five math problems related to greatest common factor and least common multiple.

Problem 1: Berkley and Bronx's Spanish Lessons

Question: Berkley goes to the class every 3 days, and Bronx attends the class once in 5 days. If they both took their lessons today, how long will it take Berkley and Bronx to take their lessons again on the same day?

Solution:
To determine when Berkley and Bronx will attend their lessons together again, we need to find the Least Common Multiple (LCM) of their attendance cycles (3 days for Berkley and 5 days for Bronx).

1. Prime Factorization:
- The prime factorization of 3 is \(3\).
- The prime factorization of 5 is \(5\).

2. LCM Calculation:
- The LCM is found by taking the highest power of all prime factors involved.
- Here, the prime factors are \(3\) and \(5\), and neither is repeated.
- Therefore, \(\text{LCM}(3, 5) = 3 \times 5 = 15\).

Answer: It will take 15 days for Berkley and Bronx to take their lessons again on the same day.

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Problem 2: Ethan's Jute Rope

Question: Ethan has two rolls of jute rope measuring 78 feet and 91 feet. If he wants to cut them into several pieces of equal length, what would be the greatest possible length of each piece?

Solution:
To find the greatest possible length of each piece, we need to determine the Greatest Common Factor (GCF) of the lengths of the two ropes (78 feet and 91 feet).

1. Prime Factorization:
- The prime factorization of 78 is \(78 = 2 \times 3 \times 13\).
- The prime factorization of 91 is \(91 = 7 \times 13\).

2. GCF Calculation:
- The GCF is found by taking the product of the common prime factors with the lowest powers.
- The only common prime factor between 78 and 91 is \(13\).
- Therefore, \(\text{GCF}(78, 91) = 13\).

Answer: The greatest possible length of each piece is 13 feet.

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Problem 3: Caryn and Delma's Alarms

Question: Caryn sets her alarm to ring every 20 minutes, and Delma sets hers to ring every 15 minutes. If both start their alarms at the same time, how long will it take for their alarms to ring simultaneously?

Solution:
To determine when Caryn's and Delma's alarms will ring simultaneously, we need to find the Least Common Multiple (LCM) of their alarm intervals (20 minutes and 15 minutes).

1. Prime Factorization:
- The prime factorization of 20 is \(20 = 2^2 \times 5\).
- The prime factorization of 15 is \(15 = 3 \times 5\).

2. LCM Calculation:
- The LCM is found by taking the highest power of all prime factors involved.
- The prime factors are \(2\), \(3\), and \(5\).
- The highest powers are \(2^2\), \(3^1\), and \(5^1\).
- Therefore, \(\text{LCM}(20, 15) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\).

Answer: Their alarms will ring simultaneously after 60 minutes.

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Problem 4: Dylan's Seashells

Question: Dylan found 30 conchs, 15 cowries, and 5 clamshells. If he plans to give them away in identical sets to his friends, how many identical sets can he make?

Solution:
To determine the maximum number of identical sets Dylan can make, we need to find the Greatest Common Factor (GCF) of the quantities of each type of seashell (30, 15, and 5).

1. Prime Factorization:
- The prime factorization of 30 is \(30 = 2 \times 3 \times 5\).
- The prime factorization of 15 is \(15 = 3 \times 5\).
- The prime factorization of 5 is \(5\).

2. GCF Calculation:
- The GCF is found by taking the product of the common prime factors with the lowest powers.
- The only common prime factor among 30, 15, and 5 is \(5\).
- Therefore, \(\text{GCF}(30, 15, 5) = 5\).

Answer: Dylan can make 5 identical sets.

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Problem 5: Cycling Session

Question: Claudia, Evelyn, and Amanda covered the distance in 12 minutes, 15 minutes, and 18 minutes respectively. They agreed to continue cycling until all three reach the finish line together. How many hours would the session continue for?

Solution:
To determine how long the session will continue, we need to find the Least Common Multiple (LCM) of their cycling times (12 minutes, 15 minutes, and 18 minutes).

1. Prime Factorization:
- The prime factorization of 12 is \(12 = 2^2 \times 3\).
- The prime factorization of 15 is \(15 = 3 \times 5\).
- The prime factorization of 18 is \(18 = 2 \times 3^2\).

2. LCM Calculation:
- The LCM is found by taking the highest power of all prime factors involved.
- The prime factors are \(2\), \(3\), and \(5\).
- The highest powers are \(2^2\), \(3^2\), and \(5^1\).
- Therefore, \(\text{LCM}(12, 15, 18) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\).

3. Convert Minutes to Hours:
- Since there are 60 minutes in an hour, we convert 180 minutes to hours:
\[
\frac{180}{60} = 3 \text{ hours}.
\]

Answer: The session will continue for 3 hours.

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Final Answers:

1. \(\boxed{15}\)
2. \(\boxed{13}\)
3. \(\boxed{60}\)
4. \(\boxed{5}\)
5. \(\boxed{3}\)