## NCERT Class VI Mathematics Chapter 11 Algebra

National Council of Educational Research and Training (NCERT) Book for Class VI

Subject: Mathematics

Chapter: Chapter 11 – Algebra

Class VI NCERT Mathematics Text Book Chapter 11 Algebra is given below.

**11.1 Introduction**

Our study so far has been with numbers and shapes. We have learnt numbers, operations on numbers and properties of numbers. We applied our knowledge of numbers to various problems in our life. The branch of mathematics in which we studied numbers is **arithmetic**. We have also learnt about figures in two and three dimensions and their properties. The branch of mathematics in which we studied shapes is **geometry**. Now we begin the study of another branch of mathematics. It is called **algebra**.

The main feature of the new branch which we are going to study is the use of letters. Use of letters will allow us to write rules and formulas in a general way. By using letters, we can talk about any number and not just a particular number. Secondly, letters may stand for unknown quantities. By learning methods of determining unknowns, we develop powerful tools for solving puzzles and many problems from daily life. Thirdly, since letters stand for numbers, operations can be performed on them as on numbers. This leads to the study of algebraic expressions and their properties.

You will find algebra interesting and useful. It is very useful in solving problems. Let us begin our study with simple examples.

**11.2 Matchstick Patterns**

Ameena and Sarita are making patterns with matchsticks. They decide to make simple patterns of the letters of the English alphabet. Ameena takes two matchsticks and forms the letter L as shown in Fig 11.1 (a).

Sarita says, “The rule is very powerful! Using the rule, I can say how many matchsticks are required to form even 100 Ls. I do not need to draw the pattern or make a table, once the rule is known”. Do you agree with Sarita?

**11.3 The Idea of a Variable**

In the above example, we found a rule to give the number of matchsticks required to make a pattern of Ls. The rule was :

**Number of matchsticks required = 2n**

Here, n is the number of Ls in the pattern, and n takes values 1, 2, 3, 4,…. Let us look at Table 1 once again. In the table, the value of n goes on changing (increasing). As a result, the number of matchsticks required also goes on changing (increasing).

**n is an example of a variable. Its value is not fixed; it can take any value 1, 2, 3, 4, … . We wrote the rule for the number of matchsticks required using the variable n.**

The word ‘variable’ means something that can vary, i.e. change. The value of a variable is not fixed. It can take different values.

We shall look at another example of matchstick patterns to learn more about variables.

**11.4 More Matchstick Patterns**

Ameena and Sarita have become quite interested in matchstick patterns. They now want to try a pattern of the letter C. To make one C, they use three matchsticks as shown in Fig. 11.2(a).

Can you complete the entries left blank in the table?

Sarita comes up with the rule :

**Number of matchsticks required = 3n**

She has used the letter n for the number of Cs; n is a variable taking on values

1, 2, 3, 4, …

Do you agree with Sarita ?

Remember 3n is the same as 3 × n.

Next, Ameena and Sarita wish to make a pattern of Fs. They make one F using 4 matchsticks as shown in Fig 11.3(a).

in the drill; here, r is a variable which stands for the number of rows and so takes on values 1, 2, 3, 4, … .

In all the examples seen so far, the variable was multiplied by a number. There can be different situations as well in which numbers are added to or subtracted from the variable as seen below.

Sarita says that she has 10 more marbles in her collection than Ameena. If Ameena has 20 marbles, then Sarita has 30. If Ameena has 30 marbles, then Sarita has 40 and so on. We do not know exactly how many marbles Ameena has. She may have any number of marbles.

But we know that, Sarita’s marbles = Ameena’s marbles + 10. We shall denote Ameena’s marbles by the letter x. Here, x is a variable, which can take any value 1, 2, 3, 4,… ,10,… ,20,… ,30,… . Using x, we write Sarita’s marbles = x + 10. The expression (x + 10) is read as ‘x plus ten’. It means 10 added to x. If x is 20, (x + 10) is 30. If x is 30, (x + 10) is 40 and so on.

2. We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?

3. Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)

4. If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)

5. The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)

6. A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)

of the sides AD or BC. Therefore,

Perimeter of a rectangle = length of AB + length of BC + length of CD

+ length of AD

= 2 × length of CD + 2 × length of BC = 2l + 2b

The rule, therefore, is that the perimeter of a rectangle = 2l + 2b where, l and b are respectively the length and breadth of the rectangle. Discuss what happens if l = b.

If we denote the perimeter of the rectangle by the variable p, the rule for perimeter of a rectangle becomes p = 2l + 2b

**Note** : Here, both l and b are variables. They take on values independent of each other. i.e. the value one variable takes does not depend on what value the other variable has taken.

In your studies of geometry you will come across several rules and formulas dealing with perimeters and areas of plane figures, and surface areas and volumes of three-dimensional figures. Also, you may obtain formulas for the sum of internal angles of a polygon, the number of diagonals of a polygon and so on. The concept of variables which you have learnt will prove very useful in writing all such general rules and formulas.

For example,

4 × 3 = 12, 3 × 4 = 12

Hence, 4 × 3 = 3 × 4

This property of numbers is known as **commutativity of multiplication** **of numbers**. Commuting (interchanging) the order of numbers in multiplication does not change the product. Using variables a and b as in the case of addition, we can express the commutativity of multiplication of two numbers as a × b = b × a

Note that a and b can take any number value. They are variables. All the special cases like

4 × 3 = 3 × 4 or 37 × 73 = 73 × 37 follow from the general rule. 5. Distributivity of numbers

Suppose we are asked to calculate 7 × 38. We obviously do not know the table of 38. So, we do the following:

7 × 38 = 7 × (30 + 8) = 7 × 30 + 7 × 8 = 210 + 56 = 266

This is always true for any three numbers like 7, 30 and 8. This property is known as** distributivity of multiplication over addition of numbers.**

By using variables, we can write this property of numbers also in a general and concise way. Let a, b and c be three variables, each of which can take any number. Then, a × (b + c) = a × b + a × c

Properties of numbers are fascinating. You will learn many of them in your study of numbers this year and in your later study of mathematics. Use of variables allows us to express these properties in a very general and concise way. One more property of numbers is given in question 5 of Exercise 11.2. Try to find more such properties of numbers and learn to express them using variables.

**EXERCISE 11.2**

5. Give expressions in the following cases.

(a) 11 added to 2m

(b) 11 subtracted from 2m

(c) 5 times y to which 3 is added

(d) 5 times y from which 3 is subtracted

(e) y is multiplied by – 8

(f) y is multiplied by – 8 and then 5 is added to the result

(g) y is multiplied by 5 and the result is subtracted from 16

(h) y is multiplied by – 5 and the result is added to 16.

6. (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.

(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.

**11.8 Using Expressions Practically**

We have already come across practical situations in which expressions are useful. Let us remember some of them.

Try to find more such situations. You will realise that there are many statements in ordinary language, which you will be able to change to statements using expressions with variables. In the next section, we shall see how we use these statements using expressions for our purpose.

**EXERCISE 11.4**

1. Answer the following:

(a) Take Sarita’s present age to be y years

(i) What will be her age 5 years from now?

(ii) What was her age 3 years back?

(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?

(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?

(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?

(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.

(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.

(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.

2. Change the following statements using expressions into statements in ordinary language.

(For example, Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)

(a) A notebook costs Rs p. A book costs Rs 3 p.

(b) Tony puts q marbles on the table. He has 8 q marbles in his box.

(c) Our class has n students. The school has 20 n students.

(d) Jaggu is z years old. His uncle is 4 z years old and his aunt is (4z – 3) years old.

(e) In an arrangement of dots there are r rows. Each row contains 5 dots.

was satisfied by n = 5. No other value of n satisfies the equation.** The value of the variable in an equation which satisfies the equation is called a solution to the equation.** Thus, n = 5 is a solution to the equation 2 n = 10.

Note, n = 6 is not a solution to the equation 2n = 10; because for n = 6, 2n = 2 × 6 = 12 and not 10.

Also, n = 4 is not a solution. Tell, why not?

Let us take the equation x – 3 = 11 (2)

This equation is satisfied by x = 14, because for x = 14,

LHS of the equation = 14 – 3 = 11 = RHS

It is not satisfied by x = 16, because for x = 16,

LHS of the equation = 16 – 3 = 13, which is not equal to RHS.

Thus, x = 14 is a solution to the equation x – 3 = 11 and x = 16 is not a solution to the equation. Also, x = 12 is not a solution to the equation. Explain,

**why not?**

Now complete the entries in the following table and explain why your answer is Yes/No.

In finding the solution to the equation 2n = 10, we prepared a table for various values of n and from the table, we picked up the value of n which was the solution to the equation (i.e. which satisfies the equation). What we used is** a trial and error method**. It is not a **direct** and **practical** way of finding a

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