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Techniques of Estimation: Mental Arithmetic- Using the Distibutive Property

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses using the distributive property. By the end of the module students should understand the distributive property and be able to obtain the exact result of a multiplication using the distributive property.

Section Overview

  • The Distributive Property
  • Estimation Using the Distributive Property

The Distributive Property

Distributive Property

The distributive property is a characteristic of numbers that involves both addition and multiplication. It is used often in algebra, and we can use it now to obtain exact results for a multiplication.

Suppose we wish to compute 3(2 + 5)3(2 + 5) size 12{3 \( "2 "+" 5" \) } {}. We can proceed in either of two ways, one way which is known to us already (the order of operations), and a new way (the distributive property).

  1. Compute 3(2 + 5)3(2 + 5) size 12{3 \( "2 "+" 5" \) } {} using the order of operations.

    3 ( 2 + 5 ) 3 ( 2 + 5 ) size 12{3 \( "2 "+" 5" \) } {}

    Operate inside the parentheses first: 2 + 5 = 72 + 5 = 7 size 12{"2 "+" 5 "=" 7"} {}.

    3 ( 2 + 5 ) = 3 ⋅ 7 3 ( 2 + 5 ) = 3 ⋅ 7 size 12{3 \( "2 "+" 5" \) =" 3 " cdot " 7"} {}

    Now multiply 3 and 7.

    3 ( 2 + 5 ) = 3 ⋅ 7 = 21 3 ( 2 + 5 ) = 3 ⋅ 7 = 21 size 12{3 \( "2 "+" 5" \) =" 3 " cdot " 7 "=" 21"} {}

    Thus, 3(2 + 5)= 213(2 + 5)= 21 size 12{3 \( "2 "+" 5" \) =" 21"} {}.

  2. Compute 3(2 + 5)3(2 + 5) size 12{3 \( "2 "+" 5" \) } {} using the distributive property.

    We know that multiplication describes repeated addition. Thus,

    3(2+5) = 2+5+2+5+2+5 ︸ 2 + 5 appears 3 times = 2+2+2+5+5+5 (by the commutative property of addition) = 3⋅2+3⋅5 (since multiplication describes repeated addition) = 6+15 = 21 3(2+5) = 2+5+2+5+2+5 ︸ 2 + 5 appears 3 times = 2+2+2+5+5+5 (by the commutative property of addition) = 3⋅2+3⋅5 (since multiplication describes repeated addition) = 6+15 = 21

    Thus, 3(2 + 5)= 213(2 + 5)= 21 size 12{3 \( "2 "+" 5" \) =" 21"} {}.

    Let's look again at this use of the distributive property.

    3(2+5) = 2+5+2+5+2+5 ︸ 2 + 5 appears 3 times 3(2+5) = 2+5+2+5+2+5 ︸ 2 + 5 appears 3 times

    3(2+5) = 2+2+2 ︸ 2 appears 3 times + 5+5+5 ︸ 5 appears 3 times 3(2+5) = 2+2+2 ︸ 2 appears 3 times + 5+5+5 ︸ 5 appears 3 times

    3 times the quantity two plus five. Arrows point from the three to both the two and the five. This is equal to three times two plus three times five.

    The 3 has been distributed to the 2 and 5.

    This is the distributive property. We distribute the factor to each addend in the parentheses. The distributive property works for both sums and differences.

Sample Set A

Example 1

4 times the quantity 6 plus 2. Arrows point from the 4 to both the 6 and the 2. This is equal to 4 times 6 plus 4 times 2. This is equal to 24 plus 8, which is equal to 32.

Using the order of operations, we get

4(6+2) = 4⋅8 = 32 4(6+2) = 4⋅8 = 32

Example 2

8 times the quantity 9 plus 6. Arrows point from the 8 to both the 9 and the 6. This is equal to 8 times 9 plus 8 times 6. This is equal to 72 plus 48, which is equal to 120.

Using the order of operations, we get

8(9+6) = 8⋅15 = 120 8(9+6) = 8⋅15 = 120

Example 3

4 times the quantity 9 minus 5. Arrows point from the 4 to both the 9 and the 5. This is equal to 4 times 9 minus 4 times 5. This is equal to 36 minus 20, which is equal to 16.

Example 4

25 times the quantity 20 minus 3. Arrows point from the 20 to both the 20 and the 3. This is equal to 25 times 20 minus 25 times 3. This is equal to 500 minus 76, which is equal to 425.

Practice Set A

Use the distributive property to compute each value.

Exercise 1

6(8 + 4)6(8 + 4) size 12{6 \( "8 "+" 4" \) } {}

Solution

6 ⋅ 8 + 6 ⋅ 4 = 48 + 24 = 72 6⋅8+6⋅4=48+24=72

Exercise 2

4(4 + 7)4(4 + 7) size 12{4 \( "4 "+" 7" \) } {}

Solution

4 ⋅ 4 + 4 ⋅ 7 = 16 + 28 = 44 4⋅4+4⋅7=16+28=44

Exercise 3

8(2 + 9)8(2 + 9) size 12{8 \( "2 "+" 9" \) } {}

Solution

8 ⋅ 2 + 8 ⋅ 9 = 16 + 72 = 88 8⋅2+8⋅9=16+72=88

Exercise 4

12(10 + 3)12(10 + 3) size 12{"12" \( "10 "+" 3" \) } {}

Solution

12 ⋅ 10 + 12 ⋅ 3 = 120 + 36 = 156 12⋅10+12⋅3=120+36=156

Exercise 5

6(11 - 3)6(11 - 3) size 12{6 \( "11 - 3" \) } {}

Solution

6 ⋅ 11 - 6 ⋅ 3 = 66 - 18 = 48 6⋅11-6⋅3=66-18=48

Exercise 6

8(9 - 7)8(9 - 7) size 12{8 \( "9 - 7" \) } {}

Solution

8 ⋅ 9 – 8 ⋅ 7 = 72 – 56 = 16 8⋅9–8⋅7=72–56=16

Exercise 7

15(30 - 8)15(30 - 8) size 12{"15" \( "30 - 8" \) } {}

Solution

15 ⋅ 30 - 15 ⋅ 8 = 450 - 120 = 330 15⋅30-15⋅8=450-120=330

Estimation Using the Distributive Property

We can use the distributive property to obtain exact results for products such as 25 ⋅ 2325 ⋅ 23 size 12{"25 " cdot " 23"} {}. The distributive property works best for products when one of the factors ends in 0 or 5. We shall restrict our attention to only such products.

Sample Set B

Use the distributive property to compute each value.

Example 5

25 ⋅ 2325 ⋅ 23 size 12{"25 " cdot " 23"} {}

Notice that 23 = 20 + 323 = 20 + 3 size 12{"23 "=" 20 "+" 3"} {}. We now write

25 times 23 equals 25 times the quantity 20 plus 3. This is equal to 25 times 20 plus 25 times 3. This is equal to 500 + 75. This is equal to 575.

Thus, 25 ⋅ 23 = 57525 ⋅ 23 = 575 size 12{"25 " cdot " 23 "=" 575"} {}

We could have proceeded by writing 23 as 30 - 730 - 7 size 12{"30 - 7"} {}.

25 times 23 equals 25 times the quantity 30 minus 7. This is equal to 25 times 30 minus 25 times 7. This is equal to 750 minus 175. This is equal to 575.

Example 6

15 ⋅ 3715 ⋅ 37 size 12{"15 " cdot " 37"} {}

Notice that 37 = 30 + 737 = 30 + 7 size 12{"37 "=" 30 "+" 7"} {}. We now write

15 times 37 equals 15 times the quantity 30 plus 7. This is equal to 15 times 30 plus 15 times 7. This is equal to 450 plus 105, which is equal to 555.

Thus, 15 ⋅ 37 = 55515 ⋅ 37 = 555 size 12{"15 " cdot " 37 "=" 555"} {}

We could have proceeded by writing 37 as 40 - 340 - 3 size 12{"40 - 3"} {}.

15 times 37 equals 15 times the quantity 40 minus 3. This is equal to 15 times 40 plus 15 times 3. This is equal to 600 minus 45, which is equal to 555.

Example 7

15 ⋅ 8615 ⋅ 86 size 12{"15 " cdot " 86"} {}

Notice that 86 = 80 + 686 = 80 + 6 size 12{"86 "=" 80 "+" 6"} {}. We now write

15 times 86 equals 15 times the quantity 80 plus 6. This is equal to 15 times 80 plus 15 times 6. This is equal to 1,200 plus 90, which is equal to 1,290.

We could have proceeded by writing 86 as 90 - 490 - 4 size 12{"90 - 4"} {}.

15 times 86 equals 15 times the quantity 90 minus 4. This is equal to 15 times 90 minus 15 times 4. This is equal to 1,350 minus 60, which is equal to 1,290.

Practice Set B

Use the distributive property to compute each value.

Exercise 8

25 ⋅ 1225 ⋅ 12 size 12{"25 " cdot " 12"} {}

Solution

2510+2=25⋅10+25⋅2=250+50=3002510+2=25⋅10+25⋅2=250+50=300 size 12{"25" left ("10"+2 right )="25" cdot "10"+"25" cdot 2="250"+"50"="300"} {}

Exercise 9

35 ⋅ 1435 ⋅ 14 size 12{"35 " cdot " 14"} {}

Solution

3510+4=35⋅10+35⋅4=350+140=4903510+4=35⋅10+35⋅4=350+140=490 size 12{"35" left ("10"+4 right )="35" cdot "10"+"35" cdot 4="350"+"140"="490"} {}

Exercise 10

80 ⋅ 5880 ⋅ 58 size 12{"80 " cdot " 58"} {}

Solution

8050+8=80⋅50+80⋅8=4,000+640=4,6408050+8=80⋅50+80⋅8=4,000+640=4,640 size 12{"80" left ("50"+8 right )="80" cdot "50"+"80" cdot 8=4,"000"+"640"=4,"640"} {}

Exercise 11

65 ⋅ 6265 ⋅ 62 size 12{"65 " cdot " 62"} {}

Solution

6560+2=65⋅60+65⋅2=3,900+130=4,0306560+2=65⋅60+65⋅2=3,900+130=4,030 size 12{"65" left ("60"+2 right )="65" cdot "60"+"65" cdot 2=3,"900"+"130"=4,"030"} {}

Exercises

Use the distributive property to compute each product.

Exercise 12

15⋅1315⋅13 size 12{"15" cdot "13"} {}

Solution

15(10+3)=150+45=19515(10+3)=150+45=195 size 12{"15" \( "10"+3 \) ="150"+"45"="195"} {}

Exercise 13

15⋅1415⋅14 size 12{"15" cdot "14"} {}

Exercise 14

25⋅1125⋅11 size 12{"25" cdot "11"} {}

Solution

25(10+1)=250+25=27525(10+1)=250+25=275 size 12{"25" \( "10"+1 \) ="250"+"25"="275"} {}

Exercise 15

25⋅1625⋅16 size 12{"25" cdot "16"} {}

Exercise 16

15⋅1615⋅16 size 12{"15" cdot "16"} {}

Solution

1520−4=300−60=2401520−4=300−60=240 size 12{"15" left ("20" - 4 right )="300" - "60"="240"} {}

Exercise 17

35⋅1235⋅12 size 12{"35" cdot "12"} {}

Exercise 18

45⋅8345⋅83 size 12{"45" cdot "83"} {}

Solution

4580+3=3600+135=37354580+3=3600+135=3735 size 12{"45" left ("80"+3 right )="3600"+"135"="3735"} {}

Exercise 19

45⋅3845⋅38 size 12{"45" cdot "38"} {}

Exercise 20

25⋅3825⋅38 size 12{"25" cdot "38"} {}

Solution

2540−2=1,000−50=9502540−2=1,000−50=950 size 12{"25" left ("40" - 2 right )=1,"000" - "50"="950"} {}

Exercise 21

25⋅9625⋅96 size 12{"25" cdot "96"} {}

Exercise 22

75⋅1475⋅14 size 12{"75" cdot "14"} {}

Solution

7510+4=750+300=1,0507510+4=750+300=1,050 size 12{"75" left ("10"+4 right )="750"+"300"=1,"050"} {}

Exercise 23

85⋅3485⋅34 size 12{"85" cdot "34"} {}

Exercise 24

65⋅2665⋅26 size 12{"65" cdot "26"} {}

Solution

6520+6=1,300+390=1,6906520+6=1,300+390=1,690 size 12{"65" left ("20"+6 right )=1,"300"+"390"=1,"690"} {} or   6530−4=1,950−260=1,6906530−4=1,950−260=1,690 size 12{"65" left ("30" - 4 right )=1,"950" - "260"=1,"690"} {}

Exercise 25

55⋅5155⋅51 size 12{"55" cdot "51"} {}

Exercise 26

15⋅10715⋅107 size 12{"15" cdot "107"} {}

Solution

15100+7=1,500+105=1,60515100+7=1,500+105=1,605 size 12{"15" left ("100"+7 right )=1,"500"+"105"=1,"605"} {}

Exercise 27

25⋅20825⋅208 size 12{"25" cdot "208"} {}

Exercise 28

35⋅40235⋅402 size 12{"35" cdot "402"} {}

Solution

35400+2=14,000+70=14,07035400+2=14,000+70=14,070 size 12{"35" left ("400"+2 right )="14","000"+"70"="14","070"} {}

Exercise 29

85⋅11085⋅110 size 12{"85" cdot "110"} {}

Exercise 30

95⋅1295⋅12 size 12{"95" cdot "12"} {}

Solution

9510+2=950+190=1,1409510+2=950+190=1,140 size 12{"95" left ("10"+2 right )="950"+"190"=1,"140"} {}

Exercise 31

65⋅4065⋅40 size 12{"65" cdot "40"} {}

Exercise 32

80⋅3280⋅32 size 12{"80" cdot "32"} {}

Solution

8030+2=2,400+160=2,5608030+2=2,400+160=2,560 size 12{"80" left ("30"+2 right )=2,"400"+"160"=2,"560"} {}

Exercise 33

30⋅4730⋅47 size 12{"30" cdot "47"} {}

Exercise 34

50⋅6350⋅63 size 12{"50" cdot "63"} {}

Solution

5060+3=3,000+150=3,1505060+3=3,000+150=3,150 size 12{"50" left ("60"+3 right )=3,"000"+"150"=3,"150"} {}

Exercise 35

90⋅7890⋅78 size 12{"90" cdot "78"} {}

Exercise 36

40⋅8940⋅89 size 12{"40" cdot "89"} {}

Solution

4090−1=3,600−40=3,5604090−1=3,600−40=3,560 size 12{"40" left ("90" - 1 right )=3,"600" - "40"=3,"560"} {}

Exercises for Review

Exercise 37

((Reference)) Find the greatest common factor of 360 and 3,780.

Exercise 38

((Reference)) Reduce 5945,1485945,148 size 12{ { {"594"} over {5,"148"} } } {} to lowest terms.

Solution

326326 size 12{ { {3} over {"26"} } } {}

Exercise 39

((Reference)) 159159 size 12{1 { {5} over {9} } } {} of 247247 size 12{2 { {4} over {7} } } {} is what number?

Exercise 40

((Reference)) Solve the proportion: 715=x90715=x90 size 12{ { {7} over {"15"} } = { {x} over {"90"} } } {}.

Solution

x=42x=42 size 12{x="42"} {}

Exercise 41

((Reference)) Use the clustering method to estimate the sum: 88 + 106 + 91 + 11488 + 106 + 91 + 114 size 12{"88 "+" 106 "+" 91 "+" 114"} {}.

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