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 reallife 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.
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.
(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 ( because 2 matchsticks are used )
(b) A pattern of letter Z as ( because 3 matchsticks are used )
(c) A pattern of letter U as ( because 3 matchsticks are used )
(d) A pattern of letter V as ( because 2 matchsticks are used )
(e) A pattern of letter E as ( because 5 matchsticks are used )
(f) A pattern of letter S as ( because 5 matchsticks are used )
(g) A pattern of letter A as ( because 6 matchsticks are used )
Answer:
Letter "T" and "v" has pattern 2n because 2 matchsticks are used in these two letters.
Answer:
cadets in a row = 5
number of rows = n
Thus, the total number of cadets = 5n
Answer:
mangoes in a box = 50
Number of boxes = b
Thus, the total number of mangoes = 50b
Answer:
pencils per student = 5
number of students = s
Thus, the total pencils needed = 5s
Answer:
Time taken by bird = t minutes
Speed of bird = 1 km per minute
Thus, distance covered by bird
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 dots.
When the number of rows is 10, then the total number of dots are dots.
Answer:
Radha's age = x years
Thus, age of Leela = (x4) years
Answer:
Number of laddus given = l
Number of laddus remaining = 5
Thus,tota number of laddus = (l+5)
Answer:
Number of oranges in one box = x
Number of box = 2
So, the total number of oranges = 2x
Remaining number of oranges = 10
Thus, the number of oranges = 2x+10
Answer:
(a) 4 matchsticks
(b) 7 matchsticks
(c) 10 matchsticks
(d) 13 matchsticks
If we remove 1 matchstick from each then it forms a table of 3 i.e.,3,6,9,12,.....
So, required equation = 3x+1 , x= number of squares
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 algebraExercise: 11.2
Answer:
The side of an equilateral triangle is .
Therefore, the perimeter of the equilateral triangle
Answer:
The side of a hexagon is l.
Therefore, the perimeter of the hexagon
Answer:
Length of one edge of cube = l
Number of edges in a cube = 12
So, total length =
Answer:
Length of diameter is double the length of the radius.
Thus,d = 2r
Question:5 To find sum of three numbers and we can have two ways:
(a) We may first add and to get and then add to it to get the total sum or
(b) We may add and to get and then add to get the sum Thus,
Answer:
According to the given condition,
Answer:
The other expressions are :
(i)
Question:4(i) Give expressions for the following cases.
Answer:
(a) added to
(b) subtracted from
(c) multiplied by
(d) divided by
(e) subtracted from
(f) multiplied by
(g) divided by
(h) multiplied by
Question:4(ii) Give expressions for the following cases.
Answer:
(e) subtracted from =
(f) multiplied by
(g) divided by
(h) multiplied by
Question:5(i) Give expressions in the following cases.
(d) times y from which is subtracted
Answer:
(a) added to
(b) subtracted from
c) times y to which is added
(d) times y from which is subtracted
(e) is multiplied by
Question:5(ii) Give expressions in the following cases.
(f) is multiplied by and then is added to the result
(g) is multiplied by and the result is subtracted from
(h) is multiplied by and the result is added to
Answer:
(f) is multiplied by and then is added to the result=
(g) is multiplied by and the result is subtracted from
(h) is multiplied by and the result is added to
Question:6(a) Form expressions using and Use not more than one number operation. Every expression must have in it.
Answer:
According to given condition,
Question:6(b) Form expressions using and . Every expression must have in it. Use only two number operations. These should be different.
Answer:
According to the given condition,
Question:1 Answer the following:
(a) Take Sarita’s present age to be years
(i) What will be her age years from now?
(ii) What was her age years back?
(iii) Sarita’s grandfather is times her age. What is the age of her grandfather?
(iv) Grandmother is years younger than grandfather. What is grandmother's age?
(v) Sarita’s father’s age is years more than times Sarita’s age. What is her father's age?
Answer:
According to the given condition,
(a) (i)
(ii)
(iii) 6y
(iv)
(v)
Answer:
Breadth = b
And length = (3b4) meters
Answer:
Height of box = h cm
Length of box = 5 times height = 5h cm
Breadth of box = 10 cm less than length = (5h10) cm
Question:1(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 steps behind. Where are Beena and Meena? The total number of steps to the hill top is less than times what Meena has reached. Express the total number of steps using
Answer:
Meena's position = s
Beena's position = 8 steps ahead = s+8
Leena's position =7 steps behind = s7
Total steps =
Answer:
Speed of bus = v km\h
Distance travelled in 5 hours = 5v km
Remaining distance = 20 km
Totsl distance =
(a) A notebook costs `.A book costs
(b) Tony puts marbles on the table. He has marbles in his box.
(c) Our class has students. The school has students.
(d) Jaggu is years old. His uncle is years old and his aunt is years old.
(e) In an arrangement of dots there are rows. Each row contains dots.
Answer:
(a) A book cost 3 times the cost of a notebook.
(b) The number of marbles in the box is 8 times marbles on the table.
(c) The total number of students in school is 20 times that in our class.
(d) Jaggu's uncle's age is 4 times the age of Jaggu. Jaggu's aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.
Question:3(a) Given Munnu’s age to be years, can you guess what may show? (Hint : Think of Munnu’s younger brother.)
Can you guess what may show? What may show?
Answer:
(a) Munnu's age = x years
Munnu’s younger brother is 2years younger than him = years
Munnu’s elder brother is 4years older than him = years
His father is 7 years more than thrice of his age =
Answer:
Her age in past =
Her age in future =
Question:3(c) Given students in the class like football, what may show? What may 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.,
NCERT solutions for class 6 maths chapter 11 algebraExercise: 11.5
Answer:
(a) It is an equation of variable as both sides are equal. The variable is x.
Answer:
(b) It is not an equation as LHS is greater than RHS.
Answer:
(c) It is an equation with no variable. We may call this a numerical equation
Answer:
(d) It is an equation with no variable. It is a numerical equation.
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.
Answer:
(e) It is an equation of variable as both sides are equal. The variable is x.
Answer:
(f) It is an equation of variable x.
Answer:
(g) It is not an equation because LHS is less than RHS.
Answer:
(h) It is an equation of variable as both sides are equal. The variable is n.
Answer:
(i) It is an equation with no variable as both sides are equal.
Answer:
(j) It is equation of variable p.
Answer:
(k) It is an equation of varaible y.
Answer:
(l) It is not an equation as LHS is less than RHS.
Answer:
(m) It is not an equation as LHS is greater than RHS.
Answer:
(n) It is an equation with no variable.
Answer:
(o) It is an equation of variable x.
Question:2 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 

(b) 
10 y = 80 
y = 8 
Yes 

(c) 
10 y = 80 
y = 5 
No 

(d) 
4l = 20 
l = 20 
No 

(e) 
4l = 20 
l = 80 
No 

(f) 
4l = 20 
l =5 
Yes 

(g) 
b + 5 =9 
b = 5 
No 

(h) 
b + 5=9 
b = 9 
Yes 

(i) 
b+5=9 
b = 4 
Yes 

(j) 
h 8 =5 
h =13 
Yes 

(k) 
h 8 =5 
h = 8 
No 

(l) 
h 8 =5 
h = 0 
No 

(m) 
p+3=1 
p = 3 
No 

(n) 
p+3=1 
p = 1 
No 

(o) 
p+3=1 
p = 0 
No 

(p) 
p+3=1 
p = 1 
Yes 

(q) 
p+3=1 
p = 2 


Answer:
(a)
Putting given values in LHS.
m = 10 is not a solution. m = 12 is a solution.
m = 5 is not a solution. m = 15 is not a solution.
(b)
Putting given values in LHS.
n = 12 is not a solution. n = 20 is not a solution.
n = 8 is a solution. n =0 is not a solution.
(c)
Putting given values in LHS.
p =0 is not a solution. p = 10 is a solution.
p = 5 is not a solution. p = 5 is not a solution
(d)
Putting given values in LHS.
q =7 is not a solution. q = 2 is not a solution.
q = 10 is not a solution. q = 14 is a solution.
(e)
Putting given values in LHS.
r =4 is a solution. r = 4 is not a solution.
r = 8 is not a solution. r = 0 is not a solution
(f)
Putting given values in LHS.
x = 2 is a solution. x = 0 is a solution.
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
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
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 using the table.
Answer:
(c)
z 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 


3 


4 


5 


6 


At z = 12 ,
Thus, z= 12 is a solution.
Question:4(d) Complete the table and find the solution to the equation
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, m7 = 3
Thus, m=10 is the solution
Question:5 Solve the following riddles, you may yourself construct such riddles.
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
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
Special number is 17.
(iv) Let the number be x.
22  x = x
22 = 2x
The number is 11.
NCERT Solutions for Class 6 Mathematics  Chapterwise
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 

Chapter  7 

Chapter  8  
Chapter  9 
CBSE NCERT solutions for class 6 maths chapter 9 Data Handling 
Chapter 10 

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 

Chapter 14 
Solutions of NCERT for class 6 maths chapter 14 Practical Geometry 
NCERT Solutions for class 6 maths 
Solutions of NCERT for class 6 science 
Happy learning!!!