NCERT Solutions for Class 6 Maths Chapter 11 Algebra

 

NCERT solutions for class 6 maths chapter 11 Algebra: This is a very important branch of mathematics. An algebraic expression contains variables, numbers, and operations. In this chapter, you will study how can we use letters or symbols to represent numbers and quantities. This chapter will give you some basics of algebra. This chapter starts with the concept of variable. It is an alphabet or symbols which can take any numerical value. In this article, you will get CBSE NCERT solutions for class 6 maths chapter 11 algebra. This chapter is very useful in solving many real-life problems using variables. In this chapter, you will learn to form and solve the equation in one variable. In higher classes (7,8,9), you will learn to solve equations in one variable and two variables using different methods. There are many questions in solutions of NCERT for class 6 maths chapter 11 algebra where you will learn to form an equation and solve them. There are some practice questions given at the end of every topic to give you conceptual clarity. In 5 exercises of this chapter, there are 27 questions given in the textbook. All these questions are explained in CBSE NCERT solutions for class 6 maths chapter 11 algebra. It will be very easy for you to understand the concept. You can get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link.

Exercise 11.1

Exercise 11.2

Exercise 11.3

Exercise 11.4

Exercise 11.5

The main topics of NCERT Solutions for Class 6 Maths Chapter 11 Algebra are 

11.1 Introduction

11.2 Matchstick Patterns

11.3 The Idea of a Variable

11.4 More Matchstick Patterns

11.5 More Examples of Variables

11.6 Use of Variables in Common Rules

11.7 Expressions with Variables

11.8 Using Expressions Practically

11.9 What is an Equation?

11.10 Solution of an Equation

Understanding the concept of variables and equations are mandatory for our higher studies. So solving all the NCERT questions are mandatory. Solve as many questions as possible from this chapter.

Here are the Solutions of NCERT for Class 6 Maths Chapter 11 Algebra:

 

NCERT solutions for class 6 maths chapter 11 algebra-Exercise: 11.1

Question:1  Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.

             (a) A pattern of letter T as .

             (b) A pattern of letter Z as .

             (c) A pattern of letter U as  .

             (d) A pattern of letter V as .

             (e) A pattern of letter E as .

             (f) A pattern of letter S as .

             (g) A pattern of letter A as .

Answer:

             (a) A pattern of letter T as =2n( because 2 matchsticks are used )

             (b) A pattern of letter Z as  =3n( because 3 matchsticks are used )

             (c) A pattern of letter U as  =3n( because 3 matchsticks are used )

             (d) A pattern of letter V as =2n( because 2 matchsticks are used )

             (e) A pattern of letter E as =5n( because 5 matchsticks are used )

             (f) A pattern of letter S as =5n( because 5 matchsticks are used )

             (g) A pattern of letter A as =6n( because 6 matchsticks are used )

 

Question: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.)

Answer:

mangoes in a box = 50

Number of boxes = b

Thus, the total number of mangoes = 50b

Question: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.)

Answer:

pencils per student = 5

 number of students = s

Thus, the total pencils needed = 5s

Question: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.)

Answer:

Time taken by bird = t minutes

Speed of bird  = 1 km per minute

Thus, distance covered by bird =speed\times time

                                               =1\times t=t \,\, \, \, km

Question:7 Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?

Answer:

Number of rows = r

Number of dots in each row  = 9 dots

Thus, the total number of dots = 9r

When the number of rows is 8, then the total number of dots are 9\times8 =72 dots.

When the number of rows is 10, then the total number of dots are 9\times 10=90 dots.

Question:Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?

Answer:

Number of laddus given = l

Number of laddus remaining = 5

Thus,tota number of laddus  = (l+5)

Question:11(b) Fig 11.7 gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

                    

                    

Answer:

(a) 3 matchsticks

(b) 5 matchsticks

(c) 7 matchsticks

(d) 9 matchsticks

If we remove 1 matchstick from each then it forms a table of 2 i.e.,  2,4,6,8............

So, required equation  = 2x+1 , x= number of triangles

 

NCERT solutions for class 6 maths chapter 11 algebra-Exercise: 11.2

Question:1 The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using 

l.

Answer:

The side of an equilateral triangle is l.

Therefore, the perimeter of the equilateral triangle =3 \times side=3l

Question:Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.

             (a)   z+1,z-1,y+17,y-17

             (b)  17y,\frac{y}{17},5z

             (c)  2y+17,2y-17

             (d)  7m,-7m+3,-7m-3

Answer:

             (a)   z+1\rightarrow addition                                         y-17\rightarrow subtraction

                  z-1\rightarrow subtraction                                     y+17\rightarrow addition

 

       

             (b)  17y\rightarrow multiplication                       \frac{y}{17}\rightarrow division

                  5z\rightarrow multiplication

              

             (c)  2y+17\rightarrow multiplication \, \, and\, \, \, addition

                  2y-17\rightarrow multiplication \, \, and\, \, \, subtraction

 

             (d)  7m\rightarrow multiplication

                  -7m+3\rightarrow multiplication\, \, and\, \, addition

                   -7m-3\rightarrow multiplication\, \, and\, \, subtraction

 

Question:4(i) Give expressions for the following cases.

            (a) 7 added to p

            (b) 7 subtracted from p

            (c) p multiplied by7

            (d) p divided by 7

            (e) 7 subtracted from -m

            (f) -p multiplied by 5

            (g) -p divided by 5

            (h) p multiplied by -5

Answer:

           (a) 7 added to p =p+7

            (b) 7 subtracted from p=p-7

            (c) p multiplied by7=p\times 7=7p

            (d) p divided by 7  =\frac{p}{7}

            (e) 7 subtracted from -m=-m-7

            (f) -p multiplied by 5=-5p

            (g) -p divided by 5  =\frac{-p}{5}

            (h) p multiplied by -5=-5p

Question:4(ii) Give expressions for the following cases.

             (e) 7 subtracted from -m

             (f) -p multiplied by 5

            (g) -p divided by 5

            (h) p multiplied by -5

Answer:

             (e) 7 subtracted from -m-m-7

             (f) -p multiplied by 5

            (g) -p divided by 5

            (h) p multiplied by -5

Question:5(i) 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

Answer:

              (a) 11 added to 2m   =2m+11

             (b) 11 subtracted from 2m =2m-11

             c) 5 times y to which 3 is added=5y+3

            (d) 5 times y from which 3 is subtracted=5y-3

            (e) y is multiplied by -8=-8y

Question:5(ii) Give expressions in the following cases.

            (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.

Answer:

           (f) y is multiplied by -8 and then 5 is added to the result=-8y+5

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

           (h) y is multiplied by -5 and the result is added to 16.=-5y+16

Question:1(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.

Answer:

Height of box  =  h cm

Length of box = 5 times height = 5h cm

Breadth of box = 10 cm less than length  = (5h-10) cm

Question:1(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.

Answer:

Speed of bus = v km\h

Distance travelled in 5 hours = 5v km

Remaining distance = 20 km

 Totsl distance = (5v+20)km

Question:3(a) Given Munnu’s age to be x years, can you guess what (x-2) may show? (Hint : Think of Munnu’s younger brother.)

                   Can you guess what (x+4) may show? What (3x+7) may show?

Answer:

(a) Munnu's age  = x years

    Munnu’s younger brother is 2years younger than him = (x-2) years

       Munnu’s elder  brother is 4years older than him = (x+4) years

  His father is 7 years more than thrice of his age = (3x+7)years

 

Question:3(c) Given n students in the class like football, what may 2n  show? What may \frac{n}{2} show? (Hint : Think of games other than football).

Answer:

Number of students like hockey is twice the students like football, i.e.,2n

Number of students like tennis is half the students like football, i.e.,\frac{n}{2}

 

NCERT solutions for class 6 maths chapter 11 algebra-Exercise: 11.5

Question:1(a) State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.

  17=x+7

Answer:

(a) It is an equation of variable as both sides are equal. The variable is x.

Question:1(e) State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.5\times 4-8=2\; x

Answer:

(e) It is an equation of variable as both sides are equal. The variable is x.

Question:1(h) State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.

 2n+1=11

Answer:

(h) It is an equation of variable as both sides are equal. The variable is n.

Question:Complete the entries in the third column of the table.

              

Answer:

S. no.

Equation

Value of variable

Equ.satisfied Yes/No

Sol. of LHS

(a)

10 y =80

  y = 10

     No

10\times 10=100

(b)

10 y = 80

  y = 8

     Yes

10\times 8=80

(c)

10 y = 80

    y = 5

      No

10\times 5=50

(d)

 4l = 20

   l = 20

       No

 4\times 20=80

(e)

 4l = 20

  l = 80

        No

4\times 80=320

(f)

 4l = 20

   l =5 

       Yes

4\times 5=20

(g)

  b + 5 =9

  b = 5

         No

5+5=10

(h)

  b + 5=9

 b = 9

          Yes

9+5=14

(i)

  b+5=9

  b = 4

        Yes

13-8=5

(j)

  h -8 =5

  h =13

         Yes

8-8=0

(k)

   h -8 =5

  h = 8

          No  

0-8=-8

(l)

   h -8 =5

  h = 0

            No

3+3=6

(m)

  p+3=1

  p = 3

            No

1+3= 4

(n)

   p+3=1

 p = 1

              No

0+3= 3

(o)

  p+3=1

  p = 0

             No

-1+3= 2

(p)

  p+3=1

  p = -1

          Yes

-2+3= 1

(q)

   p+3=1

  p = -2

 

 

 

Question:3 Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.

            (a) 5m=60                    (10,5,12,15)

            (b)n+12=20              (12,8,20,0)

             (c) p-5=5                  (0,10,5,-5)

             (d) \frac{q}{2}=7                          (7,2,10,14)

             (e) r-4=0                  (4,-4,8,0)

            (f) x+4=2                    (-2,0,2,4)

Answer:

   (a) 5m=60

Putting given values in LHS.

5\times 10=50                                            5\times 12=60

\therefore LHS\neq RHS                                  \therefore LHS= RHS

m = 10 is not a solution.                             m = 12 is a solution.

5\times 5=50                                               5\times 15=75

\therefore LHS\neq RHS                                 \therefore LHS\neq RHS

m = 5 is not a solution.                              m = 15 is not a solution.

 

    (b)n+12=20

Putting given values in LHS.

12+12=24                                            20+12=32

\therefore LHS\neq RHS                                  \therefore LHS\neq RHS

n = 12 is not a solution.                                n = 20 is not a solution. 

8+12=20                                              0+12=12

\therefore LHS= RHS                                 \therefore LHS\neq RHS

n = 8  is  a solution.                                      n  =0  is not a solution. 

  

   (c) p-5=5

Putting given values in LHS.

0-5=-5                                            10-5=5

\therefore LHS\neq RHS                                  \therefore LHS= RHS

p =0 is not a solution.                                p = 10 is a solution.

5-5=0                                               -5-5=-10

\therefore LHS\neq RHS                                 \therefore LHS\neq RHS

p = 5 is not  a solution.                              p = -5 is not a solution

 

  (d) \frac{q}{2}=7   

Putting given values in LHS.

 \frac{7}{2}                                                                      \frac{2}{2}=1

\therefore LHS\neq RHS                                  \therefore LHS\neq RHS

q =7 is not a solution.                                q = 2 is not a solution.

\frac{10}{2}=5                                                  \frac{14}{2}=7

\therefore LHS\neq RHS                                 \therefore LHS= RHS

q = 10 is not  a solution.                              q = 14 is a solution.

 

(e) r-4=0

Putting given values in LHS.

4-4=0                                              -4-4=-8

\therefore LHS= RHS                                  \therefore LHS\neq RHS

r =4  is a solution.                                r = -4 is not  a solution.

8-4=4                                                0-4=-4

\therefore LHS\neq RHS                                 \therefore LHS\neq RHS

r = 8 is not  a solution.                              r = 0 is not a solution

 

 (f) x+4=2

Putting given values in LHS.

-2+4=2                                            0+4=4

\therefore LHS= RHS                                  \therefore LHS\neq RHS

 x = -2 is a solution.                                 x = 0 is a solution.

2+4=6                                               4+4=8

\therefore LHS\neq RHS                                 \therefore LHS\neq RHS

 x = 2 is not  a solution.                               x = 4 is not a solution

 

Question:4(a) Complete the table and by inspection of the table find the solution to the equation 

m+10=16.

                

Answer:

(a)

m

1

2

3

4

5

6

7

8

9

10

11

12

13

m+10

11

12

13

14

15

16

17

18

19

20

21

22

23

   At m = 6, m+10=16

     Thus, m =6 is a solution.                

 

Question:4(b) Complete the table and by inspection of the table, find the solution to the equation  

5t=35.

                 

Answer:

(b)

t

3

4

5

6

7

8

9

10

11

12

13

14

15

16

5t

15

20

25

30

35

40

45

50

55

60

65

70

75

80

At t=7, 5t = 35

Thus, t = 7 is the solution.

 

Question:4(c) Complete the table and find the solution of the equation z/3=4 using the table.

                   

Answer:

(c)

 

z

8

9

10

11

12

13

14

15

16

17

18

19

20

\frac{z}{3}

\frac{8}{3}

3

\frac{10}{3}

\frac{11}{3}

4

\frac{13}{3}

\frac{14}{3}

5

\frac{16}{3}

\frac{17}{3}

6

\frac{19}{3}

\frac{20}{3}

At z = 12 ,\frac{z}{3}=4

Thus, z= 12 is a solution.

Question:4(d) Complete the table and find the solution to the equation  

m-7=3.

                    

Answer:

(d)

 

m

5

6

7

8

9

10

11

12

13

14

15

m - 7

-2

-1

0

1

2

3

4

5

6

7

8

 

At m=10, m-7 = 3

Thus, m=10 is  the solution

 

Question:5 Solve the following riddles, you may yourself construct such riddles.

             (i) Go round a square Counting every corner Thrice and no more! Add the count to me To get exactly thirty four!

             (ii) For each day of the week Make an upcount from me If you make no mistake You will get twenty three!

             (iii) I am a special number Take away from me a six! A whole cricket team You will still be able to fix!

             (iv) Tell me who I am I shall give a pretty clue! You will get me back If you take me out of twenty two!

Answer:

(i) Square has 4 sides.

  Counting every corner Thrice and no more!

Implies we go around square thrice,i.e. 12 times.

Add the count to me To get exactly thirty - four!

Let us assume the number as y, we get 34.

 12+y=34

 y=34-12=22

So, 22 is the number .

 

(ii) If 23 is number for Sunday.

   Then counting up, Saturday is 22, Friday is 21, Thursday is 20.

 Wednesday is 19, Tuesday is 18, Monday is 17 and Sunday is 16.

Therefore, the number considered is 16.

 

(iii) Let the number be x.

 If we take away 6 from x, we get a cricket team.

  x - 6 = 11

x=11+6=17

Special number is 17.

 

(iv) Let the number be x.

   22 - x = x

      22 = 2x

x=\frac{22}{2}=11

The number is 11.

NCERT Solutions for Class 6 Mathematics - Chapter-wise

Chapters No.

Chapters Name

Chapter - 1 

Solutions of NCERT for class 6 maths chapter 1 Knowing Our Numbers

Chapter - 2

CBSE NCERT solutions for class 6 maths chapter 2 Whole Numbers

Chapter - 3

NCERT solutions for class 6 maths chapter 3 Playing with Numbers

Chapter - 4

Solutions of NCERT for class 6 maths chapter 4 Basic Geometrical Ideas

Chapter - 5

CBSE NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes

Chapter - 6

NCERT solutions for class 6 maths chapter 6 Integers

Chapter - 7

Solutions of NCERT for class 6 maths chapter 7 Fractions

Chapter - 8

NCERT solutions for class 6 maths chapter 8 Decimals

Chapter - 9

CBSE NCERT solutions for class 6 maths chapter 9 Data Handling

Chapter -10

NCERT solutions for class 6 maths chapter 10 Mensuration

Chapter -11

Solutions of NCERT for class 6 maths chapter 11 Algebra

Chapter -12

CBSE NCERT solutions for class 6 maths chapter 12 Ratio and Proportion

Chapter -13

NCERT solutions for class 6 maths chapter 13 Symmetry

Chapter -14

Solutions of NCERT for class 6 maths chapter 14 Practical Geometry

NCERT Solutions for Class 6 - Subject wise

NCERT Solutions for class 6 maths
Solutions of NCERT for class 6 science

Benefits of NCERT solutions for class 6 maths chapter 11 algebra-

  • You will learn solutions of NCERT class 6 for this chapter very easily.
  • You can develop an algebraic expression of some real-life problems and solve them which you have learnt in this chapter. 
  • It is going to help you with your homework as you can get detailed explanations of practice questions given at the end of every topic in the CBSE NCERT solutions for class 6 maths chapter 11 algebra.  
  • You will get some short tricks and tips to solve some specific problems.
  • Tip- Mathematics required more practice to get conceptual clarity. So, you should solve all the NCERT questions including examples and practice questions on your own. 

Happy learning!!!

 

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