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The Distributive Property: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A factor on the right

Rewrite $(30 + 8) \times 7$ as a sum of two products, one for each term in the parentheses, without evaluating the products.
1. Hint 1
  Changing the order of two factors leaves their product unchanged.
2. Hint 2
  Move $7$ to the left of the parentheses, then multiply each term inside by that factor.
Answer

$30 \times 7 + 8 \times 7$ , or equivalently $7 \times 30 + 7 \times 8$ .

Full solution

Multiplication is commutative, so $(30 + 8) \times 7$ has the same value as $7(30 + 8)$ .
With the factor $7$ now on the left, distribute it to both terms.
$7(30 + 8) = 7 \times 30 + 7 \times 8$

Swapping the two factors in each product changes neither product, so the sum can be written with $7$ on the right.
$(30 + 8) \times 7 = 30 \times 7 + 8 \times 7$

Check the values: the original product is $38 \times 7$ , which is $266$ .
The two products are $210$ and $56$ , and their sum is also $266$ .

Answer

$30 \times 7 + 8 \times 7$ , or equivalently $7 \times 30 + 7 \times 8$ .

Key idea

An outside factor multiplies every term in a sum even when the factor is written on the right.
Problem 1: Mark complete
Problem 2 Completing a distributed difference

Find the whole number that belongs in the box: $8(\square - 6) = 400 - 48$ .
1. Hint 1
  The two sides must describe the same difference of two products.
2. Hint 2
  The subtracted $48$ matches $8 \times 6$ . Match $400$ with $8$ times the missing number.
Answer

$50$ .

Full solution

The factor $8$ multiplies both numbers inside the parentheses.
The second product is $8 \times 6 = 48$ , which matches the subtraction on the right.
The first product must be $400$ , so divide $400$ by the outside factor.
$400 \div 8 = 50$

Check by putting $50$ in the box.
$\begin{aligned} 8(50 - 6) &= 8 \times 44 \\ &= 352 \end{aligned}$

The right side also has value $400 - 48 = 352$ , so the equality is true.

Answer

$50$ .

Key idea

In a distributed difference, each product pairs the outside factor with one of the numbers inside the parentheses.
Problem 2: Mark complete
Problem 3 Collecting the shared factor

Write $16 + 48$ as a product with its greatest common factor outside parentheses and a sum inside.
1. Hint 1
  A positive common factor cannot exceed the smaller of the two positive terms.
2. Hint 2
  Test whether the smaller term divides the larger, then divide both terms by the factor you find.
Answer

$16(1 + 3)$ .

Full solution

The number $16$ divides itself and also divides $48$ .
No number larger than $16$ divides $16$ , so $16$ is the greatest common factor.
The two quotients are $1$ and $3$ .
$16 + 48 = 16(1 + 3)$

Check by distributing back: $16 \times 1 = 16$ and
$16 \times 3 = 48.$
The terms inside share no factor greater than $1$ .

Answer

$16(1 + 3)$ .

Key idea

When the smaller positive term divides the larger, it is their greatest common factor.
Problem 3: Mark complete
Problem 4 Pages for a print order

A print shop prints 8 manuals with 102 pages in each manual. It also prints 24 single-page notices. How many pages does it print altogether?
1. Hint 1
  Count the pages in the repeated manuals, then include the notices once.
2. Hint 2
  For mental multiplication, split $102$ into a round hundred and the amount left over.
3. Hint 3
  After finding $8 \times 100$ and $8 \times 2$ , add those results and the $24$ notice pages.
Answer

$840$ pages.

Full solution

Each manual has $100$ pages plus $2$ more.
Multiply both parts by the number of manuals.
$\begin{aligned} 8 \times 102 &= 8(100 + 2) \\ &= 800 + 16 \\ &= 816 \end{aligned}$

The notices add $24$ pages to the manual pages.
$816 + 24 = 840$

Check by counting the $800$ pages from the hundreds first.
The $16$ extra manual pages and $24$ notice pages make $40$ more pages, giving $840$ pages altogether.

Answer

$840$ pages.

Key idea

Splitting a factor near a round number makes the repeated part of a total easier to calculate.
Problem 4: Mark complete
Problem 5 Two parts of one rectangle

A rectangle is 6 cm high. A straight cut parallel to its sides, from its top edge to its bottom edge, divides it into two smaller rectangles. The left piece is 7 cm wide, and the right piece has an area of 54 square cm. Both pieces retain the full height.

Find the area of the original rectangle. Explain why adding the two piece areas gives the same result as multiplying its full height by its full width.
1. Hint 1
  The pieces cover the original rectangle without gaps or overlaps.
2. Hint 2
  The right piece has height $6$ cm, so divide its area by $6$ to find its width.
3. Hint 3
  Add the two widths for the full width, and compare the whole area with the two piece areas.
Answer

$96$ square cm; $6(7 + 9) = 6 \times 7 + 6 \times 9$ .

Full solution

A rectangle's area is its height times its width.
The right piece has height $6$ cm and area $54$ square cm, so its width is $54 \div 6 = 9$ cm.
The full width is $7 + 9 = 16$ cm.
$6 \times 16 = 96$

The left piece has area $6 \times 7 = 42$ square cm, and the right piece has area $54$ square cm.
Their sum checks the whole area.
$42 + 54 = 96$

The cut changes neither the amount of space nor the common height.
Adding the areas counts each square cm of the original rectangle exactly once.
That is why $6(7 + 9)$ equals $6 \times 7 + 6 \times 9$ .

Answer

$96$ square cm; $6(7 + 9) = 6 \times 7 + 6 \times 9$ .

Key idea

Splitting a rectangle's width separates its area into two products with the same height.
Problem 5: Mark complete
Problem 6 Counters after a class activity

A teacher has 6 sets of counters, with 32 counters in each set. After removing 7 counters from each set, the teacher puts all the remaining counters into a storage box that already holds 18 counters. How many counters are now in the storage box?
1. Hint 1
  The removal happens in every set, while the counters already in storage are counted once.
2. Hint 2
  The remaining counters from the sets are described by $6(32 - 7)$ .
3. Hint 3
  Subtract the total removed from the total originally in the sets, then include the stored counters.
Answer

$168$ counters.

Full solution

Each of the $6$ sets loses $7$ counters, so the outside factor multiplies both the starting amount and the amount removed.
$\begin{aligned} 6(32 - 7) &= 6 \times 32 - 6 \times 7 \\ &= 192 - 42 \\ &= 150 \end{aligned}$

Add the $18$ counters already in storage.
$150 + 18 = 168$

Check by finding the remainder in one set first:
$32 - 7 = 25.$
Six sets of $25$ hold $150$ counters, and the stored $18$ bring the total to $168$ counters.

Answer

$168$ counters.

Key idea

When every equal group loses the same amount, the total loss is the number of groups times that amount.
Problem 6: Mark complete
Problem 7 Joining colored tile panels

A designer uses all 12 orange tiles to make one solid rectangle and all 20 purple tiles to make another. Each square tile measures 1 cm on each side, and every tile stays whole. The rectangles must have the same height and are joined along their full heights, without gaps or overlaps.

What is the greatest possible height of the joined rectangle, and what is its width at that height?
1. Hint 1
  The shared height must allow each color's tiles to form complete rows.
2. Hint 2
  Look for the largest whole number that divides both $12$ and $20$ .
3. Hint 3
  Divide each tile count by the shared height to find the two widths, then add those widths.
Answer

Height: $4$ cm; width: $8$ cm.

Full solution

Each tile covers $1$ square cm.
A whole-number height must divide both areas.
The possible heights for the orange piece are $1$ , $2$ , $3$ , $4$ , $6$ and $12$ cm.
Of these, $1$ , $2$ and $4$ also divide $20$ , so the greatest shared height is $4$ cm.

At height $4$ cm, the orange piece is $12 \div 4 = 3$ cm wide, and the purple piece is $20 \div 4 = 5$ cm wide.
$12 + 20 = 4(3 + 5)$

The joined width is $3 + 5 = 8$ cm.
Check: a rectangle $4$ cm high and $8$ cm wide has area $4 \times 8 = 32$ square cm, matching all $32$ tiles.

Answer

Height: $4$ cm; width: $8$ cm.

Key idea

A greatest common factor can represent the greatest shared whole-number height of adjacent rectangles.
Problem 7: Mark complete
Problem 8 Checking a shortcut claim

A student knows that $7 \times 86 = 602$ and wants to find $7 \times 85$ . The student says the new product is $1$ less, since $85$ is $1$ less than $86$ . Decide whether the claim is correct, and state how much the product decreases.
1. Hint 1
  A change in the size of every equal group affects the total once for each group.
2. Hint 2
  Write $86$ as $85 + 1$ , then compare the distributed expression with $7 \times 85$ .
Answer

Incorrect; the product decreases by $7$ .

Full solution

There are $7$ equal groups.
Changing each group from $86$ to $85$ removes one from each of them.
Distribution records all seven removals.
$\begin{aligned} 7 \times 86 &= 7(85 + 1) \\ &= 7 \times 85 + 7 \times 1 \end{aligned}$

So $7 \times 85$ is $7$ less than $602$ .
$602 - 7 = 595$

Check directly: $7 \times 80 = 560$ and $7 \times 5 = 35$ , so $7 \times 85 = 595$ , which is $7$ less than $602$ .

The claim is incorrect.
The decrease is seven, matching the seven single items removed.

Answer

Incorrect; the product decreases by $7$ .

Key idea

Reducing each equal group by one reduces the total by the number of groups.
Problem 8: Mark complete
Problem 9 Boxes on shelves

There are $2$ shelves, each holding $6$ boxes, with $7$ balls in each box. A student writes the total as $2(6 \times 7)$ , then rewrites it as $(2 \times 6)(2 \times 7)$ , saying that the $2$ must multiply everything inside the parentheses. Is this rewrite valid? Explain, and give the correct total number of balls.
1. Hint 1
  Distinguish groups within groups from two amounts being added.
2. Hint 2
  Count the balls on one shelf, then multiply by the number of shelves.
3. Hint 3
  Compare how many times the factor $2$ appears before and after the rewrite.
Answer

The rewrite is invalid; $84$ balls.

Full solution

The parentheses contain a product, so the distributive rule for a sum or difference does not apply.
One shelf has $6 \times 7 = 42$ balls.
$\begin{aligned} 2(6 \times 7) &= 2 \times 42 \\ &= 84 \end{aligned}$

The proposed rewrite introduces a second factor of $2$ .
Its value is different.
$\begin{aligned} (2 \times 6)(2 \times 7) &= 12 \times 14 \\ &= 168 \end{aligned}$

Check the correct count another way: $2$ shelves of $6$ boxes make $12$ boxes, and $12 \times 7 = 84$ balls.
Regrouping the original three factors counts each shelf, box and ball once.

Answer

The rewrite is invalid; $84$ balls.

Key idea

A product can be regrouped, but multiplying both inner factors by the outside factor introduces an extra copy of that factor.
Problem 9: Mark complete
Problem 10 Two factoring decisions

To factor $40 + 72$ , Mira writes $4(10 + 18)$ and Eli writes $8(5 + 9)$ . Are both expressions equal to the original sum, and which one has the greatest common factor outside the parentheses? Justify your decisions.
1. Hint 1
  A factoring can preserve the value without taking out the greatest common factor.
2. Hint 2
  Distribute each outside factor back to check the two original terms.
3. Hint 3
  Inspect the terms still inside each pair of parentheses for a common factor greater than one.
Answer

Both equal $40 + 72$ ; Eli's expression has the greatest common factor, $8$ , outside.

Full solution

Mira's outside factor gives $4 \times 10 = 40$ and
$4 \times 18 = 72.$
Eli's gives $8 \times 5 = 40$ and
$8 \times 9 = 72.$
Both expressions therefore reproduce the original terms.

Mira has left a factor of $2$ in both inside terms.
Taking it out combines it with the outside $4$ .
$\begin{aligned} 4(10 + 18) &= 4 \times [2(5 + 9)] \\ &= 8(5 + 9) \end{aligned}$

The terms $5$ and $9$ share no factor greater than $1$ : the only factor of $5$ above $1$ is $5$ , and it does not divide $9$ .
Eli's form is therefore complete, with greatest common factor $8$ outside.

Check the values directly: $40 + 72 = 112$ , $4 \times 28 = 112$ , and
$8 \times 14 = 112.$

Answer

Both equal $40 + 72$ ; Eli's expression has the greatest common factor, $8$ , outside.

Key idea

A factored sum is complete when its inside terms share no factor greater than one.
Problem 10: Mark complete

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