# 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.

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

(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 .

(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 )

Letter "T" and "v" has pattern 2n because 2 matchsticks are used in these two letters.

cadets in a row = 5

number of rows = n

Thus, the total  number of cadets = 5n

mangoes in a box = 50

Number of boxes = b

Thus, the total number of mangoes = 50b

pencils per student = 5

number of students = s

Thus, the total pencils needed = 5s

Time taken by bird = t minutes

Speed of bird  = 1 km per minute

Thus, distance covered by bird $=speed\times time$

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

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.

Thus, age of Leela = (x-4) years

Number of laddus given = l

Number of laddus remaining = 5

Thus,tota number of laddus  = (l+5)

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

in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)

(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

(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

$l.$

The side of an equilateral triangle is $l$.

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

The side of a hexagon is l.

Therefore, the perimeter of the hexagon $=6 \times side=6l$

Length of one edge of cube = l

Number of edges in a cube  = 12

So,  total length = $12 \times l=12l$

## CBSE NCERT solutions for class 6 maths chapter 11 algebra-Exercise: 11.3

The  other expressions are :

(i)$(8\times 5)-7$

$(ii)(8+5)-7$

$(iii)(8+7)-5$

$(iv)(8\times 7)-5$

$(v) 5\times (7+8)$

$(vi) 5\times (7\times 8)$

$(vii) 5-(7+ 8)$

$(viii) 5+(8-7)$

(a)   $z+1,z-1,y+17,y-17$

(b)  $17y,\frac{y}{17},5z$

(c)  $2y+17,2y-17$

(d)  $7m,-7m+3,-7m-3$

(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 by$7$

(d) $p$ divided by $7$

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

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

(g) $-p$ divided by $5$

(h) $p$ multiplied by $-5$

(a) $7$ added to $p$ $=p+7$

(b) $7$ subtracted from $p$$=p-7$

(c) $p$ multiplied by$7$$=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 $-5$

(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 $-8$

(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 $16.$

(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$

According to given condition,

$t+4,t-4,4-t,4t,\frac{t}{4},\frac{4}{t}$

According to the given condition,

$2y+7,2y-7,7y+2,7y-2$

Height of box  =  h cm

Length of box = 5 times height = 5h cm

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

Meena's position = s

Beena's position = 8 steps ahead = s+8

Leena's position =7 steps behind = s-7

Total steps  = $4s-10$

$\inline v.$

Speed of bus = v km\h

Distance travelled in 5 hours = 5v km

Remaining distance = 20 km

Totsl distance = $(5v+20)km$

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

(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 = $(3x+7)years$

$y+7,$ $y-3,$ $y+4\frac{1}{2},$  $y-2\frac{1}{2}.$

Her age in past = $(y-3),(y-2\frac{1}{2})$

Her age in future  =$(y+7),(y+4\frac{1}{2})$

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

$17=x+7$

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

$(t-7)> 5$

(b) It is not an equation as LHS is greater than RHS.

$\frac{4}{2}=2$

(c) It is an equation with no variable. We may call this a numerical equation

$(7\times 3)-19=8$

(d) It is an equation with no variable. It is a numerical equation.

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

$x-2=0$.

(f) It is an equation of variable x.

$2m< 30$

(g) It is not an equation because LHS is less than RHS.

$2n+1=11$

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

$7=(11\times 5)-(12\times 4)$

(i) It is an equation with no variable as both sides are equal.

$7=(11\times 2)+p$

(j) It is equation of variable p.

$20=5y$

(k) It is an equation of varaible y.

$\frac{3q}{2}< 5$

(l) It is not an equation as LHS is less than RHS.

$z+12> 24$

(m) It is not an equation as LHS is greater than RHS.

$20-(10-5)=3\times 5$

(n) It is an equation with no variable.

$7-x=5$

(o) It is an equation of variable x.

 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

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

(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

$m+10=16.$

(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.

$5t=35.$

(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.

(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.

$m-7=3.$

(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

(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!

(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

## 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!!!