Elementary School Maths 
We describe the value of a digit by writing the number in a place-value chart. Our number system is called Base 10. 10 unit cubes are equal to 1 long of 10. 10 longs are equal to one flat of a hundred. 10 flats are equal to one large cube of 1000. Patterns 
Each next greater place value is 10 times more. 
Each next lesser place value is 1/10 as much. 
We can write numbers that are 10 times as much ,or 1/10 as much as any given number . 
Hund ThTen Th ThousandHundredTens Ones 
200,00020,0002,000200202
10 times as much as 20,00010 times as much as 2,000Given Number1/10 as much as 20001/10 as much as 2001/10 as much as 20
multiply by 10 divide by 10. 
As we multiply, we create greater numbers just as the place-values become greater moving left. 
Each place value column we move from right to left we multiply. 
As we divide,we create lesser numbers just as the place-values become lesser moving right. 
As we multiply by multiples of 10 we add zeros.As we divide multiples of a10 we remove zeros. 
As we divide we create lesser numbers just as the place values become lesser moving right. 
Each place value column we move to the right, we divide. 
Thousands
=5,000Hundreds
=500Tens
=50Ones
=5
5 x 10005 x 1005x10
5,000÷10 5,000÷100 5000 ÷1000
Each place value can only hold one digit. Each model of our base ten blocks is ten times more than the model to its right. We put a comma between each group of three digits starting from the right side. We can also use a place-value chart to compare values. We find or compare the value of an underlined digit. 
The digit 0 has no value. We use it as a placeholder to show an empty place-value. 
We talk about how each place-value to the left is 10 times more than the place-value to to the right. We see what number we would get if we exchanged two of its digits. 
We see what would happen if we exchanged two digits in a number and one of those digits was a zero.
Below is a Place Value Table in Millions Thousands and Ones. The value of 2 in 5,204 and 27,328 is 20. The 2 that is further to the left will have a greater value than the 2 to the right .
Compare 4,000 and 400.The value of 4 in 4000 is ten times the value of 4 in 400 .For 30 and 300, the value of 3 in 300 is ten times the value of 3 in 30. 

Thousands Ones 
Hundreds Tens OnesHunTOnes
2800
470
36000
51000

•234,098 = What number would we get if we exchanged the digits in hundred thousands place with the ten thousands place ?
•34,298. When the zero is in the left most place value we do not record it, because its not needed .
Rule Rounding to nearest underlined digit 
•0,1,2,3,4 stay the same 
•5,6,7,8,9, go up to the next number 
•51,999= 52,000
•527,829 =530,000
•27,126 = 27,100
•361,218= 400,000
A store sold $2,411 of merchandise on Monday, and $2,396 of merchandise on Tuesday .To the nearest thousand dollar, how much did the store sell on Monday and Tuesday? 
We need to add the amounts together first to make accurate estimate.
  $2,411
+$2,396
$4807 (We add them first, then round the sum).
$4807 rounds to $5000

$2,411 rounds to $2000
$2,396 rounds to $2,000
  $4000 /-
We can round a number to get an estimate .An estimate tells us how many or how much. 
•To identify place values we are rounding to 
•Look at the number to its right side
•Look if the right side is 0,1,2,3,4, yes then the number stays the same .
5th Grade Maths9.6 Word Pattern Solving find pattern Rule
We can use a strategy solve a simpler problem to help us solve a problem with patterns.
We can find a pattern using simple numbers then use the pattern that we found to predict results with greater numbers to solve the problem. We can find the rules for two sequences,then find the rule that relates the sequences to each other.
Sequence1246810
Sequence2612182430
Relating Rule :Multiply by 3 
We can use the relationship between the two sequences to find the missing terms. 
We can make a table of values of the terms to find greater numbers in the sequences. 
Sequences1246810…..20
Sequenc612182430…
  Q.Sarah made identical beaded necklaces. Each necklace used 4 green and 12 orange beads .If Sarah made 14 necklaces,how many green and orange beads did she use?
Think: We can make a table of values showing the number of beads for each necklace. 
Necklace1234…14
Green Beads48121656
OrangeBeads1224348168
First and second row multiply by 4.Second to third row multiply by 3.2ND Row: Add 4 3rd Row: Add 12 
4x3=12 14x4=56 56x3=168
Q. In a design each pattern adds 5 blocks and 80 triangles.40 Blocks have been used .How many patterns and triangles were used?

Think: 20÷5=4 and 40÷5=8

Patterns1234…….8
N of Blocks5101520…..40
Nof Triangle80160240320…..640
First two rows shows divide by 5.
  Second to third row shows multiply by 16. 
Third Row we add 80. Look at relationships between sequences going vertically. 
5x16=80 and 80÷5=16 and 40x16=640 triangles.
•Using a rule makes it easier to determine the relationship between the objects in a pattern.
•Then the rule can help us decide which operation to use. 
•The operation we chose for our rule depends on the given data. 
Q. How many patterns and triangles if 120 blocks were used? 
Nof Patterns1234…24
Nof Blocks5101520….120
Nof Triangles80160240320…1920
Rule: Second Row to first Row =Divide by 5 
  Second Row to third Row=Multiply by 16 
5Th Grade Math2.3 Model Division of 2 Digit Divisors with Base 10 Blocks 
•We can use Base 10 Blocks to understand division of whole numbers. 
•We can use hundreds,tens and ones blocks to model splitting the dividend into groups of the same size.
•We can regroup a large square of 100 for 10 tens,or a long of 1-0 for ten ones.
•(165÷11 groups=15 in each group)
•(168÷14=12 in each group).We model the dividend 168 as 1 hundred,6 tens and 8 ones.
•We make 14 groups of 10 as 140and have 28 left over.168=140+28.
•We can’t add more so we put more into each group to make each group longer. 

5th Grade MathRounding Off and Estimate to Hundred Thousand
•Rounded to the nearest hundred thousand 
•181,329 = 200,000
•328,446 =300,000
•747,692 = 700,000
•651,578= 600,000
We can use an Open number line 
100,000 138,202 150,000 200,000

•Where is the number on the number line?
•What Hundred Thousand is it between ?
•Which number is it closest to ?

138,202 is closer to 100,000 than 200,000
5th grade math 2.4Add Whole Numbers then Regroup 
Estimate then find the sum .


•72,935 +18,567 =70,000 + 20,000= 90,000 estimate
  72,935
+ 18,567
  91,502 .
Thus our answer is close to our estimate.
 Comparing whole Numbers To Compare is to decide if one number is greater than less than or equal to the other number .Each model is 10 times more than the model to its right. 
 1×10 = 10  
10 ×10=100 
100×10=1000
1000-100-10-1 
Patterns of Zeros 
Multiply by 10,add a 0 Divide by 10,remove 0
Multiply by 100 add 00Divide by 100 remove 00
Multiply by 1000 add 000 Divideby 1000,remove000.
500 is 100 times as much as 5.500has 2 more zeroes than5
6,000 is 1000 times as much as 6.
400,000is 100 times as much as 4,000.
20,000 is 1000 times as much as 20.
Complete the sentences using 10 times or 1/10 .
•4 is 1/10 as much as 40 ,so 40 is 10 times as much as 4 
•50 is 1/10 as much as 500,so 500 is 10 times as much as 50 
•200 is 10 times as much as 20, so 20 is 1/10 as much as 200.
•30,000 is 10 times as much as 3,000.So,3,000 is 1/10 as much as 30,000.
•Sasha had 200 stickers. She gave Mary 1/10 .How many stickers did she give Mary ? 
•Think : We need to find 1/10 of 200 
Of refers to a multiplication symbol. 

1/10 of 200 
​1/10 x 200= 1x200 = 200 = 20 stickers to Mary. 
5th Grade,2.5Estimate with 2-digit Divisors using Compatible Numbers 
Compatible numbers are numbers that are easy to calculate.We can use compatible numbers to estimate quotients by rounding the divisor to a multiple of 10, and rounding the dividend to the nearest multiple of the new divisor.
Then we use a basic division fact and Pattern of Zeros to complete the estimate.
2692÷28 becomes 2700÷30
Think :27÷3=9
  270÷30=9
  2700÷30=90
Basic division facts and Pattern of Zeros :
Basic Fact 24÷4=6
When we give a Zero to the dividend and divisor,the quotient stays the same.
240÷40=6
Choosing the best basic fact to use 
6,231÷80 6,231÷80
5,600÷80 6,400÷80
56÷8=7 64÷8
560÷80=7 640÷80=8
5,600÷80=70 6,400÷80=80
The actual quotient is between 70 and 80.
64÷8 is the best basic fact we use to estimate the actual quotient because 6400 is closer to 6231.
We can estimate a quotient by choosing compatible numbers using basic facts. 60 is the better estimate. We can subtract We can do 1852-1800and find the difference then 2100-1852 and whichever has the smallest difference is closest to the actual answer.
Q.Sophia has $725 to buy fabric to sew 28 shirts to sell.She plants to spend about the same amount of money for each shirt. Estimate how much she can spend for each shirt ?
Answer: We can use compatible numbers and basic facts to estimate the cost of fabric for each shirt ?

√(28&$725)=$600÷30=$20 or $900÷30=$30

She can spend between $20 and $30 on fabric for each shirt. Sophia should spend closer to $20 because $600 is closer to$725.
5th GradePlace Value of Whole Numbers 
1.2Place value Standard Form Word & Expanded 
A place- value chart contains periods as a group of three digits seperate by commas in a multi digit number.
Millions Period,Thousands Period,Ones Perio
798,421,563
HTOHTensOHTO
2800
HTO470
HTO
MillionsThousandsOnes
91,493,000
H stands for Hundreds .T stands for tens and O stands for Ones in the place value table.
The place value of the digit 9 is ten millions.
The value of 9 in 91,493,000 is 
9 x 10,000,000=90,000,000=(90 million)
Standard Form:91,493,000.
Word Form : Ninety-one million four hundred ninety-three thousand. 
ExpandedForm (9x10,000,000)+(1x1,000,000)+(9x10,000)+(3x1,000).
Each place value period contains ones Tens and Hundreds Columns.
We can read write an represent whole numbers by identifying the periods and the place values of the digits in the periods. 
We name the number and the period then place a comma between the periods starting from the right. 
Each place value period contains Ones, Tens and Hundreds columns.
5th Grade Math1.12Grouping Symbols ( ) Parentheses [ ] Brackets& Braces{ }
  When an expression or equation, has many grouping symbols we begin with the innermost se ,and work to the outermost.
When a number is directly close to a grouping symbol or variable we multiply .
2(3)= 2 x 3

 3{20-[2(5+4)]}parentheses First 
 3{20-[2(9)]} Brackets next 
3 {20-[18]} then Braces
3{2}=6 
We can use grouping symbols to write an expression that will represent the information in a word problem. Grouping symbols will help us solve the expression in the correct order. 
For 3 days, Lia ate 2 jelly beans in the morning and 4 jelly beans in the afternoon 
3(2+4) =3(6)each day =18 jelly beans in 3 days.
Rule :When following the order of operations we calculate within grouping symbols as the first step. That first step ,may contain several steps depending on the number of grouping symbols. 
Parentheses() Parentheses Please
Brackets[ ]Exponents Excuse
 Braces{ }Multiply or Divide from left to right] My
  Add or subtract from left to right] Dear 
  Aunt
Sally. 
Methods to Represent Multiplication In Algebra

Popular variables are x and n ,written in italics
 The x can be confused with the symbol x for multiplication if not written carefully. There are other methods to represent multiplication. 
2x5 can be written as 2.5,(2)5,2(5),or(2)(5).

2 x x 5 is written as 2x.5,(2x)5,2x(5) or(2x)(5)
2 x 10^3 can be written as 2.10^3 ,(2)10^3,2(10^3)
5TH Grade1.13 Number System,Subtraction with Zeros
Our number System is called Base 10
10 Unit Cubes = 1 Long 10
10 Longs = 1 Flat Hundred
10 Flats =1 Large Cube one thousand 

Each place value can only hold 1 digit .10 has two digits.We use 2 place values for 10. 
HundredsTensOnes
507

The Digit 0 has no value. We use it as a placeholder to show an empty place value. Place Value is the position of a digit in a number that tells the value of the digit. Each digit in 356,039 is written in its place on the chart .Expanded form shows the sum of values of each digit.
Expanded Form :300,000+50,000+6,000+30+9.
Rule 1 :If we stop at 0 it always becomes a 10 and when we across a o it becomes a 9 ,which is technically a 10. We borrow first ,then spend.

5th Grade Math 1.3Addition,Multiplication,Subtraction Strategies
Commutative Property Of Addition 
When the order of Two Addends is changed the sum is the same.We can add in any order. 
(4/8+2/8) +4/8 = (4/8+4/8 )+2/8 
= 12/8 = 1 2/8
To add two or more addends using mental math can 
Use the commutative and Associative property of addition. 
3+19+17=19+3+17.We can change the order using the commutative property.We can then group the addends using the Associative property to add compatible numbers first.19+(3+17)
Now we can use mental maths.3+17=20,20+19=39. 
Associative Property Of Addition 
1+(2+4)=(1+2)+4
  1+6 = 3+4
  7=7 
When the grouping of the addends is changed, the sum is the same .We can move the parentheses to different addends .
 (2/5+1/5)+4/5=2/5+(1/5+4/5)
12/5=1 2/5
Use the Associative property of Addition (the grouping property).If the grouping of the addends change the sum stays the same. 
2+(3+5)=10 (2+3)+5=10
Of Multiplication 
If the grouping of the factors change, the product stays the same.
2x(5x6)=60 (2x5)x6=60
3,928+2,480=(3,408+520)+2,480
3,408+(520+2,480)
3,408+3000=6,4083,928+80= 4,008
4,008+400=4,408
4,408+2000=6,408
Identity Property of Addition  
5+0=5
5th Grade Count up,Count Down Compensation Compatible Numbers,Properties of Multiplication 
The sum of a number and 0 is that number. Adding a 0 won’t change a number’s identity.
Of Multiplication 
The product of any number and 1 is that number .Multiplying by 1 won’t change the number’s identity.5x1=5.
Compatible numbers are numbers that are easy to compute mentally. 
They may equal a multiple of 10,like 3+17=20.
They may equal 100 ,like 4x25.
(4x9)x25=9x(4x25)
The order was changed using the commutative Property. 
The grouping was changed using Associative Property. 
Without using properties
4x9x25=36x25= ?
Using the commutative and Associative
Properties of Multiplication
9x(4x25)=9x100=900
(4x25)x9=100x9=900
Distributive Property 5x29=5x(20+9)
Multiplying a sum by a factor is the same as Multiplying each addend by that factor ,then adding their products. We use multiplication and addition .
5x(20+9)=(5x20)+(5x9) 
 = 100 +45
  =145  
The order of operations tells us to do operations within parentheses first. 
5x29=5(20+9)
We can quickly perform mental math by using properties to group numbers so that they are easier to add or multiply.
3x83 =3x80=240 then 3x3=9
  and 240+9=249(Distributive Property)
(56+19+4): (56+4)=60,then 60+19=79
  Commutative and Associative
  Properties od Addition
43x11=(43x10)=430,then(43x1)=43
And add 430+43=473
Distrivbutive and Identity Property.
Note:We cannot use the commutative properties with division and subtraction. 
We can use the equal additions method when we subtract numbers.
We add the same amount to each number so they are more compatible .
43-17=(43+3)-(17+3)

So for subtraction in compensation what we do to one side is the same we do to the other side.
86-59=(86+1)-(59+1)
  =87-60
  =27

 5th grade math 3.1.StrategiesMultiply Associative,Distributive o ,Place Value
Multiply Tens using Place Value
30 x 20 = 30 x 2 tens =60 tens 
=600 OR 60 tens
We can multiply basic facts using mental math using the number of zeroes in the factors.
30 x 20 =600 (3 x 2=6)
4 x 60 =4 x 6 tens
4 x 60 = (4 x 6 tens)
4x60= 24 x 10
4 x 6 0 =24 0
Find 4 x 600
4 x 600 =4 x 6 hundreds
4 x 600=(4x6) hundreds
4 x 600 =24 x 100
Find 4 x600 = 24 00 
 4 x 6000
4 x 6000 = (4x6) thousand
4 x 6000= 24,000 or 24 x 1000
  Place Value
8 ×359= (8×3 hundreds) +(8 × 5 tens) +(8× 9ones)
The Distributive Property 
7 × 4,056 = 7 × (4000+ 50+6)
(7 × 4000) + (7× 50)+ (7 ×6)
Distributive Property to find 3x 1275
3 x 1275=3×(1000+200+70+5)
  =(3×1000)+(3×200)
  =(3×70)+(3×5)
  = 3000+600+210+15
  =3,825
Use the Commutative Property of Multiplication states that when the order of two factors is changed the product is the same.
2 x 3 = 3x2
Use the Associative and Commutative Properties together
25 x (5 x 4) = (25 × 4 ) × 5
25 × 20 = 100 × 5
500 =500
Associative properties of Multiplication to find 8x50
8×50=(2×4)×50
  =(4x2) × 50
  =4× (2×50)
  =4 × 100 = 400
Commutative Property Of Multiplication 
4 x325 = 4 x300 +4 x25
  = 1200 +100
  =1,300 miles 
5th Grade Math Compensation Methodsadd,subtractMultiply or Divide (Mental Maths)
Addition and Subtraction Compensation Multiplication Division Strategies with Models 
Compensation By using compensation we can use mental maths to add subtract ,multiply or divide. We compensate one number with amounts from the other number to make Mental Maths easier.
Addition In Compensation (9+6)=
  =(9)+ (1+5)
=9+6 =10+5 .
Taking 1 from the 6 we add 1+9 to make a 10.Then we add 10+5 mentally.
84-28 = {84+2) – {28 +2}.We take the addend 28 and add a 2 to get 30 
Subtraction In Compensation 
43 -17 = (43 + 3 )- (17 + 3)
= 46+ 20
=86
We can use Multiplicative In Compensation 
To multiply in Compatible numbers 
68 (5) = 68 ÷2 and then ( 5)(2)
Since 68+2 =70 .We divide the first one by 2 and multiply the second one by 2 .
68 ÷2 and then 5 × 2
=34 × 10
=340
Two separate operations of dividing the Multiplicand and multiplying the multiplier.
48 (25) = 25 ×4 =100 .So we have to divide 48 by 4 and multiply 25 times 4. 
100 is more compatible than 25. 
= 48÷4 × (25 ×4)
=12 ×100
=1200
126 ÷ 18 
(126 ÷ 2 ) ÷ (18 ÷ 2)=63 ÷ 9 
Q1. 6 × 495
(We take the common factor 5
495÷5 ×(6)(5)=99 ×30
=2,970
Q2. 68 (5) = 68÷2 x (5)x(2)
68/2x(5)x(2)
  = 34 x 10 =340
Its easier to multiply by 10.
Its easier to multiply by 10 .We change the 5 to a 10 by multiplying it by 2. 
We then must change the 68 by dividing it by 2.
We can use multiplicative compensation to multiply incompatible numbers.
Q.48(25)=48/4 (25)(4)=12X100=1200.
100 is more compatible than 25. 
Two methods we divide the multiplican and , and then multiply the multiplier 
Commutative Property Of Multiplication
Multiply the friendly numbers first.
2×50 ×13 = 100 ×13 = 1300
Associative Property of Multiplication 
​4 ×(25 × 5)
(4 × 25 )×5
100 ×5=500We can use Distributive Property of Multiplication 
4 × 325 = 4 × (300 + 25)
  = (4 × 300) + (4 × 25)
  = 1200 + 100
  =1300
When multiplying mentally some numbers are easier to multiply by a 1 digit factor like 25 or 250. Numbers that have 1 non-zero digit are easier 10,20,30………..100.200,300.
5th Grade MathSubtraction Strategies
Q.Find 9125-7,985
Count Up 
7985 +15=8,000
  8000 +125= 8,125
  8,125 +1000=9,125
  Added 15+125+1,000 =1,140

Multiply tens using place value.
30 x20 = 30 x 2 tens= 60tens = 600
We can use multiplying basic facts and use mental math with the number of zeroes in factors. The number of zeroes . 
Count Down 
9125 -25=9,100
9,100-60= 9,040
9,040-900=8,140
8,140- 7000=1,140 
Use Compensation In Subtraction 
Q.9,125 - 7,985
=(9,125+15) - (7,985+15)
9,140-8,000=1,140 
The factors in the equation will help us decide which Strategy to use.
Equation Strategy 
 40 x 20 =  
  Mental math multiply basic facts and number of zeroes in the factors is the number of zeroes in the product.
12 x 20 = Halving and doubling ,this works with even numbers easily divided by 2.We multiply half of one factor then we double that product.
30 x 15 =
Associative Property 30×(5×3)
Then multiply (30 x 5)x3=150x3=450 . 
5th grade math3.2Estimate Products 
Compatible numbers are easy to compute mentally. 
10x10=100
25x2=50
25x3=75
25x4=100
50x2=100
If the rounded re greater than the actual factors the estimate will be greater than the exact answer .We will overestimate the amount. 
19 x 19 =361
20 x20= 400 (overestimate)
If the rounded factors are less than the actual factors the estimate will be less than the exact answer .We will underestimate.
21 x 21=441
20 x20 = 400 (underestimate)
If one factor is greater and another rounded factor is less than the actual factors the estimate will be close to the exact answer .
19 x21= 399
20 x20= 400
Use rounding to Estimate 7x215 .First Round 215 to the nearest hundred.

200 215 250 300
215 rounds to 200.Then multiply 7 x 200=1,400.
So ,7 x 215 is about 1,400.
Check if 2,885 x 4= 11,540 is reasonable.
First round 2,885 to the nearest thousand.2,885 
.rounds to 3,000.
Then multiply 3,000 x 4 =12,000
.So,2885 x 4 is about 12,000.11,540 is a reasonable Answer.
5th Grade Math,Estimating Products 
To estimate products you can round 3 digit numbers to nearest hundred and 4 digit numbers to nearest thousand.

Use rounding to estimate 7×215

200 215 250 300
215 rounds to 200.
Then,multiply.7×200=1,400
  So,7 ×215=is about 1400

Check if 2,885 ×4-11,540 is reasonable. 
First round 2,885 to the nearest thousand .2,885 rounded is 3000. 
Then multiply 3000 ×4= 12,000
So 2,885 ×4- 12,000.11 ,540 is a reasonable answer. Multiply by Multiples of 10,100,1000.We can use place value or the Associative Property Of Multiplication to multiply by multiples of 10,100 and 1,000.
9 ×50=9×(5×10)
  =(9×5)×10 
  =45×10
  =450
9 ×500=9×(5×100)
  =(9×5)×100
  =45×100
  =4500
9×5,000=9×(5×1000)
  =(9×5)×1000 =45×1,000 =45,000.

5th Grade MathArea Models and Partial Products for 2 digit factors 

It doesn’t matter how we break up factors because their partial products will always equal the same sum .
12 x 14 = (10 x10) +( 10 X 4)+(2X10)+(2X4)
 =100+40+20+8 =168

12 x 14 = (6x7)+(6x7)+(6x7)+(6x7)
42+ 42+ 42 +42 =168  
1 2 4 
× 3 
1 2
  6 0
  =372

We can use place value and partial products to multiply by a 1 digit factor. 
We break apart greater factor into thousands hundreds tens and ones.
Next, we can multiply each of these place values by the 1-digit factor
.Then we add the partial products to get the full products. 
When we multiply factors in vertical form ,using place value and partial products ,we begin on the left at the greatest place value. 
376
X 2 
600  
140
  12 
752 
Thus 600 ,140 and 12 are partial products. 
Add them to get product of 752 B. Use mental math and thinking 
$4 7  
X 8
 320
+ 56
$3 7 6
Estimate to check.$40 x 8 =320 ;$50 x 8 =400.$376 is between $320 and $400.So our answer is reasonable.  
 3.4 AreaModels and Partial Products 
10 4
7 = 7 ×10=707 ×4=28
Estimate 7×14 is about 7 ×10=70
7 x 10=70 ; 7x4=728
  70+28 =98
  14
  X 7 
 28 7X 4 ones +70 7x 1 tens =98 . 
Each Rectangle contains a partial product of 15 x 16. 
It doesn’t matter how we break up factors ,because their partial products will always equal the same sum.12x 14= (10x10)+(10x 4)+(2x10)+(2x4)
=100 +40+20+8 
=168
10040
208
12x14 =(6x7)+(6x7)+(6x7)+(6x7) = 42 +42 +42 +42 =168
 5th Grade Math ,3.4 Multiply Using Partial Products One and Two Digit Factors

We can use place vale and partial products to multiply by a 1 digit factor .
1)We can use place vale and partial products to multiply by a 1 digit factor .We break apart the greater factor into thousands hundreds tens and ones.
2)Next we multiply each of these place values by a 1 digit factor. 
3)Then we add the partial products to get the full product. 
When we multiply factors in vertical form using place value ,and partial products we begin on the left of the place value .

1.376
  ×2
  600
+140
  12 = 752 {We add the partial products}to get the final ,/product.
2.Estimation using Partial Products 
$47 
×8
320
  56= $376 is between $320 and $400 so our estimate is reasonable.
3.We can use the Distributive Property to break apart the greater factor to find products of 3-digit and 1-digit factors. We use place value and partial products.
4×143
40016012
400+160+12=572
143
×4 = 400  

  +160
  + 2
  = 562
We can check our answer with Estimation .143 is between 100 and 200 .4×100 =400 ; 4×200=800.572 is between 400 and 800, so our estimate is reasonable. 
•27 x 31 =(20x30)+(30x7)+(1x20)+)(1x7)
  =600 +210 + 20+ 7 =837 
•Before we multiply we can estimate the product so we know a reasonable answer
•27x 31 =30 x 30 =900
• 3 x3= 9
Multiply 2-digit factors using Partial Products 
•Multiply 23 x68 .Estimate 20 x70=1400
•1200•160
•180•24
 • 68 X23
=1200 =(20x60) so we add the partial products .First we multiply the tens then the ones and multiply to find the partial products .We multiply using Place Values 
1200 =20x60 
  160 =20 x 8 ones (tens x ones)
  180 =60x3 (ones x tens)
  24=8x3 (ones x ones)
•Multiply 39 x 24 Estimate 40x25 =1000(Compatible numbers)
  39
X24 
600 =20 x 3 tens
180 = 20 x 9 ones 
120 =4 x 3 tens 
  36 =4x9 ones 
936 (Is Reasonable because it is close to 1000.
5TH Grade7.7 Area and Mixed Numbers 
A Unit Tile is a square with side lengths of 1 unit.
The Unit maybe metric or in Standard form. It maybe a mixed number or a Fraction. 

We can use a Unit tile to find the area of a Rectangle with fractional side lengths.

We can cover the rectangle with fractional unit tiles and then multiply the number of tiles by the area of each tile to find the total area.
Area = l x w 6X4=24 SQ INCHES 
We can use square tiles with side lengths that are unit fractions to find the area of a rectangle. 

First, we need to find the area of each unit tile .
½ x ½ =1/4 ; ½ +1/2=1 inch .Each side is ½ inch but its area is ¼ th square inches.

We can review mixed numbers as fractions greater than 1 to find area.

½ x ½ = ¼ square inches 

〖15 x〗_(4=1 x 4 )^(1=15 x 1 )=15 / 4. = 3_4^3.We used 15 little squares that is ¼ sq inches.  
We can use square tiles with side lengths that are unit fractions to find the area of a rectangle.
First we need to find the area of each unit tile.

4 x 1/4=4/4=1 sq inch.
  16x 1/16=16/16=1 sq inch.(Covering the grid with smaller tiles does not change the area).We can write mixed numbers as fractions greater than 1 to find area 1 1/2=3/2.
We find the area of each section.Then we add the partial products to find the total area. 
3 +1/2+3/4+1/8 = total area.
3 +1/2 =31/2 and 3/4+1/8=6/8+1/8=7/8
3_(2=)^134/8 +7/8 as we need a common denominator
=3and 11/8=3+8/8+3/8=4_8^3 square units.
We use the same formula L X W for length times width to find the area of a rectangle whether the side lengths are whole numbers fractions or mixed numbers. 

5TH Grade6.7 Subtract mixed numbers with Renaming 
Use Renaming to find the difference of two mixed numbers. 
First write equivalent fractions with a common denominator .
Next use multiplication and addition to rename each mixed number as a fraction greater than 1.
Then subtract the fractions and write the difference in simplest form. 
By writing the mixed numbers as equivalent fractions first,it will determine if we need to use renaming. 
We look at the numerator of each fraction.
If the numerator of the minuend is less than the numerator of the subtrahend ,we,ll need to use renaming.
2_(5-)^1 1_5^3
1 <3 so we cannot subtract unless we use renaming. 
When the numerator and denominator is the same the fraction is equal to 1 whole.
1+5/5 +1/5 =6/5 
We can rename mixed numbers as fractions greater than 1 by using multiplication addition mental maths. 
1.Multiply the whole number by the denominator 
Add that product to the numerator
Use the given denominator.
We can add and subtract mixed numbers with unlike denominators by finding a common denominator for the fraction parts of the mixed numbers. 
We can use the common denominator to write Equivalent fractions with like denominators.
Then we add or subtract the fraction followed by adding or subtracting the whole numbers.
We can begin by estimating the sum / difference to be able to check our answer for reasonableness. 
3_5^2 +1_2^1 ≈3_(2 )^1 +1_2^1 =5
We can use benchmarks.
  3_5^2 x 2/2 = 3_10^4
+ 1_2^1 x5/5 =1_10^5
  =4_10^9 is very close to 5 our sum is reasonable.

We can use a common denominator to subtract mixed numbers with unlike denominators

  3_(5 )^1 x 6/6 =3_30^6
-〖 1〗_(6 x )^1 5/5 =1_30^5
  = 2_30^1 3_5^1 - 1_6^1 ≈ 3-1=2 .
2_30^1 is close to 2 so our difference is reasonable.
Model Fractions with unlike denominators to find the difference by using models.We try fitting fraction strips under difference that have an exact fit.If we need to use more than one fraction strip in the empty space they must have the same denominator as each other.
½-1/3
By finding the fewest amount of fraction strips that have an exact fit for the difference we are putting the difference into simplest form.
½-1/3-1/6.Fit larger pieces first for simplest form.  
=2 9/10
We turn one whole into10/10 and give it to the 5/10.

½ = 2,4,6,8,10,12,14
3/5=5,10,15,20,25 
2_5^1=5/5+5/5+1/5=11/5
We can rename the mixed number by grouping these fractions differently.
1+5/5+1/5= 1_5^6

 By first giving them common denominators, we see that we don’t need to use renaming ,though we would get the same answer if we did.

We can rename mixed numbers as fractions greater than 1 by using multiplication ,addition and mental math.
Multiply the whole number to the denominator and add it to the numerator. Use the given denominator.

5th Grade Math 6.6,Add &Subtract Mixed Numbers with Unlike Denominators
We can add and subtract mixed numbers with unlike
Denominators by finding a common denominator for the fraction parts of mixed numbers.
We can use the common denominator to write equivalent fractions with like denominators then write in simplest form.
Then we add or subtract the fraction followed by adding or subtracting the whole numbers. 
Make sure that our sum or difference is written in simplest form when 1 is the only common factor for the numerator and denominator.  
5th Grade 6.2 and 6.5 Add and Subtract Fractions with Unlike Denominators
We can begin by estimating the sum or difference to be able to check our answer for reasonableness. We know our answer is in simplest form when 1 is the only common factor of the numerator and denominator.
We can use a common denominator to add or subtract fractions with unlike denominators. 
We can use the product of their denominators to write equivalent fractions that have common denominators.
Then we can add or subtract and simplify.
½ and 1/3 2x3=6
We can also list the non-zero multiples of each denominator to find their least common multiple to use as their common denominator. 
1/2+1/3
We can use the numerator and denominator of our difference, to see if they can both be divided by the same number.
We can test the numerator and denominator of our difference to see if they can both be divided by the same number. 
We list their factors to find their greatest common factor then use it to divide. 
Using a common denominator, we can compare and order fractions by the amount of the numerator.
Lowest Numerator= Lowest fraction 
Greatest Numerator= Greatest Fraction 
When fractions have a common numerator 
We can compare and order fractions by their denominators .
Lowest Denominator=Greatest Fraction 
A small denominator represents larger equal parts. 
2/3>2/4
Locate 2/5,1/3 and 7/8 .
0 ½ 1
We can compare each fraction to ½.
Think : 2/5 is less than ½ ,1/10 is close to 0/10 as o. 7/8 is almost 8/8 or 1 whole.
2/5>1/10;10 is a multiple of 5.

2/5×2/5=4/10

1/10<1/2 .Benchmark Larger Denominator.


0 ¼ ½ 6/10 6/8 7/8 1
We can compare 6/8,1/4,6/10,7,10 by locating them on a number line and then comparing them.
1/4<6/10<6/8<7/8

We can test the numerator and denominator of our difference to see if they can both be divided by the same number.


We list their factors to find their greatest common factors,then use it to divide. 
6/12÷6/6=1/2
6:1,2,3,6
12:1,2,3,4,6,12 2/12÷2/2=1/2
There should be no remainder.

5thGrade 7.10 Find Unknown Fraction Lengths 
Use Strategy guess, check and revise to solve problems with fractions. To guess a Fraction, we need to replace an Unknown number and increase or decrease our guess till it is correct. 
¼ th length means ¼ of the length or ¼ x the length. 
To find the area of a Rectangle we use the formulae L × W which means length x width.


3cm

4 cm 
Area = 12 square cm 

A Square has 4 equal sides so we can use S^2
Or side times side. Area =4 square cm. 




Sarah wants to make a rectangular latch (hook rug) that has an area of 3/4 square yards.She wants the width to be 1/3 the length. What will dimensions of the rug be ? Think :The AREA OF A RECTANGLE IS LENGTH TIMES WIDTH.If the area is 3/4 square yards there are two factors that are equal to 3/4.One factor is the length and the other is the width .
We need an area of 3/4 square yards with a width that is 1/3 the length. 
LengthWidth is 1/3 length Area = L X W
¾ Yard ¾x1/3 =3/12=1/4 yard¾ x ¼ = 3/16 (TOO SMALL)
7/4 yards 7/4 x 1/3 =7/12 yd7/4 x 7/12 =49/48 (too long)
5/4 yards 5/4x1/3=5/12 yd25/48 too small


6/4 = 2/3 yd3/2 x 1/3 =3/6=1/2 yd 3/2x1/2 =3/4 (correct)
The rugs length is 3/2 yd or 11/2 yards. Its width is ½ yards. 




½ yards 


11/2 yards 
Area = ¾ square yards.

5th Grade Math 6.10,Use Properties of Addition to add Fractions
The commutative Property of Addition states that when the order of two addends is changed the sum is the same. 2+3=5and 3+2=5
The Associative Property of Addition also called the grouping property states that when the groupings of addends is changed the sum is the same.
(1+2)+3=6 and 1+(2+3)=6
By regrouping the fraction addends we can add compatible numbers using mental maths.
Using Properties of Addition we can add fractions with common denominators first. 
5th Grade Maths 0.5MetricMeasurescomparconvert M/CM/MM/GM/ML/L
(12.6)On a customary Ruler with Inches,1 foot long we can see measures along the other edge These are metric measures. The smallest lines represent millimeters.
Cutomary Units can be used when measuring length,capacity,weight
Length1 Foot (ft)=12 inches (in)
1 yard (yd)=3ft = 36in.
1 mile (mi)=1,760yd=5,280ft
Capacity1 cup©=8 fluid ounces (fl oz)
1 pint (pt)=2c=16floz
1quart(qt)=2pt=4c
1gallon(gal)=4qt=8pt=16cups
Weight1 pound(lb)=16ounces)oz)
1 ton (T)=2,000LB
Length1centimeter(cm)=10 millimeters(mm)
1meter(m)=100cm=1,000mm
1 kilometer=1,000m
Capacity and Mass1iter(L)=1,000MILLILITERS(ml)
1gram(g)=1,000milligrams(mg)
1 kilogram(kg)=1,000g

The numbers represent centimeters. 
10 millimeters =1 centimeter
10 centimeter = 1 decimeter
10 decimeter = 1 meter
10 smallest units is equal to the next great unit. 
Kilo -means 1000
Hecto-means 100
Deka-means10
Deci-means0.1(one tenth)
Centi -means 0.01(one hundredth)
1Kilometer =10 Hectometers 
1 Hectometer =10 dekameters
1 dekameter = 10 meters
1 meter =10 decimeters
1 decimeter =10 centimeters
1centimeter =10 millimeters
Metric Units use the power of 10 .The meter is the basic unit. 
Since 1 meter is 10 times longer than 1 decimeter, and 1 decimeter is 10 times longer than 1 centimeter ,1 meter is x 10 times longer than 1 centimeter. 
KmHMDkmMDECIMCMMillimeter 
Power of 10 × powerof10 ×power of 10 ×10
There are 100 centimeters in 1 meter. 10 ×10-= 100 . (MDCM).1 Meter is: 100 com.Meter is basic unit of Length. 
1 power of 10 greater than 1 decimeter.
〖10〗^█(1@)=10 decimeters. 
Meter=1 base unit used to measure length or distance .
Liter=1 base unit,used to measure liquid volume.
Gram= 1 base unit used to measure mass.
Mega-for millions
Giga-billions
Micro-millionths
Nano=billionths
We can use different metric units to describe same length. Since metric system is based on number 10 ,decimals or fractions can be used to describe metric lengths as equivalent (equal) units. 
We think of 1 meter as 1 whole.
1 meter = 10 decimeters 
Each decimeter is 1 of 10 equal parts of a meter 
1 decimeter = 1/10 of a meter. 
1 decimeter is 0.1 of a meter. 
Mass is the amount of matter in an object .

Kilogram
 (KG)Hectogram
(Hg)Dekagram
(dag) or(dkg)Gram
(G)
  ×10 × 10 ×10
We multiply each by a power of 10 .

1Kg is 3 powers of 10 ,greater than 1 Gram. 
10^3=10×10×10=1000Grams .The exponent 3 tells us how many factors of 10.

Because we need 1000 grams to have 1 Kg ,1 Gram is 1 of 1000 equal parts of a Kilogram.
1 Gram =1/1000 Kilogram ;1 Gram =0.001 Kilogram.A gram is much less than a Kilogram .
KiloHectodekaMeter,liter,gramDeclCentlmill
÷10 ÷10 ÷10
To change meters to decameters we divide by 10.
Then to change dekameters to hectometers, we divide by 10.
Finally to change hectometers to kilometers we divide by 10. 
We divide by 10 3 times .We can divide by 1,000 which is 10 x 10 x 10 to perform the conversion in one Step. 
Convert 152 milligrams to centigrams decigrams and grams.
We need to divide because milligrams are smaller and we’re going longer. 
152 mg ÷10 =15.2 cg
152mg ÷100=1.52 dg 
152mg÷1000=0.152 g

1 Kilometers=1000 Meters
1000Meters =1Km
8000Meters=1/1000 x 8000=8Km 
8000 ÷〖10〗^3 =8Km 

Convert 2.3hectoliters to dekaliters,liters and deciliters .We need to multiply ,because hectoliters are larger and we are going smaller. 


2.3hl x 10 = 23 dal

hl x100 =230 L

2.3hlx1000= 2,300dl.

Write = < >

24L ÷ 10=2.4 dal

7.6km x1000 = 7600,Conversion Rates 
FootInches
112
224
336
448
560
784
896
9108
10120
11132
12144
zYard(Yd)Foot(ft)Inches(In)
112
1336
2672
39108
412144
515180
618216
721252
824288
927324
1030360

8 inches <8 ft 5 YD ≥12 FT 3 YDS ≥90 In
Convert each unit
1 mi = 5,280 ft
1 mi= 1,760 yd 

  25 ft = 300 in 3 miles= 15,840 ft 
  ½ yd = 12 in .
Metric Units of Liquid 

Volume Kiloliter{KL}  
  Milliliter(ML)
Hectoliter(HL)
Dekaliter(daL)
Deciliter(DL)
Centiliter(CL) 1 L -=1000 ML (Milliliter)








The Liter is the basic unit for metric Liquid Volume .
Liter
(L)Deciliter
(DL)Centiliter
(CL)Milliliter
{ML
  ×10 ×10 ×10
1 Liter is 3 powers of 10 greater than 1 milliliter.
〖10〗^3=10×10×10=1,000millliters
Compare Liters and Milliliters 
We can use multiplication to convert(change) liters to milliliters.
2 liters =2×1000 milliliters=2000 milliliters
3 liters=3×1000 milliliters =3000 millliters
4 liters =4 ×1000 milliliters=4000 milliliters
Because we need 1000 milliliters to have 1 liter
1 milliliter=1/1000 liter
1 milliliter = 0.001 liter
An Eyedropper I 1 ML.A Water bottle is 1 L 
1 L =1000 ML 
Pitcher A contains 5 Liters of Liquid .It contains 5,000 ml of liquid
Pitcher B contains 3 liters of liquid.It contains 3000 ml of liquid.
The bucket contains 8 liters of liquid and the bottle contains 2 liters of liquid.How many more milliliters are in the bucket than the bottle.
Think :We need to compare 8 Liters to 2 Liters. But the difference needs to be in milliliters .
We can solve by 8 Liters and 2 Liters into milliliters and using subtraction.
8×1000=8000 milliliters
2×1000=2000 milliliters 
Subtract 6000 milliliters 
Subtracting first then converting to milliliters.
8-2=6; 6×1000=6000 milliliters.
Make sure to compare like units milliliters to milliliters and grams to grams. We need to change the larger units to the smaller units first.
A box of 1000 paper clips is about 1 KG .
1 paper clip is 1 Gram )(G).
LengthCapacity Weight
1 ft =12 in1tablespoon=3 teaspoon1pound=16 oz
1 yd= 36 in1floz=2 tablespoon1 T=2,000 LB
1 yd= 3ft1 C = 8 FL OZ
1 mi = 5,280 ft1 Pt = 2 c
1 mi=1,760 yd 1 qt = 2 pt
1 gal= 4 qt


5th Grade Math {Lesson 12.4}Units Of Capacity
1 Cup = 8 Fluid Ounces 

1 Pint =2 Cups=16 Fluid Ounces 

1 Quart 4 cups = 48 Ounces 
Ms Kim needs two gallons of iced tea for a party. Isf she made 240 fluid ounces, did she make enough ? 

A.To compare 2 gallons to 240 fluid ounces we should convert (change) 2 gallons into fluid ounces. 
1 Gallon = 16 cups = 4 quarts 
1 Cup = 8 fluid ounces
1 Gallon = 16 cups ×8 fluid ounces
1 Gallon = 4 quarts or 128 ounces
  1 Pint = 2 cups =16oz
  1 Quart =2 pints = 32 ounces
  One half gallon =64ounces
  1 Pound = 16 Ounces (oz)
  1 Ton =2000 pounds 
Conversions 
64 fl oz to Pints (smaller to larger)
64 ÷8 (fl oz ) =8 cups 
8 ÷2 cups = 4 pints 
3 gallons to pints (larger to smaller) 
3 gal x 4 qt in 1 gallon = 12 quarts
12 qt x 2 pt in 1 qt =24 pints 

2 pt = 1 qt 
30 pt is greater than 12 qt 
30 pts ÷2 pts in each quart = 15 quarts 

8 gal = 32 quart 
8 gal x 4 (qt in 1 gal)=32 quarts 

1 pt = 2 cups 1 qt = 2 pt
8 pts =16 cups 
So 4 qt is less than 18 cups 

8 fl oz =1 cup
2 cups =1 pt
8fl oz x 2cups = 16 oz in 1 pt 
16 fl oz x 4 pints = 64 fl oz. 
5th Grade Math 5.6 Divide Decimals,Decimal Divisors
When we divide a decimal by another decimal,we can multiply both the divisor and the dividend by the same power of 10 to make the divisor a whole number. 

We only need to multiply a decimal divisor by a power of 10 great enough to change it to a whole number. 

Sometimes this will change the dividend to a whole number, sometimes it won’t. This depends on the amount of hops the divisor needs to be a whole number. 

We can use multiplication if we use the correct quotient. 

Quotient x Divisor=Dividend
When we multiply a decimal divisor and a decimal dividend by the same power of 10,we change their values.but the relationship between them stays the same.  
Their decimal points must hop in the same direction for the same place value.









Q.Emma paid $1.44 for bananas and $4.34 for oranges at Mr.Lee’s store. What is the total number of items she bought 
A:We can divide $1.44by $0.24 to know how many bananas she bought,then divide $4.34 by $0.62 for the oranges .Add the items for a total. 
Mr.Lee’s Store
ItemPrice each
Banana$0.24
Lemon$0.54
Orange$0.62
Pear$0.78


5th grade math(Word Problem)Solving Addition or Subtraction 
Sophia has 2 m of fabric .he will use ½ meter to make a tote bag ,1/5 to make a dog bandana and the rest of the fabric to make a sundress. How many meters of fabric will she have to make the sundress? 



Q.Mr. Lee sold 12 pounds of apples to three customers. If the second customer bought 5_2^1 pounds,and the third customer bought 4_4^3 pounds ,how many pounds did the first customer buy ?
A. We can write an equation that fits the problem Then work backwards from the total sold, and subtract the given amounts to find the unknown amount. 
X +5_(2+)^1 4_4^3 = 12 (Total pounds sold)
Pounds for 
Three customer 

Work Backwards 
12-5_(2-)^1 4_(4 =x)^3




5thgrade10.4,MultistepMeasurementsWord problems
  1 Pound = 16 Ounces (oz)
  1 Ton =2000 pounds 
1.Mrs.Kim Bakery sells raspberry muffins.A box of one dozen weighs 4.5 pounds. How many ounces does each jumbo muffin weigh ? 
Convert 4.5 pounds to ounces.
1.4.5 or 4 ½ as 4.5 = 41/2 
X16
  270 
+45
72.0 OZ so 1 Dozen muffins weigh 72 ounces 
1 dozen = 12; we divide 72÷12=6 Oz For 1 muffin ConversionLength/Capacity/Weight
LengthCapacity Weight
1 ft =12 in1tablespoon=3 teaspoon1 lb =16 oz
1 yd= 3FT=36 in1foz=2 tablespoon1T=2,000 LB
1 yd= 3ft1 C = 8 FL OZ
1mi=5,280ft=1760 yd1Pt=2 cup=16 Fl oz
1mi=1,760yd=5280 ft 1qt=2pt=4cups
1 gal= 4 qt
2.An animal can dig upto 15 feet an hour. How many yards can a mole dig in 2 hours ?
We convert 15 feet to yards. 
15 ft ÷3 (ft in 1 yd) = 5 yards per hour .
5 yards x 2 hours = 10 yards in 2 hours. 
1 Gallon = 16 cups = 4 quarts 
1 Cup = 8 fluid ounces
1 Gallon = 16 cups ×8 fluid ounces
1 Gallon = 4 quarts or 128 ounces
  1 Pint = 2 cups =16oz

A girl used 2 quarts of orange juice to make kulfis.If a single kulfi mold has a 4 fl oz capacity, how many mould did she make ?

We need to find how many fluid ounces are in 2 quarts. Then we divide the ounces by 4 fl oz. for each mold. 
1 Cup =8 Fl Oz ,so 2x8 =16 fl oz.
2 Cups =1 pint, so 1 pint is 16 fl oz.
1 qt=2 pints,so 1 qt is 2x16(fl oz)=32 fl oz.
2(Quart) x 32 (fluid ounces)=64 fl oz)
64 ÷4 =16 pounds. 

CONVERSION TABLES
GALQTPINTC
1 GAL14816
1 QT¼12
1 PT1/8½1
1 CUP1/161

1 gal = 4 qt so 1 qt=1/4 gal
1 gal = 8 pt so 1 pt =1/8 gal ;1 gal = 16 cups ; 1 cup=1/16 gal. 
5th Grade TIMESUMS(Hours/Minutes/Seconds) Conversion 

  Minutes to Seconds =8min-sec  
  1 minute =60 sec
8min = 8/1 x 60 /1min = 480 s 
5 min 60 sec 

1 min = 60 sec 
5 min = 300 sec 

min to sec 1 min =60 sec  
7.2 min = 7.2/1 x 60 sec / 1 min =432.0 sec
Hours to Seconds 
1HR =60 MIN ; 1 MIN=60 SEC

Convert 3 H TO SEC 
3h x 60 min x 60s =3600 x 3 =10,800s
1 x 1hr x 1m

We cancel he to hr and min to min 

Convert 5400 h to s 
1 min = 60 sec , 1 Hr =60 min 

5400 sX 1 min x 1 hr  
1 X 60s x 60 min = 1.5 hr 

Convert 14,400 sec to Hours 
1 min = 60sec
1 hr =60 min 

14,400s x 1 min x 1 hr 
  1 x 60s x60 min = 4 hours 
Convert 6 h 40 mins to mins 
1 hr = 60 m 1 m =60 sec ; 1 hr = 3600 sec .  

so 6 hrs = 360 minutes +40 mins 
360 mins + 40 mins = 400 minutes. 

Rule 1 Hr =3600 sec
6 hr = 1/(3600 ) x 6 =
3600 x 6 =21,600 seconds.

  21,600 + 2,400 
= 24,000 sec .

8 hr 25 mins to seconds 
1 hr = 60 mins 
1 min=60 sec 
8 hr = 60 x 8 = 480 mins +25 mins 
  8 hr = 505 mins 
 1 min =60 sec
 505 mins =505 x 60 = 30,300 seconds. 
8 hr = 505 mins = 30,300 seconds.
Rule 1 hr 3600 sec
(30,300)/(3600 ) s=8.416=8 hrs and .4166667 x 60 =25 mins Time Sums Word Problemn 

Units Of Time
60 seconds = 1 minute
60 minutes = 1 hour
24 hours =1 day
7 days = 1 week
12 months = 1 year
365 days= 1 year
Q.Bob drinks 1 cup of coffee each day. How many pints of coffee does he drink in 4 weeks ?
There are 7 days in a week, so he drinks 7 cups each week. 
7 cups x 4 weeks =28 cups. 
28 cups ÷2 cups per pt. =14 pints of coffee in 4 weeks. 
Q: The battery for James game lasted 320 minutes before it ran out. How many hours did the battery last.
We need to convert 320 minutes into hours and minutes. 
1 hr =60 min 
320 ÷ 60 = 5 hrs and 20 minutes.Battery lasted 5 hrs and 20 mins.

  Q.Pia trained her student kathak in 3 days , 2 days to teach footwork , 8 days to teach pirouettes and 12 days to teach kathak taal .
A.First we total the days 3 days +2days+8days+12 days=25 days
7 days =1 week 
We convert the days into weeks and days 
25 ÷7 = 3 days and 4 weeks . 
Elapsed Time 
We can use a number line to find the elapsed time. A movie started at 1:30pm and was 2 hours 10 minutes long .movie end ?

1:30pm 1:40pm 2:40pm 3:40pm
1:30-1:40 = 10 minutes 
  1:40 pm -3:40 pm = 2 hours 
  A. Count the 10 minutes then 2 hours. The movie ended at 3:40pm .

AM stands for Latin word before Midday antemeridian. PM stands for Post Meridian after midday. 
Q.StartTime 2:25p.m. 
  Elapsed Time 1 hr 30 mins
The minute hand moves 30 minutes .
  2:25
+ :30
  2:55p.m. Now add 1 hour 2:55p.m. 
  + 1 hr
  3:55p.m.  
Q.Start Time:
  Elapsed time :14 hr 20 mins
  End time :10:15pm
We can work backwards on a number line .Count back minutes then hours. 

7:55 8:55 9:55
  Am am 9:55pm 10:00pm
Am  
7:55-8:55am=1 hr 
8:55-9:55 am=1 hr m
9:55pm-10:15pm =20 minutes  
Q.Aakash saw 2 videos on kathak .One video was 2 hours and 5 minutes long ,the other was 1 hour 45 minutes long .There was a 15 minute wait between the two videos. If the second movie ended at 1:00am what time did the first video start ?
2 hours5 minutesFirst video
1 hour 45 minutesSecond video
15 minutes
3 hours 65 minutes
160 minutes Regroup 
  65-60 =5 minutes. We take 60 away from minutes and give it to hours ,that is 4 hours(3+1=4 hours) . 3 +1 = 4 hours 5 minutes. 
Q. X started for his office at 7:45 am and reached after 2 hrs 20 minutes. What time did he reach office 

Reaching time is the starting time+time taken 
Starting Time is 7:45am time taken is 2 hrs 20 minutes
Reaching time is 7:45am+2 hours 20 minutes.
Q.A primary school starts at 8:40am and closes at 12.40p.m.Find the total duration for which the school remains open.Total Duration is closing time – starting time.Closing time is 12.40pm
Starting time is 8:40pm
Total duration is 12:30pm-8.40am .We transferred 60 mins to the mins column so our answer is 3 hours50 m

5th Grade Math 6.1.Model Fraction Addition with unlike denominators
We can use models to add fractions that have different denominators. 
  2/3+1/6=5/6
We use fraction strips to represent the equation .
We place fraction strips that have the same denominators as each other,below the model of ur equation for the sum.
5/10÷5/5=1/2
Factors for 5 :1,5
Factors for 10 :1,2,5,10
We need to write our sum in simplest form. 
We divide the numerator and denominator by a common factor.A fraction is in simplest form when represented by the fewest amount of possible models. 

Sometimes the fraction strips for our sum will have the same denominator as one of our addends. 
Sometimes the fraction strips for our sum will have different denominators. Than the addends. 
As we progress into Algebra we will be solving equations that use fraction bars .
We need to write our fraction bars in horizontal situation. 4x+2 =2x+1
2
  Writing fraction bars on a slant makes it more difficult to add subtract multiply or divide fractions.  

1⁄4+2⁄(4=3⁄4)
Factors of 3: 1,3
Factors of 6:1,2,3,6
Factors of 12: 1,2,3,4,6,12
Multiples of 3 :3,6,9,12,15,18,21,24
These fraction strips fit below the fraction strips for the addends ,because the denominators 3,6,12 have the common factor 3,and 6 and 12 are multiples. Of 3. 
Mrs Kim is making 2 apple pies.Each pie requires ¼ teaspoon of nutmeg and ½ teaspoon of cinnamon. What is the total number of teaspoons of nutmeg and cinnamon needed for 2 pies ? 
¼ +1/2+ ¼+1/2.We can use three ¼ fraction strips to get an equivalent denominator for 1 pie.


Q.Mrs Kim needs ¼ teaspoon of nutmeg to make a loaf of banana bread,1/2 teaspoon for a batch of oatmeal cookies and 1/8 teaspoon for a custard pie.If Mrs Kim makes one of each, how many teaspoons will she need in all?
A.1/4+1/2+1/8.We need a common denominator before we add fractions. 
1/4x2/2=2/8.4x some number equals 8, so we multiply the numerator by the same number. 
1/2 x 4/4=4⁄8

2 times some number is 8 ,So the numerator is multiplied by the same number. 

5Th Grade Math 7.5,Comapre Fractions,Factors and Products
We can compare the size of the product,to the size to the size of one factor when multiplying fractions. 
The product of 1 and any fraction will be equal to that fraction.
1x2/3=2/3or2/3=2/3.The Identity Property of Multiplication,lets the multiplier keep its Identity. 
The product of a number less than 1 and a fraction will be less than either factor. 
3/4 x (2=6)/(3=12)=1/2<2/(3 ) and 1/2<3/4
The product of a number greater than 1 and a fraction, will be greater than the fraction and less than the number greater than 1. 
1_2^1x2/3=5/4 x 2/3=10/12=5/6>2/3 5/6<1_4^1

Less than 1 multiplied xFractionProduct less than either factor
Greater than 1 xFraction Greater than the fraction ,Less than the first factor
1 xFraction The Fraction

Multiplication can be thought of resizing one number by another number.
3 x 4 will result in a product that is 3 times as great as 4. The value increases.
When we multiply a fraction by a fraction,the product will be less than either factor. 

1/4 x 1/2 means 1/ 4 as great as ½.The value decreases. 

We can use a diagram such as number line to show the relationship between the products when a fraction is multiplied or scaled(resized)by a number. 

2/3 on a number line. 1 from 0 to 1 is split into thirds.If we are at 2/3 and half of that would be half way between the 0 and the 2/3 , it would be at 1/3.
When we have1 x 2/3= 2/3 .When we have 4 times 2/3rds we skip 2/3 four times that puts us at 2_3^2.

Q. Dave bought 20 cupcakes on Monday. On Tuesday he ate some and had ¾ of the cupcakes he had on Monday. 
By Friday ,he had 2/3 of the amount he had on Tuesday. 
How many cupcakes did Bob have on Friday ?
Think : He started with 20. Tuesday was ¾ of 20 or
 20 x ¾. Then we multiply that product by 2/3.
20x 3/( 4 )= 20x3/4=60/4=15
15 x 2/3=15x2/3=30/3=10 cupcakes left on Friday. 
5th Grade math-7.8Compare mixed numbers,factors products.
Fractions greater than 1 are called improper fractions.Their numerators are greater than their denominators.
We must rename a mixed number as a fraction greater than 1.2_5^2=2x5+2 =12
  5 5
When we multiply a fraction greater than 1,by a fraction less than 1,the product will be less than the factor that is less than 1.
1/2 x1_2^1=3/4. 3/4<11/2 and 3/4>1/2.

Find the partial products then add.
We can also use a diagram such as a number line to show the relationships between the products when a fraction greater than 1 is multiplied or scaled(resized by a number).


Q.Tom, bought apples,pears and oranges. She bought an amount of apples that was 11/2 times the amount of oranges. She bought an amount of pears that were 1/3 the amount of oranges.She bought 12 oranges.
Which did she buy more of ? 
Which did she buy less of ? 
Think :she bought 12 oranges.
Apples=11_2^1 x 12
She bought more apples.
(1x12)+(1/2 x 12)
12+6=18 apples
1_2^(1of 12=18)

Pears=1/3 x 12
She bought less pears. 
1/3x12=4 pears .

5th- MathBenchmarFractions
When we multiply a fraction by a Whole number, the product will be less than the whole number because the fraction is less than 1 .The product of 6 and 1/3 will be less than the whole number, because the fraction is less than 1 whole.6×1/3=2<6.

We would need 18 ÷3=6 .We multiply the numerator by the whole number, then divide that product.

The strategy work backward can help us solve a problem with fractions that involves addition and subtraction. Use an inverse operation .
We can rename mixed numbers to subtract 
4_3^1 =3 +3/3 +1/3 =3_3^4  
or 4_3^1= 13/3
Q.Maryann has a 2 meters of a cloth. She will use ½ meter to make a tote bag, 1/5 meter to make a dog bandana ,and the rest of the cloth to make a sundress. How many meters of cloth will she have to make the sundress ?
1/2 +1/5 +m = 2 meters cloth in all .
1/(2 ) + 1/5 + m =2 We use a variable , m to take the place of the unknown amount. 
2 - 1/(2 ) - 1/5 Work backward to find the value of m .
10/(10 ) +10/(10 ) = 2 .
20/(10 )- 5/(10 ) - 2/10 m .
Give them a common denominator and rename the whole number r as a fraction greater than 1 .
15/(10 ) - 2/(10 ) = 13/10= 10/10 +3/10 =1_(10 )^3
1_10^3= m. 
5TH Grade-Subtraction of Mixed Numbers  
We can rename a mixed number to help us subtract. When the fraction in the subtrahend is greater than the fraction in the minuend, we can rename the mixed number minuend as a lesser whole number and a fraction greater than 1. We can also rename both mixed numbers as fractions greater than 1, then subtract. Be careful when subtracting the numerators, we subtract the second numerator from the first numerator as the subtrahend subtracted from the minuend. The denominators stay the same. We show how to use a number line or bar models to subtract mixed numbers with like denominators. We solve four word problems involving renaming mixed numbers as lesser whole numbers and fractions greater than one.

4 1/3=3+3/3+1/3= 3 4/3 {Minuend}
  -2 2/3 =1 2/3

The like denominator is 3 so 3/3= 1 whole 
Rule1 :We rename the minuend 4 1/3
 as a lesser whole number and a fraction greater than 1
Rule 2: We can rename both fractions .
 use the properties of addition to help us add fractions if one of the addends that has a different denominator .
  23/4+3 1/4+1 5/8=2+3+3/4+1/4+1 5/8
=5+4/4+1 5/8= 6+1 5/8
=7 5/8 {Add the compatible Fractions first } 
A
 fraction is in its simplest form when we can write it with the fewest equal parts of a whole as possible .If we can’t write it with fewer equal parts ,then the fraction cannot be simplified. All unit Fractions with 1 as the numerator are in their simplest forms. 2/8÷2/2=1/4.So 2/8 is in its simplest form ¼. 2/4 is in a simpler form because it has fewer parts than 4/8. ,but its not in simplest form because the numerator and denominator still have a 2 as a common factor. Dividing by the Greatest common factor in the beginning we find the simplest form faster and divide once instead of twice. 
Dividing Twice 4/8÷2/2=2/4 and 2/4÷2/2=1/2.
Dividing Once 4/8÷4/4=1/2 

Fractions with Common Denominators We can write a pair of fractions with common denominators by finding multiples of the denominators and using a common multiple as their common denominators. 
1/2 and 1/3
Multiples of 2 : 2,4,6,8…..
Multiples of 3 :3,6,9,12…
The 6 is common and can be used as a common denominator.
1/2×3/3=3/6
{Think what we can multiply the 2 by to get a 6}.Then multiply the numerator by the same number .
1/3×2/2=2/6
We do the same for 1/3 .We create a fraction pair with the same denominator. They will have the same number as their denominator.
We can use greater common multiples to find the common denominators ,but using the least common multiple (LCM) makes math easier.
2/3 and ¼
Multiples of 3 :3,6,9,12,15,18,21,24
Multiples of 4 :4,8,12,16,20,26,28,32
2/3×4/4=8/12 (We take the common denominator 12
2/3×8/8=16/24
Robbie has two same size pies cut into the same number of equal slices. 1/3 of the apple pie was eaten .2/5 of the cherry pie was eaten What is the least number of slices into which both pies could be sliced ? 
1/3 and 2/ 5For 1/3 multiples of 3 :3,6,9,12,15,18
  Multiples of 5:5,10,15,20,25,30
For 1/3 = 1/3×5/5=5/15
For 2/5 = 2/5×3/3=6/15
For 1/3 we think what number multiplied by 3 in the denominator will give a product of 15 .
Rule :To write a factor in its simplest form we need a common factor, To find common denominators we find common multiples. 
5th Grade Math Number SequencesRepeating Patterns  
14-1A Repeating pattern is made up of shapes or number that form a part of the repeats.  
Q1. In the Triangle Square and Trapezoid “what will be the 48th shape? What will be the 50th shape ?

〖48÷3=16〗^.Since there is no remainder the shape repeats 16 times. The 48th shape is the last shape when reach the 16th group.Thus the 50th shape is a square which is shown as 5th above. 
5th Grade 15-1Angles,AngleMeasures
A point is an exact location in space.
A Line is a straight path of points that goes on and on in opposite directions.
 A Line segments is a part of a line with two end point.

…
A Ray is a part of a line that has one endpoint and continues on forever in one direction .

<Angle ABC is a right Angle. A Right Angle forms a square corner .

<Angle DEF is an Acute Angle. An acute angle is open less than a right angle.

<Angle GH1 is an obtuse angle. An obtuse angle is open more than a right angle but less than a straight angle.

<JKL is a straight angle. A straight angle forms a straight line. 

Transversal Line is a line that intersects two lines in the same place at two different points. When a transversal runs through parallel lines they form 8 angles. We can see angles that are in the interior and exterior of the parallel lines. Or Angles that 


Corresponding Angles lie on the same side of the Transversal on the same sides of the parallel lines. Alternate Angles or Alternate Interior Angles that lie 
on the opposite sides of the transversal in the Interior. 

5thGrade Math 16.5Line Symmetry 
A figure is line symmetric if it can be folded on a line forming two
matching parts that fit exactly on top of each other.
Symmetry means having an exact match in size,shape and arrangement of parts on opposite sides of a line .Some 


Shapes have one line of symmetry ,some shapes have more than one. 

A regular polygon is a polygon in which all sides are the same length ,and all angles are the same measure.
An Irregular Polygon, have sides that are different lengths, and angles that are different from each other. 
For regular polygons the number of sides it has is the number of lines of symmetry it has . 
We can check for a shape having a vertical or horizontal line symmetry. If it has neither of these the shape has no line symmetry 
We can turn a shape to look for line symmetry .  

5TH Grade Math 11.2 Classify Triangles

We can classify triangles by the length of their sides,or by the measure of their angles.
A regular polygon has congruent sides and congruent angles.
Classify by side lengths:
A equilateral triangle has 3 congruent sides,so it is a regular polygon.
An Isosceles Triangle,has 2 congruent sides so it is not a regular polygon.
Classify by Angle Measure
An Acute Triangle has three angles that are each less than 90 degrees. An Obtuse triangle has one angle that measures greater than 90 degrees.It is not a regular polygon.

After we determine the number of congruent sides we measure each angle.

We found it was isosceles.It has a 90 degrees angle, so it’s also a right angle.


We can classify triangles by given side measures and angle measures.

Sides=22cm,14cm,12cm
Angles:110^(°,) 40^°,30^°

The sides are all different lengths and it has an angle measure greater than 90^°.
Its an obtuse scalene triangle.

Sides:6ft,3ft,6ft
Angles:75^°.30^°,75^°.
Two sides have the same length and two angles have the same measure.All angles are less than 90^°
5th Grade Math 13.1-13.5 AREA PERIMETER
We can find an unknown measure of a rectangle if we are given its area or perimeter, and the measure of one side. We can use the area or perimeter formula, and one side measure, to find the unknown side measure by writing an equation. 
  The area and perimeter formulas both use multiplication. We can use division to help us find an unknown side length. Division is the inverse, it's the opposite, of multiplication. Be careful to use the correct formula. Area is the number of square units we need to cover the surface of a figure without gaps or overlaps. Perimeter is the distance around the outside edges of a figure. We find the area and perimeter of several rectangles. We use higher order thinking skills to find the area of a square when we're given the perimeter. 

We also find the perimeter for a square when we're given an

A=B×H ; A = 40SQ CM ,B=10CM
So, 40 =10×h
W = 4 cm
Area is the number of square units we need to cover the surface of a figure without gaps or laps. 
A=B×H=B.H and is sq units like in m cm etc.

Perimeter is the distance around the outside edges of a figure 
P=(2.L)×(2.W)
Higher Thinking Skills 
1.Sofia made a square baby quilt that had a Perimeter of 12 Feet.
All the 4 sides have the same length, so we can use

 Perimeter to find the side Lengths.
P=4 ×S
12=4 ×S
S=12÷4=3 ft , so S=3Ft.

A Rectangle has 2 Pairs Of Sides of equal length.We can use the Perimeter formulae to find the unknown side length.
P=28 CM;L=10CM 
P=2×L+2×W
P= (2×L)+ (2×W)
5th grade math Quadrilaterals16.3
 A Parallelogram has 2 pairs of parallel sides and 2 pairs of sides of equal length. It is not a regular polygon .It does’nt have 4 equal sides and 4 equal angles. It has 0 lines of symmetry .










It cannot be divided in half by any lines then folded to make mirror images of the sides. 

Quadrilaterals can be classified by the angles or line segments that make their sides. Which of these quadrilaterals have only one pair of parallel sides ?
A Quadrilateral is a polygon with four sides .

Trapezoids have 1 pair of parallel lines or only one pair of parallel sides. It has 4 sides. 

Parallelograms rectangles squares and rhombus all have two pair of parallel sides.

Its sides aren’t the same.
Its angles aren’t the same. Its not a regular polygon.
 It has 1 vertical line of symmetry running from centre. 
Rectangles has 4 right angles.It is also a parallelogram .A Square has 4 right ANGLES AND ALL SIDES ARE THE SAME LENGTH .It is a parallelogram a rectangle and a Rhombus.  
A Rhombus has 2 pairs of parallel sides of equal length .But the angles are not all equal. There are two acute angles and 2 obtuse angles.A rhombus is not a regular polygon. 2 diagonal lines of symmetry . 
5th grade Math 16-1Lines 
Intersecting lines make 4 Angles.
Intersecting lines are those that cross at one point. Line AB and CD Intersect at Point X. 
Parallel Lines never Intersect.They are lines in a plane that are the same distance apart.They never cross or meet .
Perpendicular lines intersect to make right angles.Line PQ is perpendicular to line RS.






When two lines are perpendicular to a third line the two lines are parallel to each other. To be perpendicular,to a third line, the two lines must form right angles with the third line.Two rays can be parallel because a ray is a part of a line .If a Figure has a square corner symbol in one of its angle,we know that the angle is a right angle,and the lines that make the angles are perpendicular.If a figure does not have 







a square corner symbol, we don’t know if it makes really has right angles. 
An Equilateral Triangle has 3 sides and 3 lines of symmetry .Its a regular polygon
An Isosceles Triangle has only 1 Line Of Symmetry .
Isosceles triangle 2 sides are longer,1 is short it has 1 line of symmetry.
A Right Isosceles has 1 line of Symmetry 
A Right Scalene has 0 lines of symmetry
A Scalene has 0 lines of symmetry.
An Isosceles Triangle, Right Isosceles Scalene and Right Triangle are Irregular Polygons.  
5th grade math Volume Formulaes 
A formulae is a set of symbols used to state a mathematical rule. 
A= B X H {Area is the base x height } 
Use a formula to find the volume of a Rectangular Prism
V = l x w x h .Volume is written using cubic units. 
 multiply the length by the width 6 x 3 =18
Then multiply the product by the height .
18 x 5=90 〖cm〗^3
VOLUME= Base x Height
  (7 x 4 )x 2 =28 x 2 
56〖in〗^3.
Q.The volume is 30 inch by 30 in.Its height is 3 in What is the volume in the concrete of the slab ? 
V = l x w x h 
  = 30 x 30 x 3 
  = 2700 cubic inches of concrete .


1= 1 x 1x1
Volume=Base × Height = BH
If the base is 7 and height is 4 
  =(7 × 4) =28 x 2
  = 28 x 2 
  = 56 〖in〗^3.
Use the formulae for volume to find an unknown length.
Volume = length x width x height 
200= 10 x 4 
200 = 40 
 40 times a number equals 200 or 200 divided by 40 is the unknown number.
  Volume = 200 cu inches.

Sarah keeps her ponytail holders and headbands in a box. The length is twice the width the width is twice the height and the height is 3 inches .What is the volume of the box. 
Height= 3 inches Width = 2x3 =6 inches Length=2 ×6=12 inches.Volume = l x w x h =12 x 6 x 3  
=216 cubic inches
  = 216〖in〗^3.
Summary :Volume is the number of cubic units we need to fill a three dimensional space.We use length width and height as factors to find volume. 

We use a strategy make a table to compare different rectangular prisms that have the same volume.
We make a table to find all combinations of three factors whose product equals as the given volume and have different -sized bases.We begin by listing all factors for the given volume. 
5th Grade Math,11.6-11.7 EstimateVolume,Rectangular Prisms
Volume is the measure of space a solid figure occupies and is measured in cubic units.
We can use everyday objects to estimate the volume of a rectangular prism .
We can use unit cubes to find the volume of a rectangular prism. 
We find the number of unit cubes it takes to fill the base without any gaps or overlaps then multiply that number by the number of layers that make up its height. 
We can use a small rectangular prism that we know the volume as a tool to estimate the volume of a larger rectangular prism .
We can find about how many of the smaller prisms fit in the base and about how many make up the height.
We can multiply these numbers and the volume of the smaller rectangular prism to estimate the volume of the larger rectangular prism.
About 6 boxes of tissues fit in a card board box. Each box of tissues has a volume of 150 cubic inches.
If about 6 boxes of tissues fit into the card board box ,the estimated volume of the card board box can be found by multiplying 6 boxes by
150 cu in.
150 x 6 boxes =900cuin. So the volume is about 900 cu in.
Miss Cho has a box that can hold 2 dictionaries in each layer.
If the box can hold 3 layers of dictionaries, what is the estimated volume of the box ?
We know the volume of one dictionary .We multiply it by 2 for the volume of the first layer. Then we multiply that by 3 layers.
160cu in x 2 dictionaries=320 cu in.

320cu in for 1 layer x 3 layers=960 cu in. 
The volume of the box is about 960 cu.in. 
Q.If the volume of one box of cheerios is about 2,383 cubic inches what is the estimated volume of 4 boxes.
We know the volume of one box.We can multiply that volume by 4 to find the volume of all 4 boxes.
2,383x4=9,532.4 boxes of cheerios is about 9532 cu in. 
We can find volume of a rectangular prism with multiplication

3x2=6 cu units.
Volume is measured using cubic units such as 
Cu centimeters; cu inches ,cubic feet ; cu yards. 



When unit cubes are stacked,we may not see all of the cubes from our viewpoint. 
There must be a cube in the back left corner supporting the cube on top.
Each layer of a rectangular prism will have the same amount of cubes. 
To find the volume of three-dimensional figures we measure in three directions. 
For a Rectangular prism we measure its length,width and height. 



Length x width x height=volume


If each unit was centimeters the volume would be 18 cu cm.
If each unit was inches the volume would be 18cuin.
18cucm<18cuinbecause cm <in.They would have the same shape but the prism in inches would be larger. It would have greater volume.
Each cube =1 cu in
Q .Find the volume 4in x 3in x 4in=48 cu in

Q.Dave built a rectangular prism with cubes. The base of her prism has 4 inch cubes. If the prism was built with 12 inch cubes ,what is the height of her prism ?
Think : The base layer has 4 cubes.How many layers of 4 will total 12 cubes ?
We can use division to solve .
12cubes ÷4cubes per layer =3layers.The height is 3 inches.


Q.Dave and Bob estimated the volume of their boxes.Using a dictionary with a volume of 160 cubic inches,Dave can fit 3 dictionaries on the first layer and can hold 8 layers.Bob’s box can hold 6 dictionaries on the first layer and can hold 4 layers.
Dave says both boxes have the same volume.Is he correct?
A:We multiply the number of dictionaries on the first layer by the number of layers to get the volume in “dictionary units”
Daves:3x8 layers =24 dictionaries
Bob: 6x4 layers =24 dictionaries
Estimate for boxes = 160 x 24 =3,840 cu in estimate for each box.