Philip is sorting acorns he collected. He has 92 acorns: 66 still have their tops, 26 do not. Philip wants to sort the acorns into groups with the same number of acorns, including some with and without tops. He wants each group to have an equal number of acorns with tops and an equal number of acorns without tops. Which answer best shows how Philip can represent how he should sort the acorns?(1 point)

To find the best way for Philip to sort his acorns into equal groups of both types, we first need to consider the total number of acorns he has, which is 92. He has 66 acorns with tops and 26 without.

We want to find a way to create groups such that each group contains an equal number of acorns with tops and an equal number of acorns without tops.

We can represent the number of groups he wants as \( g \).

The acorns with tops for each group would be \( \frac{66}{g} \) and the acorns without tops would be \( \frac{26}{g} \).

The total number of acorns in each group would then be: \[ \frac{66}{g} + \frac{26}{g} = \frac{66 + 26}{g} = \frac{92}{g} \]

Thus, the total number of acorns per group is dependent on \( g \).

Looking at the options given:

1. 2(33 + 13): Represents 2 groups of 46.
2. 1(66 + 26): Represents 1 group of 92.
3. 24(3 + 1): Represents 24 groups of 4 (but does not divide 66 and 26 properly).
4. 6(11 + 4): Represents 6 groups of 15.

Focusing on how Philip can sort both types equally, we notice:

• When he uses 6(11 + 4), he could make 6 groups with \( 66 \div 6 = 11 \) acorns with tops in each group and \( 26 \div 6 ≈ 4.33 \) without tops (which can't happen equally).

In contrast, 2(33 + 13) provides:

• 2 groups with \( 66 \div 2 = 33 \) tops in each and \( 26 \div 2 = 13 \) without tops, perfectly matching the distribution.

Thus, the answer that best shows how Philip can represent sorting the acorns is 2(33 + 13).

Mei is playing a game with numbers. She decides to rewrite the number 78 using the Distributive Property. Which answer is a correct way to rewrite 78 using the Distributive Property of Multiplication over Addition?(1 point)
Responses

78=24+54=(6×4)+(6×9)=6(4×9)
78 equals 24 plus 54 equals left parenthesis 6 times 4 right parenthesis plus left parenthesis 6 times 9 right parenthesis equals 6 left parenthesis 4 times 9 right parenthesis

78=18+60=(6×3)+(6×10)=6(3+10)
78 equals 18 plus 60 equals left parenthesis 6 times 3 right parenthesis plus left parenthesis 6 times 10 right parenthesis equals 6 left parenthesis 3 plus 10 right parenthesis

78=22+56=(2×11)+(2×28)=2(11×28)
78 equals 22 plus 56 equals left parenthesis 2 times 11 right parenthesis plus left parenthesis 2 times 28 right parenthesis equals 2 left parenthesis 11 times 28 right parenthesis

78=11+67=(1×11)+(1×67)=1(11×67)
78 equals 11 plus 67 equals left parenthesis 1 times 11 right parenthesis plus left parenthesis 1 times 67 right parenthesis equals 1 left parenthesis 11 times 67 right parenthesis

To rewrite the number 78 using the Distributive Property of Multiplication over Addition, we need to express 78 as a sum of two numbers that can each be factored.

The correct application of the Distributive Property can be written as:

\[ a + b = c \Rightarrow c = d \cdot (e + f) \]

where \( d \cdot e = a \) and \( d \cdot f = b \).

Let's analyze each of the provided options:

1. 78 = 24 + 54 = (6 × 4) + (6 × 9) = 6(4 + 9)
   This is incorrect because \( 6(4 + 9) = 6 \times 13 = 78 \), but the factorization directly uses 6 as a multiplier and sums the terms correctly.

2. 78 = 18 + 60 = (6 × 3) + (6 × 10) = 6(3 + 10)
   This is correct because \( 6(3 + 10) = 6 \times 13 = 78 \). This expression follows the Distributive Property correctly.

3. 78 = 22 + 56 = (2 × 11) + (2 × 28) = 2(11 + 28)
   This is incorrect because \( 2(11 + 28) = 2 \times 39 = 78 \), but it uses 2 incorrectly.

4. 78 = 11 + 67 = (1 × 11) + (1 × 67) = 1(11 + 67)
   While mathematically correct, this does not showcase the Distributive Property effectively since multiplying anything by 1 does not provide new information.

The only option that effectively uses the Distributive Property is 78 = 18 + 60 = (6 × 3) + (6 × 10) = 6(3 + 10).

Thus, the correct answer is:

78 = 18 + 60 = (6 × 3) + (6 × 10) = 6(3 + 10).

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 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 .

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 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)
.

To solve the problem, we want to determine how many groups Sven can make such that each group has the same number of notebooks and the same number of binders.

Sven has:

• 52 notebooks
• 42 binders

To find the maximum number of equal groups with the same number of each item, we can find the greatest common divisor (GCD) of the two quantities:

The GCD of 52 and 42 can be determined as follows:

• The factors of 52 are: 1, 2, 4, 13, 26, 52.
• The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42.

The largest common factor is 2.

This means that Sven can make 2 groups. In each group, he would put:

• Notebooks per group: \( 52 \div 2 = 26 \)
• Binders per group: \( 42 \div 2 = 21 \)

Using the Distributive Property, this can be expressed as: \[ 2(26 + 21) \]

Therefore, 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).

Given the sum 100=64+36
, rewrite 64+36
using their GCF and multiplication.(1 point)
Responses

4(16×9)
4 left parenthesis 16 times 19 right parenthesis

2(32+18)
2 left parenthesis 32 plus 18 right parenthesis

4(16+9)
4 left parenthesis 16 plus 9 right parenthesis

1(64+36)

To rewrite \( 64 + 36 \) using their greatest common factor (GCF) and multiplication, we first need to determine the GCF of the two numbers.

1. The factors of \( 64 \) are: \( 1, 2, 4, 8, 16, 32, 64 \).
2. The factors of \( 36 \) are: \( 1, 2, 3, 4, 6, 9, 12, 18, 36 \).

The GCF of \( 64 \) and \( 36 \) is \( 4 \).

Now, we can factor out \( 4 \) from both terms:

\[ 64 + 36 = 4(16 + 9) \]

Thus, the correct answer to rewrite \( 64 + 36 \) using their GCF and multiplication is:

4(16 + 9).

Given the sum 18=6+12
, rewrite 6+12
using their GCF and multiplication. (1 point)
Responses

6(1×2)
6 times open paren 1 times 2 close paren

6(1+2)
6 times open paren 1 plus 2 close paren

1(6+12)
1 times open paren 6 plus 12 close paren

3(2+4)

To rewrite \( 6 + 12 \) using their greatest common factor (GCF) and multiplication, we first need to determine the GCF of the two numbers.

1. The factors of \( 6 \) are: \( 1, 2, 3, 6 \).
2. The factors of \( 12 \) are: \( 1, 2, 3, 4, 6, 12 \).

The GCF of \( 6 \) and \( 12 \) is \( 6 \).

Now, we can factor out \( 6 \) from both terms:

\[ 6 + 12 = 6(1 + 2) \]

Thus, the correct answer to rewrite \( 6 + 12 \) using their GCF and multiplication is:

6(1 + 2).