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In Chapter 11, Algebra – Class 6, students will be introduced to the concepts of algebra. Students will learn the use of letters to write rules and formulas to talk about any number. Algebra, Class 6 will teach students to determine the value of the unknowns which are commonly represented as letters. The chapter Algebra of Class 6 teaches students about algebraic expressions and their properties. Read through for NCERT Maths Algebra Class 6 (Chapter 11) Notes and Exercise Solutions.
Table of Contents
Multiplication and Division Word Problems
Click here to download NCERT Maths Algebra Class 6 Chapter 11 Notes and Exercise Solutions PDF
NCERT Maths Algebra Class 6 Notes – PDF Available
Check the topic-wise notes for NCERT Maths Algebra Class 6, Chapter 11 below. We have also provided a downloadable free PDF at the end of these notes so you can download and take a printout to study later when you need quick revision before going to the exam hall.
Topic 1: Variables
The word ‘variable’ means something that can vary, i.e. change. The value of a variable is not fixed. It can take different values.
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Topic 2: Rules from Geometry Using Variables
We have learned the perimeter and area formulas of the square, rectangle and triangle in the previous chapter. Now, let us learn these formulas in the form of rules using variables. This helps us to write the formulas in a more general and concise way.
- Perimeter of Rectangle Formula: The rectangle has 4 sides, having opposite sides of equal lengths.
| Perimeter of Rectangle = 2 × (length(l) + breadth(b)) = 2l × 2b |
- Perimeter of Square Formula: A square has sides of equal lengths.
| Perimeter of Square = 4 × Length of a Side (l) = 4l |
- Perimeter of Triangle Formula: Here, we will determine the perimeter of an equilateral triangle. An equilateral triangle has all sides of equal lengths.
| Perimeter of an Equilateral Triangle = 3 × Length of a Side (l) = 3l |
We can determine the length of any regular figure in a similar way if we know its number of sides. Therefore, we can determine the perimeter of a regular pentagon in a similar way.
| Perimeter of a Regular Pentagon = 5 × Length of a Side (l) = 5l |
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Topic 3: Rules from Arithematic
Here are some simple properties of addition, subtraction, and multiplication that we have studied on whole numbers. Let’s learn the rules of each of these properties using the variables here.
- Commutativity of Addition of Two Numbers:
| a + b = b + a |
- Commutativity of Multiplication of Two Numbers:
| a × b = b × a |
- Distributivity of numbers:
| a × (b + c) = a × b + a × c |
Topic 4: Equation
Any equation is a condition on variables that is satisfied only for a definite value of the variable.
- Solution to the Equation: The value of the variable in an equation that satisfies the equation is called a solution to the equation.
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NCERT Maths Algebra Class 6 Chapter 11 Exercise Solutions
Below we have provided solutions for NCERT Maths Algebra Class 6, chapter 11. Go through for answers to some important questions.
Exercise 11.1 Solutions
Q 1. 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.)
Ans. Let us assume the number of rows in the parade to be “n”.
The number of cadets in a row = 5
∴ Rule for the number of cadets in a row = 5 × n = 5n
Q 2. 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.)
Ans. Let us assume the number of boxes to be “b”.
The number of mangoes in each box = 50
∴ Rule for the total number of mangoes in terms of the number of boxes = 50 × b = 50b
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Q 3. 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.)
Ans. Let us assume the number of students to be “s”.
The number of pencils distributed per student = 5
∴ Rule for the total number of pencils needed = 5 × s = 5s
Q 4. 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. Let us assume the flying time of the bird in minutes to be “t”.
The number of kilometers covered by the bird in a minute = 1
∴ Rule for the total distance covered by the bird = 1 × t = 1t km = t km
Q 5. 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?
Ans. Let us assume the number of rows in Radha’s rangoli to be “r”.
The number of dots in a row = 9
∴ The total number of dots in r number of rows = 9 × r = 9r
- Now, given: r = 8
The total number of dots in 8 rows = 9 × 8 = 72 dots
- Given, r = 10 rows
The total number of dots in 10 rows = 9 × 10 = 90 dots
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Q 6. 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 us assume Radha’s age to be “x” years.
Given: Leela is 4 years younger than Radha.
∴ Leela’s age = (x – 4) years
Q 7. Mother has made laddus. She gives some laddoos to guests and family members; still, 5 laddus remain. If the number of laddoos mother gave away is l, how many laddoos did she make?
Ans. Let us assume the number of laddoos given away by mother to be “l”.
Given: Number of remaining laddoos = 5
∴ Total number of laddoos made by mother = l + 5 laddoos
Q 8. 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 is taken to be x, what is the number of oranges in the larger box?
Ans. Let us assume the number of oranges in the small box to be “x”.
Oranges from the larger box fill up 2 small boxes. So, the total number of oranges in smaller boxes is 2x and still 10 oranges remain out of the box.
∴ Total number of oranges in the larger box = 2x + 10 oranges
Exercise 11.2 Solutions
Q 1. The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l.
Ans. A regular hexagon has all sides of equal length. Hence, the perimeter of a regular hexagon can be determined by adding all its sides.
∴ The perimeter of the given regular hexagon = 6l
Q 2. 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 the edges of a cube.
Ans. Given, a three-dimensional cube having 6 identical square faces. Each square face has 4 edges.
The total number of uncommon edges = 12
Length of each edge = l
∴ The formula for the total length of the edges of a cube = 12l
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Q 3. The diameter of a circle is a line that joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure, AB is the diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).
Ans. A radius passes through the center of the circle and touches a point on the circle. Now, as given – the diameter of a circle is a line that joins two points on the circle and also passes through the center of the circle.
Let us assume the length of the radius of the given circle to be “r”.
∴ The length of the diameter AB = 2 × r = 2r
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Exercise 11.3 Solutions
Q 1. Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
- z +1, z – 1, y + 17, y – 17
- 17y, y/17, 5z
Ans. The answers to each of the given parts are given below.
- In (z + 1), 1 is added to z, so it is the function of addition.
In (z – 1), 1 is subtracted from z, so it is the function of subtraction.
In (y + 17), 17 is added to y, so it is the function of addition.
In (y – 17), 17 is subtracted from y, so it is the function of subtraction.
- In 17y, 17 is multiplied by y, so it is the function of multiplication.
In y/17, y is divided by 17, so it is an operation of division.
In 5z, 5 is multiplied by y, so it is the function of multiplication.
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Q 2. Give expressions for the following cases.
- 7 added to p
- 7 subtracted from p
- p multiplied by 7
- p divided by 7
- 7 subtracted from – m
- – p multiplied by 5
- – p divided by 5
- p multiplied by – 5
Ans. The expressions of each of the given cases are given below.
- 7 + p
- p – 7
- 7p
- p/7
- -m – 7
- -5p
- -p/5
- -5p
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Q 3. Give expressions in the following cases.
- 11 added to 2m
- 11 subtracted from 2m
- 5 times y to which 3 is added
- 5 times y from which 3 is subtracted
- y is multiplied by – 8
- y is multiplied by – 8 and then 5 is added to the result
- y is multiplied by 5 and the result is subtracted from 16
- y is multiplied by – 5 and the result is added to 16.
Ans. The expressions of each of the given cases are given below.
- 2m + 11
- 2m – 11
- 5y + 3
- 5y – 3
- -8y
- -8y + 5
- 16 – 5y
- -5y + 16
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Exercise 11.4 Solutions
Q 1. Take Sarita’s present age to be y years
- What will be her age 5 years from now?
- What was her age 3 years ago?
- Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
- Grandmother is 2 years younger than grandfather. What is the grandmother’s age?
- Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
Ans. Given that Sarita’s present age is “y” years. The answers are given below.
- Sarita’s age 5 years from now will be (5 + y) years.
- Sarita’s age 3 years ago was (y – 3).
- Sarita’s grandfather’s age is 6y years.
- Grandmother’s age = Grandfather’s age – 2 years = (6y – 2) years
- Sarita’s father’s age is (3y + 5) years.
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Q 2. 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. Given: the length of the breadth of the rectangular hall to be “b” metres.
∴ Length of the rectangular hall = (3b – 4) m
Q 3. A rectangular box has a height of 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.
Ans. Given: height = h cm
The length of the box is 5 times the length of height, so, length = 5h cm
The breadth of the box is 10 cm less than the length of the box, so, breadth = (5h – 10) cm
Check out Maths Class 6 Notes and Exercise Solutions for other chapters below.
FAQs
Ans: If you wish to score an absolute 100 marks in your Class 6 Maths exam, then, you must prepare for all the chapters in detail. You need to practice your exercise questions and grasp any exceptional trick to solve a particular question. Try solving your exercises 2-3 times before appearing for your maths exam.
Ans: The rule for finding the perimeter of a rectangle is (2l + 2b).
Ans: The rule for finding the perimeter of a regular triangle is 3l.
This was all about NCERT Maths Algebra Class 6, chapter 11 in which we studied the use of variables to make certain rules and that mathematical operations can be operated on variables when we need to determine an unknown value for a given quantity. Download the NCERT Maths Algebra Class 6 Notes and Exercise Solutions to ace your exam preparations. Follow the CBSE Class 6 Maths Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 6 Maths.
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