## Spectrum Math Grade 1 Answer Key Online Pdf | Spectrum Math 1st Grade Answers

Do you wanna make your child a pro in Maths? If yes, then we are here to support you by giving the best guidelines to make your child master in maths. Spectrum Math 1st Grade Answer Key is the best resource for your child to learn the basic concepts in maths. Get free access to Download Spectrum Math Answer Key for Grade 1 pdf Chapter-wise. The teachers and students can get the answers for lessons, pretests, posttests, and Mid-tests from here. Spectrum Math 1st Grade Answers are prepared by the subject experts in order to help the students.

## Spectrum Math 1st Grade Answer Key | Spectrum Math Workbook Grade 1 Answer Key

Most of the students may feel that maths is the toughest of all the subjects. But if you love the subject then it is the easiest and most interesting subject of all. This is possible from the primary level itself. So, we provided step by step explanations for each and every question in a possible way. So, that is would be for you to learn the concept of the chapter. Just tap the links provided on this page and Download Spectrum Math Workbook Grade 1 Answer Key Online pdf and prepare well for the exams.

Spectrum Math Grade 1 Answer Key Online Chapter 1 Addition and Subtraction Facts through 10

• Spectrum Math Grade 1 Chapter 1 Pretest
• Lesson 1.1 Adding Through 3
• Lesson 1.2 Subtracting from 1, 2, and 3
• Lesson 1.3 Adding 4 and 5
• Lesson 1.4 Adding to 6
• Lesson 1.5 Subtracting from 4 and 5
• Lesson 1.6 Subtracting from 6
• Lesson 1.7 Fact Families 0 Through 6
• Lesson 1.8 Problem Solving
• Lesson 1.9 Adding to 7
• Lesson 1.10 Subtracting from 7
• Lesson 1.11 Adding to 8
• Lesson 1.12 Subtracting from 8
• Lesson 1.13 Adding to 9
• Lesson 1.14 Subtracting from 9
• Lesson 1.15 Adding to 10
• Lesson 1.16 Subtracting from 10
• Lesson 1.17 Fact Families 7 Through 10
• Lesson 1.18 Addition Practice Through 10
• Lesson 1.19 Subtraction Practice Through 10
• Lesson 1.20 Adding with Money
• Lesson 1.21 Problem Solving
• Lesson 1.22 More- and Less-Than Facts Through 10
• Lesson 1.23 Using Addition for Subtraction
• Lesson 1.24 Doubles and Near-Doubles
• Spectrum Math Grade 1 Chapter 1 Posttest

Spectrum Math Grade 1 Answers Chapter 2 Place Value

• Spectrum Math Grade 1 Chapter 2 Pretest
• Lesson 2.1 Counting and Writing 10 through 14
• Lesson 2.2 Counting and Writing 15 through 19
• Lesson 2.3 Counting and Writing 20 through 24
• Lesson 2.4 Counting and Writing 25 through 29
• Lesson 2.5 Counting and Writing 30 through 49
• Lesson 2.6 Counting and Writing 50 through 69
• Lesson 2.7 Counting and Writing 70 through 99
• Lesson 2.8 Counting to 120
• Lesson 2.9 Counting Forward and Backward to 120
• Lesson 2.10 Comparing Numbers
• Spectrum Math Grade 1 Chapter 2 Posttest

Spectrum Math Grade 1 Chapters 1-2 Mid-Test

Spectrum Math 1st Grade Answer Key Chapter 3 Addition and Subtraction Facts through 20

• Spectrum Math Grade 1 Chapter 3 Pretest
• Lesson 3.1 Adding to 11
• Lesson 3.2 Subtracting from 11
• Lesson 3.3 Adding to 12
• Lesson 3.4 Subtracting from 12
• Lesson 3.5 Adding to 13
• Lesson 3.6 Subtracting from 13
• Lesson 3.7 Adding to 14
• Lesson 3.8 Subtracting from 14
• Lesson 3.9 Fact Families 11 through 15
• Lesson 3.10 Addition and Subtraction Facts through 15
• Lesson 3.11 Fact Families 16 through 20
• Lesson 3.12 Addition and Subtraction Facts through 16
• Lesson 3.13 Using Addition for Subtraction
• Lesson 3.14 Addition and Subtraction Facts through 18
• Lesson 3.15 Using Addition and Subtraction
• Lesson 3.16 More- and Less-Than Facts 11 through 20
• Spectrum Math Grade 1 Chapter 3 Posttest

Spectrum 1st Grade Math Workbook Chapter 4 Addition and Subtraction Facts through 100

• Spectrum Math Grade 1 Chapter 4 Pretest
• Lesson 4.1 Adding 2-Digit and 1-Digit Numbers
• Lesson 4.2 Adding Multiples of 10 to 2-Digit Numbers
• Lesson 4.3 Problem Solving
• Lesson 4.4 Subtracting Multiples of 10
• Lesson 4.5 Addition and Subtraction Practice through 100
• Lesson 4.6 Adding Three Numbers
• Spectrum Math Grade 1 Chapter 4 Posttest

Spectrum Math Workbook Grade 1 Answer Key Pdf Chapter 5 Measurement

• Spectrum Math Grade 1 Chapter 5 Pretest
• Lesson 5.1 Telling Time to the Hour
• Lesson 5.2 Telling Time to the Half Hour
• Lesson 5.3 Ordering Objects
• Lesson 5.4 Comparing Lengths of Objects
• Lesson 5.5 Measuring Length and Height
• Lesson 5.6 More Fewer
• Lesson 5.7 Greater Than, Less Than, and Equal To
• Lesson 5.8 Collecting Data
• Spectrum Math Grade 1 Chapter 5 Posttest

Spectrum Math 1st Grade Answers Chapter 6 Geometry

• Spectrum Math Grade 1 Chapter 6 Pretest
• Lesson 6.1 Identifying Shapes
• Lesson 6.2 Drawing Shapes
• Lesson 6.3 Finding Shapes
• Lesson 6.4 Comparing 2-D Shapes
• Lesson 6.5 Composing 3-D Shapes
• Lesson 6.6 Partitioning Shapes
• Lesson 6.7 One-Half and One-Fourth
• Spectrum Math Grade 1 Chapter 6 Posttest

Spectrum Math Grade 1 Chapters 1-6 Final Test

### What are the Advantages of Solving Spectrum Math 1st Grade Answer Key?

Teachers and parents can find many changes in students by solving the problems from Spectrum Math Grade 1 Answer Key. Have a look at the below useful points to know the importance of Spectrum Math Answer Key.

• It keeps kids at the top of their math game by improving students’ math skills, and problem-solving skills and building self-confidence in them.
• Get a Free Pdf Download of Grade 1 Spectrum Math Solution Key Chapter-wise.
• Learn simple tricks to solve problems and score maximum marks in the exams.

### Conclusion

All the details mentioned on this page are as per the latest syllabus of Spectrum Math Grade 1. Feel free to post the comments to clarify your doubts regarding Spectrum Math Grade 1 Solutions. We suggest you test yourself by solving the mid-test problems to know what you have learned from each chapter. Also, Bookmark our site ccssanswers.com to get more updates on Spectrum Math Grade 1 Answer Key.

## Properties of Arithmetic Mean | Merits and Demerits of Arithmetic Mean | AM Properties with Proof

Arithmetic Mean(AM) is the most important concept in statistics. It is one of the measures of central tendency that can be directly described as the sum of all quantities to be divided by the number of quantities. Every time we can’t apply the formula of AM to solve the problems on average or mean or arithmetic mean. So, we have explained the properties of arithmetic mean with proofs which aid students to calculate different types of questions on average with ease.

Do Check Some Other Related Articles:

## What are the Properties of Arithmetic Mean?

Properties of AM are used to solve complex problems based on mean/arithmetic mean/average. Some of the important arithmetic mean properties that are used in solving the problems based on average are mentioned here briefly. Just take a look at them and be aware of all properties to use.

• If $$\overline{x}$$ is the arithmetic mean of n observations, x1, x2, x3, …., xn, then (x1–$$\overline{x}$$)+(x2–$$\overline{x}$$)+(x3–$$\overline{x}$$)+…+(xn–$$\overline{x}$$)=0.
• The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is increased by y, the mean of the new observations is ($$\overline{x}$$+y).
• The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is decreased by y, the mean of the new observations is ($$\overline{x}$$–y).
• The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is divided by a non-zero number y, the mean of the new observations is ($$\overline{x}$$*y).
• The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is multiplied by a non-zero number y, the mean of the new observations is (y*$$\overline{x}$$).
• If all the observations in the given data set have a value say ‘y′, then their arithmetic mean is also ‘y′.

### Illustration of Arithmetic Mean Properties

Property 1:

If $$\overline{x}$$ is the arithmetic mean of n observations, x1, x2, x3, …., xn, then (x1–$$\overline{x}$$)+(x2–$$\overline{x}$$)+(x3–$$\overline{x}$$)+…+(xn–$$\overline{x}$$)=0.

Proof:

We know that

$$\overline{x}$$ = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n$$\overline{x}$$. ………………….. (i)

Hence, (x1–$$\overline{x}$$)+(x2–$$\overline{x}$$)+(x3–$$\overline{x}$$)+…+(xn–$$\overline{x}$$)

= (x1 + x2 + x3 + . . . + xn) – n$$\overline{x}$$

= (n$$\overline{x}$$ – n$$\overline{x}$$), [using (i)].

= 0.

Therefore, (x1–$$\overline{x}$$)+(x2–$$\overline{x}$$)+(x3–$$\overline{x}$$)+…+(xn–$$\overline{x}$$)=0.

Property 2:

The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is increased by A, the mean of the new observations is ($$\overline{x}$$ + A).

Proof:

$$\overline{x}$$ = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n$$\overline{x}$$. ………………….. (i)

Mean of (x1 + A), (x2 + A), …, (xn + A)

= {(x1 + A) + (x2 + A) + … + (xn+ A)}/n

= {(x1 + x2 + . . . + xn) + nA}/n

= (n$$\overline{x}$$ + nA)/n, [using (i)].

= {n($$\overline{x}$$ + A)}/n

= ($$\overline{x}$$ + A).

So, the mean of the new observations is ($$\overline{x}$$ + A).

Property 3:

The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is decreased by p, the mean of the new observations is ($$\overline{x}$$ – a).

Proof:

$$\overline{x}$$ = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n$$\overline{x}$$. ………………….. (i)

Mean of (x1 – p), (x2 – p), …., (xn – p)

= {(x1 – p) + (x2 – p) + … + (xn – p)}/n

= {(x1 + x2 + . . . + xn) – np}/n

= (n$$\overline{x}$$ – np)/n, [using (i)].

= {n($$\overline{x}$$ – p)}/n

= ($$\overline{x}$$ – p).

Hence, the mean of the new observations is ($$\overline{x}$$ + p).

Property 4:

The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is multiplied by a nonzero number p, the mean of the new observations is p$$\overline{x}$$.

Proof:

$$\overline{x}$$ = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n$$\overline{x}$$. ………………….. (i)

Mean of px1, px2, . . ., pxn,

= (px1 + px2 + … + pxn)/n

= {p(x1 + x2 + . . . + xn)}/n

= {p(n$$\overline{x}$$)}/n, [using (i)].

= p$$\overline{x}$$.

Hence, the mean of the new observations is p$$\overline{x}$$.

Property 5:

The mean of n observations x1, x2, x3, …., xn is $$\overline{x}$$. If each observation is divided by a nonzero number p, the mean of the new observations is ($$\overline{x}$$/p).

Proof:

$$\overline{x}$$ = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n$$\overline{x}$$. ………………….. (i)

Mean of (x1/p), (x2/p), . . ., (xn/p)

= (1/n) ∙ (x1/p + x2/p + …. + xn/p)

= (x1 + x2 + . . . + xn)/np

= (n$$\overline{x}$$)/(np), [using (i)].

= ($$\overline{x}$$/p).

### Merits of Arithmetic Mean

1. It is determined strictly.
2. It is calculated on the basis of all observations
3. Simple to solve and easy to understand
4. It is responsive to mathematical treatment or properties.
5. It is inspired by the value of every item in the observation series.

### Demerits of Arithmetic Mean

1. If any single observation is missing or lost then it’s unable to find the arithmetic mean of the data.
2. By inspection or graphically, it is quite difficult to find the arithmetic mean.
3. By the extreme values in the set of the data, the AM gets affected.
4. In some situations, the arithmetic mean does not exemplify the original item.

### Solved Problems on Mathematical Properties of Arithmetic Mean Proof

Example 1:
If the two variables x and y are related by 3x + 4y + 6 = 0 and x̄ = 10, then Arithmetic mean of “y” = (-6 – 3x̄) / 6, Find AM of y?

Solution:
Given x̄ = 10
Arithmetic mean of “y” = (-6 – 3x̄) / 6
Arithmetic mean of “y” = (-6 – 3×10) / 6
Arithmetic mean of “y” = (-6 – 30) / 6
Arithmetic mean of “y” = (-36) / 6
Arithmetic mean of “y” = -6

Example 2:
If a variable “x” assumes 7 observations, say 11, 22, 33, 44, 55, 66, 77 then x̄ = 44. Calculate the instance using property 1.
Solution:
Given 7 observations are 11, 22, 33, 44, 55, 66, 77 and it’s mean is 44
Now, by using property 1 ie., ∑(x – x̄) = 0.
The deviations of the observations from the arithmetic mean (x – x̄) are -33, -22, -11, 0, 11, 22, 33
Now, ∑(x – x̄) = -(33) + (-22) + (-11) + 0 +11 + 22 + 33 = 0.

## Problems Based on Average | Free & Printable Average Problems with Solutions PDF Worksheet

Problems Based on Average Worksheet PDF aid students to solve all types of easy to complex average problems with ease. Practicing from average word problems & solutions pdf can excel in the concepts of Statistics like mean, arithmetic mean, median, mode, etc.

Also, you can simply rely on the answers solved here and check out the mistakes that you have made while practicing. Access and download the Average practice questions worksheet with solutions PDF easily without any charge and prepare any time anywhere.

Do Refer:

## How To Solve Average Word Problems?

The calculation of the Average or Arithmetic Mean or Mean of a number of quantities can be done by using the formula ie., the Average of a number of quantities of the same units is equal to their sum divided by the number of those quantities.

Arithmetic average is utilized for all averages such as Average income, average profit, average age, average marks, etc. However, to solve the sum of observations, they must be in the same unit. Check out the below sections to know some quick tips & tricks to solve and practice with the questions based on average.

### Easy Average Problems Tricks | Common Average Aptitude Formulas

The list of easy tricks to solve average problems pdf is given here for your reference. Simply memorize them daily and practice well from average questions for competitive exams with solutions pdf.

1. Average = $$\frac { Sum of quantities }{ Number of quantities }$$
2. Sum of quantities = Average * Number of quantities
3. The average of first n natural numbers is $$\frac { (n +1) }{ 2 }$$
4. The average of the squares of first n natural numbers is $$\frac { (n +1)(2n+1 ) }{ 6 }$$
5. The average of cubes of first n natural numbers is $$\frac { n(n+1)2 }{ 4 }$$
6. The average of first n odd numbers is given by $$\frac { (last odd number +1) }{ 2 }$$
7. The average of first n even numbers is given by $$\frac { (last even number + 2) }{ 2 }$$
8. The average of squares of first n consecutive even numbers is $$\frac { 2(n+1)(2n+1) }{ 3 }$$
9. The average of squares of consecutive even numbers till n is $$\frac { (n+1)(n+2) }{ 3 }$$
10. The average of squares of consecutive odd numbers till n is $$\frac { n(n+2) }{ 3 }$$
11. If the average of n consecutive numbers is m, then the difference between the smallest and the largest number is 2(m-1)
12. If the number of quantities in two groups be n1 and n2 and their average is x and y respectively, the combined average is $$\frac { (n1x+n2y) }{ (n1+ n2) }$$
13. The average of n quantities is equal to x. When a quantity is removed, the average becomes y. The value of the removed quantity is n(x-y) + y
14. The average of n quantities is equal to x. When a quantity is added, the average becomes y. The value of the new quantity is n(y-x) + y

### Word Problems on Average | Average Practice Questions and Answers

Example 1:
The average mark of 50 students in a class is 70. Out of these 50 students, if the average mark of 25 students is 65, what is the average mark of the remaining 25 students?
Solution:
Average of 50 students = 70
Therefore, total marks of all 50 students = 50 x 70 =  3500
Average of 25 students = 65
Therefore, total marks of 25 students = 25 x 65 = 1625
So, the total marks of remaining 25 students = 3500 – 1625 = 1875
Hence, the average of remaining 25 students out of 50 students = 1875 / 30 = 62.

Example 2:
The mean of 15 observations is 45. If the mean of the first observations is 10 and that of the last 15 observations is 45, find the 15th observation.
Solution:
Given that,
Mean of the first 15 observations = 10
Then, Sum of the first 15 observations = 10 x 15 = 150
Mean of the last 15 observations = 45
Then, Sum of the last 15 observations = 45 x 15 = 675
Mean of the 15 observations = 45
Then, Sum of all 15 observations = 45 x 15 = 675
Now, find the 15th observation
= 150 + 675 – 675
= 150
Therefore, the 15th observation is 150.

Example 3:
The mean of 7 numbers is 35. If one of the numbers is excluded, the mean gets reduced by 5. Find the excluded number.
Solution:
Given Mean of 7 numbers = 35
Sum of the 7 numbers = 35 x 7 = 245
Mean of remaining 6 numbers = 35-5 = 30
Sum of remaining 6 numbers = 30 x 6 = 180
Now, find the excluded number to do that;
Excluded number = (sum of the given 7 numbers) – (sum of the remaining 6 numbers)
= 245 – 180
= 65
Hence, the excluded number is 65.

Example 4:
Find the average of 2, 4, 6, 8, 10, 12?
Solution:
Here, you will know that the given numbers are even in numbers ie., a total of 6 numbers. These kinds of questions are solved by looking at the difference in numbers (the constant difference is 2).

If the number were given odd in numbers means 2, 4, 6, 8, 10, then the answers will always be the middle term ie., 6. However, if the given numbers are in even count ie, six numbers like 2, 4, 6, 8, 10, 12 then the answer is always Sum of middle terms/Sum of opposite terms divided by 2 or Sum of quantities divided by total quantities.
ie., (2+12)/2 or (4+10)/2 or (6+8)/2. (the average is 7)
The answer is the same always.
As per the average formula = Sum of quantities divided by total quantities
= 2+4+6+8+10+12 / 6
= 42 / 6
= 7
Hence, the average of 2, 4, 6, 8, 10, 12 is 7.

### Some More Example Questions & Problems Based on Average

Example 5:
The average height of a group of four boys is 5.6 inches. The individual height (in inches) of three of them are 5.2, 5.4, 5.5. What is the height of the fourth boy?
Solution:
Given Average height of 4 boys = 5.6 inches
Total height of 4 boys = (5.6 x 4) in = 22.4 in
Total height of 3 boys = (5.2 + 5.4 + 5.5) in = 16.1 in
Height of the 4th boy = (total height of 4 boys) – (total height of 3 boys)
= (22.4 – 16.1) in
= 6.3 in
Hence, the height of the fourth boy is 6.3 in

Example 6:
The average of 5 numbers is 30. If each number is increased by 2, what will the new average be?
Solution:
Given that the average of 5 numbers = 30
Sum of the 5 numbers = 30 x 5 = 150
If each number is increased by 2, the total increase = 2 x 5 = 10
Then, the new sum = 150+10 = 160 and new Average = 160/10 = 16
Hence, the new average is 16.

Example 7:
The average age of four boys is 25 years and their ages are in proportion 2:4:6:8. What is the age in years of the youngest boy?
Solution:
Given, Average age of four boys = 25 years
Sum of four boys = 25 x 4 = 100
Ages = 2x + 4x + 6x + 8x
2x + 4x + 6x + 8x = 100
20x = 100
x = 100/20
x = 5 years
Age of the youngest boy = 2x years = 2(5) years = 10 years.

## Problems on Mean of Ungrouped Data | Mean of Raw Data Word Problems with Solutions PDF

Word Questions on Mean of Raw Data helps you all to find the mean (or average) of an ungrouped data set where the data is not presented in intervals. Moreover, you can learn more knowledge about the concept by practicing the example problems based on Mean of ungrouped data.

Here, we have explained various styles of word problems on Mean of arrayed data or ungrouped data for your knowledge. Take a look at them and ace up your subject preparation by answering the Mean Deviation for ungrouped data examples and cross-check the solutions here itself.

Check Related Articles:

### Mean of Ungrouped Data Example Problems PDF and Solutions

Example 1:
The monthly salary (in $) of 8 employees in a company are 6000, 8000, 4000, 9000, 5000, 3000, 8000, 6500. Find the mean of ungrouped data? Solution: Given salary (in$) for 8 employees are 6000, 8000, 4000, 9000, 5000, 3000, 8000, 6500
Mean of ungrouped data = Sum of the observations / Total number of observations
= 6000+8000+4000+9000+5000+3000+8000+6500 / 8
= 49500 / 8
= 6187.5

Example 2:
The mean age of ten girls is 25 years. If the ages of nine of them be 10 years, 12 years, 13 years, 15 years and 25 years then find the age of the tenth girl.
Solution:
Let the age of the tenth girl be x years
Now, find the mean age of 10 girls = 10 years +12 years +13 years +15 years + 25 years + x years / 10
⟹ 25 = 75 years + x years / 10 (mean age of 10 girls is 25 years from the question)
⟹ 250 = 75 + x
⟹ x = 250-75
⟹ x = 175
Hence the age of the tenth girl is 175 years.

Example 3:
Find the mean of the first four whole numbers.
Solution:
The first four whole numbers are 0, 1, 2, 3.
Hence, the mean = x1+x2+x3+x4 / 4
= 0+1+2+3/4
= 6/4
= 3/2
= 1.5

Example 4:
The mean of a sample of 5 numbers is 4. An extra value of 2.5 is added to the sample then find the new mean?
Solution:
Given that Total of original numbers =5×4=20
New total =20+2.5=22.5
The new mean of a sample = 22.5/6
= 3.75
Hence, the new mean of 6 numbers is 3.75.

Example 5:
The following table gives the points of each player scored in four games:

 Player Game 1 Game 2 Game 3 Game 4 A B C 14 0 8 16 8 11 10 6 Did not play 10 4 13

(i) Find the mean to determine C’s average number of points scored per game.
(ii) Who is the worst performer?

Solution:
(i) Mean Score of C = 8+11+13 / 3
= 32/3
= 10.6
(ii) To calculate who is the worst performer, we have to find the mean score of each player:
Mean score of A = (14 + 16 + 10 + 10) / 4 = 12.5
Mean score of B = (0 + 8 + 6 + 4) / 4 = 18/4 = 4.5
Mean Score of C = 8+11+13 / 3 = 32/3 = 10.6
Therefore, B is the worst performer.

Example 6:
A competitor scored 80%, 90%, 75%, and 60% in four subjects in an entrance test, Calculate the mean percentage of the scores achieved by his/her.
Solution:
Given observations in percentage are x1=80, x2=90, x3=75, x4=60
Hence, their mean A = x1+x2+x3+x4 / 4
= 80+90+74+60 / 4
= 304 / 4
= 76
Hence, the mean percentage of scores achieved by the competitor was 76%.

## Properties Questions on Arithmetic Mean with Answers | Word Problems on Properties of Arithmetic Mean (Average) PDF

Students who are getting confused by understanding the AM Properties with proof. If that’s the concern then have a look at this Properties Questions on Arithmetic Mean article. In this article, you will see how to solve the different example questions on average/mean/AM by using properties of arithmetic mean.

Simply follow the steps covered in the solution of Problems Based on Average & understand the explanations for solving various complex questions on the arithmetic mean properties. Also, Improve your math proficiency by answering the various example word problems on Mean/Average/Arithmetic Mean in Statistics.

Do check:

### Example Questions on Arithmetic Mean Properties with Solutions PDF

Example 1:
The mean of nine numbers is 40. If four is subtracted from each number, what will be the new average.
Solution:
Let the given numbers be x1, x2, x3,…., x9
Then, the mean of these numbers = (x1+x2+x3 + …. +x9)/9
Hence, (x1+x2+x3 + …. +x9)/9 = 40
⇒ (x1+x2+x3 + …. +x9) = 40 x 9
⇒ (x1+x2+x3 + …. +x9) = 360 ……(i)
The new numbers are (x1 – 4), (x2 – 4),…., (x9 – 4)
Mean of the new numbers = {(x1 – 4)+ (x2 – 4)+….+ (x9 – 4)} / 9
= [(x1+x2+x3 + …. +x9) – 36] / 9
= [360-36]/9, [using (i)]
= 324/9
= 36
Hence, the new mean is 36.

Example 2:
Find the average of the first five prime numbers.
Solution:
Given quantities are First five prime numbers
The first five prime numbers are 2, 3, 5, 7 and 11
The formula of Arithmetic mean or the Average = Sum of the quantities/Number of quantities
= 2+3+5+7+11 / 5
= 28/5
= 5.6

Example 3:
Find the second term of the given A.M terms $$\frac{1}{log_{3}3}$$, X , $$\frac{1}{log_{18}3}$$
Solution:
Given terms are $$\frac{1}{log_{3}3}$$, X , $$\frac{1}{log_{18}3} and this can be rewritten as log3 3, X, log3 18 ⇒ 2X = log3 3 + log3 21 = log3 3 + log3 (6×3) ⇒ 2X = 2 ( log3 3) + 2 log3 3 ⇒ 2X = 2 ( log3 3) + (2 x 1 ) ⇒ 2X = 2(1) + 2 ⇒ X = 4/2 ⇒ X = 2 Example 4: The average of 10 numbers is 5. If 2 is added to every number, what will be the new average? Solution: Let the given numbers be x1, x2, x3,…., x10 Then, the average of given numbers = x1, x2, x3,…., x10 / 10 Hence, (x1+x2+x3+…+x10)/10 = 5 ⇒ (x1+x2+x3+…+x10) = 50 …….(i) The new numbers are (x1 – 4), (x2 – 4),…., (x10 + 2) Mean of the new numbers = (x1 + 2), (x2 + 2),…., (x10 + 2) / 10 = (x1+x2+x3+…+x10) + 20 / 10 = (50 + 20)/10 [Using (i)] = 70/10 = 7. Hence, the new mean is 9. Example 5: In the exams, the mean of marks scored by 30 students was calculated as 60. Next, it was identified that the marks of one student were wrongly copied as 57 instead of 75. Find the correct mean? Solution: Given that Mean of marks = $\frac { Incorrect sum of marks of 30 students }{ 30 }$$ $$\frac { Incorrect sum of marks of 30 students }{ 30 }$$ = 60 An incorrect sum of marks of 40 students = 60 x 30 = 1800 Considering that the marks of one student were wrongly copied as 57 instead of 75, Then, the correct sum of marks of 30 students = 1800 – 57 + 75 = 1818 Finally, correct mean = 1818 / 30 = 60.6 Example 6: The mean marks of two batches of students having 60 and 40 students respectively are 35 and 65. Find the average marks of all the 100 students, taken together. Solution: Let x be the average marks of all 100 students taken together. Batch – I ($$\overline{x}$$) Marks (x1) = 35 No. of students n1 = 60 Batch – II ($$\overline{x}$$) Marks (x2)= 65 No. of students n2 = 40 $$\overline{x}$$ = $$\frac {n1[latex]\overline{x1}$$ + n2$$\overline{x2}$$}{ n1+n2 }$
= $$\frac { 60×35 + 40×65 }{ 60+40 }$$
= $$\frac { 4700 }{ 100 }$$
= 47.
Therefore, $$\overline{x}$$ = 47 marks.

## Median of Raw Data – Definition, Formula, How to Find It | Example Problems on Median of Ungrouped Data with Solutions

In statistics, Data is classified into two types ie., Grouped data and Ungrouped data. Ungrouped data is the information ie., characteristics or numbers that are not segregated into any groups or categories. Median is one of the most important measures in central tendency in statistics. Want to know more about the median of raw data? jump into the further modules and get to learn what is statistics median, the formula of Median, how to calculate median of ungrouped data, and solved examples of Median of discrete data.

Do Refer:

## Median of Raw Data – Definition

The Median of Raw Data is the middlemost number in the data set achieved after ordering the data from small to big in two equal parts. Also, it is the number that is halfway into the dataset.

For example, let’s assume the set of raw data: 6, 4, 3, 4, 2 to find the median. Now, arrange the data in ascending order: 2, 3, 4, 4, 6 and find the n observations ie., n=5 here n is odd then media of raw data is middle value i.e. 4. Therefore, 4 is the median of the given dataset. ### Median Formula for Ungrouped Data

• The formula of Median = [(n+1)/2]th observation, if n is odd.
• The formula to find median of ungrouped data if n is even is M = $$\frac { 1 }{ 2 }$$ { (n/2)th+(n/2+1)th}

### Method to Calculate Median of Ungrouped Data

Step by Step explanation of calculating the median of raw data is given here. Simply, follow the below steps without any fail and learn how to find the median of ungrouped data.

1. Firstly, arrange the given raw data in ascending or descending order.
2. Calculate the number of observations in the given data and then it is denoted by n.
3. An important part of solving the median problems is knowing the n is odd or even to find the median of the data.
4. If n is odd, the [(n+1)/2]thobservation is the median of raw data.
5. If n is even, the mean of (n/2)th observation and [(n/2)+1]th observation is the median ie.,

Median = $$\frac { 1 }{ 2 }$$ { (n/2)th+(n/2+1)th}

### Solved Examples on Calculating the Median of Discrete Data

Example 1:
Find the median of the raw data: 10, 25, 5, 15, 20.

Solution:
Given raw data is 10, 25, 5, 15, 20
Arrange the data in ascending order ie., 5, 10, 15, 20, 25
The number of observations is 5, which is odd.
Hence, Median = $$\frac { 5+1 }{ 2 }$$ = $$\frac { 6 }{ 2 }$$ = 3rd observation = 15.

Example 2:
Find the median of the following gasoline price: $1.79,$1.61, $1.96,$2.09, $1.84,$1,75, $2.11 Solution: Given gasoline prices are$1.79, $1.61,$1.96, $2.09,$1.84, $1,75 Organize the data from least to greatest ie.,$1.61, $1.75,$1.79, $1.84,$1.96, $2.09 Number of states are 6, which is even Median = $$\frac { 1 }{ 2 }$$ { (n/2)th+(n/2+1)th} = $$\frac { 1 }{ 2 }$$ { (6/2)th+(6/2+1)th} = $$\frac { 1 }{ 2 }$$ {1.79+1.84} = $$\frac { 1.79+1.84 }{ 2 }$$ = $$\frac { 3.63 }{ 2 }$$ = 1.815 Hence, the median of gasoline prices is$1.81

Example 3:
Let’s consider the data: 34, 67, 87, 23, 12, 55, 8. What is the median?

Solution:
Given data is 34, 67, 87, 23, 12, 55, 8
Arrange the data from smallest to highest i.e., 8, 12, 23, 34, 55, 67, 87
The number of variants is 7, which is odd then median = middle value
Hence, Median = 34 (middle value in the data).

### FAQs on Median

1. What is the median?

In easy words, the middle value of the given data is called the median in statistics.

2. What is the median formula for continuous data?

If the data is in continuous series then we will take the cumulative frequencies, values from the class-interval (N/2)th term. To find the median for continuous data, we use the formula: M= L – Cf-N1/f × i

3. What is the Median of grouped data formula?

For grouped data, the median of frequency distribution is calculated by the formula: Median = l + [(n/2−cf)/f] × h

4. How to find the Median class of ungrouped data?

Follow the steps given below and calculate the median class for raw data:

1. Firstly, arrange the data values in ascending order.
2. If required use the median formula c = (n + 1)/2 where n is the total number of observations.
3. Look for the value at (n + 1)/2. Use the result in step 2. If c holds a fractional half, it means the average of two values is a median.

## Problems on Median of Ungrouped Data | Median of Raw Data Questions and Answers PDF

The middle value of an ordered data set is called the median of ungrouped data. To find the median of raw data, you need to arrange the given data in ordered form (ascending or descending order) and then consider the middle number as a median. If you want to practice more Problems on Median of Ungrouped Data then go with this article. Here, we have listed some of the practice questions on calculating the median of raw data. Assess your knowledge gap by solving them on a daily basis and enhance your problem-solving & conceptual skills in the Statistics concepts.

Do Check:

## Median of Ungrouped Data Word Problems with Solutions

Example 1:
Find the Median for an odd number of values: 12; 15; 56; 23; 48; 79; 5
Solution:
Step 1: Firstly, we have to sort the values in the data set from the smallest to the largest;
5, 12, 15, 23, 48, 56, 79
Step 2: Find the number in the middle to get the median of the data. Here median is the fourth positioned value of the data set ie., 23.
Hence, the median of these odd raw data is 23.

Example 2:
Find the median of the following data set: {3; 14; 10; 14; 15; 5; 3; 10; 12; 13}
Solution:
First, we have to order the data set: 3, 3, 5, 10, 10, 12, 13, 14, 14, 15
Since the given number of observations is 10 then the median lies between the fifth and sixth place:
Median = 10+12 / 2 = 22/2 = 11
Therefore, the median of the ungrouped data set is 11.

Example 3:
The median of observation 13, 15, 17, 19, x + 3, x + 5, 31, 33, 35, 41 arranged in ascending order is 25. Find the values of x.
Solution:
The given data is in ascending order.
The number of variants of the given data is 10
Median is the average of the 5th and 6th observations ie.,
[(x + 3) + (x + 5)]/2 = 25
2x+8/2 = 25
2x+8 = 50
2x= 42
x = 42/2
x = 21
Therefore, the value of x is 21.

Example 4:
The median of a set of 7 distinct observations is 18. If each of the largest 3 observations of the set is increased by 4, then what is the median of the new set?
Solution:
Given n = 7
Median = 18
Median term = [(n+1)/2]th term = [(7+1)/2]th term = 4th term
Given the largest 3 observations are increased by 4. Considering the median is the 4th term, there will be no change in it.
Hence, Remains the same as that of the original set.

Example 5:
The following data were obtained about the time that four students took for completing a running race. Find the Median of the racing time: 8.7hrs, 4.3hrs, 3.5hrs, 5.1hrs
Solution:
First, let us sort the data in ascending order: 3.5hr, 4.3hr, 5.1hr, 8.7hr
The number of observations on the data set is 4, which is even
So, we can find the median by taking the mean of two middlemost numbers.

Median = Mean of (4.3hr, 5.1hr)
= 4.3+5.1 / 2
= 9.4/2
= 4.7
Hence, the median race time is 4.7 hrs.

Example 6:
The test scores (out of 20 points) of 25 students in a Statistics class are 12, 15, 10, 9, 8, 6, 12, 13, 19, 20, 18, 14, 15, 6, 5, 10, 12, 15, 6, 7, 9, 13, 17, 18, 20 . Find the median of the test scores.
Solution:
Given Scores are 12, 15, 10, 9, 8, 6, 12, 13, 19, 20, 18, 14, 15, 6, 5, 10, 12, 15, 6, 7, 9, 13, 17, 18, 20
Let’s arrange it in ascending order: 5, 6, 6, 6, 7, 8, 9, 9, 10, 10, 12, 12, 12, 13, 13, 14, 15, 14, 17, 18, 18, 19, 19, 20, 20.
The number of observations are 25, which is odd
Median of the test scores = n+1/2
= 25+1/2
= 26/2
= 13th Observation
= 12.
Hence, the median of the test scores is 12.

## Cube of Numbers – Definition, Formula, Examples | List of Cubes from Numbers 1 to 50

Do you know what is a cube? Have you find a cube for any number! If not, let us find how to calculate cubes for numbers. Basically, a cube is a number acquired by multiplying the number three times. Thus, the number(x) becomes x³ or x- cube. There is a difference between cube and cube root whereas cube root is the reverse process of the cube of a number and it is denoted by ∛.

Usually, in mathematics, we run mostly with squares and cubes of numbers in arithmetic operations. On this page, 5th Grade Math students will go through more about the cube of numbers, a perfect cube, a cube of negative numbers, and cubes of rational numbers, etc.

Do Check:

• Volume of Cubes and Cuboids
• Worksheet on Volume of a Cube and Cuboid

## Cube – Definition

When an integer is multiplied by three times itself and the resultant product of a number is its cube number. If you consider a number n, then the formula of a cube of a number n is n × n × n = n³ because n is a natural number. In other words, we can say a number raised its exponent by 3 is called the cube of a number. For example, to find the cube of a number 5, we write as 5 × 5 × 5 = 125 and says as the cube of number 5 is 125.

We can also write a cube of the number 6 as 6³ in the exponent form and read 6- cubed. For example, to find 7³, first, we calculate 7 × 7 = 49 and next we calculate 49 × 7 = 343. Thus, we say that 343 is the cube of a number 7. It is to be emphasized that the number obtained using a cube formula is the perfect cube number. The cube is also known as the number which is calculated by its square.

### Cube of a Number Examples

(i) 2³ = (2 × 2 × 2) = 8, here 8 is the cube of 2.
(ii) 3³ = (3 × 3 × 3) = 27, here 27 is the cube of 3.
(iii) 4 × 4 × 4 = 64, here 64 is the cube of 4.

### Perfect Cube

A perfect cube is a number that is similar to the number, multiplied by itself three times or with three same or equal integers. If m is a perfect cube of n, then m = n³. Thus, if we take the cube root of a perfect cube, we get a natural number but not any fraction number. Therefore, ∛m = n. For example, 8 is a perfect cube because ∛8 = 2.

For instance, 125 is a perfect cube because 5³ = 5 × 5 × 5 = 125 where 123 is not a perfect cube because there is no number that comes when the number multiplied three times gives the product 123. The following table shows the perfect cubes of the first 10 natural numbers. ### Cubes of Negative Integers

Cube of any negative integer will always be a negative integer, where the cube of any positive number results always a positive number, there a negative integer product will always give a product of negative integer because when a negative number is calculated by the same number three times, it results in a negative number.

Examples of a cube of negative integers when a number is calculated thrice are as such;
(-1)³ = (-1) × (-1) × (-1) = -1,
(-2)³ = (-2) × (-2) × (-2) = -8
(-3)³ = (-3) × (-3) × (-3) = -27, etc.

### Cubes of Numbers from 1 to 25 Chart

The below table gives the cube values from 1 to 25 numbers along with their notations. These numbers will help the children in solving the numerical problems accurately.

 Number Multiplying Three Times Itself Cube of a Number (x3) 1 1× 1× 1 1 2 2× 2× 2 8 3 3× 3× 3 27 4 4× 4× 4 64 5 5× 5× 5 125 6 6× 6× 6 216 7 7× 7× 7 343 8 8× 8× 8 512 9 9× 9× 9 729 10 10× 10× 10 1000 11 11× 11× 11 1331 12 12× 12× 12 1728 13 13× 13× 13 2197 14 14× 14× 14 2744 15 15× 15× 15 3375 16 16× 16× 16 4096 17 17× 17× 17 4913 18 18× 18× 18 5832 19 19× 19× 19 6859 20 20× 20× 20 8000 21 21× 21× 21 9261 22 22× 22× 22 10648 23 23× 23× 23 12167 24 24× 24× 24 13824 25 25× 25× 25 15625

### Properties of a Cube

The following are some properties of a cube of numbers where students should remember while doing any cube for an integer.
(i) The cube of even natural numbers is always even.
(ii) The cube of odd natural numbers is always odd.
(iii) The sum of the cubes of first n natural numbers is equal to the square of their sum.
(iv) Cubes of the numbers ending in digits 1, 4, 5, 6, and 9 are the numbers ending in the same digit. For example, 1³=1, 4³=64, 5³=125, 6³=216, and 9³=729.

### Cubes of Numbers 1 to 50 PDF Download

The following list helps the children while solving the problems with arithmetic operations. Children can refer to this list of cubes 1 to 50 to solve the problems accurately. ### Examples of Cube of Numbers

Example 1:
Find the cube of each of the following numbers
(i) 55    (ii) -41    (iii) 3.5
Solution:
(i) Given number is 84
To get the cube of 84, we use the formula
n³ = n × n × n
55 = 55 × 55 × 55 = 1,66,375
Hence, the cube of the number 55 is 1,66,375.

(ii) Given number is -41
The number given is a negative integer and the resultant answer is also a negative integer because it is calculated three times and gives a negative integer.
-41 = -41 × -41 × -41 = -68,921.
Thus, the cube of a number -41 is -68,921.

(iii) 3.5 is a given number
When a decimal number is cubed we convert the decimal number into a fraction number.
(3.5)³ = (35/10)³
= (7/2)³
= 7³/2³
= (7× 7× 7)/(2× 2× 2)
= 343/8.

Example 2:
What is the value of x if x³ = 64?
Solution:
The formula to calculate the cube is
x³ = x × x × x
Here, x³ = 64
x = ∛64 = 4
Therefore, x=4.

Example 3:
Find out the cube numbers 7 and 12. Also, find the sum of the cube numbers?
Solution:
Firstly, we find the cubed numbers 7 and 12.
The cube number 7 is 7 × 7 × 7 = 343.
The cube number 12 is 12 × 12 × 12 = 1728.
Now, we have to find the sum of the cubed numbers.
i.e., 7³ + 12³ = 343 + 1728 = 2071.
As a result, Sum of the cube numbers is 2,071.

### FAQ’s on Cubes

1. What is a Cube?
A cube is a number where it is multiplied thrice by itself. The symbol for the cubed is ³. For example, to get 5³ we multiply 5 three times itself i.e., 5 × 5 × 5 = 125.

2. What are the cube numbers from 1 to 10?
The cubed numbers from 1 to 10 are 1, 8, 27, 64, 125, 216, 343, 512, 729, and 1000.

3. Is 600 is a perfect cube?
To know 600 is a perfect cube or not. We have to find the multiplies of 600.
The multiplies of 600 are 2× 2× 2× 3× 5× 5 are multiplies of 600.
2³ × 3 × 5² = 600.
Therefore, 600 is not a perfect cube because all the factors are not multiple of three.

4. Which are the perfect cubes among the numbers 343, 576, 2197?
Among the given numbers 343 and 2197 are the perfect cubes because 7× 7× 7 = 343 and 13× 13× 13 =2197. Whereas 576 is the perfect square number i.e., 24× 24 = 576.

## Worksheet on Mean of Ungrouped Data | Activity Sheet on Finding Mean Practice Problems and Solutions PDF

Mean is the most commonly used central tendency measure. In Mathematics, there are various types of means but in Statistics, the mean is the sum of observations divided by the total number of observations. Also, there are other names for mean like arithmetic mean, average. Practicing Mean of Raw Data word problems can make you solve any kind of questions in exams easily. So, Answer all the questions provided in this Worksheet on Mean of Ungrouped Data PDF and gain extra knowledge on finding the arithmetic mean.

Also Check:

## Practice Question on Mean for Ungrouped Data Worksheet PDF

Example 1:

The heights of five students are 155 in, 140 in, 150 in, 160 in, and,165 in respectively. Find the mean height of the students.

Solution:

Given heights of 5 students are 155 in, 140 in, 150 in, 160 in, and,165 in
Sum of the heights of five students = (155+140+150+160+165) = 770
Using Mean Formula,
Mean = {Sum of Observation} ÷ {Total numbers of Observations}
= 770 ÷ 5
= 154 in

Example 2:

Find the mean of the following.
(i) The first five positive integers.
(ii) 4, 11, 3, 5, 10, 15, 40

Solution:

(i) The first five positive numbers are 1, 2, 3, 4, 5
Mean of the five positive numbers = Sum of five positive numbers ÷ Total number
= 1+2+3+4+5 ÷ 5
= 15 ÷ 5
= 3
Hence, the mean of the first five positive numbers is 3.
(ii) Given list of observations are 4, 11, 3, 5, 10, 15, 40
To find the mean, use the mean formula and apply the given observations;
Mean = Sum of Observations ÷ Number of Observations
= 4+11+3+5+10+15+40 ÷ 7
= 88 ÷ 7
= 12.57(approx)

Example 3:

In the annual board exams in mathematics, 5 students scored 80 marks, 8 students scored 75 marks, 10 students scored 65 marks and 2 students scored 55 marks. Find the mean of their score.

Solution:

Given that,
Number of students scored 80 marks  =  5
Number of students scored 75 marks  =  8
Number of students scored 65 marks  =  10
Number of students scored 55 marks  =  2
Mean = [5(80) + 8(75) + 10(65) + 2(55)] / (5 + 8 + 10 + 2)
= 1760 / 25
= 70.4

Example 4:

The mean weight of five complete computer stations is 167.2 pounds. The weights of four of the computer stations are 158.4 pounds, 162.8 pounds, 165 pounds, and 178.2 pounds respectively. What is the weight of the fifth computer station?

Solution:

Given Mean weight of five computer stations = 167.2 pounds
To find the weight of the fifth computer station, use the mean formula
Let the fifth weight of computer stations be x.
Mean = Sum of weights / Number of weights
167.2 = 158.4+162.8+165+178.2+x / 5
167.2*5 = 664.4 + x
664.4 + x = 836
x = 836 – 664.4
x = 171.6 pounds
Hence, the weight of the fifth computer station is 171.6 pounds.

Example 5:

The mean height of 4 members of a family is 5.5. Three of them have heights of 5.6, 6.0, and 5.2. Find the height of the fourth member.

Solution:

Given heights of family members are 5.6, 6.0, 5.2
Mean height of 4 members of a family = 5.5
Let the fourth member height would be x
To find the fourth height x, apply the mean formula
Mean = sum of observations / number of observations
5.5 = 5.6+6.0+5.2+x / 4
5.5 * 4 = 16.8+x
16.8 + x = 22
x = 22-16.8
x = 5.2
Hence, the height of the fourth member is 5.2

Example 6:

The following data represent the number of pop-up advertisements received by 8 families during the past month. Determine the mean number of advertisements received by each family during the month.
20 25 30 25 40 45 50 55

Solution:

Given data is 20 25 30 25 40 45 50 55
Mean = Sum of data / Total number of data
= 20+25+30+35+40+45+50+55 / 8
= 300/8
= 37.5

Example 7:

In a week, the temperature of a certain place is measured during winter are as follows 24ºC, 28ºC, 22ºC, 18ºC, 30ºC, 26ºC, 22ºC. Find the mean temperature of the week.

Solution:

Given temperatures are 24ºC, 28ºC, 22ºC, 18ºC, 30ºC, 26ºC, 22ºC
Mean temperature = Sum of all temperature / Number of terms
= 24ºC+28ºC+22ºC+18ºC+30ºC+26ºC+22ºC / 7
= 170 / 7
= 24.28 (approx)

## Drawing Parallel Lines with Set Squares | How to Construct Parallel Lines Using Set Square with Examples?

This article helps the students to learn about the construction of parallel lines and perpendicular lines using set squares. 5th Grade students should be aware of drawing types of lines with set squares and compass and protractor. Here, we use set squares to construct the parallel lines. Are you excited to know how to use them? Then, look at this guide without any fail.

On this page, you can find what are parallel lines and the steps followed while constructing parallel lines by using set squares and a few examples to give practical knowledge to children.

## Parallel Lines – Definition

Parallel lines are the lines that lie on the same plane and do not meet and never intersect each other and keep the same distance between them are known as parallel lines. They are also known as non-intersecting lines and they meet at infinity. If we have two lines AB and CD of the same length and equidistant to each other then we call them as the line AB is parallel to the line CD. Let us have a small representation of how the parallel lines look as follows: To represent parallel lines we use the symbol ‘ || ‘and read as the line AB is parallel to the line CD. The distance between the two lines is always the same.

### Set Squares

Set squares are of two types and they are named according to their angles. The set squares or pairs of triangles included a right angle and two angles of 45 degrees. It is known as the isosceles triangle. The other with a right angle, one of 60 degrees and another of 30 degrees. It is known as the scalene triangle. Just have a look at set squares as follows The above figure says that one is 45o set square and the other is 60o-30o set square. Set squares are used for drawing lines like parallel and perpendicular lines.

### How to Draw a Parallel Line using Set Square?

There are steps to follow to draw parallel lines using set squares. Here are the steps that are looking for:

Step 1: Draw a straight line by positioning an edge of one of the 45 degrees set squares against a ruler to which we want to draw the parallel lines. Step 2: Place the 60set square that is attached to the 45set square, and press it lightly while drawing so it doesn’t move and it fixes the exact position. Step 3: Now, fix the 60o– 30o set square and move the 45o triangular set square upwards or downwards and draw the required parallel lines. ### Examples of Drawing Parallel Lines with Set Squares

Example 1:
Draw a parallel line through a point 4 cm by using a ruler and a set square.
Solution:
We draw parallel lines with the help of set squares by following steps.
Step1:
(i) Draw a line PQ by 45 degrees set square and mark a point A on it.
(ii) Draw AB = 4 cm with the help of a ruler. Step 2:
(i) Place one set square on the line segment PQ.
(ii) Place another 60 – 30 degrees set square as shown below in the figure. Step 3:
(i) Press the 60° set square, slide the 45° set square along the other set square till the edge of the square touches point B.
(ii) Through point B, draw a line BC along the edge.
(iii) BC is the line required parallel to PQ through point B. Here, the line PQ is parallel to the line BC i.e., PQ || BC.

Example 2:
Draw a parallel line to the given line AB and passes through point X with the help of a ruler and a set square.
Solution:
Step 1:
(i) Use a set square and draw a parallel line AB and passes through point X.
(ii) Now, place the set square on line AB and place the ruler on the short side of the square. Step 2:
Just slide the set square along the ruler till the side of the set square touches the point X that was placed against the parallel line AB. Step 3:
Use the edge of the set square to draw a line through point X as shown below and name the parallel line PQ. Thus, line AB is parallel to line PQ i.e., AB || PQ.

### FAQ’s on How to Draw Parallel Lines with Set Square and Ruler

1. Why set squares are used?
Set squares or triangles are used in technical drawing and in engineering, to furnish a straightedge at a right angle to a baseline.

2. How many set squares are used while drawing?
There are two types of set squares. One set square of 45 degrees and the other is 60-30 degrees set square. The 45 degrees square also has a 90 degrees angles. These set squares are used to draw parallel lines and perpendicular lines.

3. How to find the distance between parallel lines and set squares?

By holding it with one hand and placing a set square with one arm of the right angle corresponding with the edge of the ruler. Now, you can draw the line segment together with the edge of the set square. Later, you are good to measure the distance between parallel lines and set squares.

## Worksheet on Median of Ungrouped Data | Median of Discrete & Raw Data Activity Sheet PDF

Are you wondering how to find and practice problems on the median of raw data? This is the right page for you all as it is holding with the Worksheet on Median of Ungrouped Data. Make use of this statistics median of discrete data worksheet pdf and solve all types of questions with flexibility. Also, you can bridge all your weak areas by answering the questions involved in the worksheet of the median of ungrouped data. Try to memorize all median formulas & tricks from our median of ungrouped data activity sheet pdf and solve the questions effortlessly.

Do Refer:

### Activity for Median of Ungrouped Data PDF

Example 1:
The weight of 5 ice bars in grams is 141, 152, 135, 117, 120. Find the median.

Solution:

Given data is 141, 152, 135, 117, 120
Arrange the data in ascending order: 117, 120, 135, 141, 152
The number of observations is 5, which is odd
Median = Middle Number or n+2/2
= 5+1/2
= 6/2
= 3rd observation
= 135
Hence, the median of the weight of 5 ice bars in grams is 135.

Example 2:
Find the median of the first 5 prime numbers.

Solution:

The first five prime numbers are 2, 3, 5, 7, 11. It is in an ordered form.
Next, find the number of observations ie., 5 which is odd.
Therefore, the middle value 5 is the median of the first five prime numbers.

Example 3:
Find the median of the following:
(i) The first eight even natural numbers
(ii) 8, 0, 2, 4, 3, 4, 6

Solution:

(i) The first eight even natural numbers are 2,4,6,8,10,12,14,16
The total number of observations = 8, which is even.
Therefore, the median is the mean of the 4th and 5th observations, ie.,
Median = Mean {4th + 5th variant}
= 1/2{8 + 10}
= 18/2
= 9
(ii) Given data is 8, 0, 2, 4, 3, 4, 6
Sort them in ordered form (ascending order): 0, 2, 3, 4, 4, 6, 8
Total number of observations = 7, which is odd
Hence median is the middle value( 4th variant) ie., 4.

Example 4:
Find the median of 21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22

Solution:

Given data is 21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22
First sort the given raw data in ascending order: 6,8,9,13,15,16,18,20,21,22,25
The total number of variants (n) = 11, which is odd.
Hence, median = value of 1/2(n+1)th observation.
= 1/2(11+1)th
= 1/2(12)th
= 6th observation
Therefore, the value of 6th observation is 16.
Hence, the median is 16.

Example 5:
Find the median of the following data.
Variates: 10, 11, 12, 13, 14
Frequency: 1, 2, 3, 4, 5

Solution:

First, arrange the variates in ascending order, we get
10, 11, 11, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14
The number of variates = 15, which is odd.
Hence, Median = n+1/2th variate
= 15+1/2th
= 16/2th
= 8th variate
= 13.

Example 6:
The marks obtained by 10 students in a class test are given below.

 Marks Obtained Number of Students 10 7 9 5 4 3 2 1

Calculate the median of marks obtained by the students?

Solution:

Given 10, 10, 10, 10, 7, 7, 7, 9, 9, 5
Sorting the variates in ascending order, we get
5, 7, 7, 7, 9, 9, 10, 10, 10, 10
The number of variates = 10, which is even.
Hence, median = mean of 10/2th and (10/2+1)th variate
= mean of 5th and 6th variate
= mean of 9 and 9
= 9+9/2
= 18/2
= 9.

Example 7:
Find the median of the first four odd integers. If the fifth odd integer is also included, find the difference of medians in the two cases.

Solution:

Take the first four odd integers in ascending order, we get
1, 3, 5, 7.
The number of variates = 4, which is even.
Hence, median = mean of 2nd and 3rd variate
= 1/2 (3+5)
= 8/2
= 4
When the fifth integer is included, we have (in ascending order)
1, 3, 5, 7, 9
Now, the number of variates = 5, which is odd.
Therefore, median = 5+1/2th
= 6/2th
= 3rd variate
= 5
Hence, the difference of medians in the two cases = 4 – 5 = -1.