NCERT Solutions for Class 6 Maths Chapter 11  Algebra Free PDF Download
Algebra is a branch of mathematics that involves the use of mathematical expressions to represent problems. Often considered the unifying thread of different mathematical disciplines, it uses variables such as x,y,z, etc.
Chapter 11 of Class 6 Maths inculcates the primary aspects of Algebra. With the help of NCERT Solutions for Class 6 Maths Chapter 11 Algebra PDF, students can understand the different problemsolving strategies as well as strengthen their understanding of fundamentals. The stepwise solutions are provided by subject experts as per the latest CBSE guidelines.
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NCERT Solutions for Class 6 Maths Chapter 11  Important Topics
This chapter is an essential one for 6thgrade students since it covers all essential points and sections that are needed to grow a strong understanding of numbers and equations in Algebra. The chapter 11 forms an important part of the Class 6 Maths syllabus and has 8 major sections or topics. Students must go through each topic and concept covered in the chapter very carefully to be able to solve and practise the sums in Algebra.
Introduction to Algebra
Matchstick Problems
The Idea of a Variable
Use of Variables in Common Rules
Rules from Geometry
Rules from Arithmetic
Expressions with Variables
Practical Applications of Expressions
What is an Equation?
Solution to an Equation
Students should learn each and every topic carefully and then try to attempt and practise the NCERT Solutions for Class 6 Maths Chapter 11 Algebra.
Access NCERT Solutions for Class 6 Maths Chapter 11 – Algebra
Exercise 11.1
1. Find this rule, which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write this rule.
(a). A pattern of letter T
Ans. It is observed that number of matchsticks required is $2$. Hence, the pattern of letter T is \[2n\].
(b) A pattern of letter Z
Ans. It is observed that number of matchsticks required is $3$. Hence, the pattern of letter Z is \[3n\].
(c) A pattern of letter U
Ans. It is observed that number of matchsticks required is $3$. Hence, the pattern of letter V is \[3n\].
(d) A pattern of letter V
Ans. It is observed that number of matchsticks required is $2$. Hence, the pattern of letter V is \[2n\].
(e) A pattern of letter E
Ans. It is observed that number of matchsticks required is $5$. Hence, the pattern of letter E is \[5n\].
(f) A pattern of letter S
Ans. It is observed that number of matchsticks required is $5$. Hence, the pattern of letter S is \[5n\].
(g) A pattern of letter A
Ans. It is observed that number of matchsticks required is $6$. Hence, the pattern of letter A is \[6n\].
2. We already know the rule for the pattern of letter 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?
Ans. We Observe that letter L requires $2$ matchsticks. Hence, it has pattern \[2n\]. Letters ‘T’ and ‘V’ similarly require $2$ matchsticks and hence has pattern \[2n\].
3. Cadets are marching in a parade. There are \[\mathbf{5}\]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)
Ans. Number of cadets= Number of cadets in each row × Number of rows
Number of rows= $n$
Number of cadets in each row=$5$ (∵Given)
∴ Total number of cadets= $5\times n=5n$
4. If there are \[\mathbf{50}\] 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)
Ans. Number of mangoes= Number of mangoes in each box × Number of boxes
Number of boxes= b
Number of mangoes in each box= $50$ (∵Given)
∴ Total number of mangoes= $50\times b=50b$
5. The teacher distributes \[\mathbf{5}\] pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students)
Ans. Number of pencils needed= Number of pencils distributed per student × Number of students
Number of students= s
Number of pencils distributed per student= $5$ (∵Given)
∴ Total number of pencils needed= $5\times s=5s$
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)
Ans. Distance covered by birds (in km) = speed × time
Time (in minutes) = t
Speed (in km/min) =$1$ km/min (∵Given)
∴ Total distance = $1\times t=t$km
7. Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder as in figure). She has \[\mathbf{8}\] dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are \[\mathbf{8}\] rows? If there are \[\mathbf{10}\] rows?
Ans. Number of dots = Number of dots in each row × Number of rows
Number of rows= r
Number of dots in each row= $8$ (∵Given)
∴ Total number of dots= $8\times r=8r$……(Equation 1)
For $8$ rows,
$r=8$
Total number of dots= \[8\times r=8\times 8=64\]……(∵ From Equation 1)
For $10$ rows,
$r=10$
Total number of dots= \[8\times r=8\times 10=80\]……(∵ From Equation 1)
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.
Ans. Let Radha’s age be x years
Given that, Leela is 4 years younger than Radha.
Hence, Leela’s age = \[\left( x4 \right)\] years
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?
Ans. Total laddus mother made = laddus given to guests and family + remaining laddus
Laddus given to guests and family = l
Laddus remaining = 5 (∵Given)
∴ Total laddus = l+5
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?
Ans. Number of smaller boxes = 2
Number of oranges in 1 small box = \[x\]
Oranges remaining outside = 10
Total number of oranges in larger box = (number of smaller boxes) × (number of oranges in 1 smaller box) + oranges remaining outside
∴ Total oranges in larger box =$2x+10$
11. (a) Look at the following matchstick pattern of squares. 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.)
Ans. If we remove the last vertical stick, we are left with a pattern of C.
Letter C has 3 matchsticks which gives us pattern $3n$.
Now, add the removed vertical stick. This gives us the required equation$3n+1$.
(b) Figs. Below 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.
Ans. We can see that the figures have 1 matchstick more than twice the number of triangles in the pattern.
Hence, required equation is $2n+1$. (where, n is the number of triangles)
Exercise 11.2
1. The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Ans. Side of equilateral triangle= l
Number of sides of triangle = 3
Perimeter of equilateral triangle=\[3\times l=3l\]
Hence the perimeter of the given equilateral triangle is 3l.
2. The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l.
Ans. Side of hexagon = l
Number of sides in a hexagon = 6
Perimeter of hexagon =$6\times l=6l$
Hence, the perimeter of given regular hexagon is 6l.
3. A cube is a three dimensional figure. 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 edged of cube.
Ans. Length of one edge of a cube= l
Number of edges in a cube = 12
Total length of edges of the given cube =$12\times l=12l$
Hence, the total length of edges in a cube is 12l.
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 AB is a diameter of the circle. C is its centre). Express the diameter of the circle d in terms of its radius r.
Ans. From figure, we can see that AC=BC=PC=r
Also, diameter AB (d)= AC+BC =$r+r=2r$
Hence, $d=2r$
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 \[\left( \mathbf{14}\text{ }+\text{ }\mathbf{27} \right)\text{ }+\text{ }\mathbf{13}\text{ }=\text{ }\mathbf{14}\text{ }+\text{ }\left( \mathbf{27}\text{ }+\text{ }\mathbf{13} \right)\]
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.
Ans. Let a, b and c be three variables.
Property Associativity of addition of Whole Numbers can be expressed as –
\[\left( a+b \right)+c=a+\left( b+c \right)\]
Exercise 11.3
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 \[\mathbf{5}+\left( \mathbf{8}\mathbf{7} \right),\text{ }\mathbf{5}\text{ }\left( \mathbf{8}\mathbf{7} \right),\text{ }\left( \mathbf{5}\times \mathbf{8} \right)+\mathbf{7}\]make the other expressions)
Ans. So, by using multiplication, addition and subtraction the possible expressions with the given numbers are taken as,
$5\times (7+8)$.
\[\left( 8+7 \right)5.\]
\[5\left( 7+8 \right).\]
\[5+\left( 87 \right).\]
\[(8\times 7)5\].
\[\left( 8+5 \right)7.\]
\[\left( 8\times 5 \right)7\]
2. Which out of the following are expressions with numbers only:
(a) $y+3$
(b) $7\times 208z$
(c) $5(217)+7\times 2$
(d) $5$
(e) $3x$
(f) $55n$
(g) \[\left( \mathbf{7}\times \mathbf{20} \right)\left( \mathbf{5}\text{ }\times \text{ }\mathbf{10} \right)\mathbf{45}+p\]
Ans. Options (c) and (d) as we can see no variables in these expressions.
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
Ans. z+1 → Addition
z1 → Subtraction
y +17→ Addition
y 17→ Subtraction
(b) 17y, \[\frac{\mathbf{y}}{\mathbf{17}}\], 5z
Ans. 17y → Multiplication
$\frac{y}{17}$→ Division
5z→ Multiplication
(c) 2y + 17, 2y – 17
Ans. 2y + 17→ Multiplication, Addition
2y  17→ Multiplication, Subtraction
(d) 7m, 7m + 3, 7m – 3
Ans. 7m → Multiplication
7m+3 → Multiplication, Addition
7m  3→ Multiplication, Subtraction
4. Give expressions for the following cases:
(a) 7 added to p.
Ans. $p+7$
(b) 7 subtracted from p.
Ans. $p7$
(c) p multiplied by 7.
Ans. $p\times 7=7p$
(d) p divided by 7.
Ans. $\frac{p}{7}$
(e) 7 subtracted from m.
Ans. $m7$
(f) p multiplied by 5.
Ans. $p\times 5=5p$
(g) p divided by 5.
Ans. $\frac{p}{5}$
(h) p multiplied by 5.
Ans. $p\times 5=5p$
5. Give expression in the following cases:
(a) 11 added to 2m.
Ans.$2m+11$
(b) 11 subtracted from 2m.
Ans. $2m11$
(c) 5 times y to which 3 is added.
Ans. $5y+3$
(d) 5 times y from which 3 is subtracted.
Ans. $5y3$
(e) y is multiplied by 8.
Ans. $8y$
(f) y is multiplied by 8 and then 5 is added to the result.
Ans. $8y+5$
(g) y is multiplied by 5 and result is subtracted from 16.
Ans. $165y$
(h) y is multiplied by 5 and the result is added to 16.
Ans. $5y+16$
6. (a) From expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
Ans. \[t+4,\text{ }t4,\text{ }t\times 4,\text{ }\frac{t}{4}\]
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Ans. \[2y+7,\text{ }2y7,\text{ }7y+2,7y2\]
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?
Ans. Sarita’s age 5 years from now will be \[\left( y\text{ }+\text{ }5 \right)\]years.
(ii) What was her age 3 years back?
Ans. Sarita’s age 3 years back was \[\left( y\text{ }\text{ }3 \right)\]years.
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
Ans. Age of Sarita’s grandfather is $6y$ years.
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
Ans. Age of Sarita’s grandmother is \[\left( 6y2 \right)\] years.
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
Ans. Age of Sarita’s father is \[\left( 3y+5 \right)\] years.
(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?
Ans. Length of rectangular hall = \[\left( 3b4 \right)\] meters
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth are 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
Ans. Height of Box = \[h\]cm
Length of Box = \[5h\] cm
Breadth of Box = \[\left( 5h10 \right)\] cm
(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.
Ans. Meena’s position = $s$
Beena’s position = \[s+8\]
Leena’s position = \[s7\]
Total number of steps = 4s  10
(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.
Ans. Speed of bus = $v$ km/hr
Distance travelled = speed × time = \[v\times 5=5v\] km
Distance remaining = 20 km
Total distance from Daspur to Beespur = \[\left( 5v+20 \right)\] km
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 note book costs ₹p. A book costs ₹3p.
Ans. A book cost 3 times the notebook.
(b) Tony puts q marbles on the table. He has 8q marbles in his box.
Ans. Number of marbles in box is 8 times the marbles on the table.
(c) Our class has n students. The school has 20n students.
Ans. Total number of students in the school is 20 times that of our class.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is \[\left( \mathbf{4}z\mathbf{3} \right)\] years old.
Ans. Jaggu’s uncle’s age is 4 times the age of Jaggu and Jaggu’s aunt is 3 years younger than his uncle.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Ans. Total number of dots is 5 times the number of rows.
3. (a) Given, Munnu’s age to be x years. Can you guess what \[\left( x\mathbf{2} \right)\]may show? (Hint: Think of Munnu’s younger brother). Can you guess what \[\left( x+\mathbf{4} \right)\] show? What \[\left( \mathbf{3}x+\mathbf{7} \right)\] may show?
Ans. \[\left( x2 \right)\]years will represent age of Munnu’s younger brother.
(\[x+4\]) years will represent age of Munnu after 4 years.
(\[3x+7\]) years represent age of an elder to Munnu whose age is 7 more than 3 times Munnu’s age.
(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? \[\mathbf{y}+\mathbf{7},\text{ }\mathbf{y}\mathbf{3},\text{ }\mathbf{y}+\mathbf{4}\text{ }\frac{1}{2},\text{ }\mathbf{y}\mathbf{2}\text{ }\frac{1}{2}\]
Ans. Age after 7 years = (\[y+7\]) years
Age 3 year ago = (\[y3\]) years
Age after 4 and half years = \[\left( y+4\frac{1}{2} \right)\] years
Age 2 and half year ago = $\left( y2\frac{1}{2} \right)$ years
(c) Given, n students in the class like football, what may 2n show? What may \[\frac{n}{\mathbf{2}}\] show? (Hint: Think of games other than football).
Ans. Number of students who like swimming is twice the students who like football = 2n
Number of students who like Athletics is half the students who like football =$\frac{n}{2}$
Exercise 11.5
1.State which of the following are equations (with a variable). Given reason for your answer. Identify the variable from the equations with a variable.
(a)\[\mathbf{17}=x+\mathbf{7}\]
Ans. It is an equation with variable x.
(b) \[\left( t\mathbf{7} \right)>\mathbf{5}\]
Ans. It is not an equation since there is no equal sign.
(c) \[\frac{\mathbf{4}}{\mathbf{2}}=\mathbf{2}\]
Ans. It is an equation with no variable.
(d) \[\left( \mathbf{7}\times \mathbf{3} \right)\mathbf{19}=\mathbf{8}\]
Ans. It is an equation with no variable.
(e) \[5\text{ }\times \text{ }4\text{ }\text{ }8\text{ }=\text{ }2\text{ }x\sqrt{{{a}^{2}}+{{b}^{2}}}\]
Ans. It is an equation with variable x.
(f) \[x\mathbf{2}=\mathbf{0}\]
Ans. It is an equation of variable x.
(g) \[\mathbf{2}m<\mathbf{30}\]
Ans. It is not an equation since there is no equal sign.
(h) \[\mathbf{2}n+\mathbf{1}=\mathbf{11}\]
Ans. It is an equation with variable n.
(i) \[7\text{ }=\text{ }\left( 11\text{ }\times \text{ }5 \right)\left( 12\text{ }\times \text{ }4 \right)\]
Ans. It is an equation with no variable.
(j) \[\mathbf{7}=\left( \mathbf{11}\times \mathbf{2} \right)+p\]
Ans. It is an equation with variable p.
(k) \[\mathbf{20}=\mathbf{5}y\]
Ans. It is an equation with variable y.
(l) \[\frac{\mathbf{3}q}{\mathbf{2}}<\mathbf{5}\]
Ans. It is not an equation since there is no equal sign.
(m) \[z+\mathbf{12}>\mathbf{24}\]
Ans. It is not an equation since there is no equal sign.
(n) \[\mathbf{20}\left( \mathbf{10}\mathbf{5} \right)=\mathbf{3}\times \mathbf{5}\]
Ans. It is an equation with no variable.
(o) \[\mathbf{7}x=\mathbf{5}\]
Ans. It is an equation with variable x.
2. Complete the entries of the third column of the table:
Sr. No.  Equation  Value of Variable  Equation Satisfied Yes/No 
(a)  10y = 80  y = 10  
(b)  10y = 80  y = 8  
(c)  10y = 80  y = 5  
(d)  4l = 20  l = 20  
(e)  4l = 20  l = 80  
(f)  4l = 20  l = 5  
(g)  b + 5 = 9  b = 5  
(h)  b + 5 = 9  b = 9  
(i)  b + 5 = 9  b = 4  
(j)  h – 8 = 5  h = 13  
(k)  h – 8 = 5  h = 8  
(l)  h – 8 = 5  h = 0  
(m)  p + 3 = 1  p = 3  
(n)  p + 3 = 1  p = 1  
(o)  p + 3 = 1  p = 0  
(p)  p + 3 = 1  p = 1  
(q)  p + 3 = 1  p = 2 
Ans.
Sr. No.  Equation  Value of Variable  Equation Satisfied Yes/No  Solution of L.H.S. 
(a)  10y = 80  y = 10  No  10×10=100 
(b)  10y = 80  y = 8  Yes  10×8=80 
(c)  10y = 80  y = 5  No  10×5=50 
(d)  4l = 20  l = 20  No  4×20=80 
(e)  4l = 20  l = 80  No  4×80=320 
(f)  4l = 20  l = 5  Yes  4×5=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  4+5=9 
(j)  h – 8 = 5  h = 13  Yes  13–8=5 
(k)  h – 8 = 5  h = 8  No  8–8=0 
(l)  h – 8 = 5  h = 0  No  0–8=–8 
(m)  p + 3 = 1  p = 3  No  3+3=6 
(n)  p + 3 = 1  p = 1  No  1+3=4 
(o)  p + 3 = 1  p = 0  No  0+3=3 
(p)  p + 3 = 1  p = 1  No  –1+3=2 
(q)  p + 3 = 1  p = 2  Yes  –2+3=1 
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) \[5m=\mathbf{60}\left( \mathbf{10},\text{ }\mathbf{5},\text{ }\mathbf{12},\text{ }\mathbf{15} \right)\]
Ans. \[5m=60\]
Putting the given values in L.H.S.,
5×10=50
∴ L.H.S. ≠ R.H.S.
m=10 is not the solution.
5×5=25
∴L.H.S. ≠ R.H.S.
m=5 is not the solution.
5×12=60
∴ L.H.S. = R.H.S.
m=12 is a solution.
5×15=75
∴ L.H.S. ≠ R.H.S.
m=15 is not the solution.
(b) \[n+\mathbf{12}=\mathbf{20}\left( \mathbf{12},\text{ }\mathbf{8},\text{ }\mathbf{20},\text{ }\mathbf{0} \right)\]
Ans. Putting the given values in L.H.S.,
12+12=24
∴ L.H.S. ≠ R.H.S.
n=12 is not the solution.
8+12=20
∴ L.H.S. = R.H.S.
n=8 is a solution.
20+12=32
∴ L.H.S. ≠ R.H.S.
n=20 is not the solution.
0+12=12
∴ L.H.S. ≠ R.H.S.
n=0 is not the solution.
(c) p–5=5(0, 10, 5, 5)
Ans. Putting the given values in L.H.S.,
0–5=–5
∴ L.H.S. ≠ R.H.S.
p =0 is not the solution.
10–5=5
∴ L.H.S. = R.H.S.
p=10 is a solution.
5–5=0
∴ L.H.S. ≠ R.H.S.
p=5 is not the solution.
–5–5=–10
∴ L.H.S. ≠ R.H.S.
p=5 is not the solution.
(d) \[\frac{q}{\mathbf{2}}=\mathbf{7}\left( \mathbf{7},\text{ }\mathbf{2},\text{ }\mathbf{10},\text{ }\mathbf{14} \right)\]
Ans. Putting the given values in L.H.S.,
$\frac{7}{2}=\frac{7}{2}$
∴ L.H.S. ≠ R.H.S.
q=7 is not the solution.
$\frac{2}{2}=1$
∴ L.H.S. ≠ R.H.S.
q=2 is not the solution.
$\frac{10}{2}=5$
∴ L.H.S. ≠ R.H.S.
q=10 is not the solution.
$\frac{14}{2}=7$
∴ L.H.S. = R.H.S.
q=14 is a solution.
(e) \[r\mathbf{4}=\mathbf{0}\left( \mathbf{4},\text{ }\mathbf{4},\text{ }\mathbf{8},\text{ }\mathbf{0} \right)\]
Ans. Putting the given values in L.H.S.,
4–4=0
∴ L.H.S. = R.H.S.
r=4 is a solution.
–4–4=–8
∴ L.H.S. ≠ R.H.S.
r=4 is not the solution.
8–4=4
∴ L.H.S. ≠ R.H.S.
r=8 is not the solution.
0–4=–4
∴L.H.S. ≠ R.H.S.
r=0 is not the solution.
(f) \[~x+\mathbf{4}=\mathbf{2}\left( \mathbf{2},\text{ }\mathbf{0},\text{ }\mathbf{2},\text{ }\mathbf{4} \right)\]
Ans. Putting the given values in L.H.S.,
–2+4=2
∴ L.H.S. = R.H.S.
x=2 is a solution.
0+4=4
∴ L.H.S. ≠ R.H.S.
x=0 is not the solution.
2+4=6
∴L.H.S. ≠ R.H.S.
x=2 is not the solution.
4+4=8
∴ L.H.S. ≠ R.H.S.
x=4 is not the solution.
4. (a) Complete the table and by inspection of the table find the solution to the equation \[m+\mathbf{10}=\mathbf{16}\].
M  1  2  3  4  5  6  7  8  9  10  …  …  … 
m + 10  …  …  …  …  …  …  …  …  …  …  …  …  … 
Ans.
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 $m=6,m+10=16$
is the solution.
(b) Complete the table and by inspection of the table find the solution to the equation \[\mathbf{5t}=\mathbf{35}\].
T  3  4  5  6  7  8  9  10  11  …  …  …  … 
5t  …  …  …  …  …  …  …  …  …  …  …  …  … 
Ans.
T  3  4  5  6  7  8  9  10  11  12  13  14  15 
5t  15  20  25  30  35  40  45  50  55  60  65  70  75 
∴ At $t=7$, $5t=35$
$t=7$ is the solution.
(c) Complete the table and by inspection of the table find the solution to the equation \[\frac{\mathbf{z}}{\mathbf{3}}\]=4
Z  8  9  10  11  12  13  14  15  16  …  …  …  … 
\[\frac{\mathbf{z}}{\mathbf{3}}\]  \[2\frac{2}{3}\]  3  3\[\frac{\mathbf{1}}{\mathbf{3}}\]  …  …  …  …  …  …  …  …  …  … 
Ans. $$
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
Hence, $z=12$ is the solution.
(d) Complete the table and by inspection of the table find the solution to the equation \[\mathbf{m}\mathbf{7}=\mathbf{3}\].
m  5  6  7  8  9  10  11  12  13  …  … 
\[\mathbf{m}\mathbf{7}\]  …  …  …  …  …  …  …  …  …  …  … 
Ans.
m  5  6  7  8  9  10  11  12  13  14  15 
$m7$  2  1  0  1  2  3  4  5  6  7  8 
∴ At $m=10$, $m7=3$
Hence, $m=10$is the solution.
NCERT Solutions for Class 6 Maths Chapter 11  Algebra Free PDF Download
The main characteristic of NCERT Class 6 Maths Chapter 11 is that it uses variables to deduce mathematical analogies. This will give students of Class 6 Maths Chapter 11 knowledge about using algebraic values and their applications.
NCERT Maths Book Class 6 Chapter 11 solutions are available in PDF which makes accessing the material much more convenient. The solutions are curated by an expert team of teachers.
The PDF has been developed, keeping in view the latest guidelines issued by CBSE and NCERT.
Benefits of NCERT Solutions for Class 6 Maths Chapter 11 Algebra PDF
Unlock the secrets to mastering Grade 6 Algebra effortlessly with NCERT Solutions PDF. These concise solutions provide clear explanations, organised formulas, effective approaches, and a stepwise process, making math comprehension a breeze for students.
Quick Preparation: The NCERT Solutions PDF provides straightforward answers with illustrative examples, facilitating swift preparation for the final exam.
Formula Memorization: The solutions address the common challenge of memorising formulas by presenting them in an organised manner, making them more manageable for students.
Effective Approaches: Specifically tailored for Grade 6 Algebra, the solutions offer insights into the best approaches for solving problems of varying difficulty levels.
Enhanced Understanding: Detailed solutions contribute to a better grasp of fundamental Algebra concepts, ultimately improving overall academic performance.
Stepwise Approach: The solutions adopt a stepwise approach, ensuring clarity and leaving no room for confusion during the learning process.
With the help of Algebra Class 6 NCERT Solutions, students can understand the chapter in more depth. The solutions PDF is equipped with different approaches so as to cover all the aspects of the chapter. It is aimed to offer a more fun way of learning the basics of Algebra. For more study material and CBSE updates, ensure to download the Vedantu app.
Chapter wise NCERT Solutions for Class 6 Maths
Chapter 11  Algebra
Along with this, students can also view additional study materials provided by Vedantu, for Class 6 Maths Chapter 1 Knowing Our Numbers
NCERT Solution Class 6 Maths of Chapter 11 All Exercises
Chapter 11  Algebra Exercises in PDF Format  

11 Questions with Solutions  
4 Questions with Solutions  
6 Questions with Solutions  
3 Questions with Solutions  
4 Questions with Solutions 
Conclusion
The NCERT Solutions for Class 6 Maths Chapter 11  "Algebra" offer a comprehensive and accessible resource for students to master fundamental algebraic concepts. By providing stepbystep solutions, these materials facilitate a clear understanding of algebraic expressions, equations, and operations. The chapter's emphasis on problemsolving skills and reallife applications enhances students' mathematical proficiency. Utilising these solutions, learners can build a strong foundation in algebra, setting the stage for advanced mathematical concepts. Overall, the NCERT Solutions serve as a valuable aid in fostering mathematical competence and confidence among Class 6 students.
FAQs on NCERT Solutions for Class 6 Maths Chapter 11  Algebra
1. What is Algebra? What is a Variable?
Variables can be defined as a sign or symbol of which, we are yet to find a valuation. It is generally denoted as X or Y.
An example can be; X+3= 9, where X can be the variable.
When a variable is applied, it is not just one value. It is represented in many numerical forms. Variables can be very productive in graphical illustrations.
In simple terms, variables are changeable. This is the reason why they are called variables. There are three main types of variables, namely; dependent, controlled and independent.
Algebra is a part of mathematics explaining different Greek symbols and regulations for deducing various problems into their solutions. The symbols given constitute specific valuations known as variables.
2. Is Chapter 11 Class 6 Algebra Easy?
Algebra is a branch of Mathematics with its own rules and assumptions wherein you use and manipulate mathematical symbols for solving multiple problems. The rules that govern this manipulation of symbols are also a part of Algebra whose basics the students would learn in their 6th standard along with multiple examples. The chapter becomes easy once students get familiar with the fundament concepts and important formulas.
3. How do you solve Chapter 11 Algebra in Class 6?
Solving any algebraic equation would require some basic steps, although the steps would further have their own other smaller steps depending on the question and type of algebraic expression. The basic steps are:
First understand what the question is asking you to determine.
Group similar terms together and bring them to one side.
Solve both sides simultaneously.
Be careful with the sign changes while shifting the numbers and variables.
4. How to score good marks in Class 6 Maths Algebra?
To score good marks in grade 6 Maths Algebra, students need to master the basic concepts. Prepare a separate notebook to jot down all the Algebra related formulas. Practice different types of questions to build a better understanding. Refer to the NCERT Solutions of class 6th Algebra for more indepth answers.
5. How can I download the Solutions for NCERT Class 6 math Chapter 11??
Students can download the std 6 NCERT Solutions from Vedantu’s website.
Visit the NCERT Solutions CBSE Grade 6 Algebra page.
Just click on the Download PDF option and you get offline access to the solutions PDF.