Aimathic – Distributive property with common factors

Distributive property with common factors

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Difficulty: 1 – 5

5542866

Use the distributive property to rewrite and evaluate $6\times(4+3)$.

Hints

- The factor outside the parentheses multiplies each addend inside. - Write two products before evaluating them. - Check that both products use the outside factor $6$.

Solution

1. Distribute $6$ to both addends: $6\times4+6\times3$. 2. Evaluate the products: $24+18$. 3. Add to get $42$.

Answer

$6\times4+6\times3=42$

5112996

Students are asked to evaluate $0.4 \times \frac{5}{11} + 0.7 \times \frac{5}{11}$. Luke finds the two products separately and then adds them. Maria uses the distributive property and factors out the common factor $\frac{5}{11}$. a) Find the value using Maria’s method. b) Compare the methods. What advantage does Maria’s method have?

Hints

- Look for a factor that appears in both terms. - Use the distributive property to factor out the common factor. - Add the decimals before multiplying by the fraction.

Solution

1. Factor out the common factor: $(0.4 + 0.7) \times \frac{5}{11}$. 2. Add inside the parentheses: $0.4 + 0.7 = 1.1$. 3. Write $1.1$ as a fraction: $1.1 = \frac{11}{10}$. 4. Multiply and simplify: $\frac{11}{10} \times \frac{5}{11} = \frac{1}{2}$. 5. Maria’s method uses one multiplication instead of two, and the factors simplify immediately.

Answer

a) The value is $\frac{1}{2}$, or $0.5$. b) Maria’s method is more efficient because it uses one multiplication and allows immediate simplification.

5122536

Use factoring to calculate each expression efficiently. Name the property you use. a) $7.2\times14+7.2\times6$ b) $13\times105-13\times5$ c) $4\times28+4\times22$ d) $0.25\times37-0.25\times17$

Hints

- Find the factor shared by both products in each expression. - Write that common factor outside parentheses. - Look for a sum or difference inside the parentheses that is easy to calculate.

Solution

1. Use the distributive property in reverse by factoring out the multiplier shared by both products. 2. For a), $7.2\times(14+6)=7.2\times20=144$. 3. For b), $13\times(105-5)=13\times100=1300$. 4. For c), $4\times(28+22)=4\times50=200$. 5. For d), $0.25\times(37-17)=0.25\times20=5$.

Answer

a) $144$ b) $1300$ c) $200$ d) $5$ Property: the distributive property

5170336

Factor out the common factor to evaluate each expression efficiently. a) $13 \times 6 + 7 \times 6$ b) $24 \times 5 + 6 \times 5$ c) $18 \times 3 + 12 \times 3$

Hints

- Identify the factor that appears in both terms. - Add the other two factors inside parentheses before multiplying.

Solution

1. For a), factor out $6$: $13 \times 6 + 7 \times 6 = (13 + 7) \times 6 = 20 \times 6 = 120$. 2. For b), factor out $5$: $24 \times 5 + 6 \times 5 = (24 + 6) \times 5 = 30 \times 5 = 150$. 3. For c), factor out $3$: $18 \times 3 + 12 \times 3 = (18 + 12) \times 3 = 30 \times 3 = 90$.

Answer

a) $120$ b) $150$ c) $90$

5197056

Use the distributive property to fill in each missing number. The shapes represent new unknowns in each part. a) $14 \times 7 + 14 \times 3 = \square \times (7 + 3)$ b) $48 \times 5 - 8 \times 5 = (48 - 8) \times \triangle$ c) $(\triangle + 12) \times 9 = 20 \times 9 + 12 \times \square$ d) $15 \times 11 - 15 \times \square = 15 \times (11 - 4)$

Hints

- Identify the factor repeated in each pair of products. - Compare the factored form with the expanded form. - In part c, match each factor on the left with the corresponding product on the right.

Solution

1. For a), the common factor is $14$, so $\square = 14$. 2. For b), the common factor is $5$, so $\triangle = 5$. 3. For c), compare corresponding factors in $(20 + 12) \times 9 = 20 \times 9 + 12 \times 9$. Thus, $\triangle = 20$ and $\square = 9$. 4. For d), the second factor being subtracted is $4$, so $\square = 4$.

Answer

a) $\square = 14$ b) $\triangle = 5$ c) $\triangle = 20$; $\square = 9$ d) $\square = 4$

5197276

Evaluate efficiently and name the properties you use: $38 \times 63 + 37 \times 38$

Hints

- Rewrite the factors so the repeated factor appears in the same position. - Factor out the common factor. - Add inside the grouping symbols before multiplying.

Solution

1. Use the commutative property to rewrite $37 \times 38$ as $38 \times 37$. 2. Factor out the common factor $38$: $38(63 + 37)$. 3. Evaluate: $38 \times 100 = 3800$.

Answer

$3800$; commutative and distributive properties

5198836

Factor out the common factor to evaluate efficiently: $17 \times 8 - 17 \times 3 + 5 \times 17$

Hints

- Identify the factor present in all three terms. - Rewrite factors in the same order if needed. - Factor first, then evaluate inside the grouping symbols.

Solution

1. Rewrite $5 \times 17$ as $17 \times 5$. 2. Factor out $17$: $17(8 - 3 + 5)$. 3. Evaluate: $17 \times 10 = 170$.

Answer

$17(8 - 3 + 5) = 170$

5199846

Use the distributive property to factor out the common factor and evaluate: $12 \times 17 + 12 \times 3$

Hints

- Identify the factor present in both products. - Factor it out. - Add inside the grouping symbols before multiplying.

Solution

1. Factor out $12$: $12(17 + 3)$. 2. Evaluate: $12 \times 20 = 240$.

Answer

$12(17 + 3) = 240$; distributive property

5209016

Calculate efficiently by factoring out the common factor: $125\,\text{mg} \times 18 + 0.875\,\text{g} \times 18$.

Hints

- Notice that both terms have the same factor. - Use the distributive property in reverse to factor out the common factor. - Convert the measurements inside the parentheses to the same unit.

Solution

1. Factor out $18$: $(125\,\text{mg} + 0.875\,\text{g}) \times 18$. 2. Convert: $0.875\,\text{g} = 875\,\text{mg}$. 3. Add inside the parentheses: $125\,\text{mg} + 875\,\text{mg} = 1000\,\text{mg} = 1\,\text{g}$. 4. Multiply: $1\,\text{g} \times 18 = 18\,\text{g}$.

Answer

$18\,\text{g}$

5412436

Use the greatest common factor and the distributive property to rewrite $48+30$ as a product of a whole number and a sum of two whole numbers with no common factor greater than $1$.

Hints

- Find the largest whole number that divides both addends. - Determine what remains after each addend is divided by that number. - Check that distributing the outside factor recreates the original sum.

Solution

1. The greatest common factor of $48$ and $30$ is $6$. 2. Divide each addend by $6$: $48\div6=8$ and $30\div6=5$. 3. Therefore, $48+30=6(8+5)$, and $8$ and $5$ have no common factor greater than $1$.

Answer

$6(8+5)$

5412456

Complete the equation $56+\square=8(7+3)$. Find the missing addend and explain how the distributive property determines it.

Hints

- Connect each term inside the parentheses with one addend outside. - Use the outside factor with the term that has not yet been matched. - Expand the completed product to check the entire equation.

Solution

1. Distribute $8$: $8(7+3)=8\times7+8\times3$. 2. The first product is $56$, matching the given addend. 3. The missing addend is $8\times3=24$.

Answer

$24$

5412516

Which sum is equivalent to $12(5+7)$: $48+72$, $60+84$, $60+72$, or $72+84$? Explain how the distributive property identifies the answer.

Hints

- Keep the two products separate when expanding. - Match the first inside term to the first addend and the second to the second addend. - Check the selected sum by factoring out the stated outside number.

Solution

1. Multiply $12$ by each term inside the parentheses. 2. Compute $12\times5=60$ and $12\times7=84$. 3. Therefore, the equivalent sum is $60+84$.

Answer

$60+84$

5542876

Rewrite $18+24$ as a product by factoring out the greatest common factor. Then verify your product with the distributive property.

Hints

- Find the greatest number that divides both terms. - Express each term as that common factor times another whole number. - Check by distributing the common factor back across the parentheses.

Solution

1. The GCF of $18$ and $24$ is $6$. 2. Write $18=6\times3$ and $24=6\times4$. 3. Factor: $18+24=6\times(3+4)$. 4. Distributing $6$ gives $6\times3+6\times4=18+24$, so the factorization is verified.

Answer

$6\times(3+4)$

5123056

Use factoring to calculate each expression efficiently. a) $4.2\times17+4.2\times3$ b) $\frac{2}{3}\times5.8+\frac{2}{3}\times4.2$

Hints

- Find the factor shared by the two products in each expression. - Factor that common number outside parentheses. - Check whether the remaining numbers make an especially convenient sum.

Solution

1. For a), factor out $4.2$: $4.2\times(17+3)=4.2\times20=84$. 2. For b), factor out $\frac{2}{3}$: $\frac{2}{3}\times(5.8+4.2)=\frac{2}{3}\times10=\frac{20}{3}=6\frac{2}{3}$.

Answer

a) $84$ b) $\frac{20}{3}$, or $6\frac{2}{3}$

5170346

Compare two methods for evaluating $16 \times 8 + 4 \times 8$. Method A: $16 \times 8 = 128$ $4 \times 8 = 32$ $128 + 32 = \square$ Method B: $(16 + 4) \times 8 = \square \times 8 = \square$ Complete both methods. Which method is more efficient for mental math? Explain why.

Hints

- Complete each method before comparing them. - In Method B, identify the common factor and add the remaining factors first. - Decide which intermediate numbers are easier to use mentally.

Solution

1. Method A gives $128 + 32 = 160$. 2. Method B factors out the common factor $8$: $(16 + 4) \times 8 = 20 \times 8 = 160$. 3. Method B is more efficient mentally because it creates the friendly factor $20$ before multiplying.

Answer

Method A: $160$ Method B: $160$ Method B is more efficient for mental math because factoring out $8$ creates the easier product $20 \times 8$.

5170356

For a field day, Mr. Walker buys two boxes of jump ropes. The first box contains $14$ red jump ropes, and the second contains $16$ green jump ropes. Each jump rope costs $\$4$. a) Write an expression that finds the total cost by calculating the cost of each box separately. b) Write an equivalent expression that first finds the total number of jump ropes. c) How much does Mr. Walker pay altogether?

Hints

- Write a multiplication expression for the cost of each color first. - Both groups have the same price per jump rope, so that common factor can be used once. - Add the numbers of jump ropes before multiplying in the second expression.

Solution

1. For a), calculate each box separately: $14 \times 4 + 16 \times 4$. 2. For b), factor out the common cost of $4$: $(14 + 16) \times 4$. 3. Evaluate the factored expression: $(14 + 16) \times 4 = 30 \times 4 = 120$.

Answer

a) $14 \times 4 + 16 \times 4$ b) $(14 + 16) \times 4$ c) $\$120$

5180396

Two fourth-grade classes each have $22$ students. Every student in Class 4A buys a $\$6$ popcorn, and every student in Class 4B buys a $\$4$ popcorn. Use the distributive property in both directions to find how much more Class 4A spends: a) first evaluate $22\times6-22\times4$; b) then factor the common factor $22$ and evaluate the equivalent factored expression. Explain why the two expressions must have the same value.

Hints

- Identify the factor that appears in both products. - Factoring reverses distribution: the common factor moves outside parentheses. - After rewriting, evaluate both forms and compare their values.

Solution

1. Expanded method: $22\times6-22\times4=132-88=44$. 2. Factor the common factor $22$: $22\times6-22\times4=22(6-4)$. 3. Evaluate the factored form: $22(2)=44$. 4. The distributive property guarantees that expanding $22(6-4)$ gives $22\times6-22\times4$, so the expressions are equivalent.

Answer

$\$44$. The equivalent expressions are $22\times6-22\times4$ and $22(6-4)$; they have the same value because the distributive property expands the factored form into the expanded form.

5192546

A gym has $12$ rows with $15$ adult seats in each row and $12$ rows with $25$ student seats in each row. a) Write one expanded expression and one factored expression for the total number of seats. The two expressions must be related by the distributive property. b) Evaluate both expressions and explain why they give the same total.

Hints

- Look for the factor that is common to both seat groups. - In the factored form, place that common factor outside parentheses. - Evaluate each form separately after you have written the equivalent expressions.

Solution

1. The expanded expression is $12\times15+12\times25$. 2. Factor the common factor $12$: $12\times15+12\times25=12(15+25)$. 3. The expanded form gives $180+300=480$. 4. The factored form gives $12\times40=480$. 5. They agree because the distributive property makes the two expressions equivalent.

Answer

a) Expanded: $12\times15+12\times25$; factored: $12(15+25)$ b) Both equal $480$ seats.

5196826

Use the distributive property to factor out the common factor and evaluate each expression. a) $12 \times 17 + 12 \times 11 + 12 \times 2$ b) $18 \times 24 + 16 \times 18$ c) $27 \times 15 - 15 \times 7$ d) $13 \times 29 - 13 \times 18 - 13$ e) $19 \times 14 + 19 \times 23 + 19 \times 13$ f) $22 \times 35 - 22 \times 14 - 22$

Hints

- Identify the factor shared by every term. - A term such as $13$ can be written as $13 \times 1$. - Factor first, then evaluate inside the grouping symbols.

Solution

1. For a), factor out $12$: $12(17 + 11 + 2) = 12 \times 30 = 360$. 2. For b), rewrite $16 \times 18$ as $18 \times 16$, then factor: $18(24 + 16) = 18 \times 40 = 720$. 3. For c), factor out $15$: $15(27 - 7) = 15 \times 20 = 300$. 4. For d), write $13$ as $13 \times 1$, then factor: $13(29 - 18 - 1) = 13 \times 10 = 130$. 5. For e), factor out $19$: $19(14 + 23 + 13) = 19 \times 50 = 950$. 6. For f), write $22$ as $22 \times 1$, then factor: $22(35 - 14 - 1) = 22 \times 20 = 440$.

Answer

a) $360$ b) $720$ c) $300$ d) $130$ e) $950$ f) $440$

5197076

For each equation, first use the distributive property in reverse to rewrite the left side as one product. Then find the missing value. a) $17\times5+17\times x=170$ b) $4\times y-4\times15=40$ c) $13\times12-x\times12=60$

Hints

- Identify the common factor in the two terms on the left side of each equation. - Your rewritten equation must show that common factor outside parentheses. - Only after rewriting should you use inverse operations to find the missing value.

Solution

1. For a), factor out $17$: $17\times(5+x)=170$. Since $170\div17=10$, $5+x=10$, so $x=5$. 2. For b), factor out $4$: $4\times(y-15)=40$. Since $40\div4=10$, $y-15=10$, so $y=25$. 3. For c), factor out $12$: $(13-x)\times12=60$. Since $60\div12=5$, $13-x=5$, so $x=8$.

Answer

a) $17\times(5+x)=170$; $x=5$ b) $4\times(y-15)=40$; $y=25$ c) $(13-x)\times12=60$; $x=8$

5197286

Use the distributive property to evaluate efficiently: $99 \times 17 + 17$

Hints

- Write the single $17$ as a product with factor $1$. - Identify the common factor. - Factor before calculating.

Solution

1. Rewrite $17$ as $1 \times 17$: $99 \times 17 + 1 \times 17$. 2. Factor out $17$: $(99 + 1) \times 17$. 3. Evaluate: $100 \times 17 = 1700$.

Answer

$1700$

5197296

Evaluate efficiently and name the property you use: $36 \times 12 + 14 \times 12 - 5 \times 4 \times 12$

Hints

- Simplify $5 \times 4$ first. - Identify the factor shared by all three terms. - Factor it out, then evaluate inside the grouping symbols.

Solution

1. Simplify the coefficient in the last term: $5 \times 4 = 20$. The expression becomes $36 \times 12 + 14 \times 12 - 20 \times 12$. 2. Factor out the common factor $12$: $(36 + 14 - 20) \times 12$. 3. Evaluate: $30 \times 12 = 360$.

Answer

$360$; distributive property

5197396

Examine the incorrect calculation $(12 + 9) \times 4 = 48 + 9 = 57$. Explain the error. Then correct the calculation in two ways: first by evaluating the parentheses, and second by using the distributive property.

Hints

- A factor outside parentheses applies to the entire sum. - Try evaluating the sum before multiplying. - When distributing, multiply every addend by the outside factor.

Solution

1. The factor $4$ was applied to $12$ but not to $9$. 2. Evaluate the parentheses first: $(12 + 9) \times 4 = 21 \times 4 = 84$. 3. Use the distributive property: $(12 + 9) \times 4 = 12 \times 4 + 9 \times 4 = 48 + 36 = 84$.

Answer

The factor $4$ must multiply both addends. Parentheses first: $(12 + 9) \times 4 = 84$. Distributive property: $12 \times 4 + 9 \times 4 = 84$.

5205986

A school-supply store delivers $28$ notebooks to Class 4A, $32$ to Class 4B, and $30$ to Class 4C. Each notebook costs $\$0.85$. a) Write a factored expression for the total cost using $\$0.85$ as the common factor. b) Use the distributive property to write the equivalent expanded expression. c) Evaluate the expressions to find the total cost.

Hints

- The price per notebook is the factor shared by all three class costs. - Distribution multiplies that common factor by every term inside the parentheses. - Evaluate one of the equivalent forms, then use the other as a check.

Solution

1. A factored expression is $0.85(28+32+30)$. 2. Distributing $0.85$ gives $0.85\times28+0.85\times32+0.85\times30$. 3. The factored form gives $0.85\times90=76.50$. 4. The expanded form gives $23.80+27.20+25.50=76.50$.

Answer

a) $0.85(28+32+30)$ b) $0.85\times28+0.85\times32+0.85\times30$ c) $\$76.50$

5412446

The expression $7(6+10)$ is equivalent to $42+70$, but it does not factor out the greatest common factor. Rewrite $42+70$ using the greatest common factor and explain what changes inside the parentheses.

Hints

- Find the greatest common factor of the original addends. - Divide both addends by that factor. - Compare the new inside terms with the terms in the given equivalent expression.

Solution

1. The greatest common factor of $42$ and $70$ is $14$. 2. Divide the addends by $14$: $42\div14=3$ and $70\div14=5$. 3. The greatest-common-factor form is $14(3+5)$. 4. Compared with $7(6+10)$, the outside factor is doubled and each inside term is halved; $3$ and $5$ have no common factor greater than $1$.

Answer

$14(3+5)$. The inside terms change from $6$ and $10$ to $3$ and $5$, which have no common factor greater than $1$.

5412466

A workshop prepares $54$ small tags and $72$ large tags. Write the total number of tags as the greatest possible number of identical groups times the number of small and large tags in each group. Use the distributive property.

Hints

- The group count must be a factor of both tag totals. - Maximize that shared factor before finding the contents of one group. - Match the context to the outside factor and the two terms inside.

Solution

1. The greatest common factor of $54$ and $72$ is $18$, so there can be $18$ identical groups. 2. Each group contains $54\div18=3$ small tags and $72\div18=4$ large tags. 3. The total is represented by $54+72=18(3+4)$.

Answer

$18(3+4)$: $18$ groups with $3$ small tags and $4$ large tags in each group

5412476

Fill the blank: $63+42=\square(3+2)$. Find the outside factor and verify that it is the greatest common factor of the original addends.

Hints

- Compare each original addend with its matching term inside the parentheses. - The same multiplier must connect both pairs. - Check whether any larger number could divide both original addends.

Solution

1. Since $63\div3=21$ and $42\div2=21$, the outside factor is $21$. 2. Distributing gives $21\times3+21\times2=63+42$. 3. The greatest common factor of $63$ and $42$ is $21$, so the form uses the greatest possible outside factor.

Answer

$63+42=21(3+2)$

5412486

Expand $6(11+4)$ into a sum of two whole numbers. Then determine whether $6$ is the greatest common factor of those addends.

Hints

- Multiply the outside factor by each inside term separately. - After expanding, examine whether the inside terms share another factor. - A shared inside factor would mean the outside factor was not greatest.

Solution

1. Distribute: $6(11+4)=6\times11+6\times4$. 2. The expanded sum is $66+24$. 3. Since $11$ and $4$ have no common factor greater than $1$, the greatest common factor of $66$ and $24$ is $6$.

Answer

$66+24$, and $6$ is the greatest common factor.

5412496

The sum $36+24$ can be factored using several common factors. Which of these forms uses the greatest common factor: $3(12+8)$, $4(9+6)$, $6(6+4)$, or $12(3+2)$? Explain why the other forms are not fully factored.

Hints

- All the choices may be equivalent, so equivalence alone is not enough. - Inspect the two terms inside each set of parentheses. - The fully factored form leaves relatively prime terms inside.

Solution

1. The greatest common factor of $36$ and $24$ is $12$. 2. Thus, the greatest-common-factor form is $12(3+2)$. 3. In each other form, the two inside terms still have a common factor greater than $1$, so a larger factor can be taken outside.

Answer

$12(3+2)$

5412506

Use the greatest common factor and the distributive property to decide whether $25+18$ can be rewritten with a whole-number factor greater than $1$ outside the parentheses. Explain and write its greatest-common-factor form.

Hints

- Find the prime factors of both addends. - Check whether any prime factor is shared. - Interpret what a greatest common factor of $1$ means for factoring a sum.

Solution

1. The prime factorization of $25$ is $5^2$, and the prime factorization of $18$ is $2\times3^2$. 2. The addends share no prime factor, so $\operatorname{GCF}(25,18)=1$. 3. Therefore, there is no whole-number common factor greater than $1$ to factor out. 4. The greatest-common-factor form is $1(25+18)$, which is usually left as $25+18$.

Answer

No. $\operatorname{GCF}(25,18)=1$, so the greatest-common-factor form is $1(25+18)$, usually written $25+18$.

5412526

Factor each sum using its greatest common factor: $42+30$ and $42+35$. Which sum can be arranged into more identical groups, and how does the outside factor show this?

Hints

- Find the greatest shared factor for each pair separately. - Divide the addends to obtain the terms inside each set of parentheses. - Interpret each outside factor as a possible number of equal groups.

Solution

1. The greatest common factor of $42$ and $30$ is $6$, so $42+30=6(7+5)$. 2. The greatest common factor of $42$ and $35$ is $7$, so $42+35=7(6+5)$. 3. The second sum makes more identical groups because its outside factor is $7$, compared with $6$.

Answer

$42+30=6(7+5)$ and $42+35=7(6+5)$. The second sum makes more identical groups.

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