Math News! - Colchester Public Schools · PDF file · 2015-10-305,000 x 60 ≠ 30,000 5,000 x 60 (3 zeros) (1 zero) (4 zeros ... Numerical Expression ... The Grand Theatre purchased

OBJECTIVES OF TOPIC A

Multiply multi-digit whole numbers and multiples of 10 using place value patterns and the distributive and associative properties.

Estimate multi-digit products by rounding factors to a basic fact and using place value patterns.

5th Grade Math Module 2: Multi-Digit Whole Number and Decimal Fraction Operations

Math Parent Letter

This document is created to give parents and students a better

understanding of the math concepts found in Eureka Math (©

2013 Common Core, Inc.) that is also posted as the Engage New

York material which is taught in the classroom. Grade 5 Module 2

of Eureka Math ( Engage New York) covers Multi-Digit Whole

Number and Decimal Fraction Operations. This newsletter will

discuss Module 2, Topic A.

Topic A. Mental Strategies for Multi-Digit Whole Number

Multiplication

Words to know

Product Estimate

Associative Property Factor

Commutative Property Equation

Distributive Property

Things to Remember:

Commutative Property – The word "commutative" comes

from "commute" or "move around", so the Commutative

Property is the one that refers to moving stuff around.

Example: 2 x 3 = 3 x 2

Associative Property - The word "associative" comes from

"associate" or "group"; the Associative Property is the rule

that refers to grouping. Example: 5 x 7 x 2 = (5 x 2) x 7

Distributive Property - The Distributive Property is easy to

remember, if you recall that "multiplication distributes over

addition". Example: 43 x 6 = (40 x 6) + (3 x 6)

Symbol for meaning ‘about’ - ≈

When multiplying whole numbers by multiples of 10 you

cannot always count zeros in the factors and end up with

the correct product.

5,000 x 60 ≠ 30,000 5,000 x 60 (3 zeros) (1 zero) (4 zeros) = 5 x 1,000 x 6 x 10

= (5 x 6) x (1,000 x 10) = 30 x 10,000 = 300,000

Focus Area– Topic A Module 2: Multi-Digit Whole Number and Decimal Fraction Operations

Find the product. Show your thinking

6 x 70 80 x 50

= 6 x 7 x 10 = (8 x 10) x (5 x 10)

= 42 x 10 = (8 x 5) x (10 x 10)

= 420 = 40 x 100

= 4,000

542 x 3

= (500 x 3) + (40 x 3) + (2 x 3)

= 1,500 + 120 + 6

= 1,626

Round the factors to estimate the products.

867 x 46 ≈ 900 x 50 7,231 x 25 ≈ 7,000 x 30

= 45,000 = 210,000

Determine if these equations are true or false. Defend your

answer using your knowledge of place value and the

commutative, associative and/or distributive property.

850 x 6 x 10 = 85 x 6 x 100 -- these equations are TRUE

(85 x 10) x 6 x 10 = 85 x 6 x (10 x 10)

85 x 6 x 10 x 10 = 85 x 6 x 10 x 10

77 x 30 x 10 = 770 x 3 -- these equations are FALSE

(77 x 10) x 30 = 770 x 3

770 x 30 ≠ 770 x 3

Associative

Property

Distributive

Property

Grade 5, Module 2, Topic A

Math News!

Example Problems and Answers

Laura wants to buy a new car. If the car payment each month is $367 for 5 years, about how much will the car cost her after the five

year?

$367 is about $400 --- there are 12 months in a year

$400 x 12

= (4 x 100) x 12

= (4 x 12) x 100

= 48 x 100

= 4,800

For 5 years --- $4,800 x 5

= (48 x 100) x 5

= 48 x 5 x 100

= (40 x 5) + (8 x 5) x 100

= (200 + 40) x 100

= 240 x 100

= 24,000

The car will cost her about $24,000.

Tickets to a baseball game are $23 for an adult and $12 for a student. If 37 adult tickets and 325 student tickets were bought, about

how much money would it cost for everyone to attend the baseball game?

$23 x 37 adults ≈ $20 x 40 = $800

$12 x 325 children ≈ $10 x 300 = $3,000 OR $12 x 300

$800 + $3,000 = $3,800 = 12 x (3 x 100)

= (12 x 3) x 100

= 36 x 100

= 3,600

$800 + $3,600 = $4,200

It will cost about $3,800 for everyone to attend the game. Or It will cost about $4,200 for everyone to attend the game.

This information was generously shared by LPSS, Lafayette, LA

OBJECTIVES OF TOPIC B

Connect visual models and the distributive property topartial products of the standard algorithm withoutrenaming.

Fluently multiply multi-digit whole numbers using thestandard algorithm to solve multi-step word problems.

Connect area diagrams and the distributive property topartial products of the standard algorithm with andwithout renaming.

Fluently multiply multi-digit whole numbers using thestandard algorithm to solve multi-step word problemsand using estimation to check for reasonableness of theproduct.


Math Parent Letter



2013 Common Core, Inc.) that is also posted as the Engage

New York material which is taught in the classroom. Grade

5 Module 2 of Eureka Math ( Engage New York) covers Multi-

Digit Whole Number and Decimal Fraction Operations. This

newsletter will discuss Module 2, Topic B.

Topic B. The Standard Algorithm for Multi-Digit Whole Number Multiplication

Words to Know

Area Model Product

Standard Algorithm Factor

Numerical Expression

Estimate

Decompose

Things to Remember!!!

Standard Algorithm

Step-by-step procedure to solve a problem

Numerical ExpressionA mathematical phrase involving only numbers and one or

more operational symbol Example: 11 x (6+13)

Symbol for ‘about’ ≈

ProductThe answer when two or more numbers are multiplied together.

7 x 3 = 21 Factor Factor Product

1

4 3 2 x 2 4 1

1 7 2 8 +8 6 4 0 1 0,3 6 8

1600 120 8

8000 600 40

=1,728

=8,640

Focus Area– Topic B Module 2: Multi-Digit Whole Number and Decimal Fraction Operations

Problem 1: 432 x 24

Draw using area model and then solve using the standard algorithm. Use arrows to match the partial products from the area model to the partial products of the algorithm.

To find the answer to this problem, first we represent units of 432. Decompose 432 to make finding the partial product easier.

400 + 30 + 2

How many four hundred thirty-twos are we counting? (24) Decompose 24 ( 20 + 4)

Multiply: What is the product of 4 and 2? 8 What is the product of 4 and 30? 120 Continue recording the product in the area model. Now add each row of partial products. Solve using the standard algorithm. Compare the partial products in the area model to the partial products in the algorithm.

400 + 30 + 2

4

+

20

What are 24 groups of 432? 10,368

Problem 2: 532 x 283

Estimate the product. Solve using standard algorithm. Use your estimate to check the reasonableness of the product. To estimate the product round each factor.

532 closer to 5 hundreds than 6 hundreds on the number line

283 closer to 3 hundreds than 2 hundreds on the number line

Multiply the rounded factors to estimate the product.

532 x 283

≈ 500 x 300

= 150,000

Multiply the rounded factors to estimate the product.

5 3 2 x 2 8 3

1 1

1 1 5 9 6 4 2 5 6 0 +1 0 6 4 0 0

1 5 0,5 5 6

Grade 5, Module 2, Topic B

Math News!

Problems and Answers The Grand Theatre purchased 257 new theatre seats for their auditorium at $129 each. What’s the total cost of the new theatre seats?

200 + 50 + 7

9 +

20

5, + 100

The total cost of the theatre seats is $33,153.

To find the answer to this problem, first we draw an area model. We represent the number of seats in the area model by decomposing 257 to make finding the partial product easier. Next, decompose 129 which is the cost of each seat. Record the products.

Peter has collected 15 boxes of football cards. Each box has 312 cards in it. Peter estimates he has about 6,000 cards, so he buys 10 albums that hold 600 cards each.

A. Did Peter purchase too many, not enough, or just the right amount of albums to hold his football cards? Explain your answer?

Step 1: To solve this problem, first estimate the number of cards in each box. 312 closer to 300 than 400

Multiply the number of boxes times estimated number of cards in each box. 312 x 15 ≈ 300 x 15

= (3 x 100) x 15 = (3 x 15) x 100 = 45 x 100 = 4500 Peter has about 4,500 cards.

Step 2: Find the total number of cards the 10 albums hold altogether.

600 x 10 = 6,000 The 10 albums can hold 6,000 cards.

Step 3: Peter purchased too many albums to hold his football cards. He has about 4,500 cards and ten albums would hold 6,000 cards. (Explanation could be justified by statement written in the note above.)

B. How many cards does Peter have? Use the standard algorithm to solve the problem.

3 1 2 x 1 5 1 5 6 0

+ 3 1 2 0 4,6 8 0 Peter has a total of 4,680 cards.

C. How many albums will he need for all his cards?

Peter will need 8 albums for all his cards.

1 album

600 cards

2 albums

1,200 cards

3 albums

1,800 cards

4 albums

2,400 cards 5 albums

3,000 cards

6 albums

3,600 cards 7 albums

4,200 cards

8 albums

4,800 cards

1

1,800 450 63

4,000 1,000 140

20,000 5,000 700

2 5 7 x 1 2 9

1

2 3 1 3 1 5 1 4 0

+2 5 7 0 0 3 3,1 5 3

= 2,313

= 5,140

= 25,700

Note: You may round 15 to 20 and then multiply 300 x 20 which equals 6,000. Therefore you could say that Peter has about 6,000 cards. Since both factors were rounded up, the actual number of cards is less than 6,000.


OBJECTIVES OF TOPIC C

Multiply decimal fractions with tenths by multi-digit whole numbers using place value understanding to record partial products.

Multiply decimal fractions by multi-digit whole numbers through conversion to a whole number problem and reasoning about the placement of the decimal.

Reason about the product of a whole number and a decimal with hundredths using place value understanding and estimation.


Math Parent Letter



2013 Common Core, Inc.) that is also posted as the Engage New

York material which is taught in the classroom. Grade 5 Module 2

of Eureka Math ( Engage New York) covers Multi-Digit Whole

Number and Decimal Fraction Operations. This newsletter will

address decimal multi-digit multiplication.

Topic C. Decimal Multi-Digit Multiplication

Words to know

Product Factor

Estimate

Decimal Fraction

Standard Algorithm

Things to Remember:

A decimal fraction uses a point to separate the whole number part fromthe fractional part of a number. Example: in the number 36.9 the pointseparates the 36 (the whole number part) from the 9 (the fractional part,which really means 9 tenths). So 36.9 is 36 and nine tenths.

When multiplying by a decimal fraction, you convert the decimal fraction

to a whole number by multiplying it by a power of 10 (10 or 100)

depending on the number of places after the decimal point. The

problem now resembles a whole number multiplication problem. Once

you finish multiplying, you then have to divide the answer by the same

power of 10 you multiplied by.

If the decimal fraction has one place after the decimal, you multiply by

10. The digits will then shift one place to the left. The result is a

number that is 10 times greater than the original number. If the decimal

has two places after the decimal, you multiply by 100. The digits will shift

two places to the left. The result is a number that is 100 times greater

than the original number.

When a number is divided by 10, the digits shift one place to the right.

The result is a number that is as large as the original number. When a

number is divided by 100, the digits shift two places to the right. The

result is a number that is as large as the original number.

Focus Area– Topic C Module 2: Multi-Digit Whole Number and Decimal Fraction Operations

Problem 1

Solve using standard algorithm.

54 x 3.5

3.5 35

x 54 x 54

140

+ 1750

1890

1890 ÷ 10 = 189.0

Problem 2

Round the factors to estimate the products. (Symbol ≈ means about)

Solve

7.5 x 52 17.6 x 22 95 x 3.3 ≈ 8 x 50 ≈ 18 x 20 ≈ 100 x 3

= 400 = 360 = 300

Problem 3 Estimate the product. Solve using an area model and the standard algorithm.

Solve: 4.7 x 24

4.7 x 24

≈ 5 x 20

= 100

x 10

x 10 ÷ 10 54 x 3.5 1890 189.0

Estimation

When we compare our answer (112.8) to our estimate (100), we can conclude that our answer is reasonable.

Standard Algorithm

40

Area Model

+ 7 tenths

4

+

20

160 28

800 140

= 188

= 940

1128 tenths = 112.8

4.7 x 2 4

47 x 24

188 + 940 1128 1128 ÷ 10 = 112.8

x 10

Grade 5, Module 2, Topic C

Math News!

Example Problems and Answers

Pat rides his bike a total of 6.83 miles to and from school every day. How many miles does he ride in 25 days?

6.83 miles x 25 days 6.83 (x 100) 683

x 25 x 25

3415

+ 13660

17075

17075 ÷ 100 = 170.75

Pat rides his bike a total of 170.75 miles in 25 days.

A. Courtney buys 79 protractors at $1.09 each and 32 composition notebooks at $2.19 each. About how

much money did she spend?

$1.09 per protractor x 79 protractors ≈ $1 x 80 = $80 $2.19 per notebook x 32 notebooks ≈ $2 x 30 = $60

$80 + $60 = $140

Courtney spent about $140 on protractors and notebooks.

B. How much money did she actually spend?

79 x $1.09 $1.09 (x 100) 109 32 x $2.19 $2.19 (x 100) 219 x 79 x 79 x 32 x 32

1 981 1 438 + 7630 + 6570

8611 7008

8611 ÷ 100 = $86.11 cost of protractor 7008 ÷ 100 =$70.08 cost of notebooks

$86.11 cost of protractors +$70.08 cost of notebooks

$156.19 total cost of supplies

Courtney actually spent $156.19.

A

A kitchen measures 32 feet by 17 feet. If tile cost $7.98 per square foot, what is the total cost of putting tile in the kitchen?

$7.98 (x 100) 798 32 ft x 544 x 544

3192 31920

+ 399000 17 ft 434112 434112 ÷ 100 = $4,341.12

x 100 ÷ 100

25 x 6.83 17075 170.75

32 ft x 17 ft =

544 sq. ft

Note: Area refers to the number of square units needed to cover the inside of a shape. To determine the area of this rectangle you multiply the length times the width. The formula for area is Area =length x width .

The total cost of putting tile in the kitchen is $4,341.12

width

length


OBJECTIVES OF TOPIC D

Use whole number multiplication to express equivalentmeasurements.

Use decimal multiplication to express equivalentmeasurements.

Solve two-step word problems involving measurementand multi-digit multiplication.


Math Parent Letter




New York material which is taught in the classroom. Grade 5

Module 2 of Eureka Math ( Engage New York) covers Multi-


newsletter will discuss Module 2, Topic D.

Topic D. Measurement Word Problems with Whole Number and Decimal Multiplication

Words to know

millimeter (mm) milligram (mg)

centimeter (cm) gram (g)

kilometer (km) kilogram (kg)

inch (in) ounce (oz)

foot/feet (ft) pound (lb)

yard (yd) ton

mile (mi) fluid ounce (fl oz)

cup (c) liter (L)

pint (pt) milliliter (ml)

quart (qt) kiloliter (kl)

gallon (gal) unit


When converting bigger unit to smaller unit, you multiply bythe bigger unit by whole number of smaller units.

Focus Area– Topic D Measurement Conversions through multiplication

Knowing the unit conversions

1 foot = 12 inches 1 yard = 3 feet = 36 inches

1 mile = 5,280 feet 1 mile = 1,760 yards

1 centimeter = 10 millimeter

1 meter = 100 centimeters = 1,000 millimeters

1 kilometer = 1,000 meters

1 pound = 16 pound 1 ton = 2,000 pounds

1 gram= 1,000 milligrams 1 kilogram = 1,000 grams

1 cup= 8 fluid ounces 1 pint = 2 cups

1 quart = 2 pints 1 gallon = 4 quarts

1 liter = 1,000 milliliters 1 kiloliter = 1,000 liters

Convert.

a. 15 yd =__________ ft

yards to feet: big unit to small unit - multiply

3 ft = 1 yd 15 yd x 3 ft per yd = 45 ft

b. ________g = 18 kg

kilograms to gram: big unit to small unit - multiply

1,000 g =1 kg 18 kg x 1,000 g per kg =18,000 g

c. 16 gal=_______qt =________pt

gallons to quarts to pints: big unit to small unit to

smaller unit – multiply twice

4 qt = 1 gal 1 qt =2 pt

16 gal x 4 qt per gal = 64 qt

64 qt x 2 pt per qt =128 pt

d. _________fl oz = 6.32 c

cups to fluid ounces: big unit to small unit -multiply

8 fl oz =1 cup

6.32 c x 8 fl oz per c

= 632 hundredths c x 8 fl oz per c

= 5056 hundredths fl oz

= 50.56 fl oz

e. 9.54 g = ___________mg

grams to milligrams: big unit to small unit - multiply

1,000 mg = 1 g

9.54 g x 1000 mg per g

= 954 hundredths g x 1000 mg per g

= 954,000 hundredths mg

= 9540.00 or 9540 mg

Grade 5, Module 2, Topic D

Math News!

John’s dog had 5 puppies! When John and his sister Peggy weigh all the puppies together, they weigh 4 pounds 1 ounce. Since all the puppies

are about the same size, how many ounces does each puppy weigh?

Answer: First, we need to put all of the puppies’ weight in the same units. We are looking for a final answer of ounces. So, we are converting from pounds to

ounces: big unit to small unit - multiply. 16 ounces = 1pound

4 pounds x 16 ounces per pound = 64 ounces 64 ounces + 1 ounce = 65ounces

65 ounces = 5 puppies weight in ounces

65ounces

? oz (weight of puppy)





65ounces ÷ 5 puppies = 13 ounces

Each puppy weighs 13 ounces.

Susan is training to be in the Mrs. Fitness contest. She ran 3.75 km, swam 0.76 km, and biked for 23.2 km. Susan completed this routine three

times a week. How far did Susan travel in one week while training? Express your answer in meters.

Answer: First, we will convert from km to m: big unit to small unit - multiply. 1,000 m = 1km

3.75km x 1000 m per km =3,750 m 0.76 km x 1000 m per km =760 m 23.2km x 1000m per km = 23,200m

(Susan ran) (Susan swam) (Susan biked)

3,750m

760m

+ 23,200m

27,710m (Susan’s travel for 1time)

Susan traveled a total of 83,130 meters in one week of training. Another Approach: 3.75 km 27,710m

0.76 km x __ 3 (trainings in a week)

23.20 km 83,130m (total distance in one week of training)

27.71 km x 1000 m per km = 27,710 m

Fast Mail charges $5.35 to ship a 2 lb-package. For each ounce over 2 lb, they charge an additional $0.18 per ounce. How much would it cost

to ship a package weighing 3 lb 8 oz?

Answer: First we need to see how many 2 pounds can be taken out of the total weight of the package. 3 lb 8 oz (weight of package)

- 2 lb 0 oz ($5.35 - cost for shipping 2 lb)

Now we need to convert our packages left over weight into the same unit of ounces. 1 lb 8 oz (left over weight)

Convert pounds to ounces: big unit to small unit (multiply) 16 oz = 1 lb

It will cost $9.67 to ship a package weighing 3 lb 8 oz.

27,710 m

x 3 (trainings in a week)

83,130 m

16 oz + 8 oz 24 oz ($0.18 per oz) 0.18 x 100 = 18

24 oz x 18 1 9 2 + 2 4 0 4 3 2 432 ÷ 100 = 4.32

$5.35 (cost for 2 lb) + $4.32 (cost for 24 oz) $9.67

6 5

-5

1 5

1 5

0

5

1 3

-

OBJECTIVES OF TOPIC E

Use divide by 10 patterns for multi-digit whole numberdivision.

Use basic facts to approximate quotients with two-digitdivisors.


Math Parent Letter





Module 2 of Eureka Math (Engage New York) covers Multi-


newsletter will discuss Module 2, Topic E.

Topic E. Mental Strategies for Multi-Digit Whole Number Division

Words to know

multiples dividend (whole)

quotient divide

divisor division

round estimation

approximate ( ) basic facts

Focus Area– Topic E Mental Multi-digit whole number division

Knowing the multiples of a number

2 – 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24,…

3 – 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,…

4 – 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48,…

5 – 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60,…

6 – 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,…

7 – 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,…

8 – 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96,…

9 – 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108,…

10-10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120,…

11-11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132,…

12-12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144,…

Divide. Below number disks are used to show what happens

when 400 is divided by 10.

400 ÷ 10

Answer:

÷ 10 = 40

Divide.

a. 640,000 ÷ 100 b. 420,000 ÷7,000

(shift two places to the right) = 420,000 ÷ 1,000 ÷ 7

= 6,400 =( 420,000 ÷ 1,000) ÷ 7

(shift three places to the right)

= 420 ÷ 7

= 60

c. 27,000 ÷ 90 d. 350,000 ÷ 500

=27,000 ÷10 ÷ 9 = 350,000 ÷ 100 ÷ 5

=(27,000 ÷ 10) ÷ 9 = (350,000 ÷ 100) ÷ 5

(shift one place to the right) (shift two places to the right)

= 2,700 ÷ 9 = 3,500 ÷ 5

= 300 = 700


When estimating quotients, round the divisor only.

Once the divisor is rounded, find a multiple of the first digit ofthe divisor that would create a number that is close to thedividend.Example: 835 ÷ 34 Round 34 to 30. 8 is not a multiple of 3 but 9

≈900 ÷ 30 is, so our dividend becomes 900. = 30

The dividend is referred to as the whole.

When dividing by a power of 10 (10, 100, 1000) the digits in thewhole (dividend), shift to the right. When dividing by 10, thedigits shift 1 place to the right. When dividing by 100, the digitsshift 2 places to the right and when dividing by 1,000, the digitsshift 3 places to the right.

Grade 5, Module 2, Topic E

Math News!

Estimate the quotient for the following problems.

a. 243 ÷ 56 56 rounds to 60

240 ÷ 60___ 24 is a multiple of 6,

= (240 ÷ 10) ÷ 6 so the dividend

= 24 ÷ 6 becomes 240

= 4__

b. 633 ÷ 92 92 rounds to 90

630 ÷ 90_ 63 is a multiple of 9,

= (630 ÷ 10) ÷ 9 so the dividend

= 63 ÷ 9 becomes 630

= 7

c. 483 ÷ 64 64 rounds to 60

480 ÷ 60 48 is a multiple of

=(480 ÷ 10) ÷ 6 6, so the dividend

= 48 ÷ 6 becomes 480

= 8

d. 3,924 ÷ 64 64 rounds to 60

3,600 ÷ 60 39 is not a multiple

= (3,600 ÷ 10) ÷ 6 of 6, but 36 is and it is

= 360 ÷ 6 close to 39, so the

= 60 dividend becomes 3,600

e. 5,567 ÷ 94 94 rounds to 90


= (5,400 ÷ 10) ÷ 9 of 9, but 54 is and it is



f. 2,749 ÷ 47 47 rounds to 50


= (2,500 ÷ 10) ÷ 5 of 5, but 25 is and it is



g. 8,391 ÷ 38 38 rounds to 40

8,000 ÷ 40 8 is a multiple of 4, so

= (8,000 ÷ 10) ÷ 4 the dividend becomes

= 800 ÷ 4 8,000

= 200

h. 6,438 ÷ 73 73 rounds to 70


= (6,300 ÷ 10) ÷7 of 7, but 63 is and it



i. 6,205 ÷ 27 27 rounds to 30

6,000 ÷ 30 6 is a multiple of 3,

= (6,000 ÷ 10) ÷ 3 so the dividend

= 600 ÷ 3 becomes 6,000

= 200

Mrs. Henry spent $513 buying Christmas gifts for her 21 grandchildren. If all of the gifts were the same cost, about how much did she

spend on each gift?

Problem Solving Approach: $513 (amount spent on gifts) ÷ 21 (number of grandchildren) 21 rounds to 20

$600 ÷ 20 5 is not a multiple of 2, but 6 is and it is close to 5,

= (600 ÷ 10) ÷ 2 so the dividend becomes 600

= 60 ÷ 2

= $30

Mrs. Henry spent about $30 on each gift for her 21 grandchildren.

Marcus has saved $3,345 working about 42 different home repair jobs. If he was paid about the same amount per job, about how much did

Marcus make at each different job?

Problem Solving Approach: $3,345 (Marcus’s savings) ÷ 42 (number of Marcus’ jobs) 42 rounds to 40

≈$3,200 ÷ 40 33 is not a multiple of 4, but 32 is and it is close to 33,

= (3,200 ÷ 10) ÷ 4 so the dividend becomes 3,200

= 320 ÷ 4

= $80

Marcus made about $80 at each of his different home repair jobs.


OBJECTIVES OF TOPIC F

Divide two- and three-digit dividends by multiples of 10

with single-digit quotients and make connections to a

written method.

Divide two- and three-digit dividends by two-digit

divisors with single-digit quotients and make connections

to a written method.

Divide three- and four-digit dividends by two-digit

divisors resulting in two- and three-digit quotients,

reasoning about the decomposition of successive

remainders in each place value.

5th Grade Math Module 2: Multi-Digit Whole Number and Decimal Fraction

Operations

Math Parent Letter





Module 2 of Eureka Math (Engage New York) covers Multi-


newsletter will discuss Module 2, Topic F.

Topic F. Partial Quotients and Multi-Digit Whole Number Division

Words to know

dividend (whole) estimate

divisor about (≈)

quotient multiple

remainder decompose

division algorithm


Before dividing, estimate to get an idea of about how many

groups of the divisor can be made.

Example: 84 ÷ 23

≈80 ÷ 20

= 4

(4 is placed in the ones

place of the quotient.)

The remainder has to be smaller than the divisor.

A division problem is not complete unless there is a digit

above the last digit in the dividend.

Focus Area– Topic F Multi-digit Whole Number Division

Knowing division with the standard algorithm

Can we divide 6 hundreds by 80? (No)

Since there are 10 tens in 1 hundred, we can decompose 6hundreds to 60 tens. There are already 4 tens, so there is atotal of 64 tens. Can we make a group of 80 with 64 tens?(No)

Since there are 10 ones in 1 ten, we can decompose 64 tens to640 ones. There are already 4 ones, so there is a total of 644ones. Can we make a group of 80 with 644 ones? (Yes)

So we are dividing 644 ones by 80.

*************************************************************

Application Problem and Answer: A shipment of 288 brochures was sent to the main rest areas in the state of Louisiana. Each of the 30 rest areas in the state received the same amount. After the brochures were distributed, were there any extras? If so how many extra brochures were there?

(Cannot make a group of 30 with 2 hundred, so decompose 2 hundreds to 20 tens and combine with the 8 tens you already have. Now there is a total of 28 tens. Cannot make a group of 30 with 28 tens, so 28 tens is decompose to 288 ones.)

Step 1: Estimate 288 ÷ 30

≈270 ÷ 30 = 9

Step 2: Solve

Step 3: Check 30 x 9 = 270 270 + 18 = 288

644 ÷ 80

Step 1: Estimate quotient to know where to begin. 64 is a multiple of 8

64 ÷8 = 8 so 644 becomes 640.

640÷80 = 8

Step 2: Set up the division algorithm and use the estimation to find the actual quotient.

8 0 x 8 6 4 0

640 + 4= 644

Step 3: Check by multiplying the divisor of 80 by the quotient of 8, and then add the remainder of 4. If the quotient is correct, the answer will equal the

dividend.

6 4 4 - 6 4 0 4

80 8 R 4

Answer: There will be 18 extra brochures.

28 is not a multiple of 3 but 27 is and it is close to 28.

2 8 8 -2 7 0 1 8

30

9 R 18

8 4 - 92

23

4

8 4 - 69 15

23

3 R 15

15

Grade 5, Module 2, Topic F

Math News!

There were 192 students at 4-H camp. The camp has 32 cabins. An equal number of students sleep in each cabin. How many students slept in

each cabin?

Strategy: 192 ÷ 32

Can we make a group of 32 with 1 hundred? (No)

We decompose 1 hundred to 10 tens. There are already 9 tens, so there is a total of 19 tens. Can we make a group of 32 with 19 tens? (No)

We decompose 19 tens to 190 ones. There are already 2 ones, so there is a total of 192 ones. Can we make a group of 32 with 192 ones? (Yes)

Step 1: Estimate Round 32 to 30.

192 ÷ 32

≈180 ÷ 30

= 6

Step 2: Solve

Step 3: Check 32 x 6 = 192

Bart was arranging his 823 baseball cards in a book that holds 24 cards per page. Bart divided 823 by 24 and got a quotient of 34 with a

remainder 7. Explain what the quotient and remainder represent.

Erin made 1,695 chocolate fudge candies for her Christmas gifts. She put them in bags of 36 candies per bag. How many candy bags did Erin

give out? Did she have any candies left over? If so, how many candies were left over?

Strategy: 1,695 ÷ 36

Answer: 823 is the total number of baseball cards, and the divisor, 24, is the amount of cards a page holds. Therefore 34, the quotient, is the amount of pages that will be full in his book. The remainder of 7 will be the amount of cards on the last page.

1 9 2 - 1 9 2

32 6

Answer: 6 students slept in each cabin.

8 2 3 -7 2 1 0 3

- 9 6 7

3 4 R 7 24

36 4

Can we make a group of 36 with 1 thousand? (No)

Decompose 1 thousand to 10 hundreds. There are already 6 hundreds, sothere is a total of 16 hundreds. Can we make a group of 36 with 16hundreds? (No)

Since there are 10 tens in 1 hundred, we decompose 16 hundreds to 160 tens.There are already 9 tens, so there is a total of 169 tens. Can we make a groupof 36 with 169 tens? (Yes)

First division step 169 tens ÷ 36

Estimate: ≈160 tens ÷ 4 tens = 40 (4 is placed in the tens place of the quotient)

• After subtracting, there are 25 tens left. Can we make a group of 36 with 25tens? (No) Since there are 10 ones in 1 ten, we compose 25 tens to 250 ones.There are already 5 ones, so there is a total of 255 ones. Can we make agroup of 36 with 255 ones? (Yes)

Next division step 255 ones ÷ 36

≈ 240 ones ÷ 40 = 6 (6 is placed in the ones place of the quotient)

• After subtracting, there are 39 ones left. Can we make a group of 36 with 39?(Yes) So there are 7 groups of 36 in 255 and not 6.

Answer: Erin gave out 47 bags of candy. She had 3 candies left over.

Check:

4 7 x 3 6

2 8 2 1,692 +1 4 1 0 + 3 1,6 9 2 1,695

1, 6 9 5 - 1 4 4 2 5

1, 6 9 5 - 1 4 4 2 5 5 - 2 1 6

3 9

36 4 6

1, 6 9 5 - 1 4 4 2 5 5 - 2 5 2

3

36 4 7 R 3

19 is not a multiple of 3 but 18 is and it is close to 19.


OBJECTIVES OF TOPIC G

Divide decimal dividends by multiples of 10, reasoning

about the placement of the decimal point and makingconnections to a written method.

Use basic facts to approximate decimal quotients withtwo-digit divisors, reasoning about the placement of thedecimal point.

Divide decimal dividends by two-digit divisors, estimatingquotients, reasoning about the placement of the decimalpoint, and making connections to a written method.

Focus Area– Topic G Multi-Digit Whole Number and Decimal Fraction Operations

Divide. Show division in two steps.

Let’s decompose

60 with 10 as a factor.

10 x 6 = 60

Would the quotient be affected if we divided by 6 first

then by 10?

****************************************************

Divide. Show division in two steps.

****************************************************

Estimate the quotients.

1. 4.23 ÷ 62

≈ 4.2 ÷ 60

= (4.2 ÷ 10) ÷ 6

= 0.42 ÷ 6

= 0.07

2. 53.9 ÷ 91

≈ 54 ÷ 90

= (54 ÷ 9) ÷10

= 6 ÷ 10

= 0.6


Math Parent Letter

This document is created to give parents and students a better understanding of the math concepts found in Eureka Math (© 2013 Common Core, Inc.) that is also posted as the Engage New York material which is taught in the classroom. Grade 5 Module 2 of Eureka Math (Engage New York) covers Multi-Digit Whole Number and Decimal Fraction Operations. This newsletter will discuss Module 2, Topic G.

Topic G. Partial Quotients and Multi-Digit Decimal Division

Words to know

multiple dividend (whole)

factor quotient

divisor

approximate/estimate ( )

round

decompose


The dividend is referred to as the whole.

When dividing by a power of 10 (10, 100, 1000) the digits inthe whole (dividend), shift to the right. When dividing by 10,the digits shift 1 place to the right. When dividing by 100, thedigits shift 2 places to the right and when dividing by 1,000,the digits shift 3 places to the right. This is how it would look ona place value chart.

36 3.6 .36 .036

tens ones . tenths hundredths thousandths

3 6 . 3 . 6

. 3 6

. 0 3 6

÷10 ÷10 ÷10

2.4 ÷ 60 =

2.4 ÷ 10 ÷ 6

= (2.4 ÷ 10) ÷ 6

= 0.24 ÷ 6

= 0.04

Step 1: Divide 2.4 by 10 Step 2: Divide 0.24 by 6

0.36 ÷ 90

= (0.36 ÷ 10) ÷ 9

= 0.036 ÷ 9

= 0.004

2.4 ÷ 6 ÷ 10

= (2.4 ÷ 6) ÷ 10

= 0.4 ÷ 10

= 0.04

The divisor didn’t

change so the

quotient didn’t change.

62 rounds to 60.

4.2 is a divisible by 6,

so, the dividend

becomes 4.2.

91 rounds to 90.

53 is not a multiple of 9,

but 54 is and it close to 53.

so, the dividend becomes

54.

0.36 ÷ 90

= (0.36 ÷ 9) ÷ 10

= 0.04 ÷ 10

= 0.004

84.2 ÷ 200

= (84.2 ÷ 2) ÷ 100

= 42.1 ÷ 100

= 0.421

84.2 ÷ 200

= (84.2 ÷ 100) ÷ 2

= 0.841 ÷ 2

= 0.421

OR

OR

Grade 5, Module 2, Topic G

Math News!

At times you may have to extend the dividend to tenths and hundredths.

The weight of 35 identical toy cars is 844.2 grams. What is the weight of each toy car?

Strategy: 844.2 ÷ 35

Can we make a group of 35 with 8 hundreds? (No)

Since there are 10 tens in 1 hundred, decompose 8 hundreds to 80 tens. There are

already 4 tens, so there is a total of 84 tens. Can we make a group of 35 with 84 tens? (Yes)

First division step 84 tens ÷ 35

Estimate ≈ 80 tens ÷ 40

= 2 tens or 20 (2 is placed in the tens place of the quotient.)

After subtracting, there are 14 tens left. Can we make a group of 35 with 14 tens? (No)

Since there are 10 ones in 1 ten, we decompose 14 tens to 140 ones. There are already

4 ones, so there is a total of 144 ones. Can we make a group of 35 with 144 ones? (Yes)

Next division step 144 ones ÷ 35

≈ 120 ones ÷ 40

= 3 (3 is placed in the ones place.)

After subtracting, there are 4 ones left. Can we make a group of 35 with 4 ones? (No)

Since there are 10 tenths in 1 one, we decompose 4 ones to 40 tenths. There are already

2 tenths, so there is a total of 42 tenths. Can we make a group of 35 with 42 tenths? (Yes)

Next division step 42 tenths ÷ 35

≈ 40 tenths ÷ 40

= 1 tenth (1 is placed in the tenths place.)

After subtracting, there are 7 tenths left. Can we make a group of 35 with 7 tenths? (No)

Since there are 10 hundredths in 1 tenth, we decompose 7 tenths to 70 hundredths. A zero

is added to dividend to show hundredths.

Next division step 70 hundredths ÷ 35

≈80 hundredths ÷ 40

= 2 hundredths (2 is placed in the hundredths place.)

Now check to make certain quotient is correct.

2 4 . 1 2 same as 2 4 1 2 hundredths

x 3 5 x 3 5

1 2 0 6 0

7 2 3 6 0

8 4 4 2 0 hundredths = 844.20

Each toy car weighs 24.12 grams.

A member of the cross country track team ran a total of 300.9 miles in practice over 59 days. If the member ran the same

number of miles each day, how many miles did the member run per day?

Strategy: 300.9 ÷ 59

Can we make a group of 59 with 3 hundreds? (No)

There are 10 tens in 1 hundred, so decompose 3 hundreds to 30 tens.

Can we make a group of 59 with 30 tens? (No)

There are 10 ones in 1 ten, so decompose 30 tens to 300 ones. Can we

make a group of 59 with 300 ones? (Yes)

First division step 300 ones ÷ 59

≈ 300 ones ÷ 60

= 5 (5 is placed in the ones place.)

After subtracting, there are 5 ones left. Can we make a group of 59 with

5 ones? (No)

There are 10 tenths in 1 one, so decompose 5 ones to 50 tenths. There are

already 9 tenths, so there is a total of 59 tenths. Can we make a group of 59

with 59 tenths? (Yes)

Next division step 59 tenths ÷ 59

= 1 tenth (1 is placed in the tenths place.)

The member ran 5.1 miles each day.

2 35 8 4 4 . 2 7 0 1 4

2 3 35 8 4 4 . 2 7 0 1 4 4 1 0 5 39 (We can get another group of 35 with 39; so we

can get 4 groups of 35 instead of 3 groups in 144 ones.)

2 4 35 8 4 4 . 2 7 0 1 4 4 1 4 0

4

2 4 . 1 35 8 4 4 . 2 7 0 1 4 4 1 4 0

4 2 3 5 7

2 4 . 1 2 35 8 4 4 . 2 0 7 0 1 4 4 1 4 0

4 2 3 5 7 0 7 0

5

5 9 3 0 0 . 9

2 9 5

5

5 . 1

5 9 3 0 0 . 9

2 9 5

5 9

5 9

Check: 5.1 same as 5 1 tenths

x 5 9 x 5 9 4 5 9 2 5 5 0 3 0 0 9 tenths = 300.9


OBJECTIVE OF TOPIC H

Solve division word problems involving

multi-digit division with group size unknown

and the number of groups unknown.

5th Grade Math Module 2: Multi-Digit Whole Number and Decimal Fraction

Operations

Math Parent Letter

This document is created to give parents and students a better understanding of the math concepts found in Eureka Math (© 2013 Common Core, Inc.) that is also posted as the Engage New York material which is taught in the classroom. Grade 5 Module 2 of Eureka Math (Engage New York) covers Multi-Digit Whole Number and Decimal Fraction Operations. This newsletter will discuss Module 2, Topic H. In this topic, students apply the work of Module 2 to solve multi-step word problems using multi-division.

Topic H. Measurement Word Problems with Multi-Digit Division

Words to know:

Model/Tape Diagram Reasonableness

Equation/Number Sentence

Units/Sections

Solution/Answer

Things to remember:

Tape Diagram – Drawing that looks like a segment of

tape, used to illustrate number relationships.

Example: There are 452 heart pamphlets that must be

delivered to 4 hospitals. If each hospital receives the same

amount, how many pamphlets are delivered to each hospital?

Focus Area– Topic H Measurement Word Problems with Multi-Digit Division

Example 1:

Billy is saving for a 52 inch flat screen TV that costs $1,218. He

already saved half of the money. Billy earns $14.00 per hour. How

many hours must he work in order to save the rest of the money?

Strategy Approach:

Step 1:

Step 2:

Step 3:

$1218 Draw a tape diagram to represent $1,218 which is amount he needs to buy the TV. It is divided into 2 equal units since the problem states that he already saved half of the money.

$1218

Saved

?

To find out how much he already saved, we divide 1218 by 2.

Equation: 1218 ÷ 2 = 609

$1218

Saved $609.00

If half is equal to $609.00, then the other half is equal to $609.00.

Since he makes $14.00 per hour, we want to find out how many 14s are contained in 609.

Number sentence: 609 ÷ 14 = 43.5

How many hours does he need to

work?

$609.00 $609.00

Reasonableness: 609 ÷ 14

≈ 600 ÷ 10 = 60

OR

609 ÷ 14

≈ 600 ÷ 15 = 40

Student could round 14 to 10 and 60 is a multiple of 10. Since 14 is rounded down to 10, the answer will be less than the estimate.

Student knows that 15 is a factor of 60 and 15 is very close to 14.

Solution/Answer: Billy needs to work 43.5 more hours.

452

1 hospital

The whole diagram represents the 452 pamphlets. Since there are 4 hospitals, the diagram is divided into 4 units/sections. To find the value of 1unit/ section you would divide 452 by 4.

Approach to solving a problem: Draw amodel, write an equation/number sentence,compute, and assess the reasonableness ofanswer/solution.

Grade 5, Module 2, Topic H

Math News!

Example 2: Mr. Smith has 1354.5 kilograms of potatoes to deliver in equal amounts to 18 stores. 12 of the stores are in Lafayette. How many kilograms of potatoes will be delivered to stores in Lafayette?

Strategy Approach:

Step 1:

Step 2:

Step 3:

3 11 8

The tape diagram drawn to represent the total kilograms of potatoes (1354.5 kg) that needs to be deliver to 18 stores. The three dots in the rectangle between 12 and 18 represent stores 13 to 17.

*****************************************

The model is showing 18 units (equal sections) that equal 1354.5. We have to find the value of 1 unit or one section.

Equation:

1354.5 ÷ 18 = 75.25

75.25 kg of potatoes delivered to each store.

Assess the reasonableness: Round the divisor: 18 rounds to 20 13 is not a multiple of 2 but 14 is so our whole or dividend is 1400.

1400 ÷ 20

= (1400 ÷ 2) ÷ 10

= 700 ÷ 10

= 70

We can conclude that 75.25 does make sense since it is

close to 70.

*********************************************

Now that we know the kilograms of potatoes delivered at each store, we need to multiply 75.25 kg times 12 to determine how many kilograms of potatoes were delivered to the 12 stores in Lafayette.

Equation: 75.25 x 12 = 903 kg

903 kg of potatoes were delivered to 12 stores in Lafayette.

Assess the reasonableness: When studying the model it is easy to see that more than half of the total amount of potatoes is being delivered to stores in Lafayette. 903 kg is more than half.

4

1354.5 kg of potatoes

1 2 5 6 7 9 12 18

. . .

10

Stores in Lafayette

8 3 4


1 2 5 6 7 9 12 18

. . .

10

Stores in Lafayette

11

8 3 4


1 2 5 6 7 9 12 18

. . .

10

Stores in Lafayette

11


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