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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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• Kerala Syllabus Class 10 Physics Model Question Paper Set 3
• Kerala Syllabus Class 10 Physics Model Question Paper Set 4
• Kerala Syllabus Class 10 Physics Model Question Paper Set 5

Kerala SSLC Physics Model Question Papers in Malayalam Medium

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

Old Syllabus

• Kerala SSLC Physics Question Paper March 2024 English Medium
• Kerala SSLC Physics Question Paper March 2023 English Medium
• Kerala SSLC Physics Question Paper March 2022 English Medium
• Kerala SSLC Physics Board Model Paper March 2024 English Medium
• Kerala SSLC Physics Board Model Paper March 2023 English Medium
• Kerala SSLC Physics Board Model Paper March 2022 English Medium
• Kerala SSLC Physics Question Paper March 2024 Malayalam Medium
• Kerala SSLC Physics Question Paper March 2023 Malayalam Medium
• Kerala SSLC Physics Question Paper March 2022 Malayalam Medium
• Kerala SSLC Physics Question Paper March 2021 Malayalam Medium
• Kerala SSLC Physics Question Paper March 2020 Malayalam Medium
• Kerala SSLC Physics Board Model Paper March 2024 Malayalam Medium
• Kerala SSLC Physics Board Model Paper March 2021 Malayalam Medium
• Kerala SSLC Physics Board Model Paper March 2020 Malayalam Medium

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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.

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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.

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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.