The Distributive Property for 7th Grade - Examples, Exercises and Solutions

$140-70=$

In order to simplify the resolution process, we begin by using the distributive property for 140:

$100+40-70=$

We then rearrange the exercise using the substitution property into a more practical form:

$100-70+40=$

Lastly we solve the exercise from left to right:

$100-70=30$

$30+40=70$

70

$94+72=$

In order to simplify the calculation , we first break down 94 and 72 into smaller and preferably round numbers.

We obtain the following exercise:

$90+4+70+2=$

Using the associative property, we then rearrange the exercise to be more functional.

$90+70+4+2=$

We solve the exercise in the following way, first the round numbers and then the small numbers.

$90+70=160$

$4+2=6$

Which results in the following exercise:

$160+6=166$

166

Solve the exercise:

84:4=

There are several ways to solve the following exercise,

We will present two of them.

In both ways, we begin by decomposing the number 84 into smaller units; 80 and 4.

$\frac{4}{4}=1$

Subsequently we are left with only the 80.

Continuing on with the first method, we will then further decompose 80 into smaller units; $10\times8$

We know that:$\frac{8}{4}=2$

And therefore, we are able to reduce the exercise as follows: $\frac{10}{4}\times8$

Eventually we are left with$2\times10$

which is equal to 20

In the second method, we decompose 80 into the following smaller units:$40+40$

We know that: $\frac{40}{4}=10$

And therefore: $\frac{40+40}{4}=\frac{80}{4}=20=10+10$

which is also equal to 20

Now, let's remember the 1 from the first step and add it in to our above answer:

$20+1=21$

Thus we are left with the following solution:$\frac{84}{4}=21$

21

$133+30=$

In order to solve the question, we first use the distributive property for 133:

$(100+33)+30=$

We then use the distributive property for 33:

$100+30+3+30=$

We rearrange the exercise into a more practical form:

$100+30+30+3=$

We solve the middle exercise:

$30+30=60$

Which results in the final exercise as seen below:

$100+60+3=163$

163

$63-36=$

To solve the problem, first we will use the distributive property on the two numbers:

(60+3)-(30+6)

Now, we will use the substitution property to arrange the exercise in the way that is most convenient for us to solve:

60-30+3-6

It is important to pay attention that when we open the second parentheses, the minus sign moved to the two numbers inside.

30-3 = 

27

27

$30\times39=$

To solving easier, we break down 39 into more convenient numbers, preferably round ones.

We obtain:

$30\times(40-1)=$

We multiply 30 by each of the terms in parentheses:

$(30\times40)-(30\times1)=$

We solve the exercises in parentheses and obtain:

$1,200-30=1,170$

1170

$480\times3=$

In order to simplify the resolution process, we begin by breaking down the number 480 into a smaller addition exercise:

$(400+80)\times3=$

We then multiply each of the terms within the parentheses by 3:

$(400\times3)+(80\times3)=$

Lastly we solve the exercises inside the parentheses and obtain the following:

$1200+240=1440$

1440

$35\times4=$

In order to simplify the resolution process, we divide the number 35 into a smaller addition exercise.

It is easier to choose round whole numbers, hence the following calculation:

$(30+5)\times4=$

We then multiply each of the terms inside of the parentheses by 4:

$(4\times30)+(4\times5)=$Lastly we solve the exercises inside of the parentheses:

$120+20=140$

140

$$$3\times56=$

In order to facilitate the resolution process, we break down 56 into an exercise with smaller, preferably round, numbers.

$3\times(50+6)=$

We use the distributive property and multiply each of the terms in parentheses by 3:

$(3\times50)+(3\times6)=$

We then solve each of the exercises inside of the parentheses and obtain the following result:

$150+18=168$

168

$6\times29=$

To make solving easier, we break down 29 into more comfortable numbers, preferably round ones.

We obtain:

$6\times(30-1)=$

We multiply 6 by each of the terms in parentheses:

$(6\times30)-(6\times1)=$

We solve the exercises in parentheses and obtain:

$180-6=174$

174

Solve the exercise:

=74:4

In order to simplify the resolution process we begin by breaking down the number 74 into a subtraction exercise:

We choose numbers divisible by:

$(80-6):4=$

Next we divide each of the terms inside of the parentheses by 4:

$80:4=20$

$6:4=1.5$

Lastly we subtract the result that we obtained:

$20-1.5=18.5$

18.5

Solve the following exercise

=90:5

We use the distributive property of division to separate the number 90 between the sum of 50 and 40, which facilitates the division and gives us the possibility to solve the exercise without a calculator.

Keep in mind: it is beneficial to choose to split the number according to your knowledge of multiples. In this case into multiples of 5, because it is necessary to divide by 5.

Reminder: the distributive property of division actually allows us to separate the larger term in a division exercise into the sum or the difference of smaller numbers, which facilitates the division operation and gives us the possibility to solve the exercise without a using calculator.

We use the formula of the distributive property

 (a+b):c=a:c+b:c 

$90:5=(50+40):5$

$(50+40):5=50:5+40:5$

$50:5+40:5=10+8$

$10+8=18$

Therefore, the answer is option c: 18

18

$4\times53=$

To make solving easier, we break down 53 into more comfortable numbers, preferably round ones.

We obtain:

$4\times(50+3)=$

We multiply 2 by each of the terms inside the parentheses:

$(4\times50)+(4\times3)=$

We solve the exercises inside the parentheses and obtain:

$200+12=212$

212

$3\times93=$

In order to simplify our calculation, we first break down 93 into smaller, more manageable parts. (Preferably round numbers )

We obtain the following:

$3\times(90+3)=$

We then use the distributive property in order to find the solution.

We multiply each of the terms in parentheses by 3:

$(3\times90)+(3\times3)=$

Lastly we solve each of the terms in parentheses and obtain:

$270+9=279$

279

$9\times33=$

In order to facilitate the resolution process, we first break down 33 into a smaller addition exercise with more manageable and preferably round numbers:

$9\times(30+3)=$

Using the distributive property we then multiply each of the terms in parentheses by 9:

$(9\times30)+(9\times3)=$

Finally we solve each of the exercises inside of the parentheses:

$270+27=297$

297