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NCERT Solutions for Class 6 Maths Chapter 11 Algebra | KalviGavani

NCERT Class 6 Maths Chapter 11 Algebra exercise solutions — variables expressions equations patterns, MCQ and step-by-step answers.

Updated 16 June 202619 min read

NCERT Solutions for Class 6 Maths Chapter 11 — Algebra

Complete step-by-step NCERT exercise solutions for CBSE Class 6 Maths Chapter 11 Algebra. Covers variables, algebraic expressions, equations, patterns, and word problems with detailed worked examples, common mistakes, and exam tips.


Key Concepts

ConceptDefinition / Formula
VariableA symbol (letter like x, y, n) that represents an unknown quantity; its value can change
ConstantA fixed number whose value does not change; e.g., 3, −5, 100
Algebraic ExpressionA combination of variables, constants, and operations; e.g., 2x + 3, 5y − 7
TermsParts of an expression separated by + or − signs; e.g., in 3x + 5, the terms are 3x and 5
CoefficientThe numerical factor in a term; e.g., in 7x, the coefficient is 7
Like TermsTerms with the same variable; e.g., 3x and 5x are like terms
EquationA statement that two expressions are equal; contains a variable; e.g., x + 4 = 10
Solution of an EquationThe value of the variable that makes the equation true; e.g., x = 6 solves x + 4 = 10
Patterns in AlgebraUsing variables to describe number patterns; e.g., 1, 3, 5, 7, ... can be written as 2n − 1
Using Variables in Common RulesWriting geometry and arithmetic rules using variables; e.g., perimeter of rectangle = 2(l + b)

Exercise 11.1 — Introduction to Variables

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

Solution

To make the letter T, we use 2 matchsticks.

For 1 T: 2 matchsticks

For 2 T's: 2 × 2 = 4 matchsticks

For 3 T's: 2 × 3 = 6 matchsticks

For n T's: 2 × n = 2n matchsticks

The rule is: Number of matchsticks = 2n, where n is the number of T's.

(b) A pattern of letter Z

Solution

To make the letter Z, we use 3 matchsticks.

For n Z's: 3n matchsticks.

The rule is: Number of matchsticks = 3n.

(c) A pattern of letter U

Solution

To make the letter U, we use 3 matchsticks.

For n U's: 3n matchsticks.

The rule is: Number of matchsticks = 3n.

(d) A pattern of letter V

Solution

To make the letter V, we use 2 matchsticks.

For n V's: 2n matchsticks.

The rule is: Number of matchsticks = 2n.

(e) A pattern of letter E

Solution

To make the letter E, we use 5 matchsticks.

For n E's: 5n matchsticks.

The rule is: Number of matchsticks = 5n.


Question 2

We already know the rule for the pattern of letters L, C and F. The rule for letter L is 2n, for C is 3n and for F is 4n, where n is the number of letters. The teacher asks the children to choose any other letter and give the rule for it. Look at the picture and answer:

(a) How many matchsticks are needed to form 3 S's?

Solution

Letter S requires 5 matchsticks.

For 3 S's: 5 × 3 = 15 matchsticks

(b) How many matchsticks are needed to form 5 S's?

Solution

For 5 S's: 5 × 5 = 25 matchsticks

(c) How many matchsticks are needed to form n S's?

Solution

The rule is: Number of matchsticks = 5n


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.)

Solution

Each row has 5 cadets.

Number of rows = n

Total cadets = 5 × n = 5n

The rule is: Number of cadets = 5n


Question 4

If there are 10 mangoes in a box, how many mangoes will be there in 5 boxes? In 20 boxes? In 100 boxes? In n boxes?

Solution

Mangoes in 1 box = 10

Mangoes in 5 boxes = 10 × 5 = 50 mangoes

Mangoes in 20 boxes = 10 × 20 = 200 mangoes

Mangoes in 100 boxes = 10 × 100 = 1,000 mangoes

Mangoes in n boxes = 10 × n = 10n mangoes


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.)

Solution

Pencils per student = 5

Number of students = s

Total pencils = 5 × s = 5s


Exercise 11.2 — Using Variables in Common Rules

Question 1

The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l.

Solution

A regular hexagon has 6 equal sides, each of length l.

Perimeter = 6 × l = 6l


Question 2

A cube is a three-dimensional figure as shown in the figure. It has 6 identical square faces. What is the total length of the edges of a cube?

Solution

A cube has 12 edges. Each edge has the same length, say l.

Total length of edges = 12 × l = 12l


Question 3

The diameter of a circle is a line segment that joins two points on the circle and passes through the centre. The length of the diameter is twice the length of the radius. If the radius is r, what is the length of the diameter?

Solution

Diameter = 2 × radius = 2r


Question 4

Express the following properties using variables a, b and c:

(a) Commutative property of addition

Solution

a + b = b + a

This means the order of addition does not change the sum.

(b) Commutative property of multiplication

Solution

a × b = b × a

This means the order of multiplication does not change the product.

(c) Associative property of addition

Solution

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

This means grouping of numbers in addition does not change the sum.

(d) Associative property of multiplication

Solution

(a × b) × c = a × (b × c)

This means grouping of numbers in multiplication does not change the product.

(e) Distributive property of multiplication over addition

Solution

a × (b + c) = a × b + a × c

This means multiplication can be distributed over addition.


Question 5

Use variables to express the following rules:

(a) Perimeter of an equilateral triangle = 3 × side

Solution

If side = s, then Perimeter = 3s

(b) Perimeter of a rectangle = 2 × (length + breadth)

Solution

If length = l and breadth = b, then Perimeter = 2(l + b)


Exercise 11.3 — Expressions with Variables

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 only once. You can only use the operations +, −, × and ÷.

Solution

Using 5, 7, and 8 with +, −, ×, ÷ (each number used once):

  • 5 + 7 + 8 = 20
  • 5 + 7 − 8 = 4
  • 5 − 7 + 8 = 6
  • 5 × 7 − 8 = 27
  • 5 × 7 + 8 = 43
  • 8 × 7 − 5 = 51
  • 8 × 7 + 5 = 61
  • 8 + 7 − 5 = 10
  • 5 × 8 + 7 = 47
  • 5 × 8 − 7 = 33
  • 8 × 5 ÷ 7 = 40/7
  • 7 × 8 ÷ 5 = 56/5
  • 8 ÷ 4 + 7 = 9 (not valid as 4 not available)
  • (5 + 7) × 8 = 96
  • (8 − 7) × 5 = 5
  • (8 + 7) × 5 = 75
  • (8 − 5) × 7 = 21

(Many more expressions are possible.)


Question 2

Which of these expressions have numbers only?

(a) 2 × 4 + 6 ÷ 3

Solution

This expression has only numbers and operations — no variables. It is a numerical expression.

(b) 2y + 5

Solution

This expression contains the variable y. It is an algebraic expression, not a numerical expression.

(c) 3 × 11 − 2 × 9

Solution

This expression has only numbers and operations — no variables. It is a numerical expression.


Question 3

Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions:

(a) 4x + 7

Solution

Operations: Multiplication (4 × x) and Addition (+7)

(b) 8y − 3

Solution

Operations: Multiplication (8 × y) and Subtraction (−3)

(c) 5z + 12

Solution

Operations: Multiplication (5 × z) and Addition (+12)

(d) 2a − 9

Solution

Operations: Multiplication (2 × a) and Subtraction (−9)


Question 4

Give expressions in the following cases:

(a) 5 added to p

Solution

p + 5

(b) 7 subtracted from y

Solution

y − 7

(c) m multiplied by 9

Solution

9m (or 9 × m)

(d) y divided by 4

Solution

y/4 (or y ÷ 4)

(e) 3 subtracted from −n

Solution

−n − 3

(f) 5 multiplied by y and then 7 added to the result

Solution

5y + 7

(g) y multiplied by −3

Solution

−3y

(h) p divided by 5 and then 8 subtracted from the result

Solution

(p ÷ 5) − 8 or (p/5) − 8


Question 5

If Meena's present age is x years:

(a) What will be her age 5 years from now?

Solution

Meena's present age = x years

Age 5 years from now = x + 5 years

(b) What was her age 3 years back?

Solution

Age 3 years back = x − 3 years

(c) Meena's grandfather is 6 times her age. What is the grandfather's age?

Solution

Grandfather's age = 6 × x = 6x years

(d) Meena's grandmother is 2 years younger than grandfather. What is the grandmother's age?

Solution

Grandmother's age = Grandfather's age − 2 = 6x − 2 years

(e) Meena's father's age is 5 years more than 3 times Meena's age. What is the father's age?

Solution

Father's age = 3x + 5 = 3x + 5 years


Question 6

A bus travels at a speed of v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, how far is it from Daspur?

Solution

Speed = v km/hr

Time = 5 hours

Distance = Speed × Time = v × 5 = 5v km

The bus is 5v km from Daspur.


Exercise 11.4 — Equations

Question 1

Complete the following table and find whether the value in the second column satisfies the equation or not:

EquationValueSatisfies?
x + 3 = 7x = 4?
x + 3 = 7x = 3?
x + 3 = 7x = 5?
y − 3 = 2y = 5?
y − 3 = 2y = 4?
Solution
EquationValueLHSRHSSatisfies?
x + 3 = 7x = 44 + 3 = 77Yes
x + 3 = 7x = 33 + 3 = 67No
x + 3 = 7x = 55 + 3 = 87No
y − 3 = 2y = 55 − 3 = 22Yes
y − 3 = 2y = 44 − 3 = 12No

Question 2

Solve the following equations by trial and error method:

(a) 5p + 2 = 17

Solution

Try different values of p:

  • p = 1: 5(1) + 2 = 7 ≠ 17
  • p = 2: 5(2) + 2 = 12 ≠ 17
  • p = 3: 5(3) + 2 = 17 ✓

p = 3 is the solution.

(b) 3n − 7 = 14

Solution
  • n = 4: 3(4) − 7 = 5 ≠ 14
  • n = 5: 3(5) − 7 = 8 ≠ 14
  • n = 6: 3(6) − 7 = 11 ≠ 14
  • n = 7: 3(7) − 7 = 14 ✓

n = 7 is the solution.


Question 3

Solve the following equations:

(a) x + 5 = 12

Solution

x + 5 = 12

Subtract 5 from both sides:

x = 12 − 5

x = 7

Check: 7 + 5 = 12 ✓

(b) y − 7 = 11

Solution

y − 7 = 11

Add 7 to both sides:

y = 11 + 7

y = 18

Check: 18 − 7 = 11 ✓

(c) 3p = 18

Solution

3p = 18

Divide both sides by 3:

p = 18 ÷ 3

p = 6

Check: 3 × 6 = 18 ✓

(d) m ÷ 5 = 4

Solution

m ÷ 5 = 4

Multiply both sides by 5:

m = 4 × 5

m = 20

Check: 20 ÷ 5 = 4 ✓

(e) 2x + 3 = 9

Solution

2x + 3 = 9

Subtract 3 from both sides:

2x = 9 − 3

2x = 6

Divide both sides by 2:

x = 6 ÷ 2

x = 3

Check: 2(3) + 3 = 6 + 3 = 9 ✓

(f) 4y − 5 = 15

Solution

4y − 5 = 15

Add 5 to both sides:

4y = 15 + 5

4y = 20

Divide both sides by 4:

y = 20 ÷ 4

y = 5

Check: 4(5) − 5 = 20 − 5 = 15 ✓


Exercise 11.5 — Word Problems Forming Equations

Question 1

Set up equations and solve them:

(a) Add 4 to a number to get 13. Find the number.

Solution

Let the number be n.

Equation: n + 4 = 13

n = 13 − 4

n = 9

Check: 9 + 4 = 13 ✓

(b) Subtract 5 from a number to get 8. Find the number.

Solution

Let the number be n.

Equation: n − 5 = 8

n = 8 + 5

n = 13

Check: 13 − 5 = 8 ✓

(c) Multiply a number by 3 to get 21. Find the number.

Solution

Let the number be n.

Equation: 3n = 21

n = 21 ÷ 3

n = 7

Check: 3 × 7 = 21 ✓

(d) Divide a number by 4 to get 6. Find the number.

Solution

Let the number be n.

Equation: n ÷ 4 = 6

n = 6 × 4

n = 24

Check: 24 ÷ 4 = 6 ✓


Question 2

Raju is 8 years younger than his brother. If his brother's age is x, what is Raju's age? If Raju's age is 12 years, what is his brother's age?

Solution

Raju's age = Brother's age − 8 = x − 8

If Raju's age = 12:

x − 8 = 12

x = 12 + 8 = 20

Brother's age = 20 years


Question 3

A shop has 45 notebooks. If the number of notebooks sold is s, how many notebooks are left?

Solution

Notebooks left = Total − Sold = 45 − s

If s = 18:

Notebooks left = 45 − 18 = 27 notebooks


Question 4

The length of a rectangular hall is 5 m more than its breadth. If the breadth is b metres, express the length in terms of b. If breadth is 10 m, what is the length?

Solution

Length = breadth + 5 = b + 5

If b = 10 m:

Length = 10 + 5 = 15 m


Question 5

Solve: 3x + 8 = 29

Solution

3x + 8 = 29

Subtract 8 from both sides:

3x = 29 − 8

3x = 21

Divide by 3:

x = 21 ÷ 3

x = 7

Check: 3(7) + 8 = 21 + 8 = 29 ✓


Question 6

Solve: 5y − 3 = 22

Solution

5y − 3 = 22

Add 3 to both sides:

5y = 22 + 3

5y = 25

Divide by 5:

y = 25 ÷ 5

y = 5

Check: 5(5) − 3 = 25 − 3 = 22 ✓


Question 7

Solve: 7m ÷ 2 = 14

Solution

7m ÷ 2 = 14

This means: 7m/2 = 14

Multiply both sides by 2:

7m = 28

Divide by 7:

m = 28 ÷ 7

m = 4

Check: 7(4) ÷ 2 = 28 ÷ 2 = 14 ✓


Question 8

The sum of two numbers is 45. One number is 15. What is the other number?

Solution

Let the other number be n.

n + 15 = 45

n = 45 − 15

n = 30

Check: 30 + 15 = 45 ✓


Question 9

A number when multiplied by 6 and then increased by 3 gives 33. Find the number.

Solution

Let the number be n.

Equation: 6n + 3 = 33

6n = 33 − 3

6n = 30

n = 30 ÷ 6

n = 5

Check: 6(5) + 3 = 30 + 3 = 33 ✓


Question 10

Sarita has 15 rupees. She buys some pencils costing Rs 2 each. If she is left with Rs 3, how many pencils did she buy?

Solution

Let number of pencils = n

Cost of n pencils = 2n

Money left = 15 − 2n

Equation: 15 − 2n = 3

2n = 15 − 3

2n = 12

n = 6

She bought 6 pencils.

Check: 15 − 2(6) = 15 − 12 = 3 ✓


Common Mistakes

1. Confusing expressions and equations: An expression (like 2x + 3) does NOT have an equals sign and cannot be "solved." An equation (like 2x + 3 = 7) has an equals sign and can be solved for the variable.

2. Forgetting the coefficient means multiplication: In the term 5x, the coefficient 5 is multiplied by x. Students sometimes add instead of multiplying. 5x means 5 × x, not 5 + x.

3. Not performing the same operation on both sides: When solving x + 5 = 12, if you subtract 5 from the left, you MUST subtract 5 from the right too. Always maintain balance in an equation.

4. Incorrect order in subtraction expressions: "7 subtracted from y" means y − 7, NOT 7 − y. The number being subtracted comes after the minus sign.

5. Mixing up like and unlike terms: 3x and 5x can be combined (3x + 5x = 8x), but 3x and 5y cannot be combined. They are unlike terms.

6. Wrong inverse operations: To undo addition, use subtraction. To undo multiplication, use division. For 3x = 15, divide by 3 (not subtract 3). Think: what operation "undoes" the given operation?


Exam Checklist

  • [ ] I can identify a variable in an expression and distinguish it from a constant.
  • [ ] I can write algebraic expressions from word descriptions (e.g., "5 more than x" → x + 5).
  • [ ] I know the difference between an expression and an equation.
  • [ ] I can write the rule for matchstick patterns using a variable (e.g., 2n, 3n, 4n).
  • [ ] I can express geometry formulas using variables (perimeter of rectangle, square, etc.).
  • [ ] I can write commutative, associative, and distributive properties using variables.
  • [ ] I can form an equation from a word problem by defining the unknown as a variable.
  • [ ] I can solve simple equations using inverse operations (trial and error or systematic method).
  • [ ] I always verify my answer by substituting back into the original equation.
  • [ ] I correctly interpret phrases: "added to," "subtracted from," "multiplied by," "divided by."

Quick Reference — Key Formulas

ConceptExpression / Formula
Matchstick pattern for letter T2n
Matchstick pattern for letter E5n
Perimeter of regular hexagon6l
Commutative property (addition)a + b = b + a
Commutative property (multiplication)a × b = b × a
Associative property (addition)(a + b) + c = a + (b + c)
Distributive propertya × (b + c) = a × b + a × c
Solving x + a = bx = b − a
Solving x − a = bx = b + a
Solving ax = bx = b ÷ a

These solutions follow the NCERT textbook pattern for Class 6 Maths Chapter 11 — Algebra. Practice each question step-by-step and verify your answers by substitution.

Frequently Asked Questions

A variable is a symbol (usually a letter like x, y, n) that represents an unknown quantity that can take different values.
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