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Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 11 Letter Math Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 11 Solutions Letter Math

Class 6 Kerala Syllabus Maths Solutions Chapter 11 Letter Math Questions and Answers

Letter Math Class 6 Questions and Answers Kerala Syllabus

Addition and Subtraction

Intext Questions (Page No. 162-163)

Question 1.
Now consider the polygon given in the textbook.

Do you see any relation between the number of sides of a polygon, the number of corners, the number of lines, and the number of triangles?

Answer:
First, we check the relation between the number of sides and the number of lines.

The number of lines is 3 less than the number of corners.
That is, Number of lines = Number of corners – 3
In any polygon, the number of corners is the same as the number of sides.
That means, Number of lines = Number of sides – 3

Let s be the number of sides of the polygon and l be the number of lines of the polygon.
Then we can write the relation as l = s – 3
We can say this in another two forms, that is, s = l + 3
s – l = 3
(If we add 3 to the number of lines, we get the number of sides.
The difference between the number of sides and the number of lines will results into 3.)

Now, let’s look at the relation between the number of sides of a polygon and the triangles formed within it.
The number of triangles is 2 less than the number of sides.
Let t be the number of triangles and s be the number of sides.
Then the relation between them can be written as follows.
s – t = 2
t = s – 2
s = t + 3
What is the relation between the number of triangles and the number of lines within them?
If we draw 1 lines we get 2 triangles.
If we draw 2 lines, we get 3 triangles.
That means the number of triangles is 1 more than the number of lines.
This relation can be expressed using letters in different forms, such as
l + 1 = t,
t – 1 = l,
t – l = 1
The last two relations can be explained as follows.
The number of lines is 1 less than the number of triangles.
The difference between the number of triangles and the number of lines is 1.

Addition and Subtraction (Page No. 165)

Question 1.
The pictures below show a line drawn from a point on another line:

(i) In what ways can we say the relation between the angles marked on the left and right of the top line?
(ii) If we denote the measure of the angle on the left by l and the measure of the angle on the right by r, then in what ways can we write the relation between the numbers l and r?
(iii) In all cases, l and r are between what numbers?
Answer:
(i) The sum of the angles on the left and right side of the slant line is 180°.
The angle on the right side can be obtained by subtracting the angle on the left side from 180°.
The angle on the left side can be obtained by subtracting the angle on the right side from 180°.
(ii) Angle on the left = l
Angle on the right = r
Here, l + r = 180
180 – l = r
180 – r = l
(iii) In all cases, the values of l & r are between 0° and 180°.

Question 2.
(i) Each side of a square of perimeter 40 centimetres is extended by 1 centimetre to make a larger square. By how much is the perimeter increased?
(ii) If the sides of a square of perimeter 60 centimetres are extended like this, what would be the perimeter of the large square?
(iii) Is the difference in perimeters of any square, and the larger square made by extending each side by 1 centimetre, the same number for all squares? Why?
(iv) Is it also true for rectangles that are not squares?
(v) If the perimeter of the larger rectangle is denoted by p and the perimeter of the original small rectangle is denoted by s, in what ways can we write the relation between these two numbers?
Answer:
(i) If the side is extended by 1 cm, then its perimeter also increases by 4 cm.
Perimeter = 40 cm
Perimeter of the larger square = 44 cm
Increase in the perimeter = 44 – 40 = 4
(ii) Perimeter = 60 cm
Each side is extended by 1 cm, then its perimeter is increased by 4 cm.
Perimeter of the larger square = 64 cm
(iii) Extending each side by 1 cm increases each of the 4 sides by 1 cm, so the perimeter always increases by 4 cm.
(iv) Yes. The perimeter increase is also 4cm for any rectangle when each side is increased by 1 cm.
(v) p = s + 4
s = p – 4
p – s = 4

Letter Multiplication

Intext Questions (Page No. 167)

Question 1.
Now, suppose we make triangles like this?

Write the number of triangles and the number of matchsticks in a table like this:

Answer:
In this problem, the relation between the number of triangles and the number of matchsticks is different.
First, we can complete the table:

From this, we can see that the relation between the triangles and the number of matchsticks is like this.
If one is added to twice the number of triangles, we get the number of matchsticks required to make the triangle.
This relation can be written in this form.
m = 2t + 1
t = \(\frac{m-1}{2}\)
\(\frac{m-1}{t}\) = 2
Difference in the number of matchsticks:

The number of matchsticks used to make 3 triangles in this way is 3 × 3 = 9.

If the 3 triangles are made in this form, the number of matchsticks used is 3 × 2 + 1 = 7.
The difference in the number of triangles in two different ways is 1.
The difference remains the same for any number of triangles.
4 triangles:
Method-1: 4 × 3 = 12
Method-1: 4 × 2 + 1 = 9
Difference = 3
5 triangles:
Method-1: 5 × 3 = 15
Method-1: 5 × 2 + 1 = 11
Difference = 4
Therefore, the relation is that if the triangles are made in two different ways, then the number of matchsticks needed is one less than the number of triangles.
Let’s denote the number of triangles by t and the difference in the number of matchsticks needed in the two ways by d.
That means we get d = t – 1
If 10 triangles are made in these two different ways, then
Difference in the number of matchsticks = 10 – 1 = 9
If it is 100, then the difference in the number of matchsticks = 100 – 1 = 99

Letter Multiplication (Page No. 172-174)

Question 1.
We can also make squares with matchsticks:

(i) As we go on making such squares, what is the relation between the number of squares and the number of matchsticks used?
(ii) Write this relation, denoting the number of squares by 5 and the number of matchsticks by m.
(iii) What kind of numbers are s and m in this?
(iv) For a multiple of 4, how do we calculate the number of squares that can be made with that many matchsticks?
(v) Write this calculation also using letters.
Answer:
(i)

The number of matchsticks is 4 times the number of squares.
(ii) The number of squares is denoted by s
Number of matchsticks is m = 4s
(iii) Both s and m are whole numbers.
m is always an even number, since each square requires 4 matchsticks.
m is always a multiple of 4.
(iv) For getting the number of squares, divide the number of matchsticks used by 4.
(v) If the number of squares is denoted by s
The number of matchsticks is m
s = \(\frac {m}{4}\)

Question 2.
We can make matchstick squares like this also:

(i) Write as a table the number of squares and the number of matchsticks needed to make squares like this, and find the relation between these numbers.
(ii) Explain this relation by drawing pictures showing how the squares are made.
(iii) Write this relation denoting the number of squares by s and the number of matchsticks used by m.
(iv) What kind of numbers are s and m in this?
(v) If we take a number that leaves a remainder of 1 on division by 3, how do we calculate the number of squares that can be made like this using that many matchsticks?
(vi) Write this calculation also using letters.
Ans:
(i)

(ii)

For the first square, 1 matchstick is joined with 3 matchsticks, we get 4 matchsticks
For two squares, join 3 and 1 with 3 matchsticks, we get 7 matchsticks.
For three squares, join two times 3 and 1 with 3 matchsticks, we get 10 matchsticks (That means 3 times 3 and 1).
Continuing like this.
For ten squares, join nine times 3 and 1 with 3 matchsticks, we get 30 matchsticks (that means, 10 times 3 and 1).
In general, we can say that 1 is added to 3 times the number of squares.
(iii) Number of squares be s
The number of matchsticks is m
Therefore, the relation is m = 3s + 1
(iv) Both s and m are natural numbers.
The value of m is 3, ’s are added with 4.
That means the value is the number that is divisible by 3, and the remainder is 1.
(v) If some matchsticks m leaves a remainder of 1 when divided by 3, then we can make squares like this.
That means (Number of matchsticks – 1)
(vi) s = \(\frac{m-1}{3}\)

Question 3.
(i) When an equal number of matchstick squares are made in the two different ways as in the first and second questions, what is the difference in the number of matchsticks used?
(ii) Denoting the number of squares as s and the difference in the number of matchsticks used in the two ways as d, how do we write the relation between s and d?
(iii) Explain the reason for this relation.
(iv) One kid uses 25 matchsticks to make squares in the second way. How many matchsticks would be needed to make the same number of squares first?
Answer:
(i) For making the squares in the second method, the value obtained by subtracting 1 from the number of squares in the first method can be removed from the number of matchsticks used in the first method.

To make 5 squares, subtract 4 from 20, we get 16 (20 – 4 = 16).
Matchsticks are enough for making it in the second method.
(ii) s – d = 1
(iii) In the first method, number of matchsticks is 4 times the number of squares.
In the second method, 1 is added to 3 times the number of squares.
That means, in the first method, it is in the form of: 4 + 4 + 4 + 4 + ….
In the second method, it is in the form of: 4 + 3 + 3 + 3 + …
From the second square onwards, 1 is decreased from it.
So the difference in the matchsticks is 1 less than the total number of squares.
(iv) The number of matchsticks used for making 25 squares in the second method is 25 × 3 + 1 = 76
The number of matchsticks used for making 25 squares in the first method is 25 × 4 = 100

Question 4.
See these pictures:

The first picture is a square of sides 1 centimetre. The second picture is a rectangle made by joining together two such squares. The third picture is a rectangle made by joining together three such squares.
(i) Calculate the perimeter of each.
(ii) What is the perimeter of the rectangle made by joining together four such squares like this? What about the rectangle made with five squares?
(iii) Make a table of the number of squares used to make the rectangles and the perimeter of the rectangles, starting from the first. What is the relation between the number of squares and the perimeter? Explain the reason for this relation?
(iv) Denoting the number of squares by s and the perimeter by p, write the relation between them.
(v) To get a rectangle of perimeter 1 metre, how many squares must be joined like this?
Ans:
(i) (a) 4 cm
(b) 6 cm
(c) 8 cm
(ii) 10 cm, 12 cm
(iii)

The perimeter of the square is obtained by adding 2 to twice the number of squares.
When we join squares side by side to make a rectangle, the two lengths of the rectangle increase by 2 (1 + 1 = 2)
But the width remains the same at 1 cm.
The length of the square is 1 cm.
By considering the 2 sides, we get twice the squares, and also it increases by 2 when considering the two widths of 2 cm.
Thus, we get the perimeter of the rectangle.
(iv) p = 2s + 2
(v) From this relation,
Number of squares = (perimeter – 2) ÷ 2
That is s = \(\frac{p-2}{2}\)
⇒ s = \(\frac{100-2}{2}=\frac{98}{2}\)
⇒ s = 49
49 Squares must be joined.

Class 6 Maths Chapter 11 Kerala Syllabus Letter Math Questions and Answers

Class 6 Maths Letter Math Questions and Answers

Question 1.
(i) Find the number of matchsticks used to make this hexagon.

(ii) If the hexagons are joined like this, then the number of matchsticks used to make this is.

(iii) If 10 such patterns are joined together. How many matchsticks are used to make this?
(iv) How can we say this relation? Explain this with letters.
(v) If there is 48 sticks joined together, then how many patterns are there?
(vi) How can we say this relation? Explain this with letters.
(vii) How can we explain the relation between the hexagon and the number of matchsticks used? Express it in letters also.
Answer:
(i) 6
(ii) 12
(iii) 10 × 6 = 60
(iv) The number of matchsticks used is 6 times the number of hexagons.
Let h be the number of hexagons.
m be the number of matchsticks
m = 6h
(v) 48 ÷ 6 = 8
(vi) If the number of matchsticks used is divided by 6, we get the number of hexagons.
h = \(\frac {m}{6}\)
(vii) If the number of matchsticks is divided bythe number of hexagons, we get 6.
\(\frac {m}{h}\) = 6

Question 2.
If the hexagon is arranged in this way.

(i) To form a pattern of 3 hexagons joined together, how many matchsticks are used?
(ii) If it is 10, how many matchsticks are there?
(iii) How can we say these relations? Express it in letters.
Answer:
(i) 16
(ii) 51
(iii) Let the number of hexagons be h
Number of matchsticks is m
m = 5h + 1
h = \(\frac{n-1}{5}\)
\(\frac{n-1}{h}\) = 5

Question 3.
Some portions of a circle is shaded.
(i) The angle measure of the shaded portion is 40°. What is the angle measure of the remaining portion?
(ii) If the shaded portion is 120°, what is the remaining measure?
(iii) How can we express the relation between the angles of the shaded portion and the unshaded portion?
(iv) Explain it in letters.
Ans:
(i) 360° – 40° = 320°
(ii) 360° – 120° = 240°
(iii) If the angle measure of the shaded portion is subtracted from 360°, we get the angle measure of the unshaded portion.
If the angle measure of the unshaded portion is subtracted from 360°, we get the angle measure of the shaded portion.
If we take the sum of the angle measures of both the shaded and unshaded portions, we get 360°.
(iv) If the shaded portion is s
Unshaded portion is u, we get,
360 – s = u
360 – u = s
s + u = 360

Question 4.
The image of a pentagon is given below.

(i) How many matchsticks are needed to make this pattern?
(ii) If 3 pentagons are joined together like this.

How many matchsticks are needed to make this pattern?
(iii) If the number of pentagons is 8, how many matchsticks are used?
(iv) How can we explain the relation between the number of pentagons and the number of sticks used?
(v) Express this relation using letters.
Answer:
(i) 5
(ii) 4 + 4 + 4 + 1 = 13
(iii) 8 × 4 + 1 = 33
(iv) If the number of pentagons is multiplied by 4 and 1 is added, we get the number of matchsticks used.
(v) Let the number of pentagons be p, and the number of matchsticks be n.
n = 4p + 1
p = \(\frac{n-1}{4}\)
\(\frac{n-1}{p}\) = 4

Class 6 Maths Chapter 11 Notes Kerala Syllabus Letter Math

→ If three points are on the same straight line, then the distance between the points at the two ends is equal to the sum of the distances from the end points to the point in between.

→ The triangles are formed by arranging the matchsticks in a triangular shape.

→ The number of matchsticks is equal to the number of triangles multiplied by 3.
m = 3t

→ To get the number of triangles, divide the number of matchsticks by 3.
t = \(\frac {m}{3}\)

→ If we divide the number of matchsticks by the number of triangles, we get 3.
\(\frac {m}{t}\) = 3

In the previous chapter, we discussed some of the unchanging relationships. In this chapter, we are going to discuss the unchanging relationships that can be expressed algebraically. This unit introduces the basic ideas needed to represent such relationships using letters.

Addition and Subtraction
Three points are marked on a straight line. Two points are at the ends of the line, and one point lies between them. If the total length of this line is 8 cm and the distance from the left end to the point that is between the end points are 3 cm. Then the line is divided into two parts, that is, 3 cm and 5 cm. The length of the line at the left end is 3 cm, and at the right end is 5 cm.

By adding the left and right ends, we get 8 cm. Or subtracting 5 cm (that means the right end) from 8 cm, we get 3 cm (that is the length of the left end). This can be described like this:
If the length of the left end is subtracted from 8 cm, we get the length of the right end.
Similarly, if we subtract the length of the right end from 8 cm, we get the length of the left end.
This relationship is always true, even when the position of the point that is between the endpoints is changed.
These relations can be concluded like this:
Distance from the left end = 8 – Distance from the right end
Distance from the right end = 8 – Distance from the left end.
Distance from the left end + Distance from the right end = 8 cm
We can rewrite the above mentioned relations with letters.
If we consider the distance from the left end as l and the distance from the right end as r.
Then we can write the above-mentioned relations like this.
l = 8 – r
r = 8 – l
l + r = 8
In this relation, the value of l and r can be any value, but the sum of the two values must be 8.
Now consider the line problem from the textbook.

In a 5 cm long line, mark a point 1 cm from the right. At what distance from the left end should we mark it?
5 – 1 = 4 cm
Subtracting the distance from the right end from 5 centimetres gives the distance from the left end.
5 – Distance from the right end = Distance from the left end
Let’s see how can we represent the above relation using letters,
Consider the distance from the right end = r
Distance from the left end = l
Then we get, 5 – r = l
Depends on the position of the point, the value of l and r changes, but the relationship between them remains the same.
We can say it in another way:
Distance from the right end + Distance from the left end = 5
r + l = 5
5 – Distance from the left end = Distance from the right end.
5 – l = r

In general, we can say that,
If three points are on the same straight line, then the distance between the points at the two ends is equal to the sum of the distances from the endpoints to the point in between.
For example, consider the line and the points given in it.
Total length of the line is 10 centimetres. A point is marked 4 centimetres from its right end. Then the point is 6 centimetres from its left end.
4 + 6 = 10 cm
10 – 4 = 6 cm
10 – 6 = 4 cm
If we write this using the letters we get,
Let the total length be d, we get the following relations.
d = r + l
r = d – l
l = d – r
In this, we can take l, r, and d as any three numbers, but the sum of l and r should be d.

Letter Multiplication
Let’s consider the triangular problem given in the textbook.

Here, the triangles are formed by arranging the matchsticks in a triangular shape.
Therefore, the number of matchsticks is equal to the number of triangles multiplied by 3.
Shortening this statement, using letters:
Let the number of triangles be t, and the number of matchsticks be m.
Then the relation can be written as, m = t × 3
It can also be written as m = 3t
Similarly, we can say the other relations also.
To get the number of triangles, divide the number of matchsticks by 3.
That is, t = \(\frac {m}{3}\)
If we divide the number of matchsticks by the number of triangles, we get 3.
That is \(\frac {m}{t}\)= 3

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Here we have provided SCERT Kerala SSLC English Model Question Papers 2025-26 with Answers for Class 10th Std. Students can view or download the Kerala State Board SSLC English Previous Year Question Papers and Answers for their upcoming Kerala SSLC board examinations. Students can also read Kerala Syllabus 10th Standard English Solutions.

Board	Kerala Board
Textbook	SCERT, Kerala
Class	SSLC Class 10
Subject	SSLC English
Chapter	Previous Year Question Papers, Model Papers, Sample Papers
Year of Examination	2026, 2025, 2024, 2023, 2022
Category	Kerala Syllabus Question Papers

Kerala SSLC English Previous Year Model Question Papers with Answers

These SSLC English Model Question Papers 2026 Kerala with Answers Pdf are designed according to the latest exam pattern, so it will help students to know the exact difficulty level of the question papers.

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Kerala SSLC English Model Question Papers with Answers

• Kerala Syllabus Class 10 English Model Question Paper Set 1
• Kerala Syllabus Class 10 English Model Question Paper Set 2
• Kerala Syllabus Class 10 English Model Question Paper Set 3
• Kerala Syllabus Class 10 English Model Question Paper Set 4
• Kerala Syllabus Class 10 English Model Question Paper Set 5

Old Syllabus

• Kerala SSLC English Model Question Paper First Term Evaluation
• Kerala SSLC English Question Paper March 2024
• Kerala SSLC English Question Paper March 2023
• Kerala SSLC English Question Paper March 2022
• Kerala SSLC English Question Paper March 2021
• Kerala SSLC English Question Paper March 2020
• Kerala SSLC English Question Paper March 2019
• Kerala SSLC English Board Model Paper March 2024
• Kerala SSLC English Board Model Paper March 2023
• Kerala SSLC English Board Model Paper March 2022
• Kerala SSLC English Board Model Paper March 2021
• Kerala SSLC English Board Model Paper March 2020
• Kerala SSLC English Board Model Paper March 2019

It is necessary that students will also get to know about the types of questions and their level of difficulty according to the latest exam pattern. These Model Question Papers for Class 10 English Kerala State Syllabus will help to prepare for Kerala SSLC Board Exams 2026.

We hope the given KBPE Kerala State Board Syllabus SSLC English Previous Year Model Question Papers with Answers 2025 2026 Pdf Free Download of Kerala SSLC English Previous Question Papers with Answers, SSLC English Model Question Papers with Answers for Class 10 Kerala State Board, SSLC English Question Pool 2026 will help you.

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Expert Teachers at HSSLive.Guru has created KBPE Kerala SSLC Malayalam Previous Year Model Question Papers with Answers for Class 10 Kerala State Board Syllabus 2025-2026 Pdf free download are Part of Kerala SSLC Previous Year Question Papers with Answers. By solving Kerala SSLC Malayalam Previous Question Papers with Answers, SSLC Malayalam Model Question Papers and Answers, SSLC Malayalam Question Pool 2026 will help the students to check their progress.

Here we have provided SCERT Kerala SSLC Malayalam Model Question Papers 2025-26 with Answers for Class 10th Std. Students can view or download the Kerala State Board SSLC Malayalam Previous Year Question Papers and Answers for their upcoming Kerala SSLC board examinations. Students can also read Kerala Padavali Malayalam Standard 10 Solutions and Adisthana Padavali Malayalam Standard 10 Solutions.

Board	Kerala Board
Textbook	SCERT, Kerala
Class	SSLC Class 10
Subject	SSLC Malayalam
Chapter	Previous Year Question Papers, Model Papers, Sample Papers
Year of Examination	2026, 2025, 2024, 2023, 2022
Category	Kerala Syllabus Question Papers

Kerala SSLC Malayalam Previous Year Model Question Papers with Answers

These SSLC Malayalam Model Question Papers 2026 Kerala with Answers Pdf are designed according to the latest exam pattern, so it will help students to know the exact difficulty level of the question papers.

These 10th Standard SSLC Malayalam Previous Year Question Papers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for Kerala Board Exams and Score More marks. Here is the latest Kerala SSLC Malayalam Previous Question Papers Free Download given below.

Kerala SSLC Malayalam 1 Model Question Papers with Answers (Kerala Padavali)

• Kerala Syllabus Class 10 Malayalam 1 Model Question Paper Set 1
• Kerala Syllabus Class 10 Malayalam 1 Model Question Paper Set 2
• Kerala Syllabus Class 10 Malayalam 1 Model Question Paper Set 3
• Kerala Syllabus Class 10 Malayalam 1 Model Question Paper Set 4
• Kerala Syllabus Class 10 Malayalam 1 Model Question Paper Set 5

Kerala SSLC Malayalam 2 Model Question Papers with Answers (Adisthana Padavali)

• Kerala Syllabus Class 10 Malayalam 2 Model Question Paper Set 1
• Kerala Syllabus Class 10 Malayalam 2 Model Question Paper Set 2
• Kerala Syllabus Class 10 Malayalam 2 Model Question Paper Set 3
• Kerala Syllabus Class 10 Malayalam 2 Model Question Paper Set 4
• Kerala Syllabus Class 10 Malayalam 2 Model Question Paper Set 5

Old Syllabus

Kerala Padavali

• Kerala SSLC Malayalam 1 Question Paper March 2024
• Kerala SSLC Malayalam 1 Question Paper March 2023
• Kerala SSLC Malayalam 1 Question Paper March 2022
• Kerala SSLC Malayalam 1 Question Paper March 2021
• Kerala SSLC Malayalam 1 Question Paper March 2020
• Kerala SSLC Malayalam 1 Question Paper March 2019
• Kerala SSLC Malayalam 1 Board Model Paper March 2024
• Kerala SSLC Malayalam 1 Board Model Paper March 2023
• Kerala SSLC Malayalam 1 Board Model Paper March 2022
• Kerala SSLC Malayalam 1 Board Model Paper March 2021
• Kerala SSLC Malayalam 1 Board Model Paper March 2020

Adisthana Padavali

• Kerala SSLC Malayalam 2 Question Paper March 2024
• Kerala SSLC Malayalam 2 Question Paper March 2023
• Kerala SSLC Malayalam 2 Question Paper March 2022
• Kerala SSLC Malayalam 2 Question Paper March 2021
• Kerala SSLC Malayalam 2 Question Paper March 2020
• Kerala SSLC Malayalam 2 Question Paper March 2019
• Kerala SSLC Malayalam 2 Board Model Paper March 2024
• Kerala SSLC Malayalam 2 Board Model Paper March 2023
• Kerala SSLC Malayalam 2 Board Model Paper March 2022
• Kerala SSLC Malayalam 2 Board Model Paper March 2021
• Kerala SSLC Malayalam 2 Board Model Paper March 2020

It is necessary that students will also get to know about the types of questions and their level of difficulty according to the latest exam pattern. These Model Question Papers for Class 10 Malayalam Kerala State Syllabus will help to prepare for Kerala SSLC Board Exams 2026.

We hope the given KBPE Kerala State Board Syllabus SSLC Malayalam Previous Year Model Question Papers with Answers 2025 2026 Pdf Free Download of Kerala SSLC Malayalam Previous Question Papers with Answers, SSLC Malayalam Model Question Papers with Answers for Class 10 Kerala State Board, SSLC Malayalam Question Pool 2026 will help you.

If you have any queries regarding Kerala State Board SSLC Malayalam Model Question Papers and Answers 2025 26 Pdf, drop a comment below and we will get back to you at the earliest.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 10 Number Relations Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 10 Solutions Number Relations

Class 6 Kerala Syllabus Maths Solutions Chapter 10 Number Relations Questions and Answers

Number Relations Class 6 Questions and Answers Kerala Syllabus

Digit Sum (Page Number 154-155)

Question 1.
Complete the table below:

Answer:

Question 2.
The first three digits of a four-digit number, which is a multiple of 9, are 2, 5, 7. What is the number?
Answer:
Here, the first three digits are 2, 5, 7.
The sum is 14.
So the next number should be 4, only after that the sum will be a multiple of 9.
2 + 5 + 7 + 4 = 18
The number is 2574.

Question 3.
(i) Any number, and the number got by changing the order of its digits in any way, both leave the same remainder on division by 9. Why?
(ii) If we change the order of the digits of a multiple of 9 in any way, will we again get a multiple of 9? Why?
Answer:
(i) Because the order is only changing, the sum of the digits are same.
For example, the sum of the digits of the number 3247 is 16.
The remainder when dividing by 9 is 7.
Even if the order has changed, the sum will always be 16.
(ii) Even if the order is changing, the sum is not changing, so it will always be a multiple of 9.

Question 4.
Suppose we take all five-digit numbers made using the digits 1, 2, 3, 4, 5 without repetition.
(i) Which of them are multiples of 3?
(ii) Which of them are multiples of 9?
(iii) What is the remainder on dividing any of these numbers by 9?
Answer:
(i) If we make any numbers with the digits 1, 2, 3, 4, 5 without repetition, the sum will always be a multiple of 3, because the sum of the digits of the number is 1 + 2 + 3 + 4 + 5 = 15, which is a multiple of 3.
(ii) As the sum is 15, none of them will be a multiple of 9.
(iii) When 15 is divided by 9, the remainder is 6.
The sum of the digits of the numbers that are made without repetition will give 6 as a remainder.

Question 5.
Among all the ten-digit numbers using all of the ten digits 0, 1, 2,…, 9, how many are primes? What is the reason?
Answer:
When we add these ten digits, the sum is
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
Any ten-digit number formed using all these digits will have a digit sum of 45.
Since 45 is a multiple of both 3 and 9, every such number is divisible by 3 and 9.
Therefore, none of these numbers can be prime.

Question 6.
Which is the least four-digit number without any digit repeating, which is a multiple of 9? What about the least five-digit number?
Answer:
The least four-digit number without any digit repeating is 1023. But it is not a multiple of 9.
If the number should be a multiple of 9, the sum of the digits of that number should also be a multiple of 9.
So the least four-digit number without any digit repeating, which is a multiple of 9 is 1026. And the least five-digit number is 10269.

Playing with Nine (Page Number 157-158)

Question 1.
The difference of any number and the number got by rearranging its digits in any manner is a multiple of 9. Why?
Answer:
When we rearrange the digits of a number, the sum of the digits does not change.
Any number can be written as the sum of its digits plus a multiple of 9.
Therefore, two numbers that have the same digit sum differ only by a multiple of 9.
Hence, the difference between a number and any rearrangement of its digits is always a multiple of 9.

Question 2.
The difference of any three-digit number and the number written in reverse is a multiple of 99. Why? What is the relation between the number by which 99 is multiplied and the digits of the original number?
Answer:
Let’s understand this with an example:
Consider the number 563 and its reverse, 365.
563 – 365 = 198
Write both numbers in expanded form:
563 = 500 + 60 + 3 = (9 × 55 + 5) + (9 × 6 + 6) + 3
365 = 300 + 60 + 5 = (9 × 33 + 3) + (9 × 6 + 6) + 5
Now rewrite more clearly:
563 = 9(55 + 6) + 14
365 = 9(33 + 6) + 14
When we subtract:
Only part 9(55 – 33) = 9 × 22 remains, because all other components are the same and cancel out.
Thus, the difference is always:
9 × 22 = 198 = 99 × 2
Since 22 = 2 × 11, the difference is always a multiple of: 9 × 11 = 99
For any three-digit number, the difference between the number and its reverse is always a multiple of 99.
Take the difference between the first digit and the last digit of the original number.
Original number: 893
Reversed number: 398
(8 – 3) × 99 = 5 × 99 = 495
893 – 398 = 495

Question 3.
What is the relation between the digits of the number got by multiplying 99 by a single-digit number?
Answer:
From the number getting by this, adding the first and last digit, we get the middle number.

Question 4.
Take any three-digit number, write it in reverse, and find the difference of these two numbers. Write the number got as difference in reverse and add to the difference. What number do you get? Do this with the other three-digit numbers. Do you get 1089 always? For what type of numbers do you get a different answer?
Answer:
638 – 836 = 198
198 + 891 = 1089
692 + 296 = 396
396 + 693 = 1089
603 – 306 = 297
297 + 792 = 1089
If the difference between the digits in the ones and hundreds place is 1, those numbers won’t get these answers.
392 – 293 = 99
99 + 99 = 198
544 – 445 = 99
99 + 99 = 198

Class 6 Maths Chapter 10 Kerala Syllabus Number Relations Questions and Answers

Class 6 Maths Number Relations Questions and Answers

Question 1.
Write the following numbers as the multiple of 9 and the sum of the digits.
(a) 527
(b) 4325
(c) 38
(d) 6428
Answer:
(a) 527 = 500 + 20 + 7
= (9 × 55 + 5) + (9 × 2 + 2) + 7
= (9 × 57) + 14

(b) 4325 = 4000 + 300 + 20 + 5
= (9 × 444 + 4) + (9 × 33 + 3) + (9 × 2 + 2) + 5
= (9 × 479) + 14

(c) 38 = 30 + 8
= (9 × 3 + 3) + 8
= (9 × 3) + 11

(d) 6428 = 6000 + 400 + 20 + 8
= (9 × 666 + 6) + (9 × 44 + 4) + (9 × 2 + 2) + 8
= (9 × 712) + 20

Question 2.
From the numbers given below, which of them are multiples of 9?
(a) 329
(b) 1484
(c) 3299
(d) 5238
(e) 3483
Answer:
5238 and 3483
(the number those which the sum of their digits are multiples of 9)

Question 3.
From the numbers given below, which of them are multiples of 3?
(a) 486
(b) 394
(c) 6288
(d) 5430
(e) 3841
Answer:
486, 6288, and 5430
(the number those which the sum of their digits are multiples of 3)

Question 4.
From the numbers given below, which of them are multiples of both 3 and 9?
(a) 492
(b) 648
(c) 5264
(d) 3897
(e) 5486
Answer:
648 and 3897
(multiples of 9 and also multiples of 3)

Question 5.
Find the remainder of the following numbers when divided by 9?
(a) 4238
(b) 5269
(c) 4845
(d) 9890
(e) 4241
Answer:
(a) 8
(b) 4
(c) 3
(d) 8
(e) 2

Question 6.
Find the remainder of the following numbers when divided by 3?
(a) 394
(b) 1574
(c) 3248
(d) 1375
(e) 2489
Answer:
(a) 1
(b) 2
(c) 2
(d) 1
(e) 2

Question 7.
Suppose we take all four-digit numbers made using the digits 5, 9, 4, 3 without repetition, which of them are multiples of 9? Are there any multiples of 3?
Answer:
Since the sum of the digits of the numbers is not a multiple of 9, there will be no numbers with multiples of 9.
Since the sum of the digits of the numbers are the multiple of 3, all numbers formed will be multiples of 3.

Question 8.
3284_ is a five-digit number. If this is a multiple of 9, what digit will be in the place of ones?
Answer:
The digit in the ones place is 1.
(3 + 2 + 8 + 4 + 1 = 18)

Question 9.
From the following pairs of numbers, which of them have a difference that is a multiple of 9?
(a) 327,426
(b) 1624,1382
(c) 5248,3268
Answer:
327, 426, and 5248, 3268

Class 6 Maths Chapter 10 Notes Kerala Syllabus Number Relations

→ Any natural number is equal to the sum of a multiple of nine and the sum of the digits of the number.

→ The remainder on dividing any number by 9 is the remainder on dividing the sum of the digits of the number by 9.

→ If the sum of the digits of a number is a multiple of 9, then the number itself is a multiple of 9.

→ The remainder on dividing any number by 3 is the remainder on dividing the sum of the digits of the number by 3.

→ If the sum of the digits of a number is a multiple of 3, then the number itself is a multiple of 3.

In this chapter, we discuss the special properties of the factors of 9 and 3. From this, we learn how to determine the remainders when a number is divided by 9 or 3.

Digit Sum
If we take any two-digit number, it will be a multiple of 10, or the sum of a digit less than ten is added to the multiple of 10.
30 = 3 × 10
35 = 3 × 10 + 5
In detail, we can say that a two-digit number is the sum of multiples of the first digit, with 10 added to the second digit.
74 = 7 × 10 + 4
Let us now check whether the two-digit number has any relation with multiples of 9.
If we take any two-digit number, it is the sum of the first digit, which is multiplied by 9, and the sum of those two digits.
14 = 1 × 9 + 5
23 = 2 × 9 + 5
44 = 4 × 9 + 8
65 = 6 × 9 + 11

Try to write the following numbers as multiples of nine and the sum of the digits of the number.
(1) 16
(2) 24
(3) 35
(4) 63
(5) 87
Answer:
(1) 16 = 1 × 9 + 7
(2) 24 = 2 × 9 + 6
(3) 35 = 3 × 9 + 8
(4) 63 = 6 × 9 + 9
(5) 87 = 8 × 9 + 15
Let’s now check for three-digit numbers.
100 = 9 × 11 + 1
200 = 9 × 22 + 2
800 = 9 × 88 + 8
Likewise, we can write this for any multiple of 100
250 = 9 × 27 + 7
⇒ (9 × 22 + 2) + (9 × 5) + 5 = 9 × 27 + 7
625 = 9 × 68 + 13
⇒ (9 × 66 + 6) + (9 × 2 + 2) + 5 = 9 × 68 + 13
So any three-digit number is the sum of any multiple of nine, with the sum of the digits.
How can we write 697 in this manner?
Here, the sum of the digits is 22 (6 + 9 + 7), so we can subtract 22 from 697, We get 675.
When we write 675 as a multiple of 9
9 × 75 = 675
So, 697 = 9 × 75 + 22
Now try to write 726 as the sum of any multiple of nine with the sum of its digits.
726 = 700 + 20 + 6
= (9 × 77 + 7) + (9 × 2 + 2) + 6
= (9 × 79) + 15

Can we write 549 like this?
500 + 40 + 9 = (9 × 55 + 5) + (9 × 4 + 4) + 9
= (9 × 59) + 5 + 4 + 9
= 9 × 59 + 18
549 is written as the sum of the multiples of 9 and the sum of the digits, which is 18.

Now write the following numbers as the sum of the multiples of 9 and the sum of the digits.
(1) 464
(2) 387
(3) 825
(4) 949
Answer:
(1) 464 = 400 + 60 + 4
= (9 × 44 + 4) + (9 × 6 + 6) + 4
= (9 × 50) + 4 + 6 + 4
= (9 × 50) + 14

(2) 387 = 300 + 80 + 7
= (9 × 33 + 3) + (9 × 8 + 8) + 7
= (9 × 41) + 3 + 8 + 7
= (9 × 41) + 18

(3) 825 = 800 + 20 + 5
= (9 × 88 + 41) + (9 × 2 + 2) + 5
= (9 × 90) + 8 + 2 + 5
= (9 × 90) + 15

(4) 949 = 900 + 40 + 9
= (9 × 99 + 9) + (9 × 4 + 4) + 9
= (9 × 103) + 9 + 4 + 9
= 9 × 103 + 22
Even if it is two digits, or for any number, this case is true.
4379 = 4000 + 300 + 70 + 9
= (9 × 444 + 4) + (9 × 33 + 3) + (9 × 7 + 7) + 9
= 9 × (444 + 33 + 7) + 4 + 3 + 7 + 9
= (9 × 484) + 23

Now let’s write 8246 like this.
8246 = 8000 + 200 + 40 + 6
= (9 × 888 + 8) + (9 × 22 + 2) + (9 × 4 + 4) + 6
= 9 × (888 + 22 + 4) + 8 + 2 + 4 + 6
= (9 × 914) + 20

In general, we can say that any natural number is equal to the sum of a multiple of nine and the sum of the digits of the number.

Let’s now check the peculiarity of the remainder we got after dividing any of these numbers by 9.
76 ÷ 9
76 = 9 × 7 + 13
That is 76 is a number got by adding 13 to a multiple of 9.
But we didn’t get 13 as a remainder when dividing by 9.
When we again divide 13 by 9, we get 1 and a reminder of 4.
Which means this 4 is the remainder we got by dividing the sum of the digits by 9.
Then what about dividing 365 by 9
Sum of the digits: 3 + 6 + 5 = 14
14 ÷ 9 = 1
Remainder = 5
So we get 5 as the remainder after dividing 365 by 9.

What about the remainder we got after dividing 4257 by 9.
4 + 2 + 5 + 7 = 18
18 ÷ 9
Remainder = 0
We get 0 as the remainder after dividing 4257 by 9.
From this, we understood that:
The remainder on dividing any number by 9 is the remainder on dividing the sum of the digits of the number by 9.
If the sum of the digits of a number is a multiple of 9, then the number itself is a multiple of 9.

From the numbers given below, find which of them are multiples of 9? Also, find the remainder of the numbers that are not multiples of 9.
(1) 49
(2) 378
(3) 538
(4) 1489
(5) 12489
(6) 3690
(7) 42345
(8) 52374
Answer:
(1) 49
Sum of the digits = 4 + 9 = 13
⇒ not a multiple of 9
13 ÷ 9
⇒ Remainder = 4

(2) 378
Sum of the digits = 3 + 7 + 8 = 18
⇒ multiple of 9

(3) 538
Sum of the digits = 5 + 3 + 8 = 16
⇒ not a multiple of 9
16 ÷ 9
⇒ Remainder = 7

(4) 1489
Sum of the digits = 1 + 4 + 8 + 9 = 22
⇒ not a multiple of 9
22 ÷ 9
⇒ Remainder = 6

(5) 12489
Sum of the digits = 1 + 2 + 4 + 8 + 9 = 24
⇒ not a multiple of 9
24 ÷ 9
⇒ Remainder = 6

(6) 3690
Sum of the digits = 3 + 6 + 9 + 0 = 18
⇒ multiple of 9

(7) 42345
Sum of the digits = 4 + 2 + 3 + 4 + 5 = 18
⇒ multiple of 9

(8) 52374
Sum of the digits = 5 + 2 + 3 + 7 + 4 = 21
⇒ not a multiple of 9
21 ÷ 9
⇒ Remainder = 3

Remainder on Diving by 3
Same as in the case of 9, we can find the remainder on dividing by 3 simply by adding the digits in that number.
The remainder on dividing any number by 3 is the remainder on dividing the sum of the digits of the number by 3.
In other words, we can say that if the sum of the digits of a number is a multiple of 3, then the number itself is a multiple of 3.

Check whether 1427 is a multiple of 3?
The sum of digits in 1427 is 1 + 4 + 2 + 7 = 14
Here, 14 is not a multiple of 3.
So, as the remainder we got by dividing 14 by 3 is 2, the remainder we get after dividing 1427 will also be 2.

Check whether the numbers given below is the multiple of 3, else find the remainder when dividing the number by 3?
(a) 527
(b) 648
(c) 1397
(d) 12486
Answer:
(a) 527
5 + 2 + 7 = 14
⇒ not a multiple of 3
⇒ remainder = 2

(b) 648
6 + 4 + 8 = 18
⇒ multiple of 3

(c) 1397
1 + 3 + 9 + 7 = 20
⇒ not a multiple of 3
⇒ remainder = 2

(d) 12486
1 + 2 + 4 + 8 + 6 = 21
⇒ multiple of 3

From the numbers given below, which of them are multiples of 3 and 9?
(a) 999
(b) 1647
(c) 3984
(d) 4788
(e) 5642
(f) 3864
(g) 24329
(h) 49995
Answer:
(a) 9 + 9 + 9 = 27
⇒ multiple of 3 and 9

(b) 1 + 6 + 4 + 7 = 18
⇒ multiple of 3 and 9

(c) 3 + 9 + 8 + 4 = 24
⇒ multiple of 3 but not a multiple of 9
Remainder on dividing by 9 = 6

(d) 4 + 7 + 8 + 8 = 27
⇒ multiple of 3 and 9

(e) 5 + 6 + 4 + 2 = 17
⇒ not a multiple of 3 and 9
Remainder on dividing by 3 = 2
Remainder on dividing by 9 = 8

(f) 3 + 8 + 6 + 4 = 21
⇒ multiple of 3 and not a multiple of 9
Remainder on dividing by 9 = 3

(g) 2 + 4 + 3 + 2 + 9 = 20
⇒ not a multiple of 3 and 9
Remainder on dividing by 3 = 2
Remainder on dividing by 9 = 2

(h) 4 + 9 + 9 + 9 + 5 = 36
⇒ multiple of 3 and 9

Playing with Nine
Any number is a multiple of 9 added to the sum of the digits of the number.
If we subtract from a number the sum of its digits, we get a multiple of 9.
If we subtract 19 from 487
487 – 19 = 468, which is a multiple of 9.
Also, if the sum of the digits of two numbers are same, then their difference will be a multiple of 9.
Example: The sum of the digits of the numbers 4237 and 3544 is 16.
The difference between these numbers
4237 – 3544 = 693
Since the sum of the digits of the number 693 is 18, 693 is a multiple of 9.

Which of the following differences of number pairs are multiples of 9?
(1) 3487, 4583
(2) 5684, 9284
(3) 6247, 3584
(4) 7254, 5814
Answer:
The difference in the sum of the digits of the 1st and 3rd pair is different.
So their difference is also not a multiple of 9.
The difference of the sum of the digits of the 2nd and 4th pair is the same.
So their difference is also a multiple of 9.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 7 Money Math Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 7 Solutions Money Math

Class 6 Kerala Syllabus Maths Solutions Chapter 7 Money Math Questions and Answers

Money Math Class 6 Questions and Answers Kerala Syllabus

Discount

In-text Questions (Page Number 103)

Question 1.
See this ad:

Fill in the table below by computing the deduction for each purchase and the amount to be paid.

Answer:

Interest

In-text Questions (Page Number 105)

Question 1.
See this news item:

See the amounts some people have deposited in this bank:

Calculate the interest and the total amount each would get back after one year.
Answer:
1. Ash
Deposited amount = ₹ 15000, interest = 9%
15000 = 150 × 100
Interest = 150 × 9 = ₹ 1350
Amount to be received = 15000 + 1350 = ₹ 16350

2. Suma
Deposited amount = ₹ 25000, interest = 9%
25000 = 250 × 100
Interest = 250 × 9 = ₹ 2250
Amount to be received = 25000 + 2250 = ₹ 27250

3. Niharika
Deposited amount = ₹ 37500, interest = 9%
37500 = 375 × 100
Interest = 375 × 9 = ₹ 3375
Amount to be received = 37500 + 3375 = ₹ 40875

4. Vihan
Deposited amount = ₹ 12500, interest = 9%
12500 = 125 × 100
Interest = 125 × 9 = ₹ 1125
Amount to be received = 12500 + 1125 = ₹ 13625

5. Ahi
Deposited amount = ₹ 22500, interest = 9%
22500 = 225 × 100
Interest = 225 × 9 = ₹ 2025
Amount to be received = 22500 + 2025 = ₹ 24525

Interest (Page Number 106)

Question 1.
See the board in front of the shop:

Sreesha bought clothes worth 3500 rupees from this shop. How much should she pay?
Answer:
The price of the books that Sreesha bought = ₹ 3500
Discount = 25% (25 for every 100)
3500 = 35 × 100
Discount = 35 × 25 = ₹ 875
Amount to be paid = 3500 – 875 = ₹ 2625

Question 2.
Angelo Tomy borrowed 45000 rupees from a bank. The interest is 12%. How much should he pay back after one year?
Answer:
Borrowed amount = ₹ 45000
Rate of interest = 12%
45000 = 450 × 100
Interest = 450 × 12 = ₹ 5400
Amount to be repaired = 45000 + 5400 = ₹ 50400

Question 3.
Mehul contributes 8% of his earnings each month to medical aid for the poor. His income this month is 75000 rupees. How much is his contribution this month?
Answer:
His income = ₹ 75000
Giving contribution = 8%
75000 = 750 × 100
Contribution = 750 × 8 = ₹ 6000

Question 4.
Eugene saves 15% of his income each month. His income in January was ₹ 64000 rupees. How much did he save that month?
Answer:
His income = ₹ 64000
Saving amount = 15%
64000 = 640 × 100
Saving = 640 × 15 = ₹ 9600

Question 5.
A TV manufacturer decides to reduce the price by 5% from next month. What will be the price next month of a TV that costs 45000 rupees right now?
Answer:
Current price = ₹ 45,000
Discount = 5%
45000 = 450 × 100
Discount = 450 × 5 = ₹ 2,250
Next month’s price of the TV = 45,000 – 2,250 = ₹ 42,750

Question 6.
A car manufacturer decides to increase the price by 2% from next month. What will be the price next month of a car that costs 5,60,000 rupees?
Answer:
Current price of the car = ₹ 5,60,000
Higher price = 2%
560000 = 5,600 × 100
Higher price = 5,600 × 2 = ₹ 11,200
Price of next month = 5,60,000 + 11,200 = ₹ 5,71,200.

Percent Problems (Page Number 110)

Question 1.
At the book’s Festival, all books are given 15% discount. See the prices of book Sreena and her friends bought:

Calculate how much each has to pay.
Answer:
Sreena
Price of the book = ₹ 540
540 = 500 + 40
Discount for 500 rupees = 5 × 15 = ₹ 75
Since the reduction for 10 rupees is 1 rupee 50 paise
The discount for 40 rupees = ₹ 6
Total = 75 + 6 = ₹ 81
Amount to be paid = 540 – 81 = ₹ 459
Tessy
Price of the book = ₹ 375
375 = 300 + 50 + 25
Discount for 300 rupees = 3 × 15 = ₹ 45
Discount for 50 rupees = half of 15 = ₹ 7.50
Discount for 25 rupees = half of ₹ 7.50 = ₹ 3.75
Total = 45 + 7.50 + 3.75 = ₹ 56.25
Amount to be paid = 375 – 56.25 = ₹ 318.75
Nassar
Price of the book = ₹ 630
630 = 600 + 30
Discount for 600 rupees = 6 × 15 = ₹ 90
Since the reduction for 10 rupees is 1 rupee 50 paise
Discount for 30 rupees = 3 × ₹ 1.50 = ₹ 4.50
Total = 90 + 4.50 = ₹ 94.50
Amount to be paid = 630 – 94.50 = ₹ 535.50
Jacob
Price of the book = ₹ 432
432 = 400 + 30 + 2
Discount for 400 rupees = 4 × 15 = ₹ 60
Discount for 30 rupees = 3 × ₹ 1.50 = ₹ 4.50
Discount for 2 rupees = 2 × 15 paise = 30 paise
Total = ₹ 60 + ₹ 4.50 + 30 paise = ₹ 64.80
Amount to be paid = 432 – 64.80 = ₹ 367.20

Question 2.
A person borrowed 26750 rupees from a bank, where the interest is 12% each year. How much should he pay back after one year?
Answer:
Borrowed amount = ₹ 26750
Rate of interest = 12%
26750 = 20000 + 6000 + 700 + 50
Interest for 20000 rupees = 200 × 12 = ₹ 2400
Interest for 6000 rupees = 60 × 12 = ₹ 720
Interest for 700 rupees = 7 × 12 = ₹ 84
(It can also be written as 267 × 12 = 3204 for 26700 rupees)
Interest for 50 rupees = ₹ 6
Total interest = 3204 + 6 = ₹ 3210
Amount to be paid = 26750 + 3210 = ₹ 29960

Question 3.
A company gives 8% of one month’s salary as a bonus for Onam. How much does a man, whose monthly salary is 46375 rupees, get as a bonus?
Answer:
Monthly salary = ₹ 46375
Bonus = 8%
46375 = 46300 + 50 + 25
Bonus for 46300 rupees = 463 × 8 = ₹ 3704
Bonus for 50 rupees = ₹ 4 (half of 8)
Bonus for 25 rupees = ₹ 2 (half of 4)
Total bonus = 3704 + 4 + 2 = ₹ 3710

What Percent? (Page Number 112)

Question 1.
25000 rupees deposited in a bank earned 2000 rupees as interest in a year. What is the interest rate?
Answer:
Deposited amount = 25000 rupees
Interest = 2000
25000 = 250 × 100
250 for one hundred = 2000
For one hundred = 2000 ÷ 250
= 8 × 250 ÷ 250
= 8
Interest percent = 8%

Question 2.
For books worth 480 rupees, a shop gave 72 rupees as a discount. What is the discount percent?
Answer:
The worth of the book = 480 rupees
Discount = 72 rupees
48 for ten = 72 rupees
For one ten = 72 ÷ 48
= 24 × 3 ÷ 24 × 2
= 3 ÷ 2
= \(\frac {3}{2}\)
= 1\(\frac {1}{2}\) rupees
Since for ₹ 10 it is ₹ 1.50, for ₹ 100 it is ₹ 15
This can also be calculated for one rupee.
For 480 rupees = 7,200 paisa
For 1 rupee = 7,200 ÷ 48 = 150 paise
Since for ₹ 10 it is 150 paise, for ₹ 100 it is 150 × 100 = 1,500 paise = ₹ 15
The discount = 15%

Question 3.
A mobile phone manufacturer sells all the previous year’s models at a fixed discount. A phone that cost 45,000 rupees earlier is now sold for 37800 rupees. What is the discount percent?
Answer:
Previous price = ₹ 45,000
Present price = ₹ 37,800
Discount = 45,000 – 37,800 = ₹ 7,200
For 100 rupees = 7200 ÷ 450
= 720 × 10 ÷ 45 × 10
= 80 × 9 ÷ 5 × 9
= 80 ÷ 5
= 16
Discount percent = 16%

Class 6 Maths Chapter 7 Kerala Syllabus Money Math Questions and Answers

Class 6 Maths Money Math Questions and Answers

Question 1.
A mobile phone worth ₹ 8500 is sold at a discount of 6%. What is its present price?
Answer:
Original price = ₹ 8500
Discount = 6%
It is ₹ 6 for every 100 rupees
There are 85 hundred in ₹ 8500
So, the discount for 85 hundreds = 85 × 6 = ₹ 510
Therefore, present price = 8500 – 510 = ₹ 7990

Question 2.
For ₹ 2000, 12% interest has to be paid for one year. How much interest should be paid?
Answer:
Amount borrowed = ₹ 2000
Interest = 12%
Rupees 12 for every ₹ 100.
There are 20 hundred in 2000 rupees.
So, for 20 hundreds = 20 × 12 = ₹ 240

Question 3.
The price of a watch worth ₹ 1350 increased by 8%. What is its present price?
Answer:
Original price = ₹ 1350
Increase = 8%
1350 = 1300 + 50
There are 13 hundred in 1300 rupees (13 × 100 = 1300)
Increase for 13 hundreds = 13 × 8 = ₹ 104
Increase for ₹ 50 = half of 8 = ₹ 4
Total increase = 104 + 4 = ₹ 108
Present price = 1350 + 108 = ₹ 1458

Question 4.
The price of a fan was ₹ 3275. Its price decreased by 12%. How much did it decrease?
Answer:
Price of fan = ₹ 3275
Decrease = 12%
3275 = 3200 + 50 + 25
There are 32 hundred in 3200
Decrease for 32 hundreds = 32 × 12 = ₹ 384
Decrease for ₹ 50 = half of 12 = ₹ 6
Decrease for ₹ 25 = half of 6 = ₹ 3
Total decrease = 384 + 6 + 3 = ₹ 393

Question 5.
When a product worth ₹ 1530 was bought, 18% tax was added. What is the total price of the product?
Answer:
Price of the article = ₹ 1530
Tax = 18%
1530 = 1500 + 30
Tax for one hundred = ₹ 18
Tax for rupees 1500 = 15 × 18 = ₹ 270
Since the tax is ₹ 18 for every ₹ 100, the tax for ₹ 10 is ₹ 1.80.
So, the tax for ₹ 30 is three times ₹ 1.80, which is equal to ₹ 5.40.
Total tax = 270 + 5.40 = ₹ 275.40
Therefore, total price = 1530 + 275.40 = ₹ 1805.40

Question 6.
For ₹ 12000, the interest for one year is ₹ 1800. What is the rate of interest?
Answer:
Amount = ₹ 12000
Interest = ₹ 1800
12000 = 120 × 100
For 120 hundreds = ₹ 1800
For one hundred = 1800 ÷ 120
= (15 × 120) ÷ 120
= 15
Therefore, the rate of interest = 15%

Question 7.
The price of an article worth ₹ 540 increased by ₹ 27. By what percentage did it increase?
Answer:
Original price = ₹ 540
Increase = ₹ 27
540 cannot be made exactly into hundreds.
For 54 tens, the increase is ₹ 27 (2700 paise)
Increase for one ten rupees = 2700 ÷ 54
= 27 × 100 ÷ 2 × 27
= 100 ÷ 2
= 50 paise
Increase for one hundred rupees = 10 × 50 = 500 paise = 5 rupees
Therefore, increase = 5%

Question 8.
The price of a saree decreased from ₹ 1240 to ₹ 992. What is the percentage of decrease?
Answer:
Original price = ₹ 1240
New price = ₹ 992
Decrease = 1240 – 992 = ₹ 248
1240 = 124 ten rupees
For 124 tens, the decrease is ₹ 248.
Decrease for one ten rupees = 248 ÷ 124
= 2 × 124 ÷ 124
= 2 rupees
100 rupees means 10 ten rupees.
Decrease for ₹ 100 = 10 × 2 = 20 rupees
Therefore, the percentage of decrease = 20%

Question 9.
The price of an article is ₹ 1860. Including tax, it is sold for ₹ 1953. What is the percentage of tax?
Answer:
Price without tax = ₹ 1860
Price with tax = ₹ 1953
Tax = 1953 – 1860 = ₹ 93
1860 = 186 ten rupees
Tax for 186 ten rupees = ₹ 93 = 9300 paise
Tax for one ten rupees = 9300 ÷ 186
= (93 × 100) ÷ (93 × 2)
= 100 ÷ 2
= 50
Tax for 10 ten rupees = 10 × 50 = 500 paise = ₹ 5
Therefore, tax = 5%

Question 10.
Gopi paid ₹ 960 as interest for one year on ₹ 12000 in one bank. Radha paid ₹ 1050 as interest for one year on ₹ 15000 in another bank. Who paid at a higher rate of interest?
Answer:
Gopi
Amount = ₹ 12000
Interest = ₹ 960
For 120 hundreds = ₹ 960
For one hundreds = 960 ÷ 120
= (96 × 10) ÷ (10 × 12)
= 96 ÷ 12
= 8
Interest percentage = 8%

Radha
Amount = ₹ 15000
Interest = ₹ 1050
For 150 hundreds = ₹ 1050
For one hundred = 1050 ÷ 150
= (21 × 50) ÷ (3 × 50)
= 21 ÷ 3
= 7
Interest percentage = 7%
Therefore, Gopi paid at a higher rate of interest (8%)

Class 6 Maths Chapter 7 Notes Kerala Syllabus Money Math

→ 1 Rupee = 100 Paise

→ When we buy or sell goods, we deal with price, discount, tax, profit, and loss.

→ Discount = Original price – Selling price

→ Increase = New price – Old price

→ Tax is an extra amount paid while buying goods.

→ The extra amount deposited in banks or paid when borrowing money is called interest.

→ We calculated the percentages of numbers for which we can accurately find how many hundreds they contain.

In this unit, we learn about the concept of percentage and some situations in life where it is used. The term ‘percentage’ is something we often hear in our daily life. We often hear percentages being used to express things like the number of students who achieved a certain grade, the number of people who passed, discounts or price increases, and population growth or decline. We also see them in newspapers and advertisements. What is a percentage? How is it calculated? Let us learn about it in this unit. The word ‘percentage’ is derived from Latin “per centum,” which means per hundred and represents a part of a whole. Therefore, the word percentage refers to a measure based on one hundred. The symbol % is used to denote a percentage.

Discount
We are all familiar with expressing the reduction in the price of goods as a percentage.
If an item has a 20 percent discount, it means that an article worth ₹ 100 will get a ₹ 20 reduction in price. So, if the article costs ₹ 200, the discount will be 2 × 20 = ₹ 40, and if it costs ₹ 500, the discount will be 5 × 20 = ₹ 100 (since there are two hundreds in 200 and five hundreds in 500). Thus, giving a ₹ 20 reduction for every ₹ 100 is called a 20 percent deduction. This can be written using the symbol 20%.

Write the answers for the following:

Question 1.
A shop offers a 15% deduction on goods. How much discount will a buyer get on an item worth ₹ 300?
Answer:
For ₹ 100, the deduction is ₹ 15.
For ₹ 300, the deduction is 3 × 15 = ₹ 45.

Question 2.
If an article has a 25% deduction and its price is ₹ 1000, how much should be paid to buy it?
Answer:
For ₹ 100, the deduction is ₹ 25.
For ₹ 1000, the deduction is 10 × 25 = ₹ 250

Question 3.
In a shop, there is a 20% deduction. Which of the following statements related to this are correct?
(a) From this shop, you will get goods worth ₹ 100 for ₹ 80.
(b) If you buy an item worth ₹ 100 from this shop, paying ₹ 80 will be enough.
(c) If you buy an item worth ₹ 1000 from this shop, you have to pay ₹ 980.
(d) In this shop, you will get a discount of ₹ 20 on an item worth ₹ 100.
Answer:
Statements a, b, and d are correct.

Interest
The extra amount deposited in banks or paid when borrowing money is called interest. If the interest rate is 10%, it means that for every ₹ 100 deposited, you will get ₹ 10 extra. Or, if you borrow ₹ 100, you have to pay ₹ 10 extra. So, if the interest rate is 12%, you have to pay ₹ 12 as interest for every ₹ 100.
Now, what if the amount is ₹ 1000?
There are ten hundreds in 1000. Therefore, the interest will be 10 × 12 = ₹ 120.
If ₹ 1500 is deposited, there are fifteen hundred in it.
So, the interest will be 15 × 12 = ₹ 180.
After one year, the total amount, including interest, will be ₹ 1500 + ₹ 180 = ₹ 1680.

Now write the answers to these questions:

Question 1.
A person deposited ₹ 4000 in a bank. The rate of interest is 8%. How much will he get after one year?
Answer:
Interest for ₹ 100 in one year is ₹ 8.
For ₹ 4000, there are 40 hundreds.
Interest = 8 × 40 = ₹ 320
(That is, ₹ 4000 has forty hundreds. Interest for one hundred is ₹ 8. So, for 40 hundreds, 40 × 8 = ₹ 320)
Total amount received = 4000 + 320 = ₹ 4320

Question 2.
A person borrowed ₹ 7600 from a bank. The rate of interest is 11%. How much should he repay after one year?
Answer:
Interest for ₹ 100 in one year is ₹ 11.
There are 76 hundreds in ₹ 7600 (7600 = 76 × 100)
Interest = 76 × 11 = ₹ 836
Total amount to be repaid = 7600 + 836 = ₹ 8436

Percentage means how many rupees for every 100 rupees.
So, how do we calculate a given percent?
First, we check how many hundreds are in the number.
Then we multiply that number by the given percentage.
Eg: To find 20% of ₹ 16000.
There are 160 hundred in 16000
So, 20% = 160 × 9 = 1440.

Percent Problems
We calculated the percentages of numbers for which we can accurately find how many hundreds they contain.
For example: 200, 12500, 37500, etc.
Now, let’s look at some numbers for which we cannot accurately find the number of hundreds.

If you buy goods worth ₹ 650 from a shop that offers a 20% discount, how much discount will you get?
The price of the item is ₹ 650. For every ₹ 100, you get a discount of ₹ 20.
So, for six hundred, you get 6 × 20 = ₹ 120 discount.
There is an additional ₹ 50, and since the discount is ₹ 20 for every ₹ 100, for ₹ 50, you get half of that ₹ 10 discount.
Total discount = 120 + 10 = ₹ 130
Now, if the price is ₹ 875:
Discount for ₹ 800 = 8 × 20 = ₹ 160
Discount for ₹ 50 = half of 20 = ₹ 10
Discount of ₹ 25 = half of 10 = ₹ 5
Total discount = 160 + 10 + 5 = ₹ 175

Let’s look at another calculation:
At an interest rate of 15%, how much interest should be paid for ₹ 1275 in one year?
Answer:
For 1000 rupees =10 × 15 = ₹ 150
For 200 rupees = 2 × 15 = ₹ 30
For 50 rupees = half of 15 = ₹ 7.50
For 25 rupees = half of 7.50 = ₹ 3.75
Total = 150 + 30 + 7.50 + 3.75 = ₹ 191.25

At an interest rate of 7%, how much interest should be paid for ₹ 23275 in one year?
Answer:
We can write 23275 as 20000 + 3000 + 200 + 50 + 25
Interest for 20000 rupees = 200 × 7 = ₹ 1400
Interest for 3000 rupees = 30 × 7 = ₹ 210
Interest for 200 rupees = 2 × 7 = ₹ 14
Interest for 50 rupees = half of 7 = ₹ 3.50
Interest for 25 rupees = half of 3.50 = ₹ 1.75
Total interest = 1400 + 210 + 14 + 3.50 + 1.75 = ₹ 1629.25

If an item costs ₹ 748 and you get an 8% discount, how much should you pay for it?
Answer:
748 = 700 + 40 + 8
Discount for 700 rupees = 7 × 8 = ₹ 56
Since the discount on ₹ 100 is ₹ 8, for ₹ 10 the discount is 80 paise.
So, for ₹ 40, it is 4 × 80 = 320 paise = ₹ 3.20.
For ₹ 10, the discount is 80 paise, so for ₹ 1, it is 8 paise.
For ₹ 8, it is 8 × 8 = 64 paise.
Now let’s find the total discount:
56 + 3.20 + 0.64 = ₹ 59.84
Amount to be paid = 748 – 59.84 = ₹ 688.16

What Percent?
So far, we have discussed the method of finding a fixed percentage of a number.
Now let’s look at another example.
When a person buys an item worth 100 rupees and pays 8 rupees as tax, what is the tax percentage?
Since the tax is 8 rupees for 100 rupees, the tax rate is 8%.
Therefore, including the tax, the total amount to be paid is 108 rupees.
If a person pays a tax of ₹ 40 on an item worth ₹ 400, what is the tax percentage?
The tax is calculated for every 100 rupees.
Here, the tax of 40 rupees is for 400 rupees.
To find how much it is for 100 rupees, we divide 40 by 4.
40 ÷ 4 = (4 × 10) ÷ 4 = 10 (the common factor 4 can be cancelled)
The tax is 10%.
Let’s see another example.

When a book worth ₹ 1200 was bought, a discount of ₹ 360 was given. What percentage of discount was given?
Answer:
Since it is for ₹ 1200, to find how much it is for one hundred, divide 360 ÷ 12.
360 ÷ 12 = 36 × 10 ÷ 12
= 3 × 12 × 10 ÷ 12
= 3 × 10
= 30 (the common factor 12 can be cancelled)
The discount is 30%.
A person’s daily wage was ₹ 1250. Now it has increased by ₹ 250. What is the percentage of increase?
Since ₹ 1250 cannot be made into exact hundreds, we make it 125 tens.
It increased by ₹ 250 for 125 tens.
So, to find how much it increased for one ten, divide 250 by 125.
250 ÷ 125 = 2
Since it increased by 2 for one ten, it increased by 20 for one hundred.
Therefore, the percentage of increase = 20%.

If ₹ 60 is given as interest for one year on ₹ 750, what is the rate of interest?
Answer:
Here, it is difficult to calculate for ₹ 100 or ₹ 10, so we should find how much it is for ₹ 1.
₹ 750 earns ₹ 60, that is 6000 paise.
To find how many paise for ₹ 1, divide 6000 by 750.
6000 ÷ 750 = 600 ÷ 75
600 ÷ 75 = 8 × 75 ÷ 75 = 8
Since the interest for one rupee is 8 paise, the interest for 100 rupees = 8 × 100 = 800 paise = 8 rupees
Interest 8%

When an item worth ₹ 320 was bought, a discount of ₹ 48 was given. What is the percentage of discount?
Answer:
The discount on ₹ 320 is ₹ 48 (4800 paise)
The discount for one rupee = 4800 ÷ 320
= 480 ÷ 32
= 60 × 8 ÷ 4 × 8
= 60 ÷ 4
= 15 paise
Since the discount for one rupee is 15 paise, for 100 rupees it is 1500 paise = ₹ 15
Therefore, the discount is 15%.
In this way, we can calculate how much it is for ₹ 100, ₹ 10, and ₹ 1 and then find the percentage.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 8 Unchanging Relations Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 8 Solutions Unchanging Relations

Class 6 Kerala Syllabus Maths Solutions Chapter 8 Unchanging Relations Questions and Answers

Unchanging Relations Class 6 Questions and Answers Kerala Syllabus

Unchanging Sum (Page Number 116)

Question 1.
Mentally calculate the sums below:
(i) 137 + 199
(ii) 375 + 298
(iii) 697 + 174
(iv) 1489 + 2363
Answer:
(i) 137 + 199 = 136 + 200 = 336 (1 subtracted, 1 added)
(ii) 375 + 298 = 373 + 300 = 673 (2 subtracted, 2 added)
(iii) 697 + 174 = 700 + 171 = 871 (3 added, 3 subtracted)
(iv) 1489 + 2363 = 1500 + 2352 = 3852 (11 added, 11 subtracted)

Question 2.
(i) Take any four consecutive natural numbers. Calculate the sum of the first and the last, and also the sum of the middle two numbers. Take the other four consecutive natural numbers and check. Why are the sums the same every time?
Answer:
1 + 2 + 3 + 4 → 1 + 4 = 5; 2 + 3 = 5
7 + 8 + 9 + 10 → 7 + 10 = 17; 8 + 9 = 17
10 + 11 + 12 + 13 → 10 + 13 = 23; 11 + 12 = 23
The first number and the fourth number represent the sum of the first number and the number that is 3 more than it.
The second number and the third number also mean the first number and the number that is 3 more than it.
Both sums will always be equal. (both will be the first number, add 4)
Adding 3 to 1 gives 4 (1 + 4 = 5)
Adding 1 + 2 to 1 also gives 4 (1 + 4 = 5)
Adding 3 to 10 gives 13 (10 + 13 = 23)
Adding 1 + 2 to 10 also gives 13 (10 + 13 = 23)
It can also be said like this:
If you add 3 to the first number, you get 4.
The middle numbers are the first number plus one, and the fourth number minus one.
When one number decreases by one and the next number increases by one, the total remains unchanged.

(ii) Instead of four consecutive natural numbers, take six consecutive natural numbers. The sum of the first and the last numbers is equal to which other pairs? Why is this so?
Answer:
Six consecutive natural numbers are 3, 4, 5, 6, 7, 8
3 + 8 = 4 + 7 = 5 + 6 = 11
The sixth number is five more than the first number.
Now, if we take the second and fifth numbers:
The second number is one more than the first number, and the fifth number is one less than the sixth number.
(one more than 3 is 4, one less than 8 is 7)
Next, if we take the third and fourth numbers:
The third number is two more than the first number, and the fourth number is two less than the sixth number.
(two more than 3 is 5, and two less than 8 is 6)
So, 3 + 8 = (3 + 1) + (8 – 1) that is 4 + 7 = 11
3 + 8 = (3 + 2) +(8 – 2) that is 5 + 6 = 11
Consecutive counting numbers will always follow this pattern.

(iii) What if we take five consecutive naturalnumbers?
Answer:
Five consecutive natural numbers are 1, 2, 3, 4, 5
1 + 5 = 6, 2 + 4 = 6, 3 + 3 = 6
The second number is one more than the first number, and the fourth number is one less than the fifth number.
Therefore, first number + fifth number = second number + fourth number
Now, if we take the third number:
It is two more than the first number and two less than the fifth number.
So, if we add it twice, it will be equal to the sum of the first and fifth numbers.
Example: 3 + 3 = (1 + 2) + (5 – 2) = 6

(iv) What about three consecutive natural numbers?
Answer:
Three consecutive natural numbers are 4, 5, 6
4 + 6 = 5 + 5
The middle number is one more than the first number and one less than the third number.
Therefore, the sum of the end numbers will always be twice the middle number.

Question 3.
In the calendar of any one month, mark those dates which fall on any one day of the week:

The sum of the first and last numbers is equal to the sum of which other pairs? Why?
Answer:
Let’s take the first 4 Sundays.
5, 12, 19, 26
5 + 26 = 12 + 19 = 31
The second number is 7 more than the first number, and the third number is 7 less than the fourth number.
Therefore, the sums of such pairs will not change.
When we take 5 numbers 1, 8, 15, 22, 29, the same thing happens.
The middle number is 14 less than one end number and 14 more than the other end number.
So, when it is added twice, we get the same sum as the end numbers.
Example: 1 + 29 = 8 + 22 = 15 + 15

Unchanging Difference (Page Number 118)

Question 1.
Calculate these differences in your head:
(i) 453 – 196
(ii) 388 – 189
(iii) 700 – 387
(iv) 1500 – 839
Answer:
(i) 453 – 196 = (453 + 4) – (196 + 4)
= 457 – 200
= 257

(ii) 388 – 189 = (388 + 11) – (189 + 11)
= 399 – 200
= 199

(iii) 700 – 387 = (700 + 13) – (387 + 13)
= 713 – 400
= 313

(iv) 1500 – 839 = (1500 – 1) – (839 – 1)
= 1499 – 838
= 661

(Page Numbers 119 & 120)

Question 1.
Is the difference of any two multiples of 3 again a multiple of 3? Why?
Answer:
The difference of two multiples of 3 is always a multiple of 3.
If you take (3 × one number) – (3 × another number), it can be written as 3 × (bigger number – smaller number).
Since the result is 3 multiplied by some whole number, it will always be a multiple of 3.
Example: 63 – 21
= (21 × 3) – (7 × 3)
= (21 – 7) × 3
= 14 × 3
= 42
Whenever you subtract one multiple of 3.
From another, the answer will always be a multiple of 3.

Question 2.
(i) If two numbers leave a remainder of 1 on division by 3, is their difference a multiple of 3? Why?
Answer:
The difference of two numbers that leave a remainder of 1 when divided by 3 will always be a multiple of 3.
This is because if you subtract 1 from each of those numbers, they become exact multiples of 3.
So, the remaining difference will be a multiple of 3.
3 × 10 + 1 – 3 × 7 + 1
3 × 3 + 0 = 3 × 3, multiple of 3

(ii) If two numbers leave a remainder of 2 on division by 3, is their difference a multiple of 3? Why?
Answer:
A number that leaves a remainder of 2 when divided by 3 can be written in the form 3 × 10 + 2.
Then, 3 × 10 + 2 – 3 × 4 + 2
3 × 6 + 0 = 3 × 6, multiple of 3.

Question 3.
What all things can we say if we take 4 instead of 3 in the last problem?
Answer:

• If you subtract one multiple of 4 from another multiple of 4, the result will always be a multiple of 4.
• When two numbers leave the same remainder when divided by 4, their difference will also be a multiple of 4.

Question 4.
If two numbers leave the same remainder on division by a number, what is the relation between the difference of these numbers and the dividing number?
Answer:
The difference will be the divisor multiplied by some counting number.

Unchanging Product (Page Number 123)

Question 1.
If we start with 1 and double again and again, we get the numbers 1, 2, 4, 8, 16,…
(i) Take any four consecutive numbers from these. Why is the product of the first and last numbers the same as the product of the middle two numbers?
Answer:
The second number will be the first number multiplied by two.
The third number will be the last number divided by two.
1, 2, 4, 8
1 × 8 = 2 × 4

(ii) If we take three consecutive numbers from these, what is the relation between the product of the first and last numbers and the middle number?
Answer:
The product of the first and the last numbers will be equal to the middle number multiplied by itself.
4, 8, 16
4 × 16 = 8 × 8

Question 2.
If we start with 1 and triple it again and again, we get the numbers 1, 3, 9, 27, 81,… For these, are the results of the first problem true? Why?
Answer:
That’s correct. Because the second number is the first number multiplied by 3.
The third number is the last number divided by 3.
1, 3, 9, 27, 81
1 × 27 = 3 × 9
Similarly, the second number is three times the first number.
The second number is the third number divided by 3.
3 × 27 = 9 × 9 = 81
Three multiplied by 3 is nine.
If you divide 81 by 3, you get 27.

Unchanging Quotients (Page Number 126)

Question 1.
Calculate the quotients below:
(i) 345 ÷ 15
(ii) 495 ÷ 45
(iii) 325 ÷ 25
(iv) 975 ÷ 75
(v) 875 ÷ 125
Answer:
(i) 345 ÷ 15 = 690 ÷ 30
= (69 × 10) ÷ (3 × 10)
= 69 ÷ 3
= 23

(ii) 495 ÷ 45 = 990 ÷ 90
= (99 × 10) ÷ (9 × 10)
= 99 ÷ 9
= 11

(iii) 325 ÷ 25 = 650 ÷ 50
= (65 × 10) ÷ (5 × 10)
= 65 ÷ 5
= 13

(iv) 975 ÷ 75 = 3900 ÷ 300
= (390 × 10) ÷ (30 × 10)
= 390 ÷ 30
= 13

(v) 875 ÷ 125 = 3500 ÷ 500
= (350 × 10) ÷ (50 × 10)
= 350 ÷ 50
= 7

Class 6 Maths Chapter 8 Kerala Syllabus Unchanging Relations Questions and Answers

Class 6 Maths Unchanging Relations Questions and Answers

Question 1.
Take any 4 consecutive numbers from the sequence 10, 20, 30, 40, 50, 60, …Find the pairs that give the same sum. Take any 6 numbers. Find the pairs that give the same sum. Explain why? What if we take 7 numbers?
Answer:
4 consecutive numbers are 40, 50,60, 70
40 + 70 = 50 + 60
50 is 10 more than 40, 60 is 10 less than 70.
6 consecutive numbers are 30, 40, 50, 60, 70, 80
30 + 80 = 40 + 70 = 50 + 60
40 is ten more than 30, 70 is ten less than 80.
Similarly, 50 is twenty more than 30, and 60 is twenty less than 80.
7 consecutive numbers are 10, 20, 30, 40, 50, 60, 70
10 + 70 = 20 + 60 = 30 + 50 = 40 + 40
In each case, the numbers on the two sides decrease and increase by the same amount, so the sum does not change.

Question 2.
Write 4 other multiplication expressions that are equal to 36 × 24.
Answer:
36 × 24
18 × 48
72 × 12
9 × 96
144 × 6

Question 3.
Do the following mentally:
(a) 202 + 198
Answer:
202 + 198 = (202 – 2) + (198 + 2) = 400

(b) 515 + 485
Answer:
515 + 485 = (515 – 15) + (485 + 15) = 1000

(c) 394 + 306
Answer:
394 + 306 = (394 + 6) + (306 – 6) = 700

(d) 1204 + 1296
Answer:
1204 + 1296 = (1204 – 4) + (1296 + 4) = 2500

(e) 288 – 178
Answer:
288 – 178 = (288 + 12) – (178 + 12) = 110

(f) 5000 – 1424
Answer:
5000 – 1424 = (5000 – 1) – (1424 – 1) = 3576

(g) 5012 – 3212
Answer:
5012 – 3212 = (5012 – 12) – (3212 – 12) = 1800

(h) 6008 – 1458
Answer:
6008 – 1458 = (6008 – 8) – (1458 – 8) = 4550

Question 4.
For each division problem, write 2 equivalent division problems.
(a) 160 ÷ 40
Answer:
160 ÷ 40
= 80 ÷ 20
= 40 ÷ 10
= 4 ÷ 1

(b) 144 ÷ 12
Answer:
144 ÷ 12
= 288 ÷ 24
= 72 ÷ 6
= 36 ÷ 3

(c) 500 ÷ 150
Answer:
500 ÷ 150
= 1000 ÷ 300
= 100 ÷ 30
= 10 ÷ 3

(d) 750 ÷ 25
Answer:
750 ÷ 25
= 1500 ÷ 50
= 150 ÷ 5
= 300 ÷ 10

Class 6 Maths Chapter 8 Notes Kerala Syllabus Unchanging Relations

→ The sum of two numbers is the same as the sum of the numbers got by adding a number to one of them and subtracting the same number from the other.

→ The difference of two numbers is the same as the difference of the two numbers got by adding or subtracting the same number to each of them.

→ The product of two numbers is the same as the product of the numbers got by multiplying one of them by a number and dividing the other by the same number.

→ The quotient of two numbers is the same as the quotient of these two numbers multiplied or divided by the same number.

We have been learning many types of relationships involving numbers since the lower classes. For example, the sum of two consecutive natural numbers is always an odd number. The product of two odd numbers is always an odd number. We have understood such ideas. In this unit, we explain some other relationships that remain unchanged even when the numbers change.

Unchanging Sum
A rope is 10 metres long. From it, a 2-metre piece is cut off. If we add the cut piece and the remaining piece, we get 2 + 8 = 10.
Now, if the piece cut off is 3 metres, the remaining part is 7 metres: 3 + 7 = 10
If 6 metres is cut off, the remaining part is 4 metres: 6 + 4 = 10
Thus, when the cut length changes, the remaining length also changes, but the total remains 10.

Mayookhi has 100 rupees in her money box. She took 10 rupees from it and bought a pen. How much money is left? What if she had spent 20 rupees?
In the money problems in the textbook, even if the amount spent and the amount remaining change, the total always remains 100.
10 + 90 = 100, 20 + 80 = 100, 30 + 70 = 100, and so on.

Let us look at another example. We need to draw a rectangle with a perimeter of 20 cm. To draw a rectangle, we need the length of one side, right? What should it be?
We have learned that the perimeter of a rectangle is twice the sum of its width and height.
Here, since the perimeter is 20 cm, the sum of the width and height is 10 cm.
So, if we take the width as 6 cm, the height will be 10 – 6 = 4 cm.
With the same perimeter, we can draw many other rectangles too.
Width = 1 cm, height = 9 cm, Sum = 10 cm, Perimeter = 20 cm
Width = 2 cm, height = 8 cm, Sum = 10 cm, Perimeter = 20 cm
Here, the width and height of the rectangle may change, but the total remains the same.
What did we observe here?
When the perimeter remains the same, the more the width increases, the more the height decreases.
When one piece of the rope becomes longer, the other piece becomes shorter by the same amount.
When the amount of money spent increases, the amount remaining decreases by the same amount.

Now look at another situation.
A child put 64 manchadi seeds into a box. Now he has 98 seeds left in his hand. How many seeds did he have altogether?
It is 64 + 98.
If the number put into the box decreases by one, the number left increases by one.
If the number put into the box decreases by two, the number left increases by two.
So, 64 + 98 becomes 66 + 96, which is 162.
Here, when one number increases by two and the other decreases by two, the total remains the same.
Now, consider 128 + 99.
This is safe as 127 + 100, which is 227.

How can we write 97 + 203?
(97 + 3) + (203 – 3) = 100 + 200 = 300
That is, one number increases by a certain amount, and the other number decreases by the same amount.
We can say it like this:
The sum of two numbers = (one number + a certain amount) + (the other number – the same number)
Now look at these examples.
425 + 597 = 422 + 600 (3 subtracted, 3 added) = 1022
3249 + 894 = 3243 + 900 (6 subtracted, 6 added) = 4143
395 + 876 = 400 + 871 (5 added, 5 subtracted) = 1271

Look at the numbers 5, 10, 15, 20
If you add 5 and 20, you get 25 (the end of numbers).
If you add 10 and 15, you also get 25(the middle numbers).
Why are their sums equal? Because 5 + 20 = 10 + 15
In 5 + 20, the number 5 is five less than the 10 in 10 + 15, and the number 20 is five more than the 15 in 10 + 15.
So, when one number decreases by 5 and the other number increases by 5, the total doesn’t change.
Let’s extend the sequence 5, 10, 15, 20, 25, 30.
Here, also 15 + 30 = 20 + 25
Again, one number increases by 5, and the other number decreases by 5.
So, the sum remains the same.
It can also be explained like this:
When the first number and the fourth number are added, it is the same as adding the second number and its four times.
When the second number and the third number are added, it is the same as adding twice the first number and three times the first number.
Both results are equal to five times the same number.
Similarly, take 10, 15, 20, 25: 10 + 25 = 15 + 20.
Here, 10 increases to 15 and 25 decreases to 20.
So, the sum stays the same.

In the sequence 4, 8, 12, 16, 20, 24, if you take any four consecutive numbers, will the sum of the two end numbers equal the sum of the two middle numbers? Why?
Answer:
Yes.
4, 8, 12, 16
4 + 16 = 8 + 12 = 20
If you add 4 to 4, you get 8.
If you subtract 4 from 16, you get 12.
One number increases by 4, and the other decreases by 4.
So, the total does not change.

Unchanging Difference
The age difference between two people is a relationship that never changes.
If one person is 17 years old and the other is 25 years old, the age difference now is 25 – 17 = 8 years.
After 10 years?
Difference = (25 + 10) – (17 + 10) = 35 – 27 = 8.
After 3 years?
Difference = (25 + 3) – (17 + 3) = 28 – 20 = 8 again.
What do we observe here?
When finding the difference between two numbers, if we add or subtract the same number from both the first and the second number, the difference does not change.
Examples:
195 – 98 = (195 + 2) – (98 + 2)
= 197 – 100
= 97

212 – 172 = (212 – 12) – (172 – 12)
= 200 – 160
= 40

487 – 247 = (487 + 13) – (247 + 13)
= 500 – 260
= 240

5000 – 3674 = (5000 – 1) – (3674 – 1)
= 4999 – 3673
= 1326

Now we can do this mentally:
1. 704-304
704 – 304 = (704 – 4) – (304 – 4)
= 700 – 300
= 400

2. 598 – 128
598 – 128 = (598 + 2) – (128 + 2)
= 600 – 130
= 470

3. 312 – 142
312 – 142 = (312 – 12) – (142 – 12)
= 300 – 130
= 170

4. 325 – 135
325 – 135 = (325 – 25) – (135 – 25)
= 300 – 110
= 190

5. 1524 – 324
1524 – 324 = (1524 + 26) – (324 + 26)
= 1550 – 350
= 1200

6. 4000 – 1299
4000 – 1299 = (4000 + 1) – (1299 + 1)
= 4001 – 1300
= 2701

Even and Odd Numbers
The difference between two even numbers will always be an even number. Let’s see why?
When you subtract 8 × 2 from 12 × 2, you get 4 × 2. This is an even number.
In the same way, the difference between two odd numbers is always an even number.
Here is the reason.
An odd number can be written as an even number plus one.
So, 125 – 73 can be written as (125 + 1) – (73 + 1), this difference is an even number.
This is true for any pair of odd numbers.
325 – 293 = (325 + 1) – (293 + 1) = 326 – 294; this difference is an odd number.
Let us now check the difference of multiples of 5 in the same way.
125 – 85 = (25 × 5) – (17 × 5) = (25 – 17) × 5 = 8 × 5
The difference is also a multiple of 5.
Similarly, let us check whether the difference of multiples of 9 is also a multiple of 9.
Explanation: 81 and 63 are multiples of 9
81 = 9 × 9, 63 = 7 × 9
(9 × 9) – (7 × 9) = (9 – 7) × 9 = 2 × 9, the difference is multiple of 9.

Unchanging Product
There are 36 cups. We want to arrange them in rows and columns to make a rectangle.
Possible arrangements:
9 rows and 4 columns; 9 × 4 = 36
12 rows and 3 columns; 12 × 3 = 36
18 rows and 2 columns; 18 × 2 = 36
6 rows and 6 columns; 6 × 6 = 36
Look at these examples.
Even though the numbers have changed, the product has not changed.
18 × 2 = 36
9 × 4 = 36
Look at these. What change has happened to each number?
18 became its half, which is 9.
2 became its double, which is 4
Look at the next pair, 18 × 2 = 36; 6 × 6 = 36.
Here, 18 became on third, which is 6, and 2 became three times, which is 6.
Here, also did not change.
What do we understand from this?
When two numbers are multiplied, if one number is multiplied by a certain number and the other number is divided by the same number, the product remains unchanged.
Example: 24 × 6 = 144
When 24 is multiplied by 2, we get 48.
When 6 is divided by 2, we get 3.
48 × 3 = 144 (24 × 6 = 48 × 3)
When 24 is divided by 3, the result is 8; when 6 is multiplied by 3 is 18
24 × 6 = 8 × 18 = 144
Further, 24 is divided by 12, which is 2; 6 is multiplied by 12, which is 72.
24 × 6 = 2 × 72 = 144
The result can be written like this:
24 × 6 = 48 × 3 = 8 × 18 = 2 × 72
Similarly, let us try to write some multiplication expressions equal to 72 × 40
72 × 40
36 × 80
18 × 160
9 × 320

Write some other multiplication expressions equal to the multiplication given below.
(a) 48 × 32
(b) 60 × 20
(c) 72 × 48
Answer:
(a) 48 × 32 = 24 × 64
= 12 × 128
= 6 × 256
= 3 × 512

(b) 60 × 20 = 30 × 40
= 15 × 80
= 120 × 10
= 240 × 5

(c) 72 × 48 = 36 × 96
= 18 × 192
= 9 × 384
= 144 × 24

1 × 216 and 6 × 36 are equal. Why?
Answer:
Multiplying 1 by 6 gives 6.
Dividing 216 by 6 gives 36.
If you multiply one number in 1 × 216 by 6 and divide the other number by 6, their product will still be the same as 1 × 216.

Answer:
20 × 16 = 40 × 8
42 × 30 = 7 × 180
30 × 15 = 10 × 45
55 × 3 = 11 × 15

Unchanging Quotient
If you put 5 seeds in each of 10 covers, how many seeds are needed?
5 × 10 = 50 seeds
If you put 10 seeds in each of 10 covers, how many seeds are needed?
10 × 10 = 100 seeds
If you put 20 seeds in each of 10 covers, how many seeds are needed?
20 × 10 = 200 seeds
In all these examples, there are 3 quantities: the number of seeds in one packet, the number of covers, and the total number of seeds.
Here, the number of seeds in one packet and the total number of seeds change.
The number of covers does not change.
The total number of seeds divided by the number of seeds in one packet gives the number of covers.
50 ÷ 5 = 10, 100 ÷ 10 = 10, 200 ÷ 20 = 10
Similarly, we can see the same idea in the pen problems in the textbook.
The number of pens and the total cost change.
The price per pen does not change.
40 ÷ 8 = 5, 80 ÷ 16 = 5, 50 ÷ 10 = 5, 150 ÷ 30 = 5
In the same way, in square problems too, when both numbers change in the same multiple, the quotient remains the same.
Eg: 25 ÷ 5 = 5, 50 ÷ 10 = 5, 100 ÷ 20 = 5
We also know that when we divide both numbers by the same number, the quotient does not change.
40 ÷ 10 = 4, 4 ÷ 1 = 4
Here, both numbers are divided by 10; the quotient remains the same.
This is like cancelling common factors.
40 ÷ 10 = 4 × 10 ÷ 1 × 10 = 4
75 ÷ 25 = (5 × 15) ÷ (5 × 5)
If we cancel the common factor: 15 ÷ 5
Again, cancelling the common factor: 15 ÷ 5 = 3 × 5 ÷ 1 × 5 = 3
Thus, the quotient of two numbers and the quotient of their equal multiples or equal parts will always be the same.
Eg: 16 ÷ 4 = 4
If we take the multiples of 16 and 4,
(16 × 5) ÷ (4 × 5) = 80 ÷ 20 = 4
If we take the same part of 16 and 4,
(16 ÷ 2) ÷ (4 ÷ 2) = 8 ÷ 2 = 4, quotient does not change.

Eg: If 175 is divided by 5?
175 ÷ 5 = 350 ÷ 10 = 35

Eg: 495 ÷ 15
495 ÷ 15 = 990 ÷ 30
= 99 ÷ 3
= 33

Match the ones with the same quotient.

Answer:
14 ÷ 2 = 28 ÷ 4
35 ÷ 7 = 350 ÷ 70
150 ÷ 30 = 15 ÷ 3
240 ÷ 8 = 120 ÷ 4
200 ÷ 10 = 100 ÷ 5

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 9 Lines and Angles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 9 Solutions Lines and Angles

Class 6 Kerala Syllabus Maths Solutions Chapter 9 Lines and Angles Questions and Answers

Lines and Angles Class 6 Questions and Answers Kerala Syllabus

When Lines Meet

In-text Questions (Page Number 130)

Question 1.
Some pictures of two lines that meet are shown below. The measure of one of the two angles made on either side is also given. Find the measure of the other angle:

Answer:
1. 100° – 80°
2. 45° – 135°
3. 30° – 150°
4. 80° – 100°
From this, we can understand that a line drawn from another line, or two Lines meet, the sum of the angles made thus will be 180°.
The sum of the two angles made when two lines meet is 180°.

Question 2.
From the following figure, what is the measurement of the third angle?

Answer:
Since the sum of three angles is 180°.
Here, the sum of two angles is 85° + 150 = 100°
Therefore, the third angle is 180° – 100 = 80°

Question 3.
Now look at the angles in the following figure.

Answer:
Sum of the three angles = 70° + 20° + 40° = 130°
So the fourth angle is 180° – 130° = 50°

When Lines Meet (Page Numbers 131-132)

Question 1.
In the picture below, what is the angle on the left?

Answer:
The sum of two angles = 50° + 25° = 15°
The third angle = 180° – 15° = 105°

Question 2.
In the picture below, what is the measure of the angle between the lines?

Answer:
Let’s check the corners of the set-squares.
From the corners which are joined, one is 45°, and the other is 60°.
So, adding these two angles, we get 60° + 45° = 105°
Therefore the middle angle = 180° – 105° = 75°

Question 3.
Among the three angles marked in the picture below, the angles on the left and right are equal. How much is each?

Answer:
As one angle is 30°, the sum of the angles of the other two angles will be 180° – 30° = 150°
Since those two angles are equal, one angle will be \(\frac {150}{2}\) = 75°

Question 4.
Of the two angles made when two lines meet, one angle is double the other. What are the measures of the two angles?
Answer:
We need to find two numbers whose sum is 180, and one number is double the other.
If those numbers are the same, then they are 90° each.
If we subtract 30° from one and add 30° from the other, we get
90° – 30° = 60°
90° + 30° = 120°
When they are added together, we get 180°.

When Lines Intersect (Page Number 134)

Question 1.
The pictures below show various pairs of intersecting lines. One of the four angles thus made is given in each picture. Calculate the other three angles and mark them in each picture:

Answer:
1. 45°, 135°, 45°, 135°
2. 120°, 60°, 120°, 60°
3. 90° each

(Page Number 136-137)

Question 1.
In the four angles made by two intersecting lines, the sum of the two angles is 100°. Calculate all four angles.
Answer:
As the sum of two angles is 100, they are opposite angles (if they are adjacent angles, their sum is 180).
As the opposite angles are equal, each of them is 50°. So one of its adjacent angles is 130°.
So the pair of adjacent angles of another angle is also 130°.
So the measurement of four angles is 50°, 130°, 50°, 130°.

Question 2.
In the four angles made by two intersecting lines, one angle is half of another. Calculate all four angles.
Answer:
Since one angle is half of another, the two angles must be adjacent (otherwise, vertically opposite angles would be equal).
The sum of adjacent angles on a straight line is 180°.
Given that one angle is half of the other, the angles must be 120° and 60°.
Therefore, the vertically opposite angles are also 120° and 60°.
So, the four angles are 120°, 60°, 120°, and 60°.

Question 3.
The two pictures below show lines passing through a point. The measures of some of the angles are given in each. Calculate the other marked angles in each:

Answer:

Non Intersecting Lines & Line and Slant (Page Number 144)

Question 1.
Find the pairs of parallel lines in the picture below and write them down:

Answer:
Similar pairs
1. A, F (F is 70° and the other angle is 110°)
2. B, D (F is 100° and the other angle is 80°)
3. C, E (Both are perpendicular)

Parallelograms (Page Number 146)

Question 1.
Draw the parallelograms shown below:

Answer:
1. Draw a line of 4 cm.
Draw angles of 70° and 110° on each side.
Mark a point at 3cm on each of these lines.
Join these points with a line.

2. Draw a line of 5 cm.
Draw an angle of 60° from the left side and 120° on the right side.
Mark a point at 3cm on each of these lines.
Join these points with a line.

3. Draw a line of 4 cm.
Draw an angle of 50° from the left side and 130° on the right side.
Mark a point at 2cm on each of these lines.
Join these points with a line.

4. Draw a line of 3 cm.
Draw an angle of 40° from the left side and 140° on the right side.
Mark a point at 3cm on each of these lines.
Join these points with a line.
This is a parallelogram with equal sides.

Class 6 Maths Chapter 9 Kerala Syllabus Lines and Angles Questions and Answers

Class 6 Maths Lines and Angles Questions and Answers

Question 1.
In each of the following figure give the angle marked as ‘?’

Answer:
(a) 140°
(b) 50°
(c) 60°

Question 2.
Write all the angles in the picture given.

Answer:
(a) (a) 60°, (b) 120°, (c) 60°
(b) (a) 50°, (b) 130°, (c) 50°
(c) (a) 70°, (b) 40°, (c) 70°, (d) 70°
(d) (a) 60°, (b) 30°, (c) 90°, (d) 60°

Question 3.
When two angles intersect each other to make an angle of 90°, what will be the other angles?
Answer:
90° each

Question 4.
In the figure, one angle is a right angle and the other two are equal.

Answer:
Angle 1 = 45°
Angle 2 = 45°

Question 5.
In the figure given, which of the angles are the same? Which of the angle having sum of 180°?

Answer:
Equal angles are a and c; b and d
Having sum 180° = a and b; c and d; a and d; b and c.

Question 6.
Draw a line of 8 cm and draw a parallel line of it at a distance of 3 cm.
Answer:

Draw a line of 8 cm. Place a scale perpendicularly and set a square on the line.
Mark a point 3cm away from the line.
Move the set square to the point marked and draw a line along the point.

Question 7.
Draw a line and draw another line having an angle of 70°. Draw a parallel line with the second line.
Answer:

Draw a line a mark a point.
Draw an angle of 70° from this point.
Similarly, mark a point on the left side.
Draw an angle of 70° from this point.

Question 8.
Draw a parallelogram of 6 cm and another side of 4 cm and having an angle of 45°.
Answer:

Draw a line of 6 cm.
Draw an angle of 45° on one side and 135° on another side.
Mark a point on 4 cm of each line.
Join the points with a line.

Question 9.
In this picture, which of them are opposite angles?

Answer:
1 and 3
2 and 4
If angle 1 is 110°, then angle 3 is also 110°.
Then the angle 2 is 70°, so angle 4 is also 70°.

In the four angles made by two intersecting lines,

• The sum of each pair of adjacent angles is 180°.
• Each pair of opposite angles is equal.

Question 10.
Look at the angles given in the picture. Which are adjacent angles and which are opposite angles?

If angle a = 100°, find the measurement of the other angles.
Answer:
Adjacent angles: a, b; b, c; c, d; a, d
Opposite angles: a, c; b, d
Angle a = 100°
Angle b = 80°
Angle c = 100°
Angle d = 80°

Question 11.
Which are equal pairs in the given figure? Which pairs have a sum of 180°?

Answer:
Equal pairs: 1 and 3; 2 and 4
Pairs having a sum of 180°:

• 1 and 2
• 2 and 3
• 3 and 4
• 1 and 4

Question 12.
From the following, find which of them are parallel lines?

Answer:
1. 2, 9
2. 3, 6
3. 4, 7
4. 5, 8

Question 13.
Draw the picture given below in your notebook, and draw parallel lines to the slanted lines.
Answer:

Question 14.
In the following figures, there is a measurement of one angle. Find the other angles without measuring.

Answer:

(A) 140°, 40°, 140°
(B) Each 90°
(C) 100°, 80°, 100°

Class 6 Maths Chapter 9 Notes Kerala Syllabus Lines and Angles

→ The sum of the two angles made when two lines meet is 180°

→ When two lines intersect, four angles are formed. If the lines are perpendicular to each other, all four are 90°.

→ If the lines are slanted, there are two small angles and two large angles; in these,

• Both the smaller angles have the same measure.
• Both the larger angles have the same measure.
• The sum of a small angle and a large angle is 180°.

→ In the four angles made by two intersecting lines,

• The sum of each pair of adjacent angles is 180°
• Each pair of opposite angles is equal

→ Lines that keep the same distance apart and will never meet are called parallel lines.

→ A quadrilateral with each pair of opposite sides parallel is called a parallelogram.

→ Any four-sided figure is called a quadrilateral.

In this chapter, we learn more about the lines and the shapes that are made by lines and angles. When these lines intersect, we get different shapes. We also learn about the lines that do not intersect and their peculiarities.

When Lines Meet
You have learned how to draw angles and measure them. Try to measure them.

How many degrees are they each?
A = ___________
B = ___________
C = ___________
D = ___________
Look at the angles given below.

The first angle is 90°, which is a right-angle triangle. In this type of angles two lines are always vertical.
The second angle is less than the right angle, which means its angle is less than 90°.
In this type angles the second line slants inside.
The third angle is greater than the right angle, which means its angle is greater than 90°.
In this type angles the second line slants outwards.
Now, let’s check some other points regarding angles.
We know that the angles are made by drawing a line and then drawing another line from one of its ends.

These types of angles are of three types.
1. If we draw a line and the second line is drawn straight up, we get a right angle.

2. Now, what if the second line is drawn slanted towards the left side?
The angle on the left side becomes smaller than 90°, and the angle on the right side becomes larger than 90°.
The amount by which the angle on the left side is less than 90° is the same amount by which the angle on the right side is greater than 90°.
That is, the sum of those two angles is 180°.

3. Now, what if the second line is drawn slanted towards the right side?
The angle on the right side becomes smaller than 90°, and the angle on the left side becomes larger than 90°.
The amount by which the angle on the right side is less than 90° is the same amount by which the angle on the left side is greater than 90°.
That is, the sum of those two angles is 180°.

So, in these types of angles, if we know one angle, we can find the other.
Look at the picture.

Since the smaller angle is 50°, the other angle will be 130°. Because when the two angles are added, their sum is 180°.
That means the smaller angle is 40° less than 90° and the bigger angle is 40° greater than 90°.

One angle is given; find the other one of the following angles.

Answer:

1. 160°
2. 90°
3. 120°
4. 145°
5. 20°
6. 50°

When Lines Intersect
Now let’s check the peculiarities of the angles formed by the lines that intersect.
When a line intersects another line, four angles are formed.

In this, if we add the first and second angle we get 180°.
Like this, which of the pairs of angles will have a sum of 180°?

1. 2 & 3
2. 3 & 4
3. 1 & 4

Let’s check the same thing with the measurement of angles.

Here, if one angle is 120°, then the adjacent angle will be 60°.
Like this, in these types of angles, two angles will have the same measurements.
Then what will be the measurements of the other angles of the following figure?
The angle on the right side of 30° is 150°.
The opposite angle of 30° is also 30°, and the bottom of 30° is 150°.

What if the line intersecting is straight downwards?

Then all the angles formed here will be 90°. So in general,
When two lines intersect, four angles are formed. If the lines are perpendicular to each other. All four are 90° angles.
If the lines are slanted, there are two small angles and two large angles; in these,

• Both the smaller angles have the same measure
• Both the larger angles have the same measure
• The sum of a small angle and a large angle is 180°

Angles that are next to each other are called adjacent angles. When two lines intersect, they form four angles, which can be paired in different ways.
Some of these pairs have the same measure. In each such pair, the angles are next to each other.
Look at the picture given below.

When two lines intersect like this, we can find one thing from this. One angle and its opposite angle are always the same.

Non-Intersecting Lines
These lines won’t meet however much we extend them, in any direction, because the distance between them is equal everywhere.
Lines like these, which keep the same distance apart and will never meet, are called parallel lines.

To draw a line parallel to another line, first draw a line.
Then draw perpendicular lines of equal length from both ends of the line, and join the endpoints of these perpendicular lines.
This new line will be parallel to the original line.
Look at this picture. Draw a line first.
Then, draw two perpendicular lines from both ends of the line using set squares or use a protractor to draw 90 ° angles.
Now mark the same distance on both the perpendicular lines.
Join the points marked on both the perpendicular lines.

To avoid lots of unwanted lines for drawing a neat parallel line like this, we can use setsquares and a scale.

First, draw a line. Now, place a scale perpendicular to the line firmly. Place a set square close to the line and scale.
Now move the set square by placing the scale firmly and draw a line above the line from the tip of the set square.
Now, if we draw a line, and we want to draw a parallel line that is 4 cm away from this line.

Draw a line first. Place a scale perpendicular to this line firmly and place a setsquare on the line which is close to the scale.
Now move the set square in the upward direction till the measurement reaches 4cm on the scale.
Now remove the scale and draw the line starting from the tip of the setsquare.
Now, what if we draw lots of perpendicular lines in a line?

None of these perpendicular line won’t meet if this extended.
Now we can understand that any perpendicular lines drawn in a line are parallel lines.

Line and Slant
Now draw a line that is perpendicular to and slanted to a line. The lines drawn now are not parallel.
We need to draw a line that is not perpendicular, but a line that is parallel to the line.

A line is drawn 60° slanted from a line. If we want to draw a line parallel to this line, we should draw another line with the same slant.
Since two lines have the same slant, they are parallel to each other.

From this we can understand that the lines which having same slant in a line are parallel to each other.

Method of Drawing
Draw a line and make an angle in a place. Draw another angle in some other place.

Now draw this picture in your notebook.
Draw a parallel line for the slanted line.
Since the angle of the slanted line is 80°, draw a line from a certain distance from this line, which has an angle of 80°, and they will be parallel to each other.

The angle on the right side is 100°, since the adjacent pair is 80° (since they are adjacent angles, and their sum will be 180°).

Parallelograms
Any four-sided figure is called a quadrilateral.
The following figures are quadrilaterals.

All of them have four sides.
1. Opposite sides are equal; it is a rectangle
2. All sides are equal; it is a square
Look at the 5th figure.
Their opposite sides are equal and parallel.
Since their corners are not straight, they are not a rectangle.

A quadrilateral with each pair of opposite sides parallel is called a parallelogram.
If two parallel lines are drawn as shown in this figure, we get a parallelogram.

We can draw a parallelogram using a protractor.

• Draw a line of 5 cm.
• Draw an angle of 50° on the left side.
• Mark a point on 3cm of this line.
• Draw an angle of 130° on the other end of the bottom line.
• Mark a point on 3cm of this line.
• Join the two points on these lines.

By changing the length of the sides and angles, we can draw this in different ways.

Draw 90° angles in between the sides and check what we get?

Answer:
Now the angles are straight and parallel, we call this a rectangle.