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.

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 .

(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 )

(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: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?

Answer:

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

Question: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.)

Answer:

cadets in a row = 5

number of rows = n

Thus, the total number of cadets = 5n

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: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.)

Answer:

pencils per student = 5

number of students = s

Thus, the total pencils needed = 5s

Question: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.)

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:8 Leela is Radha's younger sister. Leela is $4$ years younger than Radha. Can you write Leela's age in terms of Radha's age? Take Radha's age to be $x$ years

Answer:

Radha's age = x years

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

Question:9 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:10 Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still $10$ oranges remain outside. If the number of oranges in a small box are taken to be x. what is the number of oranges in the larger box?

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

Question:11(a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks

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

Answer:

(a) 4 matchsticks

(b) 7 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

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

(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:2 The side of a regular hexagon (Fig 11.10) is denoted by $l.$ Express the perimeter of the hexagon using $l.$ (Hint: A regular hexagon has all its six sides equal in length.)

Answer:

The side of a hexagon is l.

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

Question:3 A cube is a three-dimensional figure as shown in Fig 11.11. It has six faces and all of them are identical squares. The length of an edge of the cube is given by $l.$ Find the formula for the total length of the edges of a cube.

Answer:

Length of one edge of cube = l

Number of edges in a cube = 12

So, total length = $12 \times l=12l$

Question:4 The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.12) $AB$ is a diameter of the circle; $C$ is its centre.) Express the diameter of the circle $(d)$ in terms of its radius

$(r).$

Answer:

Length of diameter is double the length of the radius.

Thus,d = 2r

Question:5 To find sum of three numbers $14,$ $27$ and $13,$ we can have two ways:

(a) We may first add $14$ and $27$ to get $41$ and then add $13$ to it to get the total sum $54$ or

(b) We may add $27$ and $13$ to get $40$ and then add $14$ to get the sum $54.$ Thus, $(14+27)+13=14+(27+13)$

This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.

Answer:

According to the given condition,

$(a+b)+c=a+(b+c)$

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

Question:1 Make up as many expressions with numbers (no variables) as you can from three numbers $5,7$ and $8.$ Every number should be used not more than once. Use only addition, subtraction and multiplication.(Hint: Three possible expressions are $5+(8-7),\ 5-(8-7), \ (5\times8)+7$ make the other expressions.)

Answer:

The other expressions are :

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

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

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

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

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

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

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

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

Question:2 Which out of the following are expressions with numbers only?

(a) $y+3$

(b) $(7\times 20)-8z$

(d) $5$

(e) $3x$

(f) $5-5n$

(g) $(7\times 20)-(5\times 10)-45+p$

Answer:

Option (c) and (d) are with numbers only. All other expression contains alphabets

Question:3 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) $17y,\frac{y}{17},5z$

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

$5z\rightarrow multiplication$

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

(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 $p$

(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 $p$ $=p-7$

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

Answer:

According to given condition,

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

Question:6(b) Form expressions using $y,$ $2$ and $7$ . Every expression must have $y$ in it. Use only two number operations. These should be different.

Answer:

According to the given condition,

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

NCERT solutions for class 6 maths chapter 11 algebra-Exercise: 11.4

Question: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?

Answer:

According to the given condition,

(a) (i) $y+5$

(ii) $y-3$

(iii) 6y

(iv) $6y-2$

(v) $3y+5$

Question:1(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?

Answer:

Breadth = b

And length = (3b-4) meters

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(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 $s.$

Answer:

Meena's position = s

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

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

Total steps = $4s-10$

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: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 $Rs.3p.$

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

(d) Jaggu is $z$ years old. His uncle is $4\; z$ years old and his aunt is $(4\; z-3)$ years old.

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

(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 $x$ 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(b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate?

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

Answer:

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

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

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

$(t-7)> 5$

Answer:

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

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

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

Answer:

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

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

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. $5\times 4-8=2\; x$

Answer:

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

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

$x-2=0$ .

Answer:

(f) It is an equation of variable x.

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

$2m< 30$

Answer:

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

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

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

Answer:

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

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

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

Answer:

(j) It is equation of variable p.

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

$20=5y$

Answer:

(k) It is an equation of varaible y.

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

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

Answer:

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

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

$z+12> 24$

Answer:

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

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

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

Answer:

(n) It is an equation with no variable.

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

$7-x=5$

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	$10\times 10=100$
(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) $n+12=20$ $(12,8,20,0)$

(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.

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!!!

Recently Asked Questions

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

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

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

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

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

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NCERT Solutions for Class 6 Maths Chapter 11 Algebra

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

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

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

NCERT solutions for class 6 maths chapter 11 algebra-Exercise: 11.4

NCERT Solutions for Class 6 - Subject wise

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

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