CBSE conducts class X board exams in March every year. The syllabus for Class 10 English is not at all tough. If students prepare right from the beginning or read chapters of their textbooks, it will be way too easy to score well. The syllabus is designed in such a way that it enhances the students’ written as well as verbal or communicative skills. Yet, children pay heed to the subject only in the end and tend to make silly mistakes. Overnight preparation for any subject can be very risky as it leads to excessive stress, anxiety and sheer confusion. Therefore, students must devote at least a couple of months to the subject, to improve their comprehension skills as well as written and verbal.

English Class 10 Board Exam Overview:

The marking scheme of English for class X is as under: 

Section A – This section tests the reading ability of the students. Questions in this section sum to 20 marks. Students are required to read passages and answer the questions asked from the same. These are in the form of multiple-choice questions.

Section B – Section B focuses on the writing skills of the students. The entire section is of 30 marks.

Section C – Section C covers the entire literature portion, questions are asked from the prescribed textbooks of students in class 10. The entire section is of 30 marks.

Best books for class X English

If you read your textbooks thoroughly and on a regular basis, you can score good marks in the literature section easily. For the grammar section, practice is the only mantra for success. Nonetheless, keeping in mind your busy schedule of the students, the following are the best reference books that may prove very helpful to children.

• Communication English by Oswal – The book also has previous years’ question papers that students can refer to.
• Together with English is also a good alternative for all those who wish to practice grammar.
• Students can also refer to High School English Grammar and Composition Book- Wren and Martin

However, it is recommended to study NCERT books prescribed by the CBSE board thoroughly on a regular basis.

Students who prepare at the last moment may suffer due to confusion. Grammar portions at times can confuse students in terms of their format, structure, and no practice. They may face problems in expressing their feelings and thoughts in story writing or composition or may not understand unseen passages that test their reading skills.

Following are a few tips to work on when studying at the last moment or at least one week before the examination: 

• Read all the stories and poems thoroughly, remember the central idea/ theme, author/ poet and the title of the chapter.
• Read news articles or short stories at least twice a day.
• Write a para on current topics, or maintain a diary wherein you can pen down your emotions.
• Pen down formats of formal letter, subject been agreement, tenses and story writing and important topics on a blank paper, keep it on your study table so that you can go through it regularly.
• Make a schedule, devote equal time to every section.
• Keep a watch in front of you when reading a passage or writing so that you get an idea of how much time it takes for the same.
• Attempt mock tests or solve the last ten years’ paper.

Frequently Asked Questions on English Class X Examination

Q 1. Is it okay to study from last years’ paper? 

Answer. If you are preparing at the last moment, you can go through the last ten years’ papers for practice. Also, do remember the correct title of the chapter in literature and the author or poet.

Q 2. What should I do to enhance my vocabulary? 

Answer. You need to keep reading your textbooks, newspapers, or news articles. Try learning three to five new words daily along with their synonyms.

Q 3. How does a school assess students internally? 

Answer. The pattern of internal assessments depends on schools. Some schools assess on the basis of internal tests or exams while some assess on the basis of the overall performance of the students or assignments as well as an oral examination.

Q 4. Is English paper tallies only 80 marks? 

Answer. English paper sums to 80 marks in total while the remaining 20 marks rest with the school internal assessment.

The post Which Book is Best for English Class 10? appeared first on Learn CBSE.

Not getting excellent marks in your English board exam class 10 or getting below-average marks in your final examination is something that you are always scared of. Passing or failing in any subject holds a lot of importance in the way one strategizes their way to attempt the exam.

It happens very rarely that you are not nervous in any of your exams. Normally students get confused and nervous as soon as they get their exam schedule. The questions they face are what to attempt, how to attempt and the time management of completing the entire question paper— all this is a pretty tough and technical job.

Things to keep in mind while attempting English Board Exam Class 10 –

There are so many things that come to your mind the moment you sit to attempt your board’s exam paper. You just have three hours to complete the entire question paper (be it any subject). The most important thing is to manage your time accordingly. But how to do so? You might get stuck up with time management, so here are some listicles that will help you manage your English Board Exam Class 10:

• Read your Class 10 English paper very carefully. You should be thoughtful and attentive while going through your subject paper.
• Do not just rush to begin your paper. While attempting any examination, you always get 10-15 minutes extra to go through your paper. Utilize those minutes watchfully.
• Do not be hasty while reading the questions. This will help you a lot in making pointers in your mind about the answers.
• Examine the nature of the questions. For instance, are they easy to understand or not.
• Now, make pointers on the rough side for the questions that you need to attempt first.
• Make points on which questions you can attempt first.
• Begin gradually after recollecting all your answers.

English Board Exam Class 10 Board –

Pan India there are majorly two boards for class 10:

• CBSE (Central Board of Secondary Education)
• ICSE (Indian Certificate of Secondary Education)

There is always a selected set of the syllabus that is given by the school/board’s authority. For CBSE English board exam class 10, the communicative latest syllabus that has been incorporated in 2020. This will help the candidate and the students to score better marks than what they used to earlier. As per the latest guidelines issued by the CBSE board, the students’ mark sheet will be evaluated on the basis of final examinations.

The total exam sheet consists of 80 marks. Various sections are marked for the students to be able to simplify their papers.

Section A (Reading Skills)- 20 marks

• To develop reading ability in the candidates.
• Read the passage carefully and understand the main theme of the topic.
• Identify various important and hidden points/messages in the message.
• Always attempt the answers in your own words.
• Need to enhance your reading skills so that you can form your answers in your own words.

Section B (Writing and Grammar)- 30 words

• You need to be quick and neat in your writing.
• Make sure of grammatical errors, there should be none.
• Read your question at least thrice to make sure that you do not make any mistakes in writing.
• Your writing should be neat so that the examiner is able to understand your answers.
• Do not rush while writing answers.
• Format plays an important role in writing your answers. There are always extra marks on the formatting in your papers.
• Plan and organize the answers and questions in a proper sequence.
• Prepare each question carefully so that you are able to write grammatically correct answers.

Section C (Literature Textbook and Supplementary Reading Text)- 30 words

• There are four out of six Short Answer Type Questions to be answered in 20-30 words each. Keep it to the point.
• Again, four out of six Short Answer Type Questions need to be answered in 40-50 words each. Do not forget to make pointers.
• Then, one Long Answer to be answered in about 100-120 words from two options. Underline the main lines with a black pen.
• Once again, one Long Answer Type Question to be answered in about 100-120 words. Divide the answer into paras, points, and underline the main context.

Which section to attempt first?

Time management holds the most importance while giving your board examination. Do not rush or jump from one question to the third question. By doing this you will end up ruining your paper. Make use of time properly. Plan your question paper whilst you were reading your paper. By this time, you should have got your answers ready at your fingertips.

It is advised to go in the reverse order. Since the last section is Literature Textbook and Supplementary Reading Text, that is something the students are already well-versed with and takes them lesser time to attempt the answers. It is also helpful the textual answers with a fresh mind as it helps to recall everything you revised from the textbook.

After Section C, switch to Section B and attempt the grammatical questions you are well trained with all the practice you have done by attempting similar questions from previous year’s papers. Lastly, when you reach Section A, remember to have underlined the answers during the 15 minutes you got to go through the paper as this section is very time-consuming and new to the students.

FAQs on English board exam class 10

Q 1. What does Section A comprise of?

Answer. Section A of English board exam class 10 comprises two passages that test the reading, vocabulary and interpretation skills of the student. The passages hold multiple choice questions and each passage values 10 marks which makes the section 20 marks in total.

Q 2. What is Section B based on?

Answer. Section B of English board exam class 10 covers concepts and topics such as:

1. Modals
2. Tenses
3. Subject-verb concord
4. Paragraph writing
5. Reported speech
6. Formal letter

a. Commands and requests

b. Determiners

c. Statements

Q 3. What are the best-recommeneded books for English board exam class 10?

Answer. The recommended books for preparation of the English board exam class 10 are—

i. Published by NCERT, New Delhi

ii. FIRST FLIGHT – Text for Class X

The post Which Section to Attempt First in English Board Exam Class 10? appeared first on Learn CBSE.

You can Practice Various problems involving Set Operations here. Use them to learn different operations such as Union, Intersection, Complement, etc. We are sure by the end of this article you will be familiar with the Operations on Sets through the Problems Provided. Sets Questions and Answers here help you to understand the Operations concept much easier. If you need more help refer to Set Theory and learn the entire concept of Sets.

Sets Questions and Answers

1.  Let A and B be two finite sets such that n(A) = 15, n(B) = 24 and n(A ∪ B) = 30, find n(A ∩ B)?

Solution:

n(A U B) = n(A) + n(B) – n(A ∩ B)

Rearranging we get the n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

= 15+24 – 30

= 9

2. If n(A – B) = 28, n(A ∪ B) = 80 and n(A ∩ B) = 35, then find n(B)?

Solution:

Using the formula n(A∪B) = n(A – B) + n(A ∩ B) + n(B – A)

80 = 28 + 35 + n(B – A)

80 = 63 + n(B – A)

n(B – A) = 80 – 43

n(B – A) = 37

Now n(B) = n(A ∩ B) + n(B – A)

= 35 + 37

= 72

3. If n(U) = 28, n(A) = 14, n (AnB) = 10, n(B’) = 22 find n(AUB)?

Solution:

n (A U B) = n (A) + n (B) – n (A n B)

n (B) = n (U) – n (B’)

= 28 – 22

= 6

n (A U B) = n (A) + n (B) – n (A n B)

= 14+6 – 10

= 10

Thus, the value of n (A U B) = 10.

4. If A = { a, b } and B = { 3, 4 }. What is the Cartesian Product of Two Sets AxB and Bx A. Verify whether they are equal or not?

Solution:

Cartesian Product AxB = { (a, 3), (a, 4), (b, 3), (b, 4)}

BxA = { (3, a), (4, a), (3, b), (4, b)}

Therefore, AxB ≠ BxA

5. If A = {12, 13, 14, 15, 16, 17 } and B = {7, 8, 9}, then find the values of (A – B) and (B-A)?

Solution:

Given A = {12, 13, 14, 15, 16, 17 }

B = {7, 8, 9}

A-B = {12, 13, 14, 15, 16, 17 } – {7, 8, 9}

= {12, 13, 14, 15, 16, 17}

12, 13, 14, 15, 16, 17 are the elements that are present in A but not in B.

B-A = {7, 8, 9} – {12, 13, 14, 15, 16, 17}

= {7, 8, 9}

6. If If A = {10, 11, 12, 13, 14 } and B = { 10, 12, 14, 15 }. Find A ∩ B?

A ∩ B = {10, 11, 12, 13, 14 } ∩ { 10, 12, 14, 15 }

= { 10, 12, 14}

A ∩ B contains all the elements that are in both Sets A and B.

7. Let A = {2, 3, 4}, B = {3, 4, 5} , U = { 1, 2, 3, 4, 5} find the Complement of A and B?

Solution:

Given

A = {2, 3, 4}

B = {3, 4, 5}

U = {1, 2, 3, 4, 5}

A^c or A^‘ = U – A

= {1, 2, 3, 4, 5} – {2, 3, 4}

={ 1, 5}

B^c or B^‘ = U – B

= {1, 2, 3, 4, 5} – {3, 4, 5}

{ 1, 2}

The post Problems on Operation on Sets | Sets Problems and Solutions appeared first on Learn CBSE.

If you looking for different questions on Sets you have come the right way. We have curated Word Problems on Sets with Step by Step Solutions for your understanding. Make use of them while practicing Sets Problems and increase your conceptual knowledge. In this, you will understand how to Solve Sets Word Problems using Venn Diagrams easily. If you need help on different concepts of Sets refer to Set Theory and learn the representation of a set, types of sets, etc. Check out the Solved Examples provided and learn how to solve related problems during your work.

1.  Let A and B be two finite sets such that n(A) = 30, n(B) = 18 and n(A ∪ B) = 26, find n(A ∩ B)?

Solution:

Formula for n(A ∪ B) = n(A) + n(B) – n(A ∩ B).

Rearranging it we get the n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

=30+18 – 26

= 22

Therefore, n(A ∩ B) = 22.

2. If If n(A – B) = 12, n(A ∪ B) = 45 and n(A ∩ B) = 15, then find n(B)?

Solution:

n(A∪B) = n(A – B) + n(A ∩ B) + n(B – A)

45 = 12+15 +n(B-A)

n(B-A) = 45-12-15

= 45-27

= 18

n(B) = n(A ∩ B) + n(B – A)

= 15+18

= 33

3. In a group of 80 people, 37 like cold drinks and 52 like hot drinks and each person likes at least one of the two drinks. Find How many people like both coffee and tea?

Solution:

Let A = Set of people who like cold drinks.

B = Set of people who like hot drinks.

Given

(A ∪ B) = 80 n(A) = 37 n(B) = 22 then;

n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

= 37+52-80

= 89 – 80

= 9

Therefore, 9 people like both tea and coffee.

4. Let S={4, 5, 6}. Write all the possible partitions of S?

Solution:

Remember that partition of S is a collection of nonempty sets that are disjoint and their union is S.

Possible Partitions of S are

{4},{5},{6}
{4,5},{6}
{4,6},{5}
{5,6},{4}
{4,5,6}.

5. In a school, there are 30 teachers who teach Mathematics or Physics. Of these, 18 teach Mathematics and 6 teach both Physics and Mathematics. How many teach Physics only?

Solution:

Total Number of Teachers who teach Mathematics or Physics = 30

Number of Teachers who teach Mathematics n(M) = 18

Number of Teachers who teach both Mathematics and Physics n( M ∩ P) = 6

n(M) = 18 n( M ∩ P) = 6 and n (M ∪ P )= 30

n (M ∪ P ) = n(M) + n(P) – n(M ∩ P)

30 = 18 + n(P) – 6

30 = 12 + n(P)

⇒ n(P) = 30 – 12

⇒ n(P) = 18

Number of teachers teach Physics only = n( P – M)

n( P – M) = n(P) – n( M ∩ P )

= 18 – 6

n( P – M) = 12

Number of teachers who teach physics only is 12.

6. In a survey of 80 people, it was found that 35 people read newspaper H, 20 read newspaper T, 15 read the newspaper I, 5 read both H and I, 10 read both H and T, 7 read both T and I, 4 read all three newspapers. Find the number of people who read at least one of the newspapers?

Solution:

n(H) is the Number of People who read the newspaper H i.e. 35

n(I) is the Number of People who read the newspaper I i.e. 15

n(T) is the Number of People who read the newspaper T i.e. 20

n(H ∩ I) is the Number of People who read H and I i.e. 5

n(H ∩ T) is the Number of People who read H and T i.e 10

n(T ∩ I) is the Number of People who read T and I i.e. 7

n(H ∩ T ∩ I) is the Number of People who read all three Newspapers H, T, I i.e. 4

n(H ∪T ∪ I ) is the Number of people who read at least one of the newspapers

n(H ∪T ∪ I ) = n(H) + n(T) + n(I) – n(H ∩ T) – n(T ∩ I) – n(H ∩ I) + n(H ∩ T ∩ I)

n(H ∪T ∪ I )= 35+20+15-10-7-5+4

= 74-22

= 52

Therefore, the number of people who read at least one of the newspapers is 52.

7.  In a school, all pupils play either Hockey or Football or both. 400 play Football, 150 play Hockey, and 130 play both the games. Find
(i) The number of pupils who play Football only,
(ii) The number of pupils who play Hockey only,
(iii) The total number of pupils in the school.

H = Hockey and F = Football

n (H ) = 150 n (F)= 400

n ( H ∩ F) = 130

(i) The number of pupils who only play Football = n (F – H )

n (F – H ) = n(F) – n( F ∩ H )

= 400 – 130

= 270

(ii) The number of pupils who only play Hockey = n (H – F )

n (H– F ) = n(H) – n( F ∩ H )

= 150 – 130

= 20

(iii) The total number of pupils in school

= n(H) + n(F) – n (F ∩ H)

= 150 + 400 – 130

= 420

The post Word Problems on Sets | Solved Problems on Sets appeared first on Learn CBSE.

Are you looking everywhere to represent sets using Venn diagrams in different situations? Then, this is the place for you as we took several examples explaining Venn Diagrams in Different Situations. We took a few examples and drew Venn diagrams for all of those situations. To learn such similar concepts check out Set Theory and learn the concepts effectively.

1. Given the set Q = { x : 4x – 5 < 11, x is a positive integer }. Draw and label a Venn diagram to represent the set Q?

Solution:

Since an Equation is given, we need to find the value of x

4x-5<11

4x< 11+5

4x<16

x<4

Q = {1, 2, 3}

Draw a Circle and Label it Q write all the elements 1, 2, 3 in it.

2. Draw a Venn Diagram representing the Complement of Set A?

Solution:

3. Draw Venn Diagram representing the Set of R = {Monday, Tuesday, Wednesday}?

Solution:

4. U is a universal set. P and Q are subsets of U. They are also overlapping sets. U = { 0, 1, 2, 3, 4 5, 6, 7, 8, 9} P ={ 6, 3, 9} Q = {2, 4, 6 8}. P ∩ Q = {6}

• Draw a rectangle which represents a universal set.
• Draw two circles inside the rectangle which represents P and Q.
• The circles overlap.
• Write the elements of P and Q in the respective circles such that common elements are written in overlapping portion (6).
• Write the rest of the elements in the rectangle but outside the two circles.
• The figure represents P ∩ Q = {6}

Solution:

5. Draw Venn Diagram showing the Disjoint Sets S and T in which S = { 2, 4, 6, 8} T = { 1, 3, 5, 7}.

Solution:

6. U is a Universal Set and A, B are Subsets of the Universal Set. Draw Venn Diagram representing U = { 1, 2, 3, 4, 5, 6, 7, 8, 9} A = { 1, 2, 5, 6} B = { 3, 9}?

• Draw a rectangle that represents the universal set.
• Draw circles inside the rectangle which represents A, B.
• Write the elements of A, B inside the circle.
• Write the leftover elements in U that are outside the circle but inside the rectangle.

Solution:

The post Venn Diagrams in Different Situations | Solving Problems with Venn Diagrams appeared first on Learn CBSE.

Venn Diagram is a Pictorial Representation of Relation Between Sets. In general, Venn Diagrams are represented using the Symbols Circles or Ovals. The Elements of the Sets are placed within the Circle. To help you understand the Relationship in Sets using Venn Diagram we have taken a few examples and explained them. Check out the Set Theory to be familiar with various concepts of Sets.

The Following Venn Diagrams show the Relationship between Sets. The Relation between them is explained with a description below the diagram.

Union of Sets using Venn Diagrams

Union of Sets using Venn Diagrams in different cases like Disjoint Sets, A ⊂ B or B ⊂ A, neither subset of A or B is explained in the below figures.

The Above Venn Diagram is A U B when neither A is a subset of B or B is a subset of A.

The above diagram shows A U B when A is a subset of B.

A U B when A and B are disjoint sets

Intersection of Sets using Venn Diagram

Intersection of Sets using Venn Diagrams in different cases like Disjoint Sets, A ⊂ B or B ⊂ A, neither subset of A or B is explained in the below figures.

A ∩ B when neither A ⊂ B nor B ⊂ A. The Intersection Part is highlighted with a color.

In the case of disjoint sets, there will be no common elements thus

A ∩ B = ϕ No shaded part.

Difference of Two Sets using Venn Diagrams

Learn about the difference of sets in various cases like A – B when neither A ⊂ B nor B ⊂ A, when A and B are disjoint Sets in the below modules.

The above diagram depicts the relationship between sets when neither A ⊂ B nor B ⊂ A.

In the case of disjoint Sets, A-B results in A simply. That is better understood by Venn Diagrams

Relationship between Three Sets using Venn Diagrams

The Above Diagram describes the Union Operation between Three Sets i.e. A U B U C.

Intersection Operation between Three Sets A, B, C is given above. The Shaded Region includes the elements that are in Sets A, B, C.

Let’s have a synopsis of the uses of Set Theory and its Applications in Problems. If at all A is a finite set then the number of elements in Set A is given by n(A).

In the Case of Relationship Between Sets using Venn Diagrams two cases arise. Let A and B be two finite sets

a) In Case if A and B are finite sets and disjoint i.e. no common elements then Formulas for Union is as such

n(A ∪ B) = n(A) + n(B)

b) If A and B are not disjoint then

(i) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(ii) n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B)

(iii) n(A) = n(A – B) + n(A ∩ B)

(iv) n(B) = n(B – A) + n(A ∩ B)

Let A, B, C be three finite Sets then

n(A ∪ B ∪ C) = n[(A ∪ B) ∪ C]

= n(A ∪ B) + n(C) – n[(A ∪ B) ∩ C]

= [n(A) + n(B) – n(A ∩ B)] + n(C) – n [(A ∩ C) ∪ (B ∩ C)]

= n(A) + n(B) + n(C) – n(A ∩ B) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)

[We know, (A ∩ C) ∩ (B ∩ C) = A ∩ B ∩ C]

Therefore, n(A ∪B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

The post Relationship in Sets using Venn Diagram| Venn Diagrams and Sets appeared first on Learn CBSE.

This page defines the relationship between the square side length and square area via a coordinate graph. Take the square area, square side as the coordinates of a point. And plot those points on the graph and read the unknown values from the graph easily. You can get the solved example questions on how to draw a graph of area vs side of a square in the following sections.

Relation Between Square Side Length & Area

The square area is defined as the product of the length of each side with itself. Its formula is given as side_length².

So, the relationship between the square side and the area is a square graph.

Square Area A = side² = s².

Solved Example Questions

Example 1.

Draw a graph of area vs side of a square. From the graph, find the value of the area, when the side length of the square = 3.

Solution:

Square Area A = side² = s².

For different values of s, we get the corresponding value of A.

When s = 0, A = 0² = 0

When s = 1, A = 1² = 1

When s = 2, A = 2² = 4

Thus, we have the points O (0, 0), A (1, 1), B (2, 4).

Plot these points on a graph paper and join them successively to obtain the required graph given below.

Reading off from the graph of area vs. side of a square:

On the x-axis, take the point L at s = 3.

Draw LP ⊥ x-axis, meeting the given graph at P.

Clearly, PL = 9 units.

Therefore, s = 3 ⇒ A = 9.

Thus, when s = 3 units, then A = 9 sq. units

Example 2.

Draw a graph for the following.

Side of the square (in cm) 1, 2, 3, 4 and Area (in cm) 1, 4, 9, 16 Is it a linear graph?

Solution:

Area of the square A = side² = s².

Draw these square side, area on a table.

Take the side of the square on the x-axis, area on the y-axis.

Plot the points P (1, 1), Q (2, 4), R (3, 9), s (4, 16) on the graph paper.

From the graph, we can say that square area vs side does not form a linear graph. It forms a square graph.

Example 3.

(a). Consider the relation between the area and the side of a square, given by A = s². Draw a graph of the above function.

(b). From the graph, find the value of A, when s = 2.5, 3.5.

Solution:

Given that,

Square Area A = s².

For different values of s, we get the corresponding value of A.

s = 0 ⇒ A = 0² = 0

s = 0.5 ⇒ A = 0.5² = 0.25

s = 1 ⇒ A = 1² = 1

s = 1.5 ⇒ A = 1.5² = 2.25

Thus, we get the points O (0, 0), A (0.5, 0.25), B (1, 1), C 1.5, 2.25)

Plot these points on a graph paper and join them successively to obtain the required graph given below.

(b). Reading off from the graph of area vs. side of a square:

On the x-axis, take the point L at s = 2.5.

Draw LP ⊥ x-axis, meeting the given graph at P.

Clearly, PL = 6.25 cm².

Therefore, s = 2.5 ⇒ A = 6.25.

Thus, when s = 2.5 cm, then A = 6.25 cm²

On the x-axis, take the point M at s = 3.5.

Draw MQ ⊥ x-axis, meeting the given graph at Q.

Clearly, MQ = 12.25 cm².

Therefore, s = 3.5 ⇒ A = 12.25 cm²

Thus, when s = 3.5 cm, then A = 12.25 cm²

The post Graph of Area vs. Side of a Square | Draw the Graph of Area of Square Vs Side of the Square appeared first on Learn CBSE.

In Maths, a Percentage is a Number or Ratio Expressed as a Fraction Over 100. In other words, percent means parts per hundred and is given by the symbol %. If we need to calculate the percent of a number you just need to divide the number by whole and multiply with 100. Percentages can be represented in any of the forms like decimal, fraction, etc.

You can get entire information regarding Percentage like Definition, Formula to Calculate Percentage, Conversions from Percentage to another form, and vice versa in the coming modules. Learn the Percentage Difference, Increase or Decrease Concepts too that you might need during your academics or in your day to day calculations.

List of Percentage Concepts

Access the concepts that you want to learn regarding the Percentage through the quick links available. Simply tap on the concept you wish to prepare and get the concerned information explained step by step. Clarify all your queries and be perfect in the corresponding topics.

• Fraction to Percentage
• Percentage to Fraction
• Percentage to Ratio
• Ratio to Percentage
• Percentage to Decimal
• Decimal to Percentage
• Percentage of the given Quantity
• How much Percentage One Quantity is of Another?
• Percentage of a Number
• Percentage Increase
• Percentage Decrease
• Basic Problems on Percentage
• Solved Examples on Percentage
• Problems on Percentage
• Real-Life Problems on Percentage
• Word Problems on Percentage
• Application of Percentage

Percentage Formula

To make your calculations quite simple we have provided the Percentage Formula here. Make use of it during your calculations and arrive at the solution easily.

Formula to Calculate Percentage is given by = (Value/Total Value) *100

How to Calculate Percentage of a Number?

To find the Percent of a Number check the following procedure

Consider the number to be X

P% of number = X

Removing the % sign we have the formula as under

P/100*Number = X

Percentage Change

% Change = ((New Value – Original Value)*100)/Original Value

There are two different types when it comes to Percent Change and they are given as under

• Percentage Increase
• Percentage Decrease

Percentage Increase

If the new value is greater than the original value that shows the percentage change in the value is increased from the original number. Percentage Increase is nothing but the subtraction of the original number from the new number divided by the original number.

% increase = (Increase in Value/Original Value) x 100

% increase = [(New Number – Original Number)/Original Number] x 100

Percentage Decrease

When the new value is less than the original value, that indicates the percentage change in the value shows the percent decrease in the original number. Percentage Decrease is nothing but the subtraction of new number from the original number.

% Decrease = (Decrease in Value/Original Value) x 100

% Decrease = [( Original Number – New Number)/Original Number] x 100

Percentage Difference

If you need to find the Percentage Difference if two values are known then the formula to calculate Percentage Difference is given by

Percentage Difference = {|N1 – N2|/(N1+N2/2)}*100

Conversion of Fraction to Percentage

To convert fraction to percentage follow the below-listed guidelines.

• Divide the numerator with the denominator.
• Multiply the result with 100.
• Simply place the % symbol after the result and that is the required percentage value.

Conversion of Decimal to Percentage

Follow the easy steps provided to change between Decimal to Percentage. They are as such

• Obtain the decimal number.
• Simply multiply the decimal value with 100 to get the percentage value.

Solved Examples on Percentage

1. What is 50% of 30?

Solution:

Given 50% of 30

= (50/100)*30

= 1500/100

= 15

Therefore, 50% of 30 is 15.

2. Find 20% of 40?

Solution:

Given 20% of 40

= (20/100)*40

= (20*40)/100

= 800/100

= 8

3. What is 15% of 60 equal to?

Solution:

= (15/100)*60

= (15*60)/100

= 900/100

= 9

4. There are 120 people present in an examination hall. The number of men is 50 and the number of women is 70 in the examination hall. Calculate the percentage of women present in the examination hall?

Solution:

Number of Women = 70

Percentage of Women = (70/100)*120

= (70*120)/100

= 8400/100

= 84%

The Percentage of Women in the Examination Hall is 84%.

5. What is the percentage change in the rent of the house if in the month of January it was Rs. 20,000 and in the month of March, it is Rs. 15,000?

Solution:

We can clearly say that there is a decrease in the rent

Decreased Value  = 20,000 – 15, 000

= 5, 000

Percent Change = (Decreased Value/Original Value)*100

= (5000/20,000)*100

= (1/4)*100

= 25%

Hence, there is a 25% decrease in the rent.

FAQs on Percentage

1. What is meant by Percentage?

A percentage is a Number or Ratio Expressed as a Fraction Over 100.

2. What is the Formula for Percentage?

The formula for Percentage is (Value/Total Value) *100

3. What is the Symbol of Percentage?

The percentage is denoted by the symbol %.

4. What is 10% of 45?

10% of 45 is given by 10/100*45 i.e. 4.5

The post Percentage (How to Calculate, Formula and Tricks) appeared first on Learn CBSE.

Data Handling is an important topic in statistics. It deals with the collecting set of data and maintaining security, confidentiality, and the preservation of the research data. Here data is nothing but a set of numeric values. Get more useful information regarding the data handling such as the types of data, definition, and terms used in the data handling. You can check out the solved example questions in the following sections of this page.

Data Handling Definition

Data handling is the process of collecting, recording, and representing information in a way that is helpful to others. Data is a collection of all numerical figures, observations that represents a particular kind of information.

Data handling is performed depending on the types of data. Data is classified into two types, such as Quantitative Data, Qualitative Data. Quantitive data gives numerical information, while qualitative data gives descriptive information about anything.

Important Terms of Data Handling

• Data: The collection of numerical figures regarding a kind of information is called the data.
• Raw Data: The collection of observations that are gathered initially is called the raw data.
• Range: The difference between the highest and lowest values in the data collection is called the range.
• Statistics: It deals with the collection, representation, analysis, and interpretation of numerical data.

Ways to Represent Data

The following are the simple ways to represent the data. Choose anyone for a better presentation.

• Bar Graph
• Line Graph
• Pictographs
• Histograms
• Stem and Leaf Plot
• Dot Plots
• Frequency Distribution
• Cumulative Tables and Graphs

Solved Examples on Data Handling

Example 1.

Given below are the marks (out of 150) in mathematics obtained by 20 students of a class in an annual examination.

86   75   69   23   72   120   56   82   96   45

100  66   73  46   19   55     46   99  110  105

Arrange the above data in ascending order and find

(i) the lowest marks obtained,

(ii) the highest marks obtained,

(iii) the range of the given data.

Solution:

Given data is

86   75   69   23   72   120   56   82   96   45

100  66   73  46   19   55     46   99  110  105

Arranging the above data in ascending order, we get

19  23  45  46  46  55  56  66  69  72

73  75  82  86  96  99  100  105  110  120

From the above data analysis, we can say that

The lowest marks obtained is 19.

The highest marks obtained is 120.

The range of the data is 120 – 19 = 101.

Example 2.

Given below are the heights (in cm) of 15 boys of a class:

135, 128, 146, 149, 127, 130, 145, 140, 131, 125, 136, 142, 144, 133, 129

Arrange the above data in ascending order and find

(i) the height of the tallest boy,

(ii) the height of the shortest boy,

(iii) the range of the given data.

Solution:

Given data sample is

135, 128, 146, 149, 127, 130, 145, 140, 131, 125, 136, 142, 144, 133, 129

The ascending order of the data is

125, 127, 128, 129, 130, 131, 133, 135, 136, 140, 142, 144, 145, 146, 149

From the above ascending order, we can find

(i) The height of the tallest boy is 149 cm

(ii) The height of the shortest boy is 125 cm.

(iii) The range is 149 – 125 = 24.

Example 3.

The daily wages of 24 laborer who works in the construction field are given below:

280, 150, 160, 500, 520, 460, 780, 590, 860, 450, 410, 425, 360, 1000, 950, 560, 590, 620, 250, 350, 380, 515, 790, 240

Arrange the above data in ascending order and find

(i) the highest amount is taken per day,

(ii) the lowest amount is taken per day,

(iii) the range of the given data.

Solution:

Given that,

The daily wages of 24 laborer in the construction field are 280, 150, 160, 500, 520, 460, 780, 590, 860, 450, 410, 425, 360, 1000, 950, 560, 590, 620, 250, 350, 380, 515, 790, 240

The ascending order of the given data is

150, 160, 240, 250, 280, 350, 360, 380, 410, 425, 450, 460, 500, 515, 520, 560, 590, 590, 620, 780, 790, 860, 950, 1000

From the ascending order, we can find

The highest amount on daily basis is 1000

The lowest amount taken is 150

The range of the given data is 1000 – 150 = 850

FAQs on Data Handling

1. What are the uses of data handling?

Various real-time uses of data handling are for recording water levels in rivers, for doctors to keep records of patients, to record the economical income of each household, etc.

2. What is the difference between data and information?

Data means raw and unorganized facts. While information is the proper structured and organized data.

3. What are the various methods of collecting data?

The various methods of collecting data are questionnaire, interview, schedule, case study, observation, and projective case study.

The post Data Handling | Definition, Solved Example Questions on Data Handling appeared first on Learn CBSE.

The frequency distribution is a topic in statistics related to data handling. The term frequency is how often something occurs and distribution means dividing and sharing something. Get the examples, the definition of frequency distribution, and learn about the frequency distribution table. Find the solved examples that help to understand the concept clearly.

Frequency Distribution Definition

The frequency distribution describes how frequencies are distributed over the data. It is a representation either in tabular or graphical format, that displays the number of observations within the given interval.

The terms of the frequency distribution are listed here:

• Presentation of Data: After collecting the data, shortly in the tabular form in order to study its features, such arrangement is known as the data presentation.
• Observation: Each entry is collected as a numerical number in the given data is called the observation.
• Frequency: The number of times a particular observation repeat is known as the frequency.
• Frequency Distribution Table: It is one form of the representation of the data. It contains the observations and how many times those observations occur.

Steps to Draw a Frequency Distribution Table

• Arrange the given data in ascending order.
• And find the frequency of each data value.
• Take the data value and frequency as the two columns in the table.
• Write each data value frequency in the table.

Example Questions

Example 1.

Suppose the runs scored by the 11 players of the Indian cricket team in a match are given as follows:

25, 65, 03,12, 35, 46, 67, 56, 00, 31, 17

State the frequency of each player.

Solution:

Given that,

Runs scored by the 11 players of the cricket team in a match are 25, 65, 03,12, 35, 46, 67, 56, 00, 31, 17

Arranging the data in ascending order, we get the observations as

00, 03, 12, 17, 25, 31, 35, 46, 56, 65, 67

We find that

Each value has occurred only once.

We may represent the above data in a tabular form, showing the frequency of each observation. This representation is called frequency distribution.

Frequency Distribution Table

Example 2.

In a quiz, the marks obtained by 20 students out of 30 are given as:

12, 15, 15, 29, 30, 21, 30, 30, 15, 17, 19, 15, 20, 20, 16, 21, 23, 24, 23, 21

State the frequency of each student.

Solution:

Given that,

Marks obtained by 20 students in quiz out of 30 are 12, 15, 15, 29, 30, 21, 30, 30, 15, 17, 19, 15, 20, 20, 16, 21, 23, 24, 23, 21

Arranging the data in ascending order, we get the marks list as

12, 15, 15, 15, 15, 16, 17, 19, 20, 20, 21, 21, 21, 23, 23, 24, 29, 30, 30, 30

The highest marks obtained student is 30, the lowest marks are 12, the range is 30 – 12 = 18

From the data, we find that

12 marks scored by one student,

15 marks scored by 4 students,

16, 17, 19 marks scored by one student,

20 marks scored by 2 students,

21 marks scored by 3 students,

23 marks scored by 2 students,

24, 29 marks scored by one student,

30 marks scored by 3 students,

We may represent the above data in a tabular form, showing the frequency of each observation. The number of times data occurs in a data set is known as the frequency of data

This tabular form of representation is called a frequency distribution.

Example 3.

The numbers of newspapers sold at a local shop over the 10 days are:

22, 20, 18, 23, 20, 25, 22, 20, 18, 20

State the frequency.

Solution:

Given that,

The total number of newspapers sold at the local shop past 10 days are 22, 20, 18, 23, 20, 25, 22, 20, 18, 20

By arranging the newspapers numbers in ascending order is 18, 18, 20, 20, 20, 20, 22, 22, 23, 25

The frequency table for the papers sold is given here:

The post Frequency Distribution | Introduction, Examples appeared first on Learn CBSE.

Grouping of data plays an important role while dealing with a large amount of data. This information can also be displayed using a bar graph or pictograph. You can the grouping of data definition, solved examples in the below sections of this article.

What is Meant by Grouping of Data?

When the number of observations is very large, we may divide the data into several groups, by using the grouping of data concept. The data formed by arranging the individual observations of a variable into groups, so that the frequency distribution table of these groups is a convenient way of summarizing the data. The advantages of grouping data are, it improves the accuracy/ efficiency of estimation, helps to focus on the important subpopulations, and ignores irrelevant ones.

Steps to Draw Frequency Distribution Table for Grouped Data

• From the given data, divide the data into some groups.
• Arrange the given observations in ascending order.
• Get the frequency of each observation.
• Write the frequency, group name in the frequency distribution table.

Example Questions & Answers

Example 1.

The mass of 40 students in a class is given below. The measurement of the weight will be in kgs.

55, 70, 57, 73, 55, 59, 64, 72, 60, 48, 58, 54, 69, 51, 63, 78, 75, 64, 65, 57, 71, 78, 76, 62, 49, 66, 62, 76, 61, 63, 63, 76, 52, 76, 71, 61, 53, 56, 67, 71

State the frequency distribution table for the grouped data.

Solution:

Given that,

The weight of 40 students in a class are 55, 70, 57, 73, 55, 59, 64, 72, 60, 48, 58, 54, 69, 51, 63, 78, 75, 64, 65, 57, 71, 78, 76, 62, 49, 66, 62, 76, 61, 63, 63, 76, 52, 76, 71, 61, 53, 56, 67, 71

The ascending order of the students weight is 48, 49, 51, 52, 53, 54, 55, 55, 56, 57, 57, 58, 59, 60, 61, 61, 62, 62, 63, 63, 63, 64, 64, 65, 66, 67, 69, 70, 71, 71, 71, 72, 73, 75, 76, 76, 76, 76, 78, 78

The range = 78 – 48 = 30

The intervals should separate the scale into equal parts. We could choose intervals of 5. We then begin the scale with 45 and end with 79.

Frequency Distribution Table for Grouped Data is

Example 2.

The marks obtained by 40 students of Class VII in an examination are given below:
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 2, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23

State the frequency distribution table for the grouped data.

Solution:

Given data,

The marks scored by 40 students is 16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 2, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23

The ascending order of the marks is 0, 1, 2, 2, 3, 3, 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 10, 10, 12, 12, 12, 13, 13, 13, 15, 16, 16, 16, 17, 18, 18, 18, 19, 20, 21, 21, 23, 23, 23, 24

Range is = 24 – 0 = 24

The intervals should separate the scale into equal parts. We could choose intervals of 5. We then begin the scale with 0 and end with 24.

Frequency Distribution Table for Grouped Data is

Example 3.

The height (in cms) of 35 persons are given below:

140, 125, 128, 126, 130, 134, 134, 146, 125, 140, 140, 127, 125, 137, 132, 134, 126, 128, 128, 129, 150, 151, 131, 138, 128, 129, 126, 127, 129, 130, 135, 132, 134, 140, 152

State the frequency distribution table for the grouped data.

Solution:

Given that,

The height of 35 persons are 140, 125, 128, 126, 130, 134, 134, 146, 125, 140, 140, 127, 125, 137, 132, 134, 126, 128, 128, 129, 150, 151, 131, 138, 128, 129, 126, 127, 129, 130, 135, 132, 134, 140, 152

Ascending order of the 35 persons height is 125, 125, 125, 126, 126, 126, 127, 127, 128, 128, 128, 128, 129, 129, 129, 130, 130, 131, 132, 132, 134, 134, 134, 134, 135, 137, 138, 140, 140, 140, 140, 146, 150, 151, 152

The range is 152 – 125 = 27

The intervals should separate the scale into equal parts. We could choose intervals of 4. We then begin the scale with 125 and end with 152.

Frequency Distribution Table for Grouped Data is

Frequently Asked Questions on Grouping of Data

1. What is grouped and ungrouped data?

Grouped data is the data given in intervals whereas ungrouped data without a frequency distribution.

2. What is a group of data called?

Grouped data is used in data analysis. The frequency table is also called the grouped data.

3. How can we convert ungrouped data to grouped data?

The first step of the conversion is to determine how many classes you have and find the range of data. And then divide the number of classes into groups.

The post Grouping of Data | Definition, Solved Example Questions appeared first on Learn CBSE.

Set Theory is a branch of mathematics and is a collection of objects known as numbers or elements of the set. Set theory is a vital topic and lays stronger basics for the rest of the Mathematics. You can learn about the axioms that are essential for learning the concepts of mathematics that are built with it. For instance, Element a belongs to Set A can be denoted by a ∈ A and a ∉ A represents the element a doesn’t belong to Set A.

{ 3, 4, 5} is an Example of Set. In this article of Introduction to Set Theory, you will find Representation of Sets in different forms such as Statement Form, Roster Form, and Set Builder Form, Types of Sets, Cardinal Number of a Set, Subsets, Operations on Sets, etc.

• Representation of a Set
• Types of Sets
• Finite Sets and Infinite Sets
• Power Set
• Problems on Union of Sets
• Problems on Intersection of Sets
• Difference of two Sets
• Complement of a Set
• Problems on Complement of a Set
• Problems on Operation on Sets
• Word Problems on Sets
• Venn Diagrams in Different Situations
• Relationship in Sets using Venn Diagram
• Union of Sets using Venn Diagram
• Intersection of Sets using Venn Diagram
• Disjoint of Sets using Venn Diagram
• Difference of Sets using Venn Diagram
• Examples on Venn Diagram

Set Definition

Set can be defined as a collection of elements enclosed within curly brackets. In other words, we can describe the Set as a Collection of Distinct Objects or Elements. These Elements of the Set can be organized into smaller sets and they are called the Subsets. Order isn’t that important in Sets and { 1, 2, 4} is the same as { 4,2, 1}.

Examples of Sets

• Odd Numbers less than 20, i.e., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
• Prime Factors of 15 are 3, 5
• Types of Triangles depending on Sides: Equilateral, Isosceles, Scalene
• Top two surgeons in India
• 10 Famous Engineers of the Society.

Among the Examples listed the first three are well-defined collections of elements whereas the rest aren’t.

Important Sets used in Mathematics

N: Set of all natural numbers = {1, 2, 3, 4, …..}

Q: Set of all rational numbers

R: Set of all real numbers

W: Set of all whole numbers

Z: Set of all integers = {….., -3, -2, -1, 0, 1, 2, 3, …..}

Z+: Set of all positive integers

Representation of a Set

Set can be denoted using three common forms. They are given along the following lines by taking enough examples

• Statement Form
• Roaster Form or Tabular Form
• Set Builder Form

Statement Form: In this representation, elements of the set are given with a well-defined description. You can see the following examples for an idea

Example:

Consonants of the Alphabet

Set of Natural Numbers less than 20 and more than 5.

Roaster Form or Tabular Form: In Roaster Form, elements of the set are enclosed within a pair of brackets and separated by commas.

Example:

N is a set of Natural Numbers less than 7 { 1, 2, 3, 4, 5, 6}

Set of Vowels in Alphabet = { a, e, i, o, u}

Set Builder Form: In this representation, Set is given by a Property that the members need to satisfy.

{x: x is an odd number divisible by 3 and less than 10}

{x: x is a whole number less than 5}

Size of a Set

At times, we are curious to know the number of elements in the set. This is called cardinality or size of the set. In general, the Cardinality of the Set A is given by |A| and can be either finite or infinite.

Types of Sets

Sets are classified into many kinds. Some of them Finite Set, Infinite Set,  Subset, Proper Set, Universal Set, Empty Set, Singleton Set, etc.

Finite Set: A Set containing a finite number of elements is called Finite Set. Empty Sets come under the Category of Finite Sets. If at all the Finite Set is Non-Empty then they are called Non- Empty Finite Sets.

Example: A = {x: x is the first month in a year}; Set A will have 31 elements.

Infinite Set: In Contrast to the finite set if the set has infinite elements then it is called Infinite Set.

Example: A = {x : x is an integer}; There are infinite integers. Hence, A is an infinite set.

Power Set: Power Set of A is the set that contains all the subsets of Set A. It is represented as P(A).

Example:  If set A = {-5,7,6}, then power set of A will be:

P(A)={ϕ, {-5}, {7}, {6}, {-5,7}, {7,6}, {6,-5}, {-5,7,6}}

Sub Set: If Set A contains the elements that are in Set B as well then Set A is said to be the Subset of Set B.

Example:

If set A = {-5,7,6}, then Sub Set of A will be:

P(A)={ϕ, {-5}, {7}, {6}, {-5,7}, {7,6}, {6,-5}, {-5,7,6}}

Universal Set:

This is the base for all the other sets formed. Based on the Context universal set is decided and it can be either finite or infinite. All the other Sets are Subsets of Universal Set and is given by U.

Example: Set of Natural Numbers is a Universal Set of Integers, Real Numbers.

Empty Set: 

There will be no elements in the set and is represented by the symbol ϕ or {}. The other names of Empty Set are Null Set or Void Set.

Example: S = { x | x ∈ N and 9 < x < 10 } = ∅

Singleton Set:

If a Set contains only one element then it is called a Singleton Set.

Example: A = {x : x is an odd prime number}

Operations on Sets

Consider Two different sets A and B, they are several operations that are frequently used

Union: Union Operation is given by the symbol U. Set A U B denotes the union between Sets A and B. It is read as A union B or Union of A and B. It is defined as the Set that contains all the elements belonging to either of the Sets.

Intersection: Intersection Operation is represented by the symbol ∩. Set A ∩ B is read as A Intersection B or Intersection of A and B. A ∩ B is defined in general as a set that contains all the elements that belong to both A and B.

Complement: Usually, the Complement of Set A is represented as A^c or A^‘ or ~A. The Complement of Set A contains all the elements that are not in Set A.

Power Set: The power set is the set of all possible subsets of S. It is denoted by P(S). Remember that Empty Set and the Set itself also comes under the Power Set. The Cardinality of the Power Set is 2ⁿ in which n is the number of elements of the set.

Cartesian Product: Consider A and B to be Two Sets. The Cartesian Product of the two sets is given by AxB i.e. the set containing all the ordered pairs (a, b) where a belong to Set A, b belongs to Set B.

Representation of Cartesian Product A × B = {(a, b) | a ∈ A ∧ b ∈ B}.

The cardinality of AxB is N*M where N is the cardinality of A and M is the Cardinality of B. Remember that AxB is not the same as BxA.

The post Introduction to Set Theory – Basics, Definition, Representation of Sets appeared first on Learn CBSE.

There are certain basic operations that can be performed on Sets. Similar to Addition, Subtraction, Multiplication Operations which we come across in Maths we have Union, Intersection, Difference in Set Theory. These operations are performed on two or more sets to result in a New Set depending on the Operation Performed. In the case of Union of Sets, all the elements are included in the result whereas in Intersection only common elements between the Sets are included.

At times you might get confused between Union of Sets and Universal Set. There is a slight difference between Union and Universal Set. Union of two or more sets is an Operation performed on them and includes the elements that are in both sets. On the other hand, Universal Set itself is a Set and has all the elements of other sets along with its set.

Union of Sets Definition

Union of Two Sets A and B is the set of elements that are present in Set A, Set B, or in both Sets A and B. The Union of Sets Operation can be denoted as such

A ∪ B = {a: a ∈ A or a ∈ B}

For instance Let A = { 3, 4, 5} B = { 4, 5, 6, 7} then A U B = { 3, 4, 5, 6, 7}

Go through the further sections to know How to represent the Union of Sets using Venn Diagrams.

Union of Sets Venn Diagram

Let us dive deep into the article to know about the representation of Union of Sets using the Venn Diagram. You can visualize the Operation of Sets with the Diagrammatic Representation.

Consider a Universal Set U in which you have the Subsets A and B. Union of Two Sets A and B is nothing but the elements in Set A and B or both the elements in A and B together. The Union of Sets is denoted by the symbol U.

A U B is the Union of Sets A and B. It is read as A Union B or Union of A and B. The Notation representing the Union of Sets A and B is given as follows

A U B = {x : x ∈ A or x ∈ B}.

It is evident that, x ∈ A U B

⇒ x ∈ A or x ∈ B

In the same way, if x ∉ A U B

⇒ x ∉ A or x ∉ B

Thus, from the definition of Union of Sets, we can say that A ⊆ A U B, B ⊆ A U B.

From the above Venn diagram, we can infer certain theorems

(i) A ∪ A = A (Idempotent theorem)

(ii) If A ⊆ B, then A ⋃ B = B

(iii) A ⋃ U = U (Theorem of ⋃) U is the universal set.

(iv) A ⋃ A’ = U (Theorem of ⋃) U is the universal set.

(v) A ∪ ϕ = A (Theorem of identity element, is the identity of ∪)

(vi) A ∪ B = B ∪ A (Commutative theorem)

Union of Sets Venn Diagram Examples with Solutions

1. If A = { 2, 5, 9, 15, 19} B = { 8, 9, 10, 13, 15, 17}. Find A U B using Venn Diagram?

Solution:

Given Sets are A = { 2, 5, 9, 15, 19} B = { 8, 9, 10, 13, 15, 17}

Let us draw Venn Diagram for the given questions considering the given sets.

A U  B is clearly {2, 5, 8, 9, 10, 13, 15, 17, 19} all the elements in both the Sets.

A U B = {2, 5, 8, 9, 10, 13, 15, 17, 19}

2. If P = { 3, 6, 9, 12, 15, 18} Q = { 2, 4, 6, 8, 10, 12, 14, 16, 18}, Find P U Q and the represent the same using Venn Diagram?

Solution:

Given Sets are P = { 3, 6, 9, 12, 15, 18} Q = { 2, 4, 6, 8, 10, 12, 14, 16, 18}

On Drawing the Venn Diagram for the given sets we have the P U Q as under

It is clearly evident from the Venn Diagram that P U Q = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18}

Thus, P U Q = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18}

The post Union of Sets using Venn Diagram | Venn Diagram Examples with Solutions appeared first on Learn CBSE.

Set Operations are performed on two or more sets to obtain the combination of elements based on the Operation. There are three major types of Operations performed on Sets like Intersection, Union, Difference in Set Theory. Check out Representation of Intersection of Sets using Venn Diagram, Properties of Intersection of Sets, Solved Examples in the later modules.

Intersection of Sets Definition

Intersection of Sets A and B is the Set that includes all the Elements that are Common to Sets A and B. Intersection is represented using the symbol ‘∩’. All the elements that belong to both A and B denote the Intersection of A and B.

A ∩ B = {x : x ∈ A and x ∈ B}

If you have n sets i.e. A₁, A₂, A₃…..A_n all are Subsets of Universal Set U the intersection is the set of elements that are in common to n sets.

Intersection of Sets Venn Diagram

Consider Two Sets A and B and their Intersection is depicted pictorially using the following Venn Diagram. A, B are subsets in the Universal Set. Intersection of Sets is all those elements that belong to both the Sets A and B. Shaded Portion denotes the Intersection of Sets A and B. Intersection of Sets A and B is represented as A ∩ B and is read as A Intersection B or Intersection of A and B.

A ∩ B = {x : x ∈ A and x ∈ B}.

Clearly, x ∈ A ∩ B

⇒ x ∈ A and x ∈ B

Thus, from the Definition of Intersection, we can conclude that A ∩ B ⊆ A, A ∩ B ⊆ B.

Properties of Intersection of Sets

(i) A ∩ A = A (Idempotent theorem)

(ii)  A ∩ B = B ∩ A (Commutative theorem)

(iii) A ∩ U = A (Theorem of Union)

(iv) A ∩ ϕ = ϕ (Theorem of ϕ)

(v) A ∩ A’ = ϕ (Theorem of ϕ)

(vi) If A ⊆ B, then A ∩ B = A.

Solved Examples on Intersection of Sets using Venn Diagram

1.  If A = {a, b, d, e, g, h} B = {b, c, e, f, h, i, j}. Find A ∩ B using the Venn Diagram?

Solution:

Given Sets are A = {a, b, d, e, g, h} B = {b, c, e, f, h, i, j}

Draw the Venn Diagram for the given Sets and then find the Intersection of Sets.

Intersection is nothing but the elements that are common in both Sets A and B.

A ∩ B = { b, e, h}

2. If C = { 3, 5, 7} D = { 7, 9, 11}. Find C ∩ D using Venn Diagram?

Solution:

Given Sets are C = { 3, 5, 7} D = { 7, 9, 11}

Let us represent the given sets in the diagrammatic representation of sets

Intersection is nothing but the elements that are common in both Sets C and D.

C ∩ D = {7}

The post Intersection of Sets using Venn Diagram | Solved Examples on Intersection appeared first on Learn CBSE.

Disjoint Sets are the Sets whose intersection with each other results in a Null Set. In Set Theory if two or more sets have no common elements between them then the Intersection of Sets is an Empty Set or Null Set. This kind of Sets is Called Disjoint Sets. Go through the entire article to know about Disjoint Sets Definition, Disjoint Sets using Venn Diagram, Pairwise Disjoint Sets, etc.

Disjoint Sets Definition

Two Sets are said to be disjoint if they have no common elements between them. If elements in two sets are common then they are said to be Non-Disjoint Sets. Condition for Disjointness is just that intersection of the entire collection needs to empty.

For Example, A = { 4, 5, 6} B = { 7, 8, 9} then A and B are said to be Disjoint Sets since they have no common elements between them. Some Sets can have null set as Intersection without being Disjoint.

Disjoint Sets using Venn Diagram

Two sets A and B are disjoint sets if the intersection of two sets A and B is either a null set or an empty set. In other words, we can say the Intersection of Sets is Empty.

i.e. A ∩ B = ϕ

Pairwise Disjoint Sets

Definition of Disjoint Sets can be proceeded to any group of sets. Collection of Sets is said to be pairwise disjoint if it has any two sets disjoint in the collection. These are also called as Mutually Disjoint Sets.

Consider P is a set of any collection of Sets and A and B. i.e. A, B ∈ P. Then, P is known as pairwise disjoint if and only if A ≠ B. Therefore, A ∩ B = ϕ.

Example:

{ {7}, {3, 4}, {5, 6, 8} }

Solved Examples on Disjoint Sets Venn Diagrams

1. Determine whether the following Venn Diagram represents Disjoint Sets or not?

Solution:

S = {2, 4, 6 8} T = {1, 3, 5, 7}

Sets S, T doesn’t have any common elements between them. Thus, Sets S, T are said to be Disjoint.

2. If A = { 1, 2, 3, 4, 5, 6, 7, 8, 9} B = {14, 15, 16, 17, 18, 19}. Check whether the following are Disjoint Sets are not?

Solution:

Given Sets are A = { 1, 2, 3, 4, 5, 6, 7, 8, 9} B = {14, 15, 16, 17, 18, 19}

Sets A, B doesn’t have any common elements between them. Thus, Sets A, B are said to be Disjoint.

The post Disjoint Sets using Venn Diagram | Pairwise Disjoint Sets & Examples appeared first on Learn CBSE.

Principal Operations on Sets include Intersection, Union, Difference of Sets, Complement of a Set, etc. To Visualise Operations of Sets we use Venn Diagrams. Difference Operation in Set Theory is a fundamental and important operation along with Union, Intersection Operations. Get to know about the Difference of Sets Definition, and how to find Difference of Sets, Difference of Sets Diagrammatic Representation in the further modules.

Difference of Sets Definition

The difference between Sets A and B is the Set of elements present in A but not in B. It is represented as A -B. The region shaded in orange denotes A -B and the one shaded in violet represents the difference between B and A i.e. B-A.

How to find the Difference of Sets using the Venn Diagram?

Let us consider Two Sets A and B that are Subsets of Universal Set U.

To find the difference between Sets A and B simply write the elements of A and take away the elements that are also present in Set B.

The difference of sets A, B is represented as such

A – B = {x : x ∈ A and x ∉ B}

A-B is the set of all the elements that are present in Set A but don’t belong to Set B.

A – B = {x : x ∈ A and x ∉ B} or A – B = {x ∈ A : x ∉ B}

Thus, we can say that x ∈ A – B

⇒ x ∈ A and x ∉ B

The Same Logic applies to B-A i.e. it contains all the elements that are included in Set B but don’t belong to Set A.

We need to be cautious about the way we compute the difference of sets. Since A-B is not the same as B-A and order does matter in the Sets Difference. Thus, we can say that Difference of Sets Operation isn’t commutative.

Identities involving Difference of Sets

There are quite a few operations that include Difference and Complement of Sets. We have stated some of the important identities related to the Difference of Sets and they may include operations such as Union, Intersection in between. They are in the following fashion

• A – A =∅
• A – ∅ = A
• ∅ – A = ∅
• A – U = ∅
• (A^C)^C = A
• DeMorgan’s Law I: (A ∩ B)^C = A^C ∪ B^C
• DeMorgan’s Law II: (A ∪ B)^C = A^C ∩ B^C

Solved Examples on Difference of Sets

1. If A = {1, 2, 3, 4, 5, 6, 7, 8, 9} B = {2, 4, 6, 8, 10, 12, 14, 16, 18}. Find Difference of Sets A-B and represent the Same using Venn Diagram?

Solution:

Given Sets A = {1, 2, 3, 4, 5, 6, 7, 8, 9} B = {2, 4, 6, 8, 10, 12, 14, 16, 18}

A-B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8, 10, 12, 14, 16, 18}

= {1, 3, 5, 7, 9}

A-B is the Set of Elements that are present in Set A and doesn’t belong to Set B.

Therefore, A-B = {1, 3, 5, 7, 9}

2. If A = {1, 2, 3, 4} B = { 3, 4, 5, 6}. Find Difference of Sets A and B using Venn Diagram?

Solution:

A = {1, 2, 3, 4} B = { 3, 4, 5, 6}

A-B = { 1, 2, 3, 4} – { 3, 4, 5, 6}

A-B = {1, 2}

A-B denotes the Elements in Set A but doesn’t belong to Set B. The Difference of Sets A-B is shaded for your reference.

3.  Let us consider Set A = {blue, green, red} & Set B = {red, orange, yellow}. Find the Difference of Sets A and B?

Solution:

Given Set A = {blue, green, red} & Set B = {red, orange, yellow} are

A-B = {blue, green, red} – {red, orange, yellow}

= {blue, green}

A-B denotes the colors that belong to Set A and don’t belong to Set B. A-B is shaded so that you can understand easily.

The post Difference of Sets using Venn Diagram | How to find Difference of Sets? appeared first on Learn CBSE.

If you are looking everywhere to find Solved Questions on Venn Diagrams then you have come the right way. We use Venn Diagrams to Visualize Set Operations in Set Theory. Refer to Solved Questions of Venn Diagrams and learn how to find Union, Intersection, Complement, etc. using the Venn Diagrams. Use the Practice Problems provided and get a good grip on the concepts involving Sets easily. You can use the below existing questions as a quick reference to solve any kind of problem-related to Sets using Venn Diagrams.

1. From the following Venn diagram, find the following sets.

(i) A

(ii) B

(iii) ξ

(iv) A’

(v) B’

(vi) C’

(vii) C – A

(viii) B – C

(ix) A – B

(x) A ∪ B

(xi) B ∪ C

(xii) A ∩ C

(xiii) B ∩ C

(xiv) (B ∪ C)’

(xv) (A ∩ B)’

(xvi) (A ∪ B) ∩ C

(xvii) A ∩ (B ∩ C)

Solution:

Given Sets are A = {1, 2, 3, 4, 6, 9, 10}, B = {1, 3, 4, 9, 13, 14, 15}, C= {1, 2, 3, 6, 9, 11, 12, 14, 15}, ξ or U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

(i) A = {1, 2, 3, 4, 6, 9, 10}

(ii) B = {1, 3, 4, 9, 13, 14, 15}

(iii) ξ or U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

(iv) A’

A’ = U -A

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} – {1, 2, 3, 4, 6, 9, 10}

= { 5, 7, 8, 11, 12, 13, 14, 15}

(v) B’

B’ = U -B

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} – {1, 3, 4, 9, 13, 14, 15}

= { 2, 5, 6, 7, 8, 10, 11, 12}

(vi) C’

C’ = U – C

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} – {1, 2, 3, 6, 9, 11, 12, 14, 15}

= {4, 5, 7, 8, 10, 13}

(vii) C – A

C-A = {1, 2, 3, 6, 9, 11, 12, 14, 15} – {1, 2, 3, 4, 6, 9, 10}

= {11, 12, 14, 15}

C- A is the Elements that are in Set C but doesn’t belong to Set A.

(viii) B – C

B-C = {1, 3, 4, 9, 13, 14, 15} – {1, 2, 3, 6, 9, 11, 12, 14, 15}

= {4, 13}

(ix) A – B

A-B = {1, 2, 3, 4, 6, 9, 10} – {1, 3, 4, 9, 13, 14, 15}

= {2, 6, 10}

(x) A ∪ B

A ∪ B = {1, 2, 3, 4, 6, 9, 10} ∪ {1, 3, 4, 9, 13, 14, 15}

= {1, 2, 3, 4, 6, 9, 10, 13, 14, 15}

(xi) B ∪ C

B U C = {1, 3, 4, 9, 13, 14, 15} U {1, 2, 3, 6, 9, 11, 12, 14, 15}

= {1, 2, 3, 4, 6, 9, 11, 12, 13, 14, 15}

(xii) A ∩ C

A ∩ C = {1, 2, 3, 4, 6, 9, 10} U {1, 2, 3, 6, 9, 11, 12, 14, 15}

= { 1, 2, 3, 6, 9}

(xiii) B ∩ C

B ∩ C = {1, 3, 4, 9, 13, 14, 15} ∩ {1, 2, 3, 6, 9, 11, 12, 14, 15}

= { 1, 3, 9, 14, 15}

(xiv) (B ∪ C)’

(B ∪ C)’ = U – (B U C)

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} – {1, 2, 3, 4, 6, 9, 11, 12, 13, 14, 15}

= { 5, 7, 8, 10}

(xv) (A ∩ B)’

Firstly, find the (A ∩ B) i.e. {1, 2, 3, 4, 6, 9, 10} ∩ {1, 3, 4, 9, 13, 14, 15}

= {1, 4, 9}

(A ∩ B)’ = U – (A ∩ B)

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} – {1, 4, 9}

= {2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15}

(xvi) (A ∪ B) ∩ C

(A ∪ B) ∩ C = {1, 2, 3, 4, 6, 9, 10, 13, 14, 15} ∩ {1, 2, 3, 4, 6, 9, 11, 12, 13, 14, 15}

= { 1, 2, 3, 4, 6, 9, 13, 14, 15}

(xvii) A ∩ (B ∩ C)

A ∩ (B ∩ C) = {1, 2, 3, 4, 6, 9, 10} ∩ { 1, 3, 9, 14, 15}

= {1, 3, 9}

2. Find the following sets from the given Venn Diagram?

(i) F

(ii) H

(iii) B

(iv) F U H

(v) B ∩ F

(vi) F U H U B

Solution:

(i) F = {9, 12, 13, 15}

(ii) H = {12, 14, 15}

(iii) B = {13, 14, 15, 20}

(iv) F U H

F U H = {9, 12, 13, 15} U {12, 14, 15}

= {9, 12, 13, 14, 15}

(v) B ∩ F

B ∩ F = {13, 14, 15, 20} ∩ {9, 12, 13, 15}

= { 13, 15}

(vi) F U H U B

F U H U B = (F U H) U B

= {9, 12, 13, 14, 15} U {13, 14, 15, 20}

= { 9, 12, 13, 14, 15, 20}

The post Examples on Venn Diagrams with Answers | Solved Problems on Venn Diagrams appeared first on Learn CBSE.

Cube and Cube Root of numbers can be found easily using the simplest and quickest methods. Check the complete details of the How to Find the Cube and Cube Root of a Number? We have covered everything like the definition of cube and Cubes Relation with Cube Numbers, Perfect Cube, etc. For better understanding, we even jotted the solved examples explained in detail.

• Cube
• To Find if the Given Number is a Perfect Cube
• Cube Root
• Method for Finding the Cube of a Two-Digit Number
• Table of Cube Roots
• Worksheet on Cube
• Worksheet on Cube and Cube Root
• Worksheet on Cube Root

Cube

The cube of a number is calculated by multiplying a number itself by 3 times. If you consider a number n, then the cube of a number n is n. Here, n is the natural number.

Example:

1, 8, 27 are the cube number of the numbers 1, 2, and 3 respectively.

Cube of 9 = 9 × 9 × 9 = 729
Cube of 8 = 8 × 8 × 8 = 512
Cube of 6 = 6 × 6 × 6 = 216

Cubes Relation with Cube Numbers

In mathematics, a cube is defined as a solid figure where all edges are of the same sizes and each edge is perpendicular to other edges.

Example:

If you take cubes of 4 units, then you can form a bigger cube of 64 units. Or else, if you take cubes of 3 units, then you can form a bigger cube of 27 units.

Perfect Cube Cube Numbers

The product of the three same numbers will give you a cube of a number (perfect cube).

Example:

The cube of a number 2 is 2 × 2 × 2 = 8.
8 is a perfect cube.

Properties of Cube Numbers

1. The cube of an even number is always an even number.

Example:
(i) Find the cube of a number 2?
2 × 2 × 2 = 8
8 is an even number.
(ii) Find the cube of a number 4?
4 × 4 × 4 = 64
64 is an even number.
(iii) Find the cube of a number 6?
6 × 6 × 6 = 216

2. The cube of an odd number is always an odd number.

Example:
(i) Find the cube of a number 3?
3 × 3 × 3 = 27
27 is an odd number.
(ii) Find the cube of a number 5?
5 × 5 × 5 = 125
125 is an odd number.
(ii) Find the cube of a number 7?
7 × 7 × 7 = 343
343 is an odd number.

Units Digits in Cube Numbers

If a number is even or odd, its cube is even or odd respective to the given number. The cube of a unit’s digit always shows the below results.

(i) Cube of 1 = 1 × 1 × 1 = 1;
The Units Digits of Cube of 1 is 1.
(ii) Cube of 2 = 2 × 2 × 2 = 8
The Units Digits of Cube of 2 is 8.
(iii) Cube of 3 = 3 × 3 × 3 = 27
The Units Digits of Cube of 3 is 7.
(iv) Cube of 4 = 4 × 4 × 4 = 64
The Units Digits of Cube of 4 is 4.
(v) Cube of 5 = 5 × 5 × 5 = 125
The Units Digits of Cube of 5 is 5.
(vi) Cube of 6 = 6 × 6 × 6 = 216
The Units Digits of Cube of 6 is 6.
(vii) Cube of 7 = 7 × 7 × 7 = 343
The Units Digits of Cube of 7 is 3.
(viii) Cube of 8 = 8 × 8 × 8 = 512
The Units Digits of Cube of 8 is 2.
(ix) Cube of 9 = 9 × 9 × 9 = 729
The Units Digits of Cube of 9 is 9.

Cube roots

Cube Root of a Number is the inverse of finding the cube of a number. If the cube of a number 3 is 27, then the cube root of 27 is 3.

How to Find the Cube Root of a Number by Prime Factorisation Method?

The Prime Factorisation of any Number Cube Root can be calculated by grouping the triplets of the same numbers. Multiply the numbers by taking each one from each triplet to provide you the Cube Root of a Number.

Example:
Cube Root of 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 × 3 = 6
6 is the cube root of 216.

FAQs on Cube and Cube Roots

1. Find the cube of 3.4?
The cube of a number can be calculated by multiplying it three times.
Cube of 3.4 = 3.4 x 3.4 x 3.4 = 39.304

2. Is 288 a perfect cube? If not, find the smallest natural number by which 288 should be multiplied so that the product is a perfect cube.
The prime factorization of 288 is
288 = 2 x 2 x 2 x 6 x 6
Since we can see number 6 cannot be paired in a group of three. Therefore, 288 is not a perfect cube.
To make it a perfect cube, we have to multiply the 6 by the original number.
Thus, 2 x 2 x 2 x 6 x 6 x 6 = 1728, which is a perfect cube.
Hence, the smallest natural number which should be multiplied to 288 to make a perfect cube is 6.

3: Find the smallest number by which 256 must be divided to obtain a perfect cube.
The prime factorization of 256 is
256 = 2×2×2×2×2×2×4
Now, if we group the factors in triplets of equal factors,
256 = (2×2×2)×(2×2×2)×4
Here, 4 cannot be grouped into triples of equal factors.
Therefore, we will divide 256 by 4 to get a perfect cube.

4. Michael makes a cuboid of plasticine of sides 3 cm, 2 cm, 3 cm. How many such cuboids will he need to form a cube?
Given that the sides of the cube are 3 cm, 2 cm, and 3 cm.
Therefore, volume of cube = 3×2×3 = 18
The prime factorization of 18 = 3×2×3
Here, 2, 3, and 3 cannot be grouped into triplets of equal factors.
Therefore, we will multiply 18 by 2×2×3 = 12 to get a perfect square.
Hence, 12 cuboids are needed.

The post Cube and Cube Roots | Perfect Cube, How to Find the Cube Root, Properties appeared first on Learn CBSE.

Hello Aspirants!! Are you one of those candidates who are preparing for competitive exams? If yes, Don’t worry. We are here to help in all the possible ways we can. Know the important questions of direct variation and inverse variation problems. Go through the below sections to know the study material and step by step procedure to solve problems.

Direct Variation and Inverse Variation Examples

Question 1.

y varies directly with x. If x=3 when y=12, what is y when x=9?

A. 42

B. 36

C. 40

D. 38

Solution: B(36)

Explanation:

From the variation equation of y=kx

12=k(3)

Dividing both sides by 3

12/3=k(3)/3

4=k

Substituting the value of k=4 in the equation y=kx

y=4(9)

y=36

Question 2.

y varies inversely with x. If x = 4 when y = 48, what is the value of y when x is 8?

A. 24

B. 36

C. 48

D. 26

Solution: A (24)

Explanation:

From the variation equation of y=k/x

48/1 = k/4

k=192

Substitute the value in the equation y=k/x

y=192/8

y= 48.4/8 = 8*6*4/8 = 24

1. x=4
2. y=48

Multiplying the 1st equation by 2

and dividing the 2nd equation by 2

Therefore, x=8, y=24.

Question 3.

y varies jointly with x and z. If y=36 and when x=2 and z=3. Find the value of y when x=4 & z=6?

A. 50

B. 80

C. 134

D. 144

Solution: D(144)

Explanation:

From the given question, if x increases then y increases, and if z increases then y increases.

As given, x= 2 and z=3

Multiplying both the equations by 2

x=4 and z=6

As x and z are directly proportional to y. x and z doubles, therefore y also doubles.

Therefore, the value of y = 36

Multiplying by 4, the value is 144

y=144

Direct and Indirect Variation Study Material

Question 4.

y varies jointly with x and z. If y=36 when x=2 and z=3. Find the value of y when x=4 and z=6?

A. 14

B. 156

C. 122

D. 142

Solution: A(144)

Explanation:

From the variation equation y=kxz

36=k(2)(3)

36=6K

K=6

Substitute k value in the equation y=kxz

y=(4)(6)

y=24(6)

y=120

y=120+24

y=144

Question 5.

y varies inversely with the cube of x. If y is 108 when x is 2, find the value of y when x is 6?

A. 2

B. 3

C. 4

D. 6

Solution:

Explanation: C(4)

From the variation of the equation y=k/x²

As given in the question

1. x=2
2. y=108

Multiply the 1st equation by 3

Divide the equation by 27

Therefore, x=6, y=4

108=k/(2)<sup3

k=108(8)

k=864

y=8(108)/(6)³

y=2³(27.4)/6³

y=2³.3³.4/6³

y=6³.4/6³

Therefore, y=4

Question 6.

The price for 75 basketballs is 1143.75 rupees. Find the price of 26 basketballs?

396. 50 rupees
397. 20 rupees
398. 60 rupees
399. 40 rupees

Solution: A (396.50 rupees)

Explanation:

Let “m” be the price of 26 basketballs

As given In the question,

The price for 75 basket ball’s is 1143.775 rupees

The price for 26 basketballs is “m”

Since, the question belongs to direct variations (For less number of basketballs, the price will be less)

We can apply cross multiplication shortcut

75*m=26*1143.75

m=(26*1143.75)/75

m=396.50

Therefore, the cost of 26 balls is 396.50 rupees.

Question 7.

A book consists of 120 papers and each paper has 35 lines. How many papers will the book consists of if each paper has 24 lines per paper?

1. 150 papers
2. 175 papers
3. 200 papers
4. 120 papers

Solution: B (175 papers)

Explanation:

Let “m” be the no of required papers

As given in the question,

There are 35 lines and 120 papers in a book

There are 24 lines and m papers in a book

Since, it is an inverse variation (for fewer lines, there are more pages)

We have to apply “straight multiplication” as it is inverse variation.

120*35=m*24

(120.35)/24 = m

175 = m

Therefore, if every paper has 24 lines, the book will contain 175 pages.

Question 8.

A truck covers some distance in 3 Hours with a speed of 60 miles per hour. Consider that the speed is increased by 30 miles per hour, find the time taken to cover the same distance by the truck?

1. 4 Hours
2. 5 Hours
3. 6 Hours
4. 2 Hours

Solution: D (2 Hours)

Explanation:

If the given speed 60 mph is increased by 30 mph, then the new speed is 90 mph

Let “m” be the required time.

In 3 hours of time, the truck covers 60 mph

For m hours of time, the truck covers 90 mph

Since it is an inverse variation (for more speed it takes less time), therefore we have to apply the “straight multiplication” shortcut.

3*60 = m*90

(3*60)/90 = m

2=m

Therefore, if the speed is increased by 30mph, then the time taken by truck is 2 Hours.

Question 9.

In 36.5 weeks, Nandu raised 2,372.50 rupees for cancer research. If he works for 20 weeks, how much money he will raise?

1. 1300 rupees
2. 1400 rupees
3. 1600 rupees
4. 1100 rupees

Solution: A (1300 rupees)

Explanation:

Let “m” be the required amount of money

From the given question, for 36.5 weeks, Nandu raised 2,372.50 rupees. Suppose, for 20 weeks, Nandu raised m rupees.

Since, it is a direct variation(for less number of weeks, the amount raised will be less), we have to apply the “cross multiplication” shortcut.

36.5*m = 20*2372.50

m = (20*2372.50)/36.5

m=1300

Therefore, the money raised in 20 weeks is 1300 rupees

Question 10.

John takes 15 days to lose 30 kgs of his weight by doing 30 minutes of exercise per day. How many days will he need if continues the exercise for 1 Hour 30 minutes per day?

1. 10 Days
2. 5 Days
3. 2 Days
4. 8 Days

Solution: B (5 Days)

Explanation:

Let “m” be the required no of days.

For 15 Days, John takes 30 minutes, suppose that in John takes 90 minutes in m days

As, from the given question, it is an inverse variation (for more minutes per day, fewer days are required to reduce weight), therefore we have to apply “straight multiplication” shortcut.

15*30=m*90

(15*30)/90=m

5=m

Therefore, if John does exercise for 1 Hour 30 minutes per day, he takes 5 days to reduce 30 kilograms of weight.

Hope the Practice Test on Direct Variation and Inverse Variation, will help you to know the various models and questions asked. With the help of the given questions, you can easily score better marks. If you need any further clarifications, then you can comment us below. Moreover, stay tuned to our site to know more shortcuts.

The post Practice Test on Direct Variation and Inverse Variation appeared first on Learn CBSE.

Do you want to calculate the highest common factor of polynomials by using the division method? If yes then stay on this page. Here we are providing the detailed step by step procedure to find the G.C.F of polynomials by division method. Along with the steps you can also check some solved example questions from the below sections.

How to find the Highest Common Factor of Polynomials by Division Method?

We are using the division method to find the G.C.F of polynomials when the polynomials have the highest factor and it is difficult to compute the HCF. Then follow the steps and instructions provided below and get the answer easily.

• Let us take two polynomials f(x), g(x).
• Divide the polynomials f(x) / g(x) to get f(x) = g(x) * q(x) + r(x). Here the degree of g(x) > degree of r(x).
• If the remainder r(x) is zer0, then g(x) is the highest common factor of polynomials.
• If the remainder is not equal to zero, then again divide g(x) by r(x) to obtain g(x) = r(x) * q(x) + r1(x). Here if r1(x) is zero then required H.C.F is r(x).
• If it is not zero, then continue the process until we get zero as a remainder.

Solved Examples on GCF of Polynomials

Example 1.

Find the H.C.F of x⁴ + 4x³ + x – 10 and x² + 3x – 5 by using the division method?

Solution:

Given polynomials are f(x) = x⁴ + 4x³ + x – 10, g(x) = x² + 3x – 5

Arranging the polynomials in the descending order f(x) = 1x⁴ + 4x³ + 0x² + x – 10

By using the division method,

The remainder is zero.

So, the required H.C.F of x⁴ + 4x³ + x – 10 and x² + 3x – 5 is x² + 3x – 5.

Example 2.

Calculate the greatest common factor of 8x³ – 10x² – x + 3 and x – 1 by using the division method?

Solution:

Given two polynomials are 8x³ – 10x² – x + 3 and x – 1

Let us take f(x) = 8x³ – 10x² – x + 3, g(x) = x – 1

Divide f(x) by g(x)

The remainder is zero.

So, the required H.C.F of 8x³ – 10x² – x + 3 and x – 1 is x – 1.

Example 3.

Find the HCF of the following pairs of polynomials using the division algorithm

2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12?

Solution:

Given two polynomials are 2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12

Let f(x) = 2 x³ + 2 x² + 2 x + 2 , g(x) = 6 x³ + 12 x² + 6 x + 12

Take 2 common from the first polynomial.

f(x) = 2(x³ + x² + x + 1)

Take 6 common from the second polynomial.

g(x) = 6(x³ + 2x² + x + 2)

Divide f(x) by g(x)

The remainder is not zero. So, we have to repeat this long division once again.

Then divide g(x) by r1(x). i.e x³ + 2x² + x + 2 by x² + 1

The remainder is zero.

The H.C.F of 2, 6 is 2.

So, required H.C.F of 2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12 is 2 ( x² + 1).

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