Important Questions for CBSE Class 6 Maths Chapter 11 - Algebra

VSAT 2022

CBSE Class 6 Maths Important Questions Chapter 11 - Algebra - Free PDF Download

Class 6 algebra is one of the preliminary stage algebra, which is instrumental in making the perfect knowledge-base for the higher classes. Therefore, they must have crystal clear concepts of the fundamental topics. The best way to prepare algebra is to practice sample questions of each of the topics. The Maths experts at Vedantu have compiled a comprehensive list of questions for the topics covered in chapter 11. In Class 6 Maths Important Questions for Chapter 11, you will get solved sums covering all the topics. You can download this set of questions for free from the Vedantu. Students face a lot of difficulties while solving problems of the NCERT textbook, as these require logic as well as an understanding of concepts. CBSE solutions (NCERT) give the best possible and easy methods to solve problems with the help of solved questions present in it for each type of question. You can also register Online for NCERT Class 6 Science tuition on Vedantu.com to score more marks in CBSE board examination.

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Access IMPQ Solutions for Class 6 Maths Chapter 11- Algebra

Very Short Answer Questions 1 Marks

1. Express algebraically: $\mathbf{3}$ more than $\mathbf{x}$.

Ans: The algebraic expression of the given statement is $x+3$.


2. What is the exponential form of $\mathbf{x}\,\times \,\mathbf{x}\,\times \,\mathbf{x}\,\times \,\mathbf{x}\,\times ....\mathbf{15}\ \mathbf{times}$?

Ans: The given expression $x\times x\times x\times x\times ...15\ times$.

Recall that, $x\times x={{x}^{2}}$,

$x\times x\times x={{x}^{3}}$,

$x\times x\times x\times x={{x}^{4}}$, etc.

Therefore, the number of times $x$ is multiplied with itself equals $x$ to the power that number.

Thus, $x\times x\times x\times x\times ...15\ times={{x}^{15}}$.


3. What is the product form of ${{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}^{\mathbf{7}}}$?

Ans: The given algebraic expression is ${{x}^{2}}{{y}^{7}}$.

Recall that, the number of times $x$ is multiplied with itself equals $x$ to the power that number.

Therefore, ${{x}^{2}}=x\times x$.

Similarly, ${{y}^{7}}=y\times y\times y\times y\times y\times y\times y$.

Thus, ${{x}^{2}}{{y}^{7}}=x\times x\times y\times y\times y\times y\times y\times y\times y$.


4. What is the coefficient of $\mathbf{b}$ in $\mathbf{-8abc}$?

Ans: The given algebraic expression is $-8abc$.

Recall that, the coefficient of any term in an algebraic expression is the multiplication of other terms in that algebraic expression.

Therefore, the coefficient of the term $b$ in the expression $-8abc$ is $-8ac$.


5. What is the numerical coefficient of $\mathbf{4}{{\mathbf{a}}^{\mathbf{2}}}$?

Ans: The given expression is $4{{a}^{2}}$.

It is known that, the numerical coefficient in an algebraic expression is the number multiplied in that expression.

Therefore, the numerical coefficient of $4{{a}^{2}}$ is $4$.


6.  What is the constant term of $\mathbf{2}{{\mathbf{a}}^{\mathbf{2}}}-\mathbf{9}$?

Ans: The constant term of the expression $2{{a}^{2}}-9$ is $-9$.


7. If $\dfrac{\mathbf{y}}{\mathbf{5}}=\mathbf{1}$, then $\mathbf{y}=$?

Ans: The given equation is

$\dfrac{y}{5}=1$.

Multiply $5$ both sides of the equation.

$5\times \dfrac{y}{5}=1\times 5$

$\Rightarrow y=5$.


Short Answer Questions 2 Marks

1. Express algebraically the following statement:

$\mathbf{8}$ times a number $\mathbf{x}$ is less than a number $\ge $.

Ans: The required algebraic expression of the given statement is

$8x=z-y$.


2. What are the terms in the following algebraic expression?

$\mathbf{4a}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{+3}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{-5ab+9}$.

Ans: The terms of in the given expression are 

$4a{{b}^{2}},\,3{{c}^{2}},-5ab,$ and $9$.


3. Solve the following equation and evaluate $\mathbf{x}$. Also, justify the result.

$\mathbf{x}-\mathbf{8}=\mathbf{17}$.

Ans: The given equation is

$x-8=17$

Add $8$ both sides of the equation.

$ x-8+8=17+8 $

$ \Rightarrow x=25$ 

Justification:

$ LHS=x-8 $

$=25-8$

$ =17$ 

$ =RHS $

Thus, Left-hand-side $=$ Right-hand-side.


4. Solve the following equation and evaluate $\mathbf{x}$. Also, justify the result.

$\dfrac{\mathbf{4}}{\mathbf{6}}\mathbf{x=18}$.

Ans: The given equation is 

$\dfrac{4}{6}x=18$

Multiply $6$ both sides of the equation.

$6\times \dfrac{4}{6}x=6\times 18$

$\Rightarrow 4x=18\times 6$

Divid both sides of the equation by $4$.

$\dfrac{4x}{4}=\dfrac{18\times 6}{4}$

$\Rightarrow x=27$.

Justification:

$ LHS=\dfrac{4}{6}x $

$ =\dfrac{4}{6}\times 27 $

$ =18 $

$=RHS $

Hence, Left-hand-side $=$ Right-hand-side.


Long Answer Questions 3 Marks

1. Sanjit scored $\mathbf{70}$ marks in mathematics and $\mathbf{''x''}$ marks in English. Find the total score in both the subjects?

Ans: He got marks in mathematics $=70$.

He got marks in English $=x$.

Therefore, the total marks in mathematics and English $=x+70$.


2. Let $\mathbf{p}=\mathbf{1},\,\,\mathbf{q}=-\mathbf{1},$ and $\mathbf{r}=\mathbf{3}$. What is the value of ${{\mathbf{p}}^{\mathbf{3}}}\mathbf{+}{{\mathbf{q}}^{\mathbf{3}}}\mathbf{+}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{-3pqr}$?

Ans: The given algebraic expression is 

${{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr$                                             …… (i)

Substituting $p=1,q=-1,$ and $r=3$ into the equation (i) gives

$ {{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr$

$ ={{\left( 1 \right)}^{3}}+{{\left( -1 \right)}^{3}}+{{\left( 3 \right)}^{3}}-3\left( 1 \right)\left( -1 \right)\left( 3 \right) $

 $=1-1+27+9$ 

 $=27+9 $ 

 $ =36 $


3. Choose the similar terms from the following expressions.

$\mathbf{xyz,}\,\,\mathbf{x}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{z,}\,\,\mathbf{xz}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{,}\,\,{{\mathbf{z}}^{\mathbf{2}}}\mathbf{xy,}\,\,{{\mathbf{y}}^{\mathbf{2}}}\mathbf{xz,}\,\,{{\mathbf{x}}^{\mathbf{2}}}\mathbf{yz,}\,\,\mathbf{zx}{{\mathbf{y}}^{\mathbf{2}}}$

Ans:

The like terms are those which have exactly one square term involved.

Therefore, the similar terms are

$x{{y}^{2}}z,\,\,xz{{y}^{2}},\,\,{{y}^{2}}xz,$ and $zx{{y}^{2}}$.


4. Solve the equation $\mathbf{12x}-\mathbf{30}=\mathbf{6}$.

Ans:

The given equation is 

$12x-30=6$

Add $30$ both sides of the equation.

$12x-30+30=6+30$

$\Rightarrow 12x=36$

Divide $12$ both sides of the equation.

$\dfrac{12x}{12}=\dfrac{36}{12}$

$\Rightarrow x=3$.

Hence, the solution is $x=3$.


5. Verify whether $\mathbf{y}=\mathbf{4}$ satisfy the equation $\mathbf{3y}-\mathbf{5}=\mathbf{7}$.

Ans: The given equation is $3y-5=7$.

Then,

$ LHS=3\left( 4 \right)-5 $

$=12-5$

$ =7 $

$ =RHS $

Thus, Left-hand-side $=$ Right-hand-side.

Hence, $y=4$ satisfy the given equation.


6. Solve the equation $\mathbf{6x}+\mathbf{10}=\mathbf{26}-\mathbf{2x}$ and evaluate $\mathbf{x}$ by using the method of transposition. Then justify the answer.

Ans: The given equation is

$6x+10=26-2x$

$\Rightarrow 6x=26-2x-10$, by the method of transposition.

$\Rightarrow 6x=16-2x$

$\Rightarrow 6x+2x=16$, by the method of transposition.

$\Rightarrow 8x=16$

Divide $8$ both sides of the equation.

$\dfrac{8x}{8}=\dfrac{16}{8}$

$\Rightarrow x=2$.

Justification:

$ LHS=6x+10$

$=6\times 2+10 $ 

$ =12+10 $

$=22$

Also,

$ RHS=26-2x $

 $=26-2\times 2 $

$ =26-4$ 

 $=22 $

Thus, Left-hand-side $=$ Right-hand-side.

Hence, $x=2$ is the solution of the given equation.


Long Answer Questions 4 or 5 Marks 

1. Find the algebraic expression of the statements given below.

(a) $\mathbf{6}$ less than the quotient of $\mathbf{x}$ and $\mathbf{y}$.

Ans. The required algebraic expression is 

$\dfrac{x}{y}-6$.

(b) The sum of the quotient of $\mathbf{p}$, $\mathbf{q}$ and the product of $\mathbf{p},\,\,\mathbf{q}$.

Ans: The quotient of $p,q$ is $\dfrac{p}{q}$.

The product of $p,q$ is $pq$.

Therefore, the required algebraic expression is given by

$\dfrac{p}{q}+pq$.

(c) One fourth of the sum of $\mathbf{x}$ and $\mathbf{y}$.

Ans: The sum of $x$ and $y$ is $x+y$.

Therefore, the required algebraic expression is given by

$\dfrac{1}{4}\left( x+y \right)$.

(d) $\mathbf{8}$ removed from twice the term $\mathbf{x}$.

Ans: Twice the term $x$ is $2x$.

Therefore, the required algebraic expression is $2x-8$.


2. Classify monomials, binomials, and trinomials from the following algebraic expressions.

$\mathbf{y}+\mathbf{5},\,\,\,\mathbf{6}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{y},\,\,\,-\mathbf{8},\,\,\,-\mathbf{3pqr},\,\,\,{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{z}}^{\mathbf{2}}}$

Ans: Recall that, sum of two terms is called binomials, sum of three terms is called trinomials and single algebraic term is called monomials.

The following table shows the required classifications.

Monomials

Binomials

Trinomials

$6{{x}^{3}}y$

$y+5$

${{a}^{2}}+{{b}^{2}}+{{z}^{2}}$

$-8$



$-3pqr$




3. Determine the solution of $\mathbf{3x}=\mathbf{12}$, by using the method of trial and errors.

Ans: The given equation is $3x=12$.

First substitute $x=1$ into the given equation.

$3x=3\times 1=3\ne 12$.

So, $x\ne 1$.

Then, substitute $x=2$ into the equation.

$3x=3\times 2=6\ne 12$.

So, $x\ne 2$.

Again, substitute $x=3$ into the given equation.

$3x=3\times 3=9\ne 12$.

Therefore, $x\ne 3$.

Now, substitute $x=4$ into the given equation.

Then $3x=3\times 4=12$.

Thus, $x=4$ is the solution of the given equation.


4. Solve the following equation and evaluate $\mathbf{x}$ by using the method of isolation.

$\mathbf{a}-\mathbf{7}=\mathbf{5}+\dfrac{\mathbf{a}}{\mathbf{2}}$.

Ans: The given equation is 

$a-7=5+\dfrac{a}{2}$

Add $7$ on both sides of the equation.

$a-7+7=5+7+\dfrac{a}{2}$

$\Rightarrow a=12+\dfrac{a}{2}$

Add $-\dfrac{a}{2}$  on both sides of the equation.

$a+\left( -\dfrac{a}{2} \right)=12+\dfrac{a}{2}+\left( -\dfrac{a}{2} \right)$

$ \Rightarrow a-\dfrac{a}{2}=12+\dfrac{a}{2}-\dfrac{a}{2} $

$ \Rightarrow \dfrac{2a-a}{2}=12 $ 

$\Rightarrow \dfrac{a}{2}=12 $

Multiply $2$ on both sides of the equation.

$\dfrac{a}{2}\times 2=12\times 2$

$\Rightarrow a=24$.

Hence, the required solution is $a=24$.


5. Solve the following equation and justify the result.

$\mathbf{3}\left( \mathbf{a+3} \right)\mathbf{+3}\left( \mathbf{a-1} \right)=\mathbf{5}\left( \mathbf{a+5} \right)$.

Ans: The given equation is 

$\text{3}\left( \text{a+3} \right)\text{+3}\left( \text{a-1} \right)=\text{5}\left( \text{a+5} \right)$

$\Rightarrow 3a+9+3a-3=5a+25$, multiplying the terms.

$\Rightarrow 6a+6=5a+25$, adding the like terms.

$\Rightarrow 6a-5a=25-6$, by the method of transposition.

$\Rightarrow a=19$.

Justification:

$ LHS=3\left( a+3 \right)+3\left( a-1 \right) $

$=3\left( 19+3 \right)+3\left( 19-1 \right)$

$ =3\times 22+3\times 18 $

$ =66+54$

$=120 $

Also, 

$ RHS=5\left( a+5 \right) $

$ =5\left( 19+5 \right) $ 

$=5\times 24 $ 

$=120 $

Thus, Left-hand-side $=$ Right-hand-side.

Hence, the required solution of the given equation is $a=19$.


6. Determine two numbers so that one is greater by $\mathbf{21}$ than another and sum of them is $\mathbf{91}$.

Ans: Let $x$ be one number.

Then, another number is $x+21$.

By the given conditions,

$x+\left( x+21 \right)=91$

$\Rightarrow 2x+21=91$

Add $-21$ on both sides of the equation.

$2x+21+\left( -21 \right)=91+\left( -21 \right) $ 

$\Rightarrow 2x+21-21=91-21 $ 

$ \Rightarrow 2x=70 $

Divide $2$ on both sides of the equation.

$ \dfrac{2x}{2}=\dfrac{70}{2} $

$\Rightarrow x=35 $

Hence, the required numbers are $35$ and $35+21$, that is $35,\,\,56$.


Important Questions for Class 6 Maths Chapter 11

CBSE Class 6 Maths Chapter 11 - Algebra PDF Free Download

To prepare for any mathematics concept, the best way to do it is to practice. You must have a set of important questions for class 6 maths chapter 11 to get your concepts cleared about the chapter. The students can download the PDF version of this question set from the Vedantu website onto their system for free. They need to have an account on the website. They can also enroll in some tuition classes to get accustomed to the topics. The students can download these class 6 maths chapter 11 important questions and use them in their preparation before any examinations.


Why is it Important to Practice Important Questions for Algebra?

Every teacher will recommend students practice class 6 maths important questions for chapter 11, and they have valid points to do so. Let us look at some of the reasons why class 6 maths chapter 11 important questions will benefit the students.


More Questions Mean More Practice

Mathematics has always been one of the rare subjects which require a lot of practice. Unlike other subjects where reading the concepts can help the students in their preparation, they have to first know about the concepts and then practice in maths. To help the students in their practice, the set of important questions for class 6 maths chapter 11 has been published. In this set of questions, all the basic principles of algebra are covered. The students will start from scratch and gradually can soak up the increasing complexity of the subject.


The Proper Introduction to a New Topic

Secondly, algebra will be a new chapter for the students in class 6. Chapter 11 will pose another issue for the students. The students will be confused about the type of questions that they will encounter in the examination. By solving these class 6 maths chapter 11 important questions, the students will know the type of questions. They can consider these questions to be the blueprint of what is coming in the examination and plan their preparation accordingly.


Proper Time Management While Solving Questions

Although the questions are not arranged in the exam format, the students can plan their preparation keeping the format in mind. That means they can make different clusters of questions and have a proper plan for managing time while they solve them. This practice will increase their proficiency in proper time management for examinations. They will know how to allocate time to each question. If they can develop these skills, they will not face any problem during the actual examination.


Growth in Confidence

Proper time management and constant practice also play a major role in building confidence in the students. As the students practice solving these class 6 maths chapter 11 important questions, they will grow their confidence in answering such questions. They will also be confident about encountering some unknown questions, as they know the right approach for such questions. Therefore, the students will face no problem when they see any unknown questions in the final exam paper.


Fine-Tuning the Preparation Process

When you start to know about a new topic in maths, you practice a lot to build the concept. You will solve all types of questions related to the topic. However, as you develop your concepts, you will focus more on the important aspects of the topics. Class 6 maths chapter 11 important questions bring you the right set of questions to fine-tune your preparation. As you focus more on these important aspects, your preparation will become more exam-oriented. You will focus more on questions that come in the examination. Therefore, you will know how to approach the questions that you will encounter in the examination.


Last-Minute Exam Preparation

For any other subject, the last-minute exam preparation will involve revision of the important topics. However, Math is such a subject that can only be revised through practice. For the last-minute preparation, you would prefer to look at the problems that are most likely to come in the examination. You might also prefer to solve them to revise the techniques. These questions will be ideal for last-minute preparation before any examination.


Important Questions on Maths Chapter 11 Algebra

Q. What is the algebraic representation of six more than x?

Ans. The problem mentioned above can be best represented as x + 6.

Q. What is the algebraic representation of nine less than three times of x?

Ans. The problem mentioned above can be best represented as 3x - 9. 

Q. What is the algebraic representation of four more than half of x?

Ans. The problem mentioned above can be best represented as (x/2) + 4.

Q. In a horticultural garden, rose plants are arranged in rows. There are seven rose plants in each row. If the total number of rows dedicated to role plants is equal to n, how many rose plants are there in the garden?

Ans. In the problem mentioned above, the following information is obtained.

The number of rose plants in a single row = 7.

The number of rows in the horticultural garden dedicated to rose plants = n.

The total number of rose plants in the horticultural garden = total number of rose plants in each row X total number of rows dedicated to rose plants in the horticultural garden.

Therefore, the total number of rose plants in the horticultural garden = 7n.

Q. In a fruit store, there are different boxes designated for different fruits. There are n boxes of mangoes, each containing 15 mangoes. Apples are present in 4 boxes with x number of apples in each box. Additionally, there are 6 boxes of guavas, each containing 20 guavas. How many fruits are present in the store in total?

Ans. In the problem mentioned above, the following information is obtained.

Number of mangoes in each box = 15 mangoes

Number of boxes containing mangoes = n boxes

The total number of mangoes in the fruit store = number of mangoes in each box X Number of boxes containing mangoes.

Therefore, the total number of mangoes in the fruit store = 15n.

Number of apples in each box = x apples

Number of boxes containing apples = 4 boxes

The total number of apples in the fruit store = number of apples in each box X Number of boxes containing apples.

Therefore, the total number of apples in the fruit store = 4x.

Number of guavas in each box = 20 guavas

Number of boxes containing guavas = 6 boxes

The total number of guavas in the fruit store = number of guavas in each box X Number of boxes containing guavas.

Therefore, the total number of guavas in the fruit store = 120 guavas.

Finally, the total number of fruits present in the fruit store = total number of mangoes + total number of apples + the total number of guavas

Total number of fruits in the fruit store = 15n + 4x +120.

Q. The speed of a car is x km/ hr. If the car covers a distance of y km by going with that constant speed, how much time will it take to complete the distance?

Ans. In the problem mentioned above, the following information is obtained.

Speed of the car = x km/hr.

Distance covered by the car at that constant speed = y km.

Time-taken by the car to cover the distance = Distance covered by the car at that constant speed/ Speed of the car.

Therefore, Time-taken by the car to cover the distance = y/x.

Q. Manoj has x number of chocolates. Rahim gave him 4 more. How many chocolates does Manoj have now?

Ans. In the problem mentioned above, the following information is obtained.

The number of chocolates Manoj has = x chocolates.

The number of chocolates Rahim gave to Manoj = 4 chocolates.

Total number of chocolates that Manoj have = number of chocolates Manoj had previously + number of chocolates Rahim gave to Manoj,

Therefore, the total number of chocolates Manoj has now = x+4.


Conclusion

Class 6 Maths important questions for Chapter 11 are an essential toolkit for many students. Since the subject is algebra, the students will require a lot of practice, especially because this is their first introduction to the subject. The question set also comes with the solution set for the students to analyze their performances. The PDF file is available for free from the Vedantu website to anyone who has an account with Vedantu.

FAQs (Frequently Asked Questions)

1. How to score good marks in Chapter 11 of NCERT Solutions for Class 6 Maths?

Ans: Chapter 11 is important in Class 6 Mathematics. Once students cover the core areas, it will always be easier for them to solve the exercise wise problems. Solving the problems related to this chapter will improve the analytical thinking and logical thinking skills of students. Choosing the right study material plays a vital role in scoring good marks in the Class 6 exam. The main aim of creating NCERT Solutions is to help students analyze the concepts which they are lagging behind and work on them for a better score.

2. Does the Vedantu’s NCERT Solutions for Class 6 Maths Chapter 11 provide accurate solutions as per the CBSE syllabus?

Ans: When it comes to Mathematics, accuracy is an important factor and regular practice is required to achieve that. As there are various study materials available online, selecting the correct one is always a challenging task. By opting for solutions that completely focus on the CBSE syllabus, students will be able to understand the concepts which are important from the exam perspective.

3. What are the terms of an expression Chapter 11 of NCERT Solutions for Class 6 Maths?

Ans: Parts of an expression that are formed separately first and then added or subtracted, are known as terms. In the given example, terms 2x and 5 are added to form the expression (2x+5).

4. How many exercises are there in Chapter 11 ‘Algebra’ of Class 6 Maths?


Ans: There are particularly five exercises in Chapter 11 ‘Algebra’ of Class 6 Maths. The exercises are enlisted as; 

Exercise 11.1- In this exercise there are ten questions. 

Exercise- 11.2- This exercise comprises five questions

Exercise 11.3- It includes six questions

Exercise 11.4- This exercise contains three long type answer type questions

Exercise 11.5- It includes five questions. 

To get answers to these textbook exercise questions, visit Vedantu’s official website (vedantu.com). One can find important questions on the Chapter 11 ‘Algebra’ of Class 6 Maths where you can download the solutions to textbook questions absolutely free of cost.


5. Will Vedantu help me to get decent scores in Chapter 11 ‘Algebra’ of Class 6 Maths?


Ans: Yes, Vedantu will help you to get decent scores in Chapter 11 ‘Algebra’ of Class 6 Maths. It is one of the best online learning platforms that enables students to understand educational concepts better. It provides numerous study materials like notes, a summary of the chapter, answers to the textbook questions mentioned in exercises, sample papers, video lectures, etc, so that you can understand the important concepts of the chapter clearly and score decent marks in your class tests as well as in your exams.

6. What are the benefits of practising important questions of Chapter 11 ‘Algebra’ of Class 6 Maths? 


Ans: It is important to go through the important questions of Chapter 11 ‘Algebra’ of Class 6 Maths as solving these questions will enable you to clarify your doubts and strengthen your concepts. If you practise them then you can test your skills and analyze your performance. This analysis will help you to know how much you have understood the chapter. With this, you can also detect your strong and weak topics. Practising questions of Chapter 11 ‘Algebra’ of Class 6 Maths will store concepts in your brain for the long term. And hence, you’ll be able to revise this chapter in less time. 


7. Is Chapter 11 ‘Algebra’ of Class 6 Maths easy? 


Ans: Chapter 11 ‘Algebra’ of Class 6 Maths is neither easy nor difficult. Its difficulty level is medium as some questions and examples of this chapter are easy to understand whereas you may find difficulty in understanding the rest of the questions. However, students who are good at their studies will comprehend concepts easily. While average and weak students have to struggle a bit to understand this chapter. But, if they practise this chapter continuously then they’ll find it easy. 


8. What concepts are students going to understand in Chapter 11 ‘Algebra’ of Class 6 Maths? 


Ans: The concepts that students are going to understand in Chapter 11 ‘Algebra’ of Class 6 Maths are as follows. 

  1. Understanding patterns with matchsticks

  2. Definition, facts, and examples of a variable

  3. Understanding more examples of patterns with matchsticks

  4. Other examples and applications of variables

  5. Application of variables in common rules

  6. Understanding regulations of geometry and arithmetic

  7. Expressions with variables

  8. Practical uses of expressions

  9. Definition, facts and examples of an equation

  10.  Solution of an equation 

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