Greatest Common Factor Word Problems

A car dealer has 16 Ford Mustangs and 24 Chevrolet Camaros on the lot. The dealer wants to arrange the cars in sections so that each section has the same number of Ford Mustangs and Chevrolet Camaros. What is the greatest number of sections that can be formed?

• 14

• 16

• 8

• 12

Answer explanation

To find the greatest number of sections that can be formed, you need to find the greatest common factor (GCF) of 16 and 24. The prime factorization of 16 is 2 × 2 × 2 × 2, and the prime factorization of 24 is 2 × 2 × 2 × 3. The common prime factors are 2, 2, and 2. So, the GCF is 2 × 2 × 2 = 8. By dividing both the number of Ford Mustangs and Chevrolet Camaros by the GCF, we get 16 ÷ 8 = 2 Ford Mustangs and 24 ÷ 8 = 3 Chevrolet Camaros per section. Therefore, the greatest number of sections that can be formed is 8, with each section having 2 Ford Mustangs and 3 Chevrolet Camaros.

Emma wants to divide her collection of 24 marbles and 36 buttons equally among her friends. What is the largest number of friends she can have if each friend should receive the same number of marbles and buttons?

• 14

• 16

• 18

• 12

Answer explanation

To find the largest number of friends Emma can have, you need to find the greatest common factor (GCF) of 24 and 36. The prime factorization of 24 is 2 × 2 × 2 × 3, and the prime factorization of 36 is 2 × 2 × 3 × 3. The common prime factors are 2, 2, and 3, so the GCF is 2 × 2 × 3 = 12. Therefore, Emma can have a maximum of 12 friends. By dividing both numbers by the GCF, you get 24 ÷ 12 = 2 marbles and 36 ÷ 12 = 3 buttons for each friend. 

A grocery store has 60 bags of apples and 72 bags of oranges. The store wants to arrange them on display tables so that each table has the same number of apples and oranges. What is the greatest number of tables that can be formed?

• 12

• 16

• 24

• 18

Answer explanation

 To find the greatest number of tables that can be formed, you need to find the greatest common factor (GCF) of 60 and 72. The prime factorization of 60 is 2 × 2 × 3 × 5, and the prime factorization of 72 is 2 × 2 × 2 × 3 × 3. The common prime factors are 2, 2, and 3. So, the GCF is 2 × 2 × 3 = 12. By dividing both the number of bags of apples and oranges by the GCF, you get 60 ÷ 12 = 5 bags of apples and 72 ÷ 12 = 6 bags of oranges per table. Therefore, the greatest number of tables that can be formed is 12, with each table having 5 bags of apples and 6 bags of oranges.

John wants to share his 30 baseball cards and 45 basketball cards equally with his friends. What is the maximum number of friends John can have if each friend should receive the same number of baseball cards and basketball cards?

• 13

• 15

• 19

• 25

Answer explanation

To find the maximum number of friends John can have, you need to find the GCF of 30 and 45. The prime factorization of 30 is 2 × 3 × 5, and the prime factorization of 45 is 3 × 3 × 5. The common prime factors are 3 and 5, so the GCF is 3 × 5 = 15. Therefore, John can have a maximum of 5 friends. By dividing both numbers by the GCF, you get 30 ÷ 15 = 2 baseball cards and 45 ÷ 15 = 3 basketball cards for each friend.

Mark has a 96-inch long rope and a 120-inch long rope. He wants to cut them into equal-sized pieces. What is the largest length of each piece that Mark can have?

• 36

• 48

• 10

• 24

Answer explanation

To find the largest length of each piece that Mark can have, you need to find the greatest common factor (GCF) of 96 and 120. The prime factorization of 96 is 2 × 2 × 2 × 2 × 2 × 3, and the prime factorization of 120 is 2 × 2 × 2 × 3 × 5. The common prime factors are 2, 2, 2, and 3, so the GCF is 2 × 2 × 2 × 3 = 24. Therefore, Mark can have pieces of length 24 inches.