The Distributive Property Quick Check

Question

3 of 53 of 5 Items
Question
Sven is making a school supply display. Sven has 52 notebooks and 42 binders. Sven wants to group the supplies in the display so that there are an equal number of supplies in each group. If we know that Sven puts the same number of notebooks and the same number of binders into each group, how many groups did he display and how many of each item went into each group? Represent your answer using the Distributive Property of Multiplication over Addition.(1 point)
Responses

Sven made 1 group. He put all 52 notebooks and all 42 binders into that one group. Using the Distributive Property, the answer can be shown as 1(52+42).
Sven made 1 group. He put all 52 notebooks and all 42 binders into that one group. Using the Distributive Property, the answer can be shown as 1 left parenthesis 52 plus 42 right parenthesis .

Sven made 1 group. He put all 52 notebooks and all 42 binders into that one group. Using the Distributive Property, the answer can be shown as 1(52×42).
Sven made 1 group. He put all 52 notebooks and all 42 binders into that one group. Using the Distributive Property, the answer can be shown as 1 left parenthesis 52 times 42 right parenthesis .

Sven made 2 groups. He put 26 notebooks and 21 binders into each group. Using the Distributive Property, the answer can be shown as 2(26×21).
Sven made 2 groups. He put 26 notebooks and 21 binders into each group. Using the Distributive Property, the answer can be shown as 2 left parenthesis 26 times 21 right parenthesis .

Sven made 2 groups. He put 26 notebooks and 21 binders into each group. Using the Distributive Property, the answer can be shown as 2(26+21).
Sven made 2 groups. He put 26 notebooks and 21 binders into each group. Using the Distributive Property, the answer can be shown as 2 left parenthesis 26 plus 21 right parenthesis .

Answer 1

To solve the problem, we first need to determine the greatest common divisor (GCD) of the number of notebooks (52) and the number of binders (42) that Sven has. This will help us find out how many groups he can make with the same number of notebooks and binders in each group.

1. Finding the GCD of 52 and 42:

   • The factors of 52 are 1, 2, 4, 13, 26, and 52.
   • The factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42.
   • The common factors are 1 and 2.
   • Therefore, the GCD is 2.

2. Calculating the number of groups:

   • Since the GCD is 2, Sven can make 2 groups.

3. Calculating the number of notebooks and binders per group:

   • For notebooks: \( \frac{52}{2} = 26 \)
   • For binders: \( \frac{42}{2} = 21 \)

4. Using the Distributive Property:

   • The total number of supplies in each group is \( 26 + 21 = 47 \).
   • The equation to represent the distribution of the supplies can be written as \( 2(26 + 21) \), where 2 is the number of groups, and \( 26 + 21 \) represents the supplies per group.

Thus, the correct response is:

Sven made 2 groups. He put 26 notebooks and 21 binders into each group. Using the Distributive Property, the answer can be shown as 2(26 + 21).