Class Limits | How to find the Class Limits from Data? | Difference between Class Limits and Class Boundaries

Class Limits

Searching for help regarding the concept of Class Limits in Statistics? If so, you have come the right way and you will get a complete idea of the entire concept by going through this article. Check class limits definition, types, the procedure to find the class limits from data in the forthcoming modules. Get to know about the lower and upper-class limits along with the steps to solve the class limits problems. Refer to the step-by-step procedure for solving questions related to class limits. Refer to all the definitions involved in it.

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Class Limits – Definition

To find the class limits there are numerous ways and methods to find the exact solution. There are two concepts involved in class limits. They are namely

1. Lower Class Limit

2. Upper-Class Limit

The class limits and the data values have the same accuracy rate and also have the same data values as the same number of decimal places. In this concept, the first-class interval extreme upper value and next class interval lower extreme value will not be equal. Class limits are considered as the maximum and minimum value of the class interval. Lower Class Limit is the minimum value of the interval and Upper-Class Limit is the maximum value of the interval.

How to find Class Limits?

While dealing with Class Limits in Statistics you will have two scenarios one is for overlapping groups and nonoverlapping groups. Refer to the following sections to get a complete idea of it.

1. Let the class intervals for some grouped data 10 – 15, 15 – 20, 20 – 25, 25 – 30, etc. In this case, the class intervals are overlapping and the distribution is continuous. 10, 15 are called the limits for the interval 10-15. However, 10 is the lower limit and 15 is the upper limit of the class.

In the same way, 15 and 20 are the lower and upper-class limits for the respective class interval 15-20. It is clear that the upper-class interval is the same as the lower limit for the next class interval in the case of overlapping groups.

2. Now, let us consider class intervals of grouped data to be 1-4, 5-8, 9 – 12, etc. in this the class intervals are non-overlapping and the distribution is discontinuous. 1, 4 are known as the class limits for the class interval 1-4 in which 1 is the lower class limit and 4 is the upper-class limit.

Similarly, 5 is the lower limit and 9 is the upper limit for the next interval 5-9. In this we can clearly observe that the upper limit of the class interval is not the same as a lower limit in the next class interval for nonoverlapping groups.

Class Mark or Mid Value or Midpoint

With respect to the class interval, it is defined as the average of two class boundaries or class limits. In other words, we can define it as the total or arithmetic mean of both class boundaries and class limits. Therefore, we have

Midpoint = LCL + UCL/2

= LCL + UCB/2

Class Boundaries

The class interval’s actual class limit is called the class boundary. For the classification of overlapping or classification of mutually exclusive which excludes some of the upper-class limits like 30-40,20-30,10-20 etc i.e., where the class limits and class boundaries coincide. Class boundaries are generally done for continuous variables. These are applicable for discrete variables that were mutually inclusive and non-overlapping classification which has the class limits like 20-29, 10-19, 0-9, etc.

LCB = LCL – (D/2)

UCB = UCL + (D/2)

where D is defined as the difference between the lower class limit of the next class interval and the upper-class limit of the given class interval.

Frequency Distribution

The frequency distribution divides the data and shows the number of data values that are present in each class.

Class Width

To find the class width, greatest data value – lowest data value /desired number of classes. If the value is in the decimal value, then round that value to the nearest convenient number.

Data Range for each class

LCL (Lower Class Limit) is the lowest data value that fits in the class

UCL (Upper-Class Limit) is the upper data that fits in the class

Frequency in the class

The number of values that fall in class is the frequency of the particular class.

Class Limits Example Problems with Solutions

Problem 1:

Data: 110, 122, 133 etc

ClassFrequencyClass LimitsClass BoundariesClass MarkClass size
80 – 99280, 9979.5, 99.589.520
100 – 1195100, 11999.5, 119.5109.520
120 – 13912120, 139119.5, 138.5129.520
140 – 1596140, 159138.5, 158.5149.520
25

Problem 2:

Data: 20.6, 11.7, 12.8 etc

ClassFrequencyClass LimitsClass BoundariesClass MarkClass size
19.6 – 24.51019.6, 24.519.55, 24.5522.055
24.6 – 34.52024.6, 34.524.55, 34.5539.5510
34.6 – 54.53034.6, 54.534.55, 54.5549.5520
44.6 – 64.52544.6, 64.544.55, 64.5559.5510
85

Math Tables 11 to 20 | Learn Multiplication Tables from 11 to 20 | Tables from Eleven to Twenty

Math Tables 11 to 20

Multiplication Tables from 11 to 20 help you learn the patterns and multiplication facts effortlessly. Math Tables for 11 to 20 can be quite essential for solving math problems on a quick basis. Tables of 11 to 20 are quite important for enhancing your math skills and arithmetic skills together. We have compiled the Multiplication Times Table from Eleven to Twenty both in the image and tabular format for free of cost.

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Tables from 11 to 20

For the sake of your comfort, we have attached the Multiplication Tables from 11 to 20 in Tabular Format. Use them as quick references and solve the math problems much efficiently and quickly. Learning the Multiplication Tables from Eleven to Twenty boosts your confidence and builds problem-solving skills in you. Make the most out of them and learn the solve the problems involving multiplication, division much simply.

Table of 11 to 15

Table of 11Table of 12Table of 13Table of 14Table of 15
11 ×‌‌ 1 = 1112 ×‌ 1 = 1213 ×‌ 1 = 1314 ×‌ 1 = 1415 ×‌ 1 = 15
11 ×‌‌ 2 = 2212 ×‌ 2 = 2413 ×‌ 2 = 2614 ×‌ 2 = 2815 ×‌ 2 = 30
11 ×‌‌ 3 = 3312 ×‌ 3 = 3613 ×‌ 3 = 3914 ×‌ 3 = 4215 ×‌ 3 = 45
11 ×‌ 4 = 4412 ×‌ 4 = 4813 ×‌ 4 = 5214 ×‌ 4 = 5615 ×‌ 4 = 60
11 ×‌ 5 = 5512 ×‌ 5 = 6013 ×‌ 5 = 6514 ×‌ 5 = 7015 ×‌ 5 = 75
11 ×‌ 6 = 6612 ×‌ 6 = 7213 ×‌ 6 = 7814 ×‌ 6 = 8415 ×‌ 6 = 90
11 ×‌ 7 = 7712 ×‌ 7 = 8413 ×‌ 7 = 9114 ×‌ 7 = 9815 ×‌ 7 = 105
11 ×‌ 8 = 8812 ×‌ 8 = 9613 ×‌ 8 = 10414 ×‌ 8 = 11215 ×‌ 8 = 120
11 ×‌ 9 = 9912 ×‌ 9 = 10813 ×‌ 9 = 11714 ×‌ 9 = 12615 ×‌ 9 = 135
11 ×‌ 10 = 11012 ×‌ 10 = 12013 ×‌ 10 = 13014 ×‌ 10 = 14015 ×‌ 10 = 150

Table of 16 to 20

Table of 16Table of 17Table of 18Table of 19Table of 20
16 ×‌ 1 = 1617 ×‌ 1 = 1718 ×‌ 1 = 1819 ×‌ 1 = 1920 ×‌ 1 = 20
16 ×‌ 2 = 3217 ×‌ 2 = 3418 ×‌ 2 = 3619 ×‌ 2 = 3820 ×‌ 2 = 40
16 ×‌ 3 = 4817 ×‌ 3 = 5118 ×‌ 3 = 5419 ×‌ 3 = 5720 ×‌ 3 = 60
16 ×‌ 4 = 6417 ×‌ 4 = 6818 ×‌ 4 = 7219 ×‌ 4 = 7620 ×‌ 4 = 80
16 ×‌ 5 = 8017 ×‌ 5 = 8518 ×‌ 5 = 9019 ×‌ 5 = 9520 ×‌ 5 = 100
16 ×‌ 6 = 9617 ×‌ 6 = 10218 ×‌ 6 = 10819 ×‌ 6 = 11420 ×‌ 6 = 120
16 ×‌ 7 = 11217 ×‌ 7 = 11918 ×‌ 7 = 12619 ×‌ 7 = 13320 ×‌ 7 = 140
16 ×‌ 8 = 12817 ×‌ 8 = 13618 ×‌ 8 = 14419 ×‌ 8 = 15220 ×‌ 8 = 160
16 ×‌ 9 = 14417 ×‌ 9 = 15318 ×‌ 9 = 16219 ×‌ 9 = 17120 ×‌ 9 = 180
16 ×‌ 10 = 16017 ×‌ 10 = 17018 ×‌ 10 = 18019 ×‌ 10 = 19020 ×‌ 10 = 200

Multiplication Table for 11 to 20

Below is the multiplication times table chart for tables 11 to 20. Use it as a reference to learn from 11 Times Table to 20 Times Table easily. It is as such

Multiplication Table Chart for 11 to 20

Tables 11 to 20 in PDF’s

Please find the below-attached Tables for 11 to 20 through the quick links provided below. Simply tap on the links and learn the entire table in no time.

Multiplication Tables for Eleven to Twenty
11 Times Table12 Times Table
13 Times Table14 Times Table
15 Times Table16 Times Table
17 Times Table18 Times Table
19 Times Table20 Times Table

FAQs on Tables from Eleven to Twenty

1. How to Memorize Tables from 11 to 20 Easily?

You can memorize the Tables from 11 to 20 easily by following the below-listed guidelines

  • Write down the Math Tables on a piece of paper.
  • Learn them orally and speak out in a loud voice.
  • Solve Problems on Multiplication as much as possible.

2. What is the Table of 13?

Table of 13 is written as follows: 13 x 1 = 13, 13 x 2 = 26, 13 x 3 =39, 13 x 4 =52, 13 x 5 =65, 13 x 6= 78, 13 x7 =91, 13 x 8 =104, 13 x9 = 117, 13 x10 = 130.

3. How to learn Multiplication Table of 15 orally?

You can learn the Multiplication Table of 15 by reading as such

Fifteen ones are 15, Fifteen twos are 30, Fifteen threes are 45, Fifteen four’s are 60, Fifteen fives are 75, Fifteen sixes are 90, Fifteen seven’s are 105, Fifteen eights are 120, Fifteen nines are 135 and Fifteen ten’s are 150.

Math Tables 1 to 12 | Printable Multiplication Chart 1 to 12 | Maths Multiplication Tables 1 to 12 PDF Download

Math Tables 1 to 12

Students are advised to go through the Multiplication Tables from 1 to 12 for faster math calculations. These Math Tables are the basic ones and help you to do mental math calculations efficiently and quickly. We have provided Multiplication Tables for 1 to 12 both in image and PDF Format. We don’t charge any amount and you can download them for free and start practicing them. Learning these Multiplication Times Tables helps you to increase your speed of solving problems.

Do Check: Math Tables 1 to 20

Multiplication Table Chart for One to Twelve

Maths Times Tables 1 to 12 help you to learn and practice the multiplication facts easily. Multiplication Tables for 1 to 12 can be of extreme help in performing your math calculations. For your convenience, we have added the Math Tables for 1 to 12 in image format which you can download for free of cost and prepare every now and then. Students who learn these Math Tables can solve complex problems too easily.

Times Tables 1 to 12

Math Tables from 1 to 12

Here is the list of Tables from 1 to 12 in tabular format. Primary School Students are advised to go through these Multiplication Charts for One to Twelve to solve mathematical problems involving multiplication and division much easily. These Math Tables are the foundation blocks for many arithmetic calculations.

Table of 1 to 6

Table of 1Table of 2Table of 3Table of 4Table of 5Table of 6
1 × 1 = 12 × 1 = 23 × 1 = 34 × 1 = 45 × 1 = 56 x 1 = 6
1 × 2 = 22 × 2 = 43 × 2 = 64 × 2 = 85 × 2 = 106 x 2 = 12
1 × 3 = 32 × 3 = 63 × 3 = 94 × 3 = 125 × 3 = 156 x 3 = 18
1 × 4 = 42 × 4 = 83 × 4 = 124 × 4 = 165 × 4 = 206 x 4 = 24
1 × 5 = 52 × 5 = 103 × 5 = 154 × 5 = 205 × 5 = 256 x 5 = 30
1 × 6 =62 × 6 = 123 × 6 = 184 × 6 = 245 × 6 = 306 x 6 = 36
1 × 7 = 72 × 7 = 143 × 7 = 214 × 7 = 285 × 7 = 356 x 7 = 42
1 × 8 = 82 × 8 = 163 × 8 = 244 × 8 = 325 × 8 = 406 x 8 = 48
1 × 9 = 92 × 9 = 183 × 9 = 274 × 9 = 365 × 9 = 456 x 9 = 54
1 × 10 = 102 × 10 = 203 × 10 = 304 × 10 = 405 × 10 = 506 x 10 = 60

Table of 7 to 12

Table of 7Table of 8Table of 9Table of 10Table of 11Table of 12
7 × 1 = 78 × 1 = 89 × 1 = 910 × 1 = 1011 x 1 = 1112 x 1 = 12
7 × 2 = 148 × 2 = 169 × 2 = 1810 × 2 = 2011 x 2 = 2212 x 2 = 24
7 × 3 = 218 × 3 = 249 × 3 = 2710 × 3 = 3011 x 3 = 3312 x 3 = 36
7 × 4 = 288 × 4 = 329 × 4 = 3610 × 4 = 4011 x 4 = 4412 x 4 = 48
7 × 5 = 358 × 5 = 409 × 5 = 4510 × 5 = 5011 x 5 = 5512 x 5 = 60
7 × 6 = 428 × 6 = 489 × 6 = 5410 × 6 = 6011 x 6 = 6612 x 6 = 72
7 × 7 = 498 × 7 = 569 × 7 = 6310 × 7 = 7011 x 7 = 7712 x 7 = 84
7 × 8 = 568 × 8 = 649 × 8 = 7210 × 8 = 8011 x 8 = 8812 x 8 = 96
7 × 9 = 638 × 9 = 729 × 9 = 8110 × 9 = 9011 x 9 = 9912 x 9 = 108
7 × 10 = 708 × 10 = 809 × 10 = 9010 × 10 = 10011 x 10 = 11012 x 10 = 120

Multiplication Tables Chart for 1 to 12

Below is the Multiplication Table Chart for One to Twelve Tables and they are as such

× (Times)123456789101112
1123456789101112
224681012141618202224
3369121518212427303336
44812162024283236404448
551015202530354045505560
661218243036424854606672
771421283542495663707784
881624324048566472808896
9918273645546372819099108
10102030405060708090100110120

Printable PDF’s of 1 to 12 Tables

Maths Times Tables for One to Twelve
1 Times Table2 Times Table
3 Times Table4 Times Table
5 Times Table6 Times Table
7 Times Table8 Times Table
9 Times Table10 Times Table
11 Times Table12 Times Table

FAQs on Tables from 1 to 12

1. How to Write the Table of 12?

Table of 12 is written as follows

12 x 1 =12

12 x 2 = 24

12 x 3 = 36

12 x 4 = 48

12 x 5 = 60

12 x 7 = 84

12 x 8 = 96

12 x 9 = 108

12 x 10 = 120

2. How to read 5 Times Table?

One time five is 5, two times five is 10, three times five is 15, four times five is 20, five times five is 25, six times five is 30, seven times five is 35, eight times five is 40, nine times five is 45 and ten times five is 50.

3. What is the 11 times table trick?

The trick for multiplying a single digit by 11 is to repeat the digit. For instance, to multiply 9 by 11 repeat the digit of 9 i.e. you will get 99.

Learn Math Tables 1 to 20 | Printable Multiplication Tables 1 to 20 | Tips to Memorize Tables of One to Twenty

Math Tables 1 to 20

Memorizing Multiplication Tables from 1 to 20 help you to related math calculations involving division, multiplication, fractions, algebra, taught in elementary school much simply. Without properly learning Math Tables you will feel difficulty in solving the math problems. Boost up your problem-solving skills and logical ability by memorizing the simple Math Multiplication Tables of 1 to 20 available here. Check out the Tips & Tricks to Memorize the Maths Times Tables provided in the later modules.

Multiplication Tables for One to Twenty

Boost up your math skills altogether by learning the Tables of 1 to 20 provided here. You can avail the Multiplication Tables from One to Twenty provided below in image format and download them free of cost. Stick it on your walls and recite it before going to bed and memorize it regularly.

Multiplication Table Charts for 1 to 10

Multiplication Tables for 11 to 20

Math Tables 1 to 20

Memorizing the Multiplication Tables one can boost their self-confidence and keep the information at one’s fingertips. Build memory in you and also enhances your problem-solving abilities. On Mastering the Multiplication Tables from 1 to 20 your speed of solving the Math Problems increases.

Tables of 2 to 10 are the most basic ones and play a crucial role in performing the arithmetic operations. If you are strong enough with the Math Tables of 2 to 10 you can recall or memorize the Tables from 11 to 20 much simply. It helps you to solve complex problems too easily and can save you a great deal of time. Thus, you are advised to learn them by heart so that you can do fundamental estimations.

While learning the Math Tables you will get to see some examples like 4×5 = 20, 5×4 = 20. On seeing such examples you can get to know the patterns and understand the logic like a number multiplied by another number will result in the same product if the numbers are multiplied the other way.

Table of 1Table of 2Table of 3Table of 4Table of 5
1 × 1 = 12 × 1 = 23 × 1 = 34 × 1 = 45 × 1 = 5
2 × 1 = 22 × 2 = 43 × 2 = 64 × 2 = 85 × 2 = 10
3 × 1 = 32 × 3 = 63 × 3 = 94 × 3 = 125 × 3 = 15
4 × 1 = 42 × 4 = 83 × 4 = 124 × 4 = 165 × 4 = 20
5 × 1 = 52 × 5 = 103 × 5 = 154 × 5 = 205 × 5 = 25
6 × 1 =62 × 6 = 123 × 6 = 184 × 6 = 245 × 6 = 30
7 × 1 = 72 × 7 = 143 × 7 = 214 × 7 = 285 × 7 = 35
8 × 1 = 82 × 8 = 163 × 8 = 244 × 8 = 325 × 8 = 40
9 × 1 = 92 × 9 = 183 × 9 = 274 × 9 = 365 × 9 = 45
10 × 1 = 102 × 10 = 203 × 10 = 304 × 10 = 405 × 10 = 50

Table of 6 to 10

Table of 6Table of 7Table of 8Table of 9Table of 10
6 × 1 = 67 × 1 = 78 × 1 = 89 × 1 = 910 × 1 = 10
6 × 2 = 127 × 2 = 148 × 2 = 169 × 2 = 1810 × 2 = 20
6 × 3 = 187 × 3 = 218 × 3 = 249 × 3 = 2710 × 3 = 30
6 × 4 = 247 × 4 = 288 × 4 = 329 × 4 = 3610 × 4 = 40
6 × 5 = 307 × 5 = 358 × 5 = 409 × 5 = 4510 × 5 = 50
6 × 6 = 367 × 6 = 428 × 6 = 489 × 6 = 5410 × 6 = 60
6 × 7 = 427 × 7 = 498 × 7 = 569 × 7 = 6310 × 7 = 70
6 × 8 = 487 × 8 = 568 × 8 = 649 × 8 = 7210 × 8 = 80
6 × 9 = 547 × 9 = 638 × 9 = 729 × 9 = 8110 × 9 = 90
6 × 10 = 607 × 10 = 708 × 10 = 809 × 10 = 9010 × 10 = 100

Table of 11 to 15

Table of 11Table of 12Table of 13Table of 14Table of 15
11 × 1 = 1112 × 1 = 1213 × 1 = 1314 × 1 = 1415 × 1 = 15
11 × 2 = 2212 × 2 = 2413 × 2 = 2614 × 2 = 2815 × 2 = 30
11 × 3 = 3312 × 3 = 3613 × 3 = 3914 × 3 = 4215 × 3 = 45
11 × 4 = 4412 × 4 = 4813 × 4 = 5214 × 4 = 5615 × 4 = 60
11 × 5 = 5512 × 5 = 6013 × 5 = 6514 × 5 = 7015 × 5 = 75
11 × 6 = 6612 × 6 = 7213 × 6 = 7814 × 6 = 8415 × 6 = 90
11 × 7 = 7712 × 7 = 8413 × 7 = 9114 × 7 = 9815 × 7 = 105
11 × 8 = 8812 × 8 = 9613 × 8 = 10414 × 8 = 11215 × 8 = 120
11 × 9 = 9912 × 9 = 10813 × 9 = 11714 × 9 = 12615 × 9 = 135
11 × 10 = 11012 × 10 = 12013 × 10 = 13014 × 10 = 14015 × 10 = 150

Table of 16 to 20

Table of 16Table of 17Table of 18Table of 19Table of 20
16 × 1 = 1617 × 1 = 1718 × 1 = 1819 × 1 = 1920 × 1 = 20
16 × 2 = 3217 × 2 = 3418 × 2 = 3619 × 2 = 3820 × 2 = 40
16 × 3 = 4817 × 3 = 5118 × 3 = 5419 × 3 = 5720 × 3 = 60
16 × 4 = 6417 × 4 = 6818 × 4 = 7219 × 4 = 7620 × 4 = 80
16 × 5 = 8017 × 5 = 8518 × 5 = 9019 × 5 = 9520 × 5 = 100
16 × 6 = 9617 × 6 = 10218 × 6 = 10819 × 6 = 11420 × 6 = 120
16 × 7 = 11217 × 7 = 11918 × 7 = 12619 × 7 = 13320 × 7 = 140
16 × 8 = 12817 × 8 = 13618 × 8 = 14419 × 8 = 15220 × 8 = 160
16 × 9 = 14417 × 9 = 15318 × 9 = 16219 × 9 = 17120 × 9 = 180
16 × 10 = 16017 × 10 = 17018 × 10 = 18019 × 10 = 19020 × 10 = 200

Important Points to Remember Regarding the Math Tables 1 to 20

Below are the key points to be remembered regarding the Multiplication Tables One to Twenty. They are as under

  • Each and Every Number in the Multiplicaton Table from 1 to 20 is a Whole Number.
  • A number multiplied by itself results in the square of a number.
  • Adding a number n times is the same as multiplying it with n. Adding 5 10 times is the same as multiplying 5 by 10 and gives the result 50.

Multiplication Table Chart

Below is the Multiplication Chart for Tables 1 to 10. They are in the following fashion

× (Times)12345678910
112345678910
22468101214161820
336912151821242730
4481216202428323640
55101520253035404550
66121824303642485460
77142128354249566370
88162432404856647280
99182736455463728190
10102030405060708090100

Benefits of learning Multiplication Table Charts for 1 to 20

Learning Maths Multiplication Tables 1 to 20 provides numerous advantages and boosts your learning abilities. Some of them are outlined as follows

  • Helps you to solve math problems much quicker.
  • You can avoid mistakes while doing calculations in mind.
  • Reciting 1 to 20 Multiplication Tables helps you to understand the patterns among multiples of a number.

Printable Multiplication Tables for 1 to 20 PDF Download

For the sake of your comfort, we have provided the Maths Times Tables from 1 to 20 via quick links available. Simply, tap on them and learn entirely regarding the particular table in no time.

Maths Times Tables for One to Twenty
1 Times Table2 Times Table
3 Times Table4 Times Table
5 Times Table6 Times Table
7 Times Table8 Times Table
9 Times Table10 Times Table
11 Times Table12 Times Table
13 Times Table14 Times Table
15 Times Table16 Times Table
17 Times Table18 Times Table
19 Times Table20 Times Table

FAQs on Multiplication Tables

1. How do you memorize multiplication tables up to 20?

The fastest way to memorize the Multiplication Tables from 1 to 20 is to master the tips & tricks for each and individual table accordingly. Another way to remember the Tables is through Addition. The number of times a number is multiplied by another number it means that it has been added to itself for the same number of times. For Example, 3 Times 3 is 3+3+3

2. How to Learn Math Tables easily?

Prepare a Multiplication Chart for each and every table and paste it on your walls of the room and try to recite it regularly so that you can remember them for a long time.

3. Why is it important to Learn Multiplication Tables?

  • Students are advised to learn Math Tables to perform their mental math calculations quickly.
  • Enhances your Problem Solving Abilities and helps you to solve math problems much faster.
  • Boosts your arithmetic capabilities of a student.

Properties of Division- Closure, Commutative, Associative, Distributive | Basic Division Properties with Examples

Properties of Division

Properties of Division definition is here. Check the formulae, various properties of division, and how they work on various problems. Know the basics regarding the division and also the division property of equality. Follow the various operators with examples and concepts. Get the expression form and also step by step procedure to solve the problems. Division rule follows many properties and those are important in solving various problems. Check the below sections to know the complete details regarding properties of division, formulae, rules, examples, etc.

Also, Read:

Properties of Division

Of the four basic arithmetic operations, the division is the one. In the operation of division, we distribute or share the number or a group of things into equal parts. Division operation defines the fair result of sharing. It is the inverse property of multiplication. Division operation has five properties which are discussed in the below sections. The division is defined as the most complicated part of the arithmetic functions. But will be easy if you have a clear idea of all the methods, concepts, rules, and formulae along with a clear understanding of its usage.

Representation of Division Operator

The notation of the division operator is a short horizontal line with 2 dots one above the line and the other below the line.

Notation:

The division is represented with the notation “÷”

Basic Terms Used in Division

Various parts involving in the division rule have a special name.

Dividend – Dividend is the term that is being divided.

Divisor – The term which is being divided by the dividend is called the divisor

Quotient – The term quotient is defined as the result that is obtained in the division process

Remainder – The term remainder is defined as the leftover portion after the division process

Rules of Division

  • The first division rule is when the number is divided by zero, then the result is always 0. For suppose, 0 ÷ 4 = 0, i.e., 0 chocolates are shared among 4 pupils and each one gets 0 chocolates.
  • No number can be divided with zero, the result gives the undefined value. For suppose, 4 ÷ 0. You have 4 chocolates but no pupil to distribute it, hence you cannot divide it by 0.
  • On dividing the number with 1, the result is the same number with which you are dividing. For suppose, 4 ÷ 1 = 4. 4 chocolates divided among one pupil.
  • If you divide the number by 2, it means that you are halving the number. For suppose, 4 ÷ 2 = 2. 4 chocolates dividing among two pupils, each gets 2 chocolates.
  • On dividing the same number, the result value will always be one. For suppose, 4 ÷ 4 = 1. If 4 chocolates are divided among 4 pupils, then each gets one chocolate.
  • The dividend rule must be applied in a proper way because if we interchange the numbers, the result value changes. For suppose, 20 ÷ 4 = 5 and 4 ÷ 20 = 0.2. Hence, the division rule must be applied in the correct order.
  • The fractions like ¼, ½, ¾ are known as the division sums. ¼ is nothing but 1 ÷ 4, i.e., 1 chocolate is divided among 4 pupils.

Division Properties

There are various properties of a division operation. They are explained in detail by considering few examples and they are as under

Closure Property

In general closure property states that, the resultant value will be always an integer. But when it comes to the division operation, the resultant value of the division need not be an integer value always. Hence, division fundamental operation does not follow closure property. i.e., a ÷ b is not an integer always. Therefore a ÷ b does not follow closure property.

Example: 7 ÷ 3 is not an integer

If we divide 7 with 3, then the resultant value is 2.33 which is not an integer. Thus, it is proved that closure property is not applicable for division operation.

Commutative Property

In general commutative property states that, even after swapping or shifting of numbers, the resultant value will be the same. When it comes to division operation, it gives the different resultant value when the operators are shifted or swapped. Hence, division operation does not follow the commutative property. i.e., a ÷ b ≠ b ÷ a. Therefore, a ÷ b does not follow the commutative property.

Example: 10 ÷ 5 ≠ 5 ÷ 10

If we divide 7 with 3, the resultant value is 2.33. If we divide 3 with 7, the resultant value is 0.42. Therefore, both the values are not equal. Thus, it is proved that commutative property is not applicable for division operation.

Associative Property

In general associative property states that, even if the parentheses of the expression are rearranged, the resultant will not be changed. When it comes to the division operation, it gives a different value when the parentheses are rearranged. Hence, division operation does not follow the associative property. i.e., a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c. Therefore, a ÷ (b ÷ c) does not follow the associative property.

Example: (16 ÷ 4) ÷ 2 ≠ 16 ÷ (4 ÷ 2)

If we solve (16 ÷ 4) ÷ 2, the resultant value is 2 and if we solve 16 ÷ (4 ÷ 2), the resultant value is 8. Therefore, both the values are not the same. Thus, it is proved that associative property is not applicable for division operation.

Distributive Property

In general distributive property states that, the resultant value is the same, even if the sum of two or more addends are multiplied or each addend multiplied separately, and then the products to be added together. When it comes to division operation, it gives different results when the addends are multiplied separately. Hence, division operation does not follow the distributive property. Therefore a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).

Example: 12 ÷ (4+ 2) ≠ (12 ÷ 4) + (12 ÷ 2)

If we solve the equation 12 ÷ (4+ 2), we get the resultant value as 2 and if we solve the equation (12 ÷ 4) + (12 ÷ 2), we get the resultant value as 9. Therefore, both the values are not similar. Thus, it is proved that commutative property is not applicable for division operation.

Division by 1

Any number that is divided by 1 gives the resultant value as the same number.

Example: 

5 ÷ 1 = 5

Example Problems on Division Properties

Problem 1: 

There are 80 chocolates. Each packet must be packed with 5 chocolates. How many packets do we need in total?

Solution:

Total number of chocolates = 80

Toffees that are to be packed in 1 packet = 5

Packets needed to pack 80 toffees = 80 ÷ 5

= 16

Therefore, we require 16 packets to pack 80 chocolates

Problem 2:

There are 100 donuts. They are equally packed in 10 packets. How many donuts are there in each box?

Solution:

Total number of donuts = 100

Total number of packets = 10

Number of donuts in each packet = 100 ÷ 10

= 10

Therefore, there are 10 donuts in each box

Problem 3:

50 bottles are placed in 5 equal trays. Find the number of bottles in each tray?

Solution:

Total no of bottles = 50

No of trays = 5

Number of bottles in each tray = 50 ÷ 5

= 10

Therefore, there are 10 bottles in each tray

Worksheet on Math Relation | Relations and Functions Worksheets with Answers

Worksheet on Math Relation

Students who are searching to get Math Relation problems can check the Worksheet on Math Relation. Our Math Relation Worksheets available improves your preparation level and are very helpful in your practice. It included various models of questions on Math Relations. Therefore, practice all the given examples and check out the answers to cross-check your method of solving. Practice different questions related to Ordered Pair, Cartesian Product of Two Sets, Relation, Domain, and Range of a Relation in Math Relation Worksheet.

See More: Sets

Relations and Functions Questions and Answers

1. Find the values of x and y, if (x + 4, y – 8) = (8, 1).

Solution:

Given that (x + 4, y – 8) = (8, 1)
Compare the elements of the given ordered pairs.
Firstly, compare the first components of the given ordered pairs.
x + 4 = 8
x = 8 – 4 = 4
So, x = 4.
Now, compare the second components of the given ordered pairs.
y – 8 = 1
y = 8 + 1 = 9
So, y = 9.

Therefore, the value of x = 4 and y = 9.


2. If (x/5 + 3, y – 5/7) = (4, 5/14), find the values of x and y.

Solution:

Given that (x/5 + 3, y – 5/7) = (4, 5/14)
Compare the elements of the given ordered pairs.
Firstly, compare the first components of the given ordered pairs.
x/5 + 3 = 4
x/5 = 4-3
Therefore, x/5 = 1
x = 1 * 5 = 5
So, x = 5.
Now, compare the second components of the given ordered pairs.
y – 5/7 = 5/14
y = 5/14 + 5/7 = 15/14
So, y = 15/14.

Therefore, the value of x = 5 and y = 15/14.


3. If X = {m, n, o} and Y = {u, v}, find X × Y and Y × X. Are the two products equal?

Solution:

Given that X = {m, n, o} and Y = {u, v},
Let’s find the X × Y
X × Y = {(m, u); (m, v); (n, u); (n, v); (o, u); (o, v)}
Now, find Y × X.
Y × X = {(u, m); (u, n); (u, o); (v, m); (v, n); (v, o)}
Compare the elements of the given ordered pairs X and Y.
X × Y not equal to Y × X

Therefore, it is clearly stated that the two products are not equal.


4. If P × Q = {(x, 7); (x, 9); (y, 7); (y, 9); (z, 7); (z, 9)}, find P and Q.

Solution:

Given that P × Q = {(x, 7); (x, 9); (y, 7); (y, 9); (z, 7); (z, 9)},
We know that P is a set of all first entries in ordered pairs in P × Q.
Q is a set of all second entries in ordered pairs in P × Q.
Therefore, P = {x, y, z}
Q = {7, 9}

Therefore, the final answer is P = {x, y, z} and Q = {7, 9}


5. If M and N are two sets, and M × N consists of 6 elements: If three elements of M × N are (8, 4) (7, 3) (6, 3). Find M × N.

Solution:

Given that M and N are two sets, and M × N consists of 6 elements: If three elements of M × N are (8, 4) (7, 3) (6, 3).
We know that M is a set of all first entries in ordered pairs in M × N.
N is a set of all second entries in ordered pairs in M × N.
Therefore, M = {8, 7, 6}, and N = {4, 3}
Now, M × N = {(8, 4); (8, 3); (7, 4); (7, 3); (6, 4); (6, 3)}

Thus, M × N contains six ordered pairs.


6. If A × B = {(m, 3); (m, 7); (m, 4); (n, 3); (n, 7); (n, 4)}, find B × A.

Solution:

Given that A × B = {(m, 3); (m, 7); (m, 4); (n, 3); (n, 7); (n, 4)},
We know that A is a set of all first entries in ordered pairs in A × B.
B is a set of all second entries in ordered pairs in A × B.
Therefore, A = {m, n}, and B = {3, 7, 4}
Now, B × A = {(3, m); (3, n); (7, m); (7, n); (4, m); (4, n)}

Therefore, the final answer is B × A = {(3, m); (3, n); (7, m); (7, n); (4, m); (4, n)}


7. If P = { 2, 1, 9} and Q = {4, 5}, then
Find: (i) P × Q (ii) Q × P (iii) P × P (iv) (Q × Q)

Solution:

Given that P = { 2, 1, 9} and Q = {4, 5}
(i) P × Q = {(2, 4); (2, 5); (1, 4); (1, 5); (9, 4); (9, 5)}
(ii) Q × P = {(4, 2); (4, 1); (4, 9); (5, 2); (5, 1); (5, 9)}
(iii) P × P = {(2, 2); (2, 1); (2, 9); (1, 2); (1, 1); (1, 9); (9, 2); (9, 1); (9, 9)}
(iv) (Q × Q) = {(4, 4); (4, 5); (5, 4); (5, 5)}


8. If P = {3, 5, 7} and Q = {2, 3, 6}, state which of the following is a relation from P to Q.
(a) R₁ = {(3, 5); (6, 7); (7, 2)} (b) R₂ = {(3, 3); (7, 6)}
(c) R₃ = {(3, 2); (5, 6); (6, 7)} (d) R₄ = {(7, 2); (7, 6); (5, 3); (3, 3), (5, 2), (5, 7)}

Solution:

Given that P = {3, 5, 7} and Q = {2, 3, 6}
Note: Every element of set P is associated with a unique element of set Q. No element of P must have more than one image.
(a) f(1) = 3 and f(1) = 5 are not possible. so, this relation is not mapping from P to Q.
(b) R₂ = {(3, 3); (7, 6)}. Every element of set P is associated with a unique element of set Q. hence, it is relation from P to Q.
(c) R₃ = {(3, 2); (5, 6); (6, 7)} it’s not relation from P to Q.
(d) R₄ = {(7, 2); (7, 6); (5, 3); (3, 3), (5, 2), (5, 7)} Every element of set P is associated with a unique element of set Q. hence, it is relation from P to Q.

Therefore, the final answer is (b) R₂ = {(3, 3); (7, 6)} and (d) R₄ = {(7, 2); (7, 6); (5, 3); (3, 3), (5, 2), (5, 7)}


9. Write the domain and range of the following relations.
(a) R₁ = {(5, 4); (7, 9); (5, 9); (1, 8); (8, 6); (1, 9)}
(b) R₂ = {(p, 3); (q, 4); (r, 3); (p, 4); (s, 5); (q, 5)}

Solution:

Given that (a) R₁ = {(5, 4); (7, 9); (5, 9); (1, 8); (8, 6); (1, 9)}
(b) R₂ = {(p, 3); (q, 4); (r, 3); (p, 4); (s, 5); (q, 5)}
(a) R₁ = {(5, 4); (7, 9); (5, 9); (1, 8); (8, 6); (1, 9)}
From the given information, the Domain = {1, 5, 7, 8} and Range = {4, 6, 8, 9}
(b) R₂ = {(p, 3); (q, 4); (r, 3); (p, 4); (s, 5); (q, 5)}
From the given information, the Domain = {p, q, r, s} and Range = {3, 4, 5}


10. Let P = {3, 4, 5, 6, 7, 8}. Define a relation R from A to A by R = {(x, y) : y = x + 1}.

  • Depict this relation using an arrow diagram.
  • Write down the domain and range of R.
Solution:

Given that P = {3, 4, 5, 6, 7, 8}. Define a relation R from A to A by R = {(x, y) : y = x + 1}.
If x = 3, y = x + 1 = 3 + 1 = 4.
x = 4, y = x + 1 = 4 + 1 = 5.
x = 5, y = x + 1 = 5 + 1 = 6.
x = 6, y = x + 1 = 6 + 1 = 7.
x = 7, y = x + 1 = 7 + 1 = 8.
x = 8, y = x + 1 = 8 + 1 = 9.
R = {(3, 4); (4, 5); (5, 6); (6, 7); (7, 8)} where P = {3, 4, 5, 6, 7, 8}.
Worksheet on Math Relation
Domain = Set of all first elements in a relation = {3, 4, 5, 6, 7}
Range = Set of all second elements in a relation = {4, 5, 6, 7, 8}


11. Adjoining figure shows a relationship between the set P and Q. Write this relation in the roster form. What are its domain and range?
Domain and Range Problems

Solution:

The relation mentioned in the figure shows, P a domain and Q as a range.
Let the relation be R.
In roster form R = {(3, 6); (6, 12); (9, 18)}
Domain = Set of all first elements in a relation = {3, 6, 9}
Range = Set of all second elements in a relation = {6, 12, 18}


12. In the given ordered pairs (2, 4); (4, 16); (5, 7); (1, 3); (6, 36); (2, 9); (1, 1), find the following relationship:
(a) Is a factor of ….
(b) Is a square root of …..
Also, find the domain and range in each case.

Solution:

Given that the ordered pairs (2, 4); (4, 16); (5, 7); (1, 3); (6, 36); (2, 9); (1, 1).
(a) Is a factor of ….
Let’s find out the factor of …. from the given order pars.
(2, 4); (4, 16); (1, 3); (6, 36); (1, 1)
Domain =  Set of all first elements in a relation = {1, 2, 4, 6}
Range = Set of all second elements in a relation = {1, 3, 4, 16, 36}
(b) Is a square root of …..
Let’s find out the square root of …. from the given order pars.
(2, 4); (4, 16); (6, 36).
Domain =  Set of all first elements in a relation = {2, 4, 6}
Range = Set of all second elements in a relation = {4, 16, 36}


13. Draw the arrow diagrams to represent the following relations.
(a) R₁ = {(2, 2); (2, 7); (2, 8); (6, 9); (7, 4)}
(b) R₂ = {(5, 11); (5, 14); (5, 17); (6, 14); (7, 17)}
(c) R₃ = {(3, 4); (4, 6); (5, 8); (6, 10); (7, 12)}
(d) R₄ = {(a, x); (a, y); (b, p); (b, z); (c, y)}

Solution:

(a) Given that R₁ = {(2, 2); (2, 7); (2, 8); (6, 9); (7, 4)}
Let the two sets are P and Q.
The required diagram is
Math Relation Worksheet
(b) Given that R₂ = {(5, 11); (5, 14); (5, 17); (6, 14); (7, 17)}
Let the two sets are P and Q.
The required diagram is
Math Relation Worksheets
(c) Given that R₃ = {(3, 4); (4, 6); (5, 8); (6, 10); (7, 12)}
Let the two sets are P and Q.
The required diagram is
Math Relation Worksheet problems
(d) Given that R₄ = {(a, x); (a, y); (b, p); (b, z); (c, y)}
Let the two sets are P and Q.
The required diagram is
Math Relation Worksheet Questions


14. Represent the following relation in the roster form.
(a) Math Relation Worksheet Questions and answers
(b) Worksheet on Math Relation Problems
(c) Math Relation Worksheet Question and answers
(d) Math Relation Worksheet Solved Examples

Solution:

(a) R = {(a, x) (a, z) (b, y) (c, x) (c, q) (d, z)}
(b) R = {(3, 7) (3, 9) (4, 7) (4, 10) (5, 9) (3, 11)}
(c) R = {(2, 2) (5, 3) (10, 4) (17, 5)}
(d) R = {(11, 3) (11, 6) (13, 3) (13, 4) (13, 5) (16, 4) (16, 6) (26, 6)}


Conversion of Numbers to Roman Numerals – Rules, Chart, Examples | How to Convert Numbers to Roman Numerals?

Conversion of Numbers to Roman Numerals

Looking for ways on how to convert from Numbers to Roman Numerals? If so, look no further as this web page gives you entire information regarding the basics like Roman Numerals Definition, Frequently Used Roman Numerals. Furthermore, you will get acquainted with the details like Procedure on How to Convert Numbers to Roman Numerals, Rules involving the Roman Numerals Conversion, their Applications in day to day lives, etc. Also, check out the Solved Examples on Changing between Numbers to Roman Numerals for a better understanding of the concept.

Also, Read:

Roman Numerals – Definition

Romans used a special kind of numerical notations that contains Latin alphabets which signifies values. Roman alphabets are English alphabets except for J, U, and w. These are used to represent roman numbers.
For Example:
1 is written as I
2 is written as II
3 is written as III
The other letters V, X, I, C, M are easy to understand.

Commonly used Roman Numerals are listed below.

NumberRoman Numeral
5V
10X
50L
100C
500D
1000M

How to Convert Numbers into Roman Numerals?

Follow the below-listed step-by-step process to change between Numbers to Roman Numerals easily. You can get the results easily by following the Numbers to Roman Numerals Conversion procedure. They are as follows

  • Break the number into thousands, hundreds, tens, and ones and write down each conversion.
  • Remember a Letter can only be repeated three times.

Roman Numerals Chart for 1-100 Numbers

Roman Numerals Chart for 1-100 Numbers

Rules for Converting Numbers to Roman Numerals

We can convert any number to a Roman numeral and vice versa. However, to do so we have to follow certain rules and they are explained in detail below by considering a few examples. Primary Rules for Reading and Writing the Roman Numerals are given here. They are as follows

  • A Letter can be repeated only thrice and not more than that. For instance XXX = 30, etc.
  • When a smaller numeral is placed after a larger or (equal) one it has the effect of addition. i.e., a smaller number is added to the larger number.
    For Example: VIII=V+I+I+I=5+1+1+1=8
    CXX=C=+X+X=100+10+10=120
  • When a smaller numeral is placed before a larger one it has the effect of subtraction. i.e., the smaller number is subtracted from the larger number.
    For Example: IX=X-I=10-1=9
    LIX=50-1+10=59
  • A bar placed on top of a letter or string of letters increases the numeral’s value by 1,000 times.

Solved Examples on Conversion of Numbers to Roman Numerals

1. Convert 1789 to Roman Numerals?

Solution:

Break the number into thousands, hundreds, tens, and ones and perform the individual conversion.

Given Number 1789 broken down into place values are as follows

1000=M
700=DCC
80=LXXX
9=IX
Therefore, 1789 converted MDCCLXXXIX

2. Convert 1674 to Roman Numerals?

Solution:

Break down the number into thousands, hundreds, tens, and ones and perform conversion each.

Given Number 1674 broken down into place values are as follows

1000 = M
600 = DC
70 = LXX
4 = IV

Therefore, 1674 converted to Roman Numerals is MDCLXXIV

Uses of Roman Numerals

Roman Numerals have various applications and are used in plenty of scenarios that we come across in our day-to-day lives. They are in the following fashion

  • They are used in chapters of the book, movie titles, television programs, and videos.
  • Roman Numbers used for the names of pope ships.
  • They are also used for displaying hours on analog clocks and watches.

FAQ’S on Conversion of Numbers to Roman Numerals

1. How do we write 1000 in Roman Numbers?

1000 Written in Roman Numerals is represented by the Letter ‘M’.

2. How to Convert Numbers to Roman Numbers?

You can convert numbers to Roman Numerals by simply breaking down the Thousands, Hundreds, Tens, Ones and perform each conversion adhering to the roman numeral conversion rules.

3. What does the Roman Number XV equal to?

Roman Numeral XV is expressed in numbers as 95.

Eureka Math Kindergarten Module 3 Lesson 10 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 10 Answer Key

Eureka Math Kindergarten Module 3 Lesson 10 Homework Answer Key

Question 1.
Eureka Math Kindergarten Module 3 Lesson 10 Homework Answer Key 1
The golf ball is as heavy as _______ pennies.
Answer:
The golf ball is as heavy as 6 pennies.

Explanation:
In the above picture in the left side of the balance their is a golfball and in the right side of the balance there are 6 pennies.The balance shows equal weight on both sides.Therefore, 1 golfball is as heavy as 6 pennies.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 10 Homework Answer Key 2
The toy train is as heavy as _______ pennies.
Answer:
The toy train is as heavy as 9 pennies.

Explanation:
In the above picture in the left side of the balance their is a Toytrain and in the right side of the balance there are 9 pennies.The balance shows equal weight on both sides.Therefore, 1 toytrain is as heavy as 9 pennies.

Question 3.
Eureka Math Kindergarten Module 3 Lesson 10 Homework Answer Key 3
Draw in the pennies so the carrot is as heavy as 5 pennies.
Answer:

Explanation:
In the above picture i drew 5 pennies as given that the carrot is as heavy as 5 pennies.

Question 4.
Eureka Math Kindergarten Module 3 Lesson 10 Homework Answer Key 4
Draw in the pennies so the book is as heavy as 10 pennies.
Answer:

Explanation:
In the above picture i drew 10 pennies as given that the book is as heavy as 10 pennies.

Eureka Math Kindergarten Module 3 Lesson 9 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 9 Answer Key

Eureka Math Kindergarten Module 3 Lesson 9 Homework Answer Key

Draw something inside the box that is heavier than the object on the balance.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 9 Homework Answer Key 1
Answer:

Explanation:
A bunch of flowers is heavier than a feather.So, i drew a bunch of flowers in the box.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 9 Homework Answer Key 2
Answer:

Explanation:
A watermelon is heavier than an apple.So, i drew a watermelon in the box.

Draw something lighter than the object on the balance.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 9 Homework Answer Key 3
Answer:

Explanation:
A balloon is lighter than a ball.So, i drew a balloon in the box.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 9 Homework Answer Key 4
Answer:

Explanation:
A tomato is lighter than a pumpkin.So, i drew a tomato in the box.

Eureka Math Kindergarten Module 3 Lesson 8 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 8 Answer Key

Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key

Which is heavier? Circle the object that is heavier than the other.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key 1
Answer:

Explanation:
I circled book because it is heavier than a scissors.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key 2
Answer:

Explanation:
I circled pen because it is heavier than a paper.

Question 3.
Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key 3
Answer:

Explanation:
I circled bear because it is heavier than a teddybear.

Question 4.
Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key 4
Answer:

Explanation:
I circled shoe because it is heavier than a socks.

Question 5.
Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key 5
Answer:

Explanation:
I circled ball because it is heavier than a balloon.

Question 6.
Eureka Math Kindergarten Module 3 Lesson 8 Problem Set Answer Key 6
Answer:

Explanation:
I circled watermelon because it is heavier than an apple.

Eureka Math Kindergarten Module 3 Lesson 8 Homework Answer Key

Draw an object that would be lighter than the one in the picture.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 8 Homework Answer Key 7
Answer:

Explanation:
A balloon will be lighter than a ball.So, i drew a balloon.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 8 Homework Answer Key 8
Answer:

Explanation:
An orange will be lighter than a pineapple.So, i drew an orange.

Question 3.
Eureka Math Kindergarten Module 3 Lesson 8 Homework Answer Key 9
Answer:

Explanation:
A chair will be lighter than a table.So, i drew a chair.

Question 4.
Eureka Math Kindergarten Module 3 Lesson 8 Homework Answer Key 10
Answer:

Explanation:
A book will be lighter than a bag.So, i drew a book.

Eureka Math Kindergarten Module 3 Lesson 7 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 7 Answer Key

Eureka Math Kindergarten Module 3 Lesson 7 Problem Set Answer Key

These boxes represent cubes.
Eureka Math Kindergarten Module 3 Lesson 7 Problem Set Answer Key 1

Question 1.
Eureka Math Kindergarten Module 3 Lesson 7 Problem Set Answer Key 2
Answer:

Explanation:
I colored 2 cubes red and 3 cubes green.There are 5 color cubes.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 7 Problem Set Answer Key 3
Answer:

Explanation:
I colored 1 cube red and 4 cubes green.There are 5 color cubes.

Question 1.
Trace a 6-stick. Find something the same length as your 6-stick.
Draw a picture of it here.
Answer:

Explanation:
I drew a 6-stick and drew a chalkpiece which is same length as 6-stick.

Question 2.
Trace a 7-stick. Find something the same length as your 7-stick.
Draw a picture of it here.
Answer:

Explanation:
I drew a 7-stick and drew a cflowervase which is same length as 7-stick.

Question 3.
Trace an 8-stick. Find something the same length as your 8-stick.
Draw a picture of it here.
Answer:

Explanation:
I drew a 8-stick and drew a marker which is same length as 8-stick.

Eureka Math Kindergarten Module 3 Lesson 7 Homework Answer Key

These boxes represent cubes.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 7 Homework Answer Key 4
Color 2 cubes green. Color 3 cubes blue.
Together, my green 2-stick and blue 3-stick are the same length as 5 cubes.
Answer:

Explanation:
I colored 2 cubes green, 3 cubes blue.
Together, my green 2-stick and blue 3-stick are the same length as 5 cubes.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 7 Homework Answer Key 5
Color 3 cubes blue. Color 2 cubes green.
Together, my blue 3-stick and green 2-stick are the same length as ___ cubes.
Answer:

Explanation:
I colored 3 cubes blue and 2 cubes green.
Together, my blue 3-stick and green 2-stick are the same length as 5 cubes.

Question 3.
Eureka Math Kindergarten Module 3 Lesson 7 Homework Answer Key 6
Color 1 cube green. Color 4 cubes blue.
How many did you color? ________
Answer:

Explanation:
I colored 1 cube green and 4 cubes blue.
Altogether i colored 5 cubes.

Question 4.
Eureka Math Kindergarten Module 3 Lesson 7 Homework Answer Key 7
Color 4 cubes green. Color 1 cube blue.
How many did you color? ________
Answer:

Explanation:
I colored 4 cubes green and 1 cube blue.
Altogether i colored 5 cubes.

Question 5.
Eureka Math Kindergarten Module 3 Lesson 7 Homework Answer Key 8
Color 2 cubes yellow. Color 2 cubes blue.
Together, my 2 yellow and 2 blue are the same as _____.
Answer:

Explanation:
I colored 2 cubes yellow, 2 cubes blue.
Together, my 2 yellow and 2 blue are the same as 4 cubes.

Eureka Math Kindergarten Module 3 Lesson 6 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 6 Answer Key

Eureka Math Kindergarten Module 3 Lesson 6 Problem Set Answer Key

In the box, write the number of cubes there are in the pictured stick. Draw a green circle around the stick if it is longer than the object. Draw a blue circle around the stick if it is shorter than the object.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 6 Problem Set Answer Key 1
Answer:

Explanation:
There are 6 cubes in the above cube.I circled the cube with blue as the stick is shorter tahn the object.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 6 Problem Set Answer Key 2
Answer:

Explanation:
There are 9 cubes in the above stick.I circled it green as the stick is longer than the object.

Question 1.
Make a 3-stick. In your classroom, select a crayon, and see if your crayon is longer than or shorter than your stick.
Trace your 3-stick and your crayon to compare their lengths.
Answer:

Explanation:
I drew a 3-stick and a crayon.I traced them to compare the lengths.The crayon is longer than the 3-stick.

Question 2.
In your classroom, find a marker, and make a stick that is longer than your marker.
Trace your stick and your marker to compare their lengths.
Answer:

Explanation:
I drew a marker and a stick longer than the marker.I traced them to comapre their lengths.The stick is longer than the marker.

Question 3.
Make a 5-stick. Find something in the classroom that is longer than your 5-stick.
Trace your 5-stick and the object to compare their lengths.
Answer:

Explanation:
I drew a 5-stick and a chair from my classroom.I traced them to compare their lengths.The chair is longer than the 5-stick.

Eureka Math Kindergarten Module 3 Lesson 6 Homework Answer Key

Color the cubes to show the length of the object.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 6 Homework Answer Key 3
Answer:

Explanation:
I colored the stick upto 7cubes as the object is 7-stick long.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 6 Homework Answer Key 4
Answer:

Explanation:
I colored the stick upto 4 cubes as the object is 4-stick long.

Question 3.
Eureka Math Kindergarten Module 3 Lesson 6 Homework Answer Key 5
Answer:

Explanation:
I colored the stick upto 3 cubes as the object is 3-stick long.

Question 4.
Eureka Math Kindergarten Module 3 Lesson 6 Homework Answer Key 6

 

 

 

Answer:

Explanation:
I colored the ctick upto 4 cubes as the object is 4-stick long.

Eureka Math Kindergarten Module 3 Lesson 5 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 5 Answer Key

Eureka Math Kindergarten Module 3 Lesson 5 Problem Set Answer Key

Circle the stick that is shorter than the other.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 5 Problem Set Answer Key 1
Answer:

Explanation:
I circled the  shorter stick in the above picture.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 5 Problem Set Answer Key 2
Answer:

Explanation:
I circled the shorter stick in the above picture.

Circle the stick that is longer than the other.
Eureka Math Kindergarten Module 3 Lesson 5 Problem Set Answer Key 3

Answer:

Explanation:
I circled the longer stick in the above picture.

My _____ -stick is longer than my _____ -stick.
Answer:
My 5 -stick is longer than my 4 -stick.

My _____ -stick is shorter than my ____ -stick.
Answer:
My 4-stick is shorter than my 5-stick.

Circle the stick that is shorter than the other stick.
Eureka Math Kindergarten Module 3 Lesson 5 Problem Set Answer Key 4

Answer:

Explanation:
I circled the shorter stick in the above picture.

My _____ -stick is longer than my _____ -stick.
Answer:
My 9-stick is longer than my 7-stick.

My _____ -stick is shorter than my _____ -stick.
Answer:
My 7-stick is shorter than my  9-stick.

Eureka Math Kindergarten Module 3 Lesson 5 Homework Answer Key

Circle the stick that is shorter than the other.
Eureka Math Kindergarten Module 3 Lesson 5 Homework Answer Key 5

Answer:

Explanation:
I circled the shorter stick in the above picture.

My _____ -stick is shorter than my _____ -stick.
Answer:
My 8-stick is shorter than my 9-stick.

My _____ -stick is longer than my _____ -stick.
Answer:
My 9-stick is longer than my 8-stick.

Eureka Math Kindergarten Module 3 Lesson 4 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 4 Answer Key

Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key

Circle the shorter stick.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 1
How many linking cubes are in the shorter stick? Write the number in the box.
Answer:

Explanation:
I circled the shorter stick and it has 4 cubes in it.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 2
How many linking cubes are in the shorter stick? Write the number in the box.
Answer:

Explanation:
I circled the shorter stick and it has 3 cubes in it.

Circle the longer stick.

Question 1.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 3
How many linking cubes are in the longer stick? Write the number in the box.
Answer:

Explanation:
I circled the longer stick and it has 6 cubes in it.

Question 2.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 4
How many linking cubes are in the longer stick? Write the number in the box.
Answer:

Explanation:
I circled the longer stick and it has 7 cubes in it.

Draw a stick shorter than my 5-stick.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 5
Answer:

Explanation:
I drew a 4 cube stick which is shorter than the 5 cube stick that is given in the above picture.

Draw a stick longer than mine.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 6
Answer:

Explanation:
I drew a 8 cube stick which is longer than the stick given in the above picture.

Draw a stick shorter than mine.
Eureka Math Kindergarten Module 3 Lesson 4 Problem Set Answer Key 7
Answer:

Explanation:
I drew a 5 cube stick which is shorter than the stick given in the picture.

Eureka Math Kindergarten Module 3 Lesson 4 Homework Answer Key

Use a red crayon to circle the sticks that are shorter than the 5-stick.
Eureka Math Kindergarten Module 3 Lesson 4 Homework Answer Key 8
Answer:

Explanation:
I circled the sticks that are shorter than the 5-stick with a red crayon.

Use a blue crayon to circle the sticks that are longer than the 5-stick.
Eureka Math Kindergarten Module 3 Lesson 4 Homework Answer Key 9
Answer:

Explanation:
I circled the sticks that are longer than the 5-stick with a blue crayon.

Eureka Math Kindergarten Module 3 Lesson 3 Answer Key

Engage NY Eureka Math Kindergarten Module 3 Lesson 3 Answer Key

Eureka Math Kindergarten Module 3 Lesson 3 Homework Answer Key

Take out a new crayon. Circle objects with lengths shorter than the crayon blue. Circle objects with lengths longer than the crayon red.
Eureka Math Kindergarten Module 3 Lesson 3 Homework Answer Key 1
Answer:

Explanation:
I circled the cars with red that are longer than a cayon and circled the cars with blue that are shorter than a crayon.

On the back of your paper, draw some things shorter than the crayon and longer than the crayon. Draw something that is the same length as the crayon.
Answer:

Explanation:
An eraser is shorter than a crayon, pencil is longer than a crayon and key is same length of crayon.