Class 5

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 12 Tables and Graphs Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 12 Solutions Tables and Graphs

Class 5 Maths Chapter 12 Tables and Graphs Questions and Answers Kerala State Syllabus

Tables and Graphs Class 5 Questions and Answers Kerala Syllabus

Question 1.
The population of the various districts of our state, according to the 2011 census, is given in the table below:

Rewrite the table with populations in the ascending order.
Answer:

District	Population
Wayanad	817420
Idukki	1108974
Pathanamthitta	1197412
Kasaragod	1307375
Kottayam	1974551
Alappuzha	2127789
Kannur	2523003
Kollam	2635375
Palakkad	2809934
Kozhikode	3086203
Thrissur	3121200
Ernakulam	3282388
Thiruvananthapuram	3301427
Malappuram	4112920

Question 2.
Calculate the total runs secured in each over and shorten the table like this.

Calculate the total runs secured in each over and shorten the table like this.

i) How many runs were scored in the first ten overs?
ii) In how many overs were more than 6 runs scored?
iii) In how many overs were less than 6 runs scored?
iv) What other information can we get from this shortened table?
Answer:
Over 1: 2 + 0 + 0 + 2 + 0 + 0 = 4
Over 2: 1 + 4 + 0 + 2 + 0 + 0 = 7
Over 3: 0 + 0 + 4 + 0 + 1 + 1 = 6
Over 4: 4 + 0 + 2 + 0 + 1 + 0 = 7
Over 5: 0 + 0 + 4 + 1 + 0 + 3 = 8
Over 6: 2 + 0 + 2 + 2 + 0 + 0 = 6
Over 7: 1 + 0 + 4 + 0 + 2 + 1 = 8
Over 8: 2 + 0 + 2 + 0 + 3 + 0 = 7
Over 9: 0 + 0 + 6 + 2 + 2 + 0 = 10
Over 10: 6 + 0 + 0 + 4 + 0 + 2 = 12

i) The total runs scored = 4 + 7 + 6 + 7 + 8 + 6 + 8 + 7 + 10 + 2 = 75
So, the total runs scored in the first ten overs is 75.

ii) Over 2-7 runs, Over 4 -7 runs, Over 5 – 8 runs, Over 7 – 8 runs, Over 8 – 7 runs, Over 9 – 10 runs, Over 10 – 12 runs
There are 7 overs with more than 6 runs scored.

iii) Overs with less than 6 Runs Scored Over 1-4 runs
iv) There is 1 over with less than 6 runs scored.
Other information We Can Get from the Shortened Table,

• Average runs per over.
• Identification of the highest-scoring over (Over 10, with 12 runs).
• Consistency of the scoring rate (fluctuation of runs over the overs).

Question 3.
The graph below shows the number of cars a company manufactured in the past few years:

i) In which year was the most number of cars made? How many cars were manufactured?
ii) And in which year was the least number of cars manufactured? How many cars were manufactured?
iii) How many cars less than those made in 2018 were made in 2023?
Answer:
i) The year with the most cars manufactured is 2019
No of cars made =14,500 cars

ii) The year with the least cars manufactured is 2020
No of cars made = 10,500 cars

iii) No of car made in 2018 = 13,000
No of car made in 2023 = 12,500

The difference is 13,000 – 12,500 = 500, which means 500 less cars were made in 2023 compared to 2018.

Question 4.
The graph below shows the units’of electricity consumed during various months of a year:

i) In which months was the maximum electricity used?
ii) In which months was it the minimum?
iii) What is the difference between the maximum and minimum?
iv) Are there any months in which the same units of electricity was used?
v) What is the total number of units used in this whole year?
Answer:
i) The maximum electricity consumption occurred in March and April, each.
ii) The minimum electricity consumption occurred in September, October, November, and December
iii) Maximum unit of electricity consumed =340 Minimum unit of electricity consumed =290
The difference between the maximum and minimum = 340 – 290 = 50 units
iv) Yes, September,October,November and December.
v) 310 + 340 + 330 + 300 + 290 + 290 = 1860 units.
The total electricity used in the whole year =1860 × 2 = 3720 unit

Question 5.
The graph below shows the number of girls and boys in classes 1-7 in a school:

Answer:
i) Class 3, Difference = 75 -70 = 5 ;
ii) Difference between the number of girls and boys the most = Class 1
Difference between the number of girls and boys the least = Class 2, 3 and 6
iii) Total number of girls in all these classes = 65 + 65 + 70 + 70 + 100 + 115 + 95 = 580
iv) Total number of students = 1100

Intext Questions And Answers

Question 1.
All the information we get from the table can be found in the picture also. Moreover, we can see the ups and downs of scores at one glance. Such a picture is called a bar graph.

(i) In which over was the most runs scored?
Answer:
11th Over

(ii) How many runs were scored?
Answer:
19 runs

(iii) In which over was the least runs scored?
Answer:
13th,14th,15th

(iv) How many runs were scored?
Answer:
3

(v) In how many overs were more than 10 runs scored?
Answer:
5 Overs

(vi) Which are those overs?
Answer:
4th, 8th, 9th, 10th, 11th

(vii) In how many overs were less than 6 runs scored?
Answer:
6 overs

(viii) Which are those overs?
Answer:
12th, 13th, 14th ,15th, 16th,17th

Question 2.
We can also compare the batting performance of the teams by drawing the graphs in different colours and combining them

Can you answer these questions using the above graph?
(i) In which overs did India score more runs than England?
Answer:
3, 5, 6, 8, 9, 10, 11, 19 Overs

(ii) In which overs did both teams score the same number of runs?
Answer:
Over 2: Both teams scored 9 runs.
Over 7: Both teams scored 6 runs.
Over 12: Both teams scored 5 runs.
Over 14: Both teams scored 3 runs.
Over 16: Both teams scored 5 runs.

(iii) Which team scored more at the end of the first ten overs?
Answer:
England’s runs in the first 10 overs: 13 + 9 + 8 + 16 + 6 + 3 + 6 + 7 + 6 + 5 = 79
India’s runs in the first 10 overs: 8 + 9 + 9 + 12 + 10 + 6 + 6 + 16 + 13 + 14 = 103
India scored more at the end of the first ten overs, with a total of 103 runs compared to England’s 79 runs.

Class 5 Maths Chapter 12 Kerala Syllabus Tables and Graphs Questions and Answers

Question 1.
The picture below shows the number of children in a class who got various grades m mathematics in the half-yearly examination.

Complete the table using the picture

Grades	Number of students
A	8
B	________
C	9
D	________
E	________

Answer:

Grades	Number of students
A	8
B	10
C	9
D	5
E	1

Question 2.
The bar graph below represents the books read by students of class 5.

Answer the following questions.
a) The students of which class read less books.
b) Total how many books read by 5th standard students all together.
c) Which divisions students together read 50 books.
Answer:
a) 5 B

b) 5 A = 10
5 B = 5
5C = 20
5D = 30
Total = 65

c) 5C = 20
5D = 30
Total = 50

Question 3.
Rewrite the table in the ascending order.

DISTRICT	POPULATION
Kottayam	1974551
Alappuzha	2127789
Pathanamthita	1197412
Kollam	2635375
Thiruvananthapuram	3301427

Answer:

DISTRICT	POPULATION
Pathanamthita	1197412
Kottayam	1974551
Alappuzha	2127789
Kollam	2635375
Thiruvananthapuram	3301427

Class 5 Maths Chapter 12 Notes Kerala Syllabus Tables and Graphs

In this chapter, we will learn how to understand and use tables and graphs to show information clearly. Here are the main ideas:

• Learn how to look at tables and find answers by comparing information.
• Understand how to show data using pictures like charts, graphs, and diagrams.
• See how bar graphs use bars to compare different groups or amounts.

By the end, you will be able to read tables and create simple graphs to show information in a clear way.

Tables
The given table provides a general overview of Class 5’s performance in the math exam by grouping students according to their grades.

While it doesn’t offer the detailed explanation found in the larger table, which would list each student’s individual grade, it still allows us to understand important aspects of the class performance. The general information that we get from this table is as follows,

• B is the grade that most students got.
• E is the grade that the least number of students got.
• The number of students who got A or B is more than the number of students who got C, D or E.
• The number of students who got A, B or C is four times the number of students who got D or E.

The table confirms there are 35 students in total, which is the basis for these percentages and observations.

We can combine all three tables into one


The information that we get from the completed table,

This table helps in assessing and comparing the performance across different grades and identifying which students might need more support or attention.

• It also highlights the general trend of student performance, which seems strong in higher grades, with a few students who may benefit from additional resources.
• Grade A is most common in Class 7, with 12 students achieving it. Class 6 follows closely with 10 students, and Class 5 has 9 students.
• Grade B is quite consistent across the three classes, though Class 6 has the most students (12) with this grade.
• The lowest grades, D and E, are generally less frequent, indicating fewer students in all classes are struggling significantly.

Data Pictures
A “Data Picture” is a visual representation of data that helps to communicate information in an easily understandable way. Data pictures can include various types of charts, graphs, and diagrams that display numerical or categorical data visually.

Common examples of data pictures include bar charts, pie charts, line graphs, histograms, scatter plots, and infographics.

A bar graph, is a visual representation of data that uses rectangular bars to show the frequency, count,- or other measures for different categories or groups. Each bar’s length or height is proportional to the value it represents, making it easy to compare data across categories. ,
England batted first in a T20 match against India. The runs they scored in each over is given in this table :

• Along a horizontal line at the bottom, numbers from 1 to 20 are marked at the same distance apart. This is to show the number of overs.
• Along a vertical line on the left, numbers from 1 to 21 are marked at the same distance apart. This is to show the number of runs scored.
• Then at the position showing each over, rectangles are drawn with heights up to the number on the vertical line showing the number of runs scored in that over. All these rectangles are of the same width.

⇒ A “Data Picture” is a visual representation of data that helps to communicate information in an easily understandable way.

⇒ Data pictures can include various types of charts, graphs, and diagrams that display numerical or categorical data visually.

⇒ Bar graph, is a visual representation of data that uses rectangular bars to show the frequency, count, or other measures for different categories or groups.

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 11 Within Numbers Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 11 Solutions Within Numbers

Class 5 Maths Chapter 11 Within Numbers Questions and Answers Kerala State Syllabus

Within Numbers Class 5 Questions and Answers Kerala Syllabus

Question 1.
Now in each pair of numbers given below, check whether the first is a multiple of the second:
i) 7,91
Answer:

This shows 91 = 13 × 7
That is, 91 is a multiple of 7.

ii) 9, 127
Answer:

This shows 127 = (14 × 9) + 1
That is, 91 is not a multiple of 7.

iii) 12, 136
Answer:

This shows
136 = (11 × 12) + 4
That is, 136 is not a multiple of 12.

iv) 15, 225
Answer:

This shows
225 = 15 × 15
That is, 225 is a multiple of 15.

Question 2.
Now write the relation between each pair of numbers below like this, as multiplication-multiple and division-factor.
i) 9,72
Answer:

ii) 12, 156
Answer:

iii) 13, 169
Answer:

iv) 25, 375
Answer:

Question 3.
See if you can mentally calculate the products below:
i) 12 × 25
Answer:
12 × 25 = 4 × 3 × 25
= 3 × (4 × 25)
= 3 × 100
= 300

ii) 35 × 18
Answer:
35 × 18 = 35 × 9 × 2
= 9 × (2 × 35)
= 9 × 70
= 630

iii) 125 × 8
Answer:
125 × 8 = 125 × 2 × 2 × 2
= 250 × 2 × 2
= 500 × 2
= 1000

iv) 125 × 24
Answer:
125 × 24 = 125 × 4 × 6
= 500 × 6
= 3000

Question 4.
Now try these problems:
i) 90 ÷ 15
Answer:
Writing 15 as the product of factors, that is 15 = 3 × 5
Now, 90 ÷ 3 = 30
Then, 30 ÷ 5 = 6
That is, 90 ÷ 15 = 6

ii) 900 ÷ 18
Answer:
Writing 18 as the product of factors, that is 15 = 3 × 6
Now, 900 ÷ 3 = 300
Then, 300 ÷ 6 = 50
That is, 900 ÷ 18 = 50

iii) 160 ÷ 32
Answer:
Writing 32 as the product of factors, that is 32 = 4 × 8
Now, 160 ÷ 4 = 40
Then, 40 ÷ 8 = 5
That is, 160 ÷ 32 = 5

iv) 168 ÷ 24
Answer:
Writing 24 as the product of factors, that is 24 = 3 × 8
Now, 168 ÷ 3 = 56
Then, 56 ÷ 8 = 7
That is, 168 ÷ 24 = 7

v) 210 ÷ 42
Answer:
Writing 42 as the product of factors, that is 24 = 6 × 7
Now, 210 ÷ 7 = 30
Then, 30 ÷ 6 = 5
That is, 210 ÷ 42 = 5

Question 5.
Now find the quotients below by removing common factors:
i) 600 ÷ 150
Answer:
600 = 4 × 2 × 3 × 25
150 = 2 × 3 × 25
Common factors: 2, 3, 25
Thus, 600 ÷ 150 = 4

ii) 900 ÷ 180
Answer:
900 = 4 × 9 × 5 × 5
180 = 4 × 9 × 5
Common factors: 4, 9, 5
Thus, 900 ÷ 180 = 5

iii) 225 ÷ 75
Answer:
225 = 3 × 3 × 25
75 = 3 × 25
Common factors: 3, 25
Thus, 225 ÷ 75 = 3

iv) 420 ÷ 105
Answer:
420 = 4 × 3 × 5 × 7
105 = 3 × 5 × 7
Common factors: 3, 5, 7
Thus, 420 ÷ 105 = 4

Intext Questions And Answers

Question 1.
Now let’s see how we can check the number is multiple or not.
(i) Is 215 the multiple of 5?
Answer:
Let’s check this by dividing 215 by 5, that is,

This shows
215 = 43 × 5
Thus, 215 is indeed a multiple of 5.

(ii) Let’s divide 168 by 5,

This shows that
168 = (33 × 5) + 3
That is, 168 is larger than 33 × 5 and smaller than 34 × 5.
Thus, 168 is not a multiple of 5.
In general, We can say
If a number is divisible by another number, the first number is a multiple of the second number; if there is a remainder on division, it is not a multiple.

Class 5 Maths Chapter 11 Kerala Syllabus Within Numbers Questions and Answers

Question 1.
Now. in each pair of numbers given below, check whether the first is a multiple of the second:
i) 15, 120
Answer:
This shows
120 = 15 × 8
That is, 120 is a multiple of 15.

ii) 7,50
Answer:
This shows
50 = (7 × 7) + 1
That is, 50 is not a multiple of 7.

iii) 12, 160
Answer:
This shows
160 = (12 × 13) + 4
That is, 160 is not a multiple of 12.

Question 2.
Now write the relation between each pair of numbers below like this, as multiplication-multiple and division-factor.
i) 20, 240
Answer:

ii) 15, 120
Answer:

Question 3.
Now find the quotients below by removing common factors:
i) 630 ÷ 126
Answer:
630 = 2 × 9 × 5 × 7
126 = 2 × 9 × 7
Common factors: 2, 9, 7
Thus, 630 ÷ 126 = 5

ii) 420 ÷ 105
Answer:
420 = 4 × 3 × 5 × 7
105 = 3 × 5 × 7
Common factors: 3, 5, 7
Thus, . 420 ÷ 105 =4

iii) 300 ÷ 75
Answer:
300 = 4 × 3 × 25
75 = 3 × 25
Common factors: 3, 25
Thus, 300 ÷ 75 = 4

Class 5 Maths Chapter 11 Notes Kerala Syllabus Within Numbers

In this chapter, we will explore the exciting world of numbers, focusing on how they relate to each other through multiplication and division. Understanding multiples and factors helps us uncover the hidden patterns within numbers, which makes problem-solving easier and more fun! Let’s dive into the concepts that form the foundation of this topic.

• Multiples of a natural number are the results we get by multiplying that number by 1,2, 3, and so on.
• If a number can be divided exactly by another number (with no remainder), then the first number is a multiple of the second. If there’s a remainder, it’s not a multiple.
• Factors of a number are the numbers that divide it exactly, leaving no remainder.
• If two numbers have the same factors, they are called common factors of those two numbers.

This chapter will help you better understand how numbers relate to each other through these concepts, making it easier to work with large numbers and solve problems efficiently. Let’s start discovering!

Multiplication And Multiples
Let’s take a closer look at the numbers in the table.

First rows and columns: Natural numbers
Second rows and columns: Twice the natural numbers.
In the language of mathematics, we get the numbers by multiplying the numbers 1, 2, 3, ……… by 2.
In short, this is multiples of 2.
Third rows and columns: Thrice the natural numbers In short, this is multiples of 3.

In general,
By the multiples of a natural number, we mean the numbers got by multiplying 1, 2, 3, ……. by that number.
Therefore, we can conclude that in the table…
numbers in the 1st row are multiples of 1, i.e. natural numbers, numbers in the 2nd row are multiples of 2. numbers in the 3 rd row are multiples of 3. numbers in the 4th row are multiples of 4.

Division And Factors
We can check whether one number is a multiple of another by division.
For example,
84 ÷ 6 = 14
84 is a multiple of 6.
That is,
6 is a factor of 84.

In general,
By the factors of a number, we mean those numbers by which this number is divisible.
Let’s now explore the relationship between numbers, expressed through multiplication and division, in terms of multiples and factors.

Factors In Multiplication And Division
Splitting a number into products of factors will make some computations easier. For example,
Calculate 14 × 26
Writing 14 = 7 × 2 as a product of factors.
Thus,
14 × 26 = 7 × 2 × 26
= 7 × (2 × 26)
= 7 × 52
= 364

Now, let’s look at an example of how using factors can make division easier.
120 chocolates are to be shared equally among 15 children. How many chocolates will each child get?
Instead of direct division, let’s do this:
Writing 15 = 3 × 5 as the product of factors.
Then, 120 ÷ 3 = 40 chocolates for each group.
Now, to find the number of chocolates each child will get, 40 ÷ 5 = 8 chocolate.

Common Factors
Now make division using factors using another method:

Let’s look what are the steps done for dividing 360 by 24:
1) First, divide 360 by 2 factor of 24
Since 2 is a factor of 360, we got a quotient of 180 with no reminder.

2) Now divide 180 by 12, here we remove 2 which is a factor for both 360 and 24.

• For this, first divide 180 by the factor 3 of 21.
• Since 3 is a factor of 180, also we got the quotient 60 with no reminder.

3) Now divide 60 by 4, here we remove 3 which is a factor for both 180 and 12.

4) This gives a quotient 15 Thus, we can conclude this:
Remove one by one, the number 2, 3, 4 which are the factors of both 360 and 24.
That is,
Numbers which are the factors of both the numbers are called common factors of these two numbers.
In dividing one number by another, common factors of both can be removed. Thus, we get the quotient.
For example, by dividing 84 by 12
84 = 7 × 3 × 4
12 = 3 × 4
Removing the common factors 3 and 4, we get the quotient as 7.

• By the multiples of a natural number, we mean the numbers got by multiplying 1, 2, 3, …… by that number.
• If a number is divisible by another number, the first number is a multiple of the second number; if there is a remainder on division, it is not a multiple.
• By the factors of a number, we mean those numbers by which this number is divisible.
• Numbers which are the factors of both the numbers are called common factors of these two numbers.

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 9 Number Relations Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 9 Solutions Number Relations

Class 5 Maths Chapter 9 Number Relations Questions and Answers Kerala State Syllabus

Number Relations Class 5 Questions and Answers Kerala Syllabus

Question 1.
Find which of the numbers below are divisible by 2, 4, 5 or 10. For the others, find the remainder on division by each of these:
i) 3624
ii) 3625
iii) 3626
iv) 3630
Answer:
i) 3624

• The number is divisible by 2. Because its last digit is 4, which is divisible by 2.
• The number is divisible by 4. Because the number formed by last two digit is 24, which is divisible by 4.
• The number is not divisible by 5. Because, only numbers ending with 0 or 5 are divisible by 5.
• The number is not divisible by 10. Because, only numbers ending with 0 are divisible by 10.

ii) 3625

• The number is not divisible by 2. Because its last digit 5 is not divisible by 2.
• The number is not divisible by 4. Because the number formed by last two digit is 25, which is not divisible by 4.
• The number is divisible by 5. Because, numbers ending with 5 are divisible by 5.
• The number is not divisible by 10. Because, only numbers ending with 0 are divisible by 10.

iii) 3626

• The number is divisible by 2. Because its last digit is 6, which is divisible by 2.
• The number is not divisible by 4. Because the number formed by last two digit is 26, which is not divisible by 4.
• The number is not divisible by 5. Because, only numbers ending with 0 or 5 are divisible by 5.
• The number is not divisible by 10. Because, only numbers ending with 0 are divisible by 10.

iv) 3630

• The number is divisible by 2. Because its last digit is 0, which is divisible by 2.
• The number is not divisible by 4. Because the number formed by last two digit is 30, which is not divisible by 4.
• The number is divisible by 5. Because, numbers ending with 0 are divisible by 5.
• The number is divisible by 10. Because, numbers ending with 0 are divisible by 10.

Question 2.
In any five consecutive natural numbers, one of them will be divisible by 5. Explain why this is so.
Answer:
In any group of five consecutive numbers, one of them will always be a multiple of 5.
Eg:
Consider the numbers 7, 8, 9, 10, and 11. Here, 10 is divisible by 5.

Question 3.
For any number with three or more digits, the remainder on division by 8 is the remainder on dividing the number by the last 3 digits of the number by 8. Explain why this is so.
Answer:
When you divide a big number by 8, you only need to look at the last three digits of that number. This works because 1,000 is divisible by 8. So, for any number that is 1,000 or more, the big part doesn’t affect the remainder when dividing by 8 only the last three digits matter.
Eg:
If the number is 2,567, you just need to look at 567.
Divide 567 by 8, and the remainder will be the same as if you divided the whole number, 2,567, by 8.
i.e, 569 ÷ 8 = (70 × 8)+ 7
2567 ÷ 8 = (320 × 8) + 7
In both the case reminder is same that is 7

Intext Questions And Answers

Question 1.
Consider the below problem:
(i) How much money is needed to buy 12 pens at 8 rupees each?
Answer:
This problem can be solved by the following ways.
If we do it mentally, 80 rupees for 10 pens and 16 rupees for 2 pens, altogether 96 rupees for 12 pens.
By mathematical calculation,
12 × 8 = (10 + 2) × 8
= (10 × 8) + (2 × 8)
= 80 + 16
= 96

The product of a difference is the difference of the products.

(ii) How much money is needed to buy 19 pens at 8 rupees each?
Answer:
19 × 8 = (20 – 1) × 8
= (20 × 8) – (1 × 8)
= 160 – 8
= 152

Question 2.
Now see whether you can do these problems:
i) 15 × 6
ii) 18 × 7
iii) 24 × 9
iv) 29 × 8
v) 99 × 6
Answer:
i) 15 × 6 = (10 + 5) × 6
= (10 × 6) + (5 × 6)
= 60 + 30
= 90

ii) 18 × 7 = (20 – 2) × 7
= (20 × 7) – (2 × 7)
= 140 – 14
= 126

iii) 24 × 9 = (20 + 4) × 9
= (20 × 9) + (4 × 9)
= 180 + 36
= 216

iv) 29 × 8 = (30 – 1) × 8
= (30 × 8) – (1 × 8)
= 240-8
= 232

v) 99 × 6 = (100 – 1) × 6
= (100 × 6) – (1 × 6)
= 600 – 6
= 594

Question 3.
Consider the following problem:
(i) How many pens at 6 rupees each can be bought with 78 rupees?
Answer:
Mentally we can calculate as, with 60 rupees 10 pen can be bought and 3 more pens can be bought with 18 rupees.
So, altogether 13 pens can be bought with 78 rupees.
Mathematically, we get the solution by dividing 78 by 6.
60 ÷ 6 = 10
18 ÷ 6 = 3
78 ÷ 6 = 13
Or,
78 ÷ 6 = (60 ÷ 18) ÷ 6
= (60 ÷ 6) + (18 ÷ 6)
= 10 + 3
= 13
We can describe this using pictures.
First split 60 into 10 parts of 6 each:

Then split 18 into 3 parts of 6 each and add it to the first picture.

Thus we can split 78 into 13 parts of 6 each.
So, to divide to numbers, we can divide each and add.
Note: This method doesn’t work when there are remainders.
Difference can also be divided like this.

(ii) How much pens at 6 rupees each can be bought with 108 rupees?
Answer:
108 ÷ 6 = (120 – 12) ÷ 6
= (120 ÷ 6) – (12 ÷ 6)
= 20 – 2
= 18
First split 120 into 20 parts of 6 each.

If we remove 2 parts , the total is reduced by 12 to become 108 and the number of parts reduced to 18.

Question 4.
Do the below problems mentally:
i) 39 ÷ 3
ii) 52 ÷ 4
iii) 125 ÷ 5
iv) 396 ÷ 4
v) 135 ÷ 15
Answer:
i) 39 ÷ 3 = (30 + 9) ÷ 3
= (30 ÷ 3) + (9 ÷ 3)
= 10 + 3
= 13

ii) 52 ÷ 4 = (60 – 8) ÷ 4
= (60 ÷ 4) – (8 ÷ 4)
= 15 – 2
= 13

iii) 125 ÷ 5 = (120 + 5) ÷ 5
= (120 ÷ 5) + (5 ÷ 5)
= 24 + 1
25

iv) 396 ÷ 4 = (400 – 4) ÷ 4
= (400 ÷ 4) – (4 ÷ 4)
= 100 – 1
= 99

v) 135 ÷ 15 = (120 + 15) ÷ 15
= (120 ÷ 15) + (15 ÷ 15)
= 8 + 1
= 9

Class 5 Maths Chapter 9 Kerala Syllabus Number Relations Questions and Answers

Question 1.
Do the below problems in your head.
i) 12 × 7
ii) 21 × 6
iii) 37 × 9
iv) 43 × 5
v) 98 × 8
vi) 31 × 5
vii) 29 × 4
viii) 47 × 7
Answer:
i) 12 × 7 = (10 + 2) × 7
= (10 × 7) + (2 × 7)
= 70 + 14
= 84

ii) 21 × 6 = (20 + 1) × 6
= (20 × 6) + (1 × 6)
= 120 + 6
= 126

iii) 37 × 9 = (40 – 3) × 9
= (40 × 9) – (3 × 9)
= 360 – 27
= 333

iv) 43 × 5 = (40 + 3) × 5
= (40 × 5) + (3 × 5)
= 200 + 15
= 215

v) 98 × 8 = (100 – 2) × 8
= (100 × 8) – (2 × 8)
= 800 – 16
= 784

vi) 31 × 5 = (30 + 1) × 5
= (30 × 5) + (1 × 5)
= 150 + 5
= 155

vii) 29 × 4 = (30 – 1) × 4
= (30 × 4) – (1 × 4)
= 120 – 4
= 116

viii) 47 × 7 = (50 – 3) × 7
= (50 × 7) – (3 × 7)
= 350 – 21
= 329

Question 2.
Do the below problems mentally:
i) 36 ÷ 3
ii) 56 ÷ 4
iii) 225 ÷ 5
iv) 304 ÷ 4
v) 165 ÷ 15
vi) 105 ÷ 5
vii) 44 ÷ 2
viii) 90 ÷ 3
Answer:
i) 36 ÷ 3 = (30 + 6) ÷ 3
= (30 ÷ 3) + (6 ÷ 3)
= 10 + 2
= 12

ii) 56 ÷ 4 = (60 – 4) ÷ 4
= (60 ÷ 4) – (4 ÷ 4)
= 15 – 1
= 14

iii) 225 ÷ 5 = (200 + 25) ÷ 5
= (200 ÷ 5) + (25 ÷ 5)
= 40 + 5
= 45

iv) 304 ÷ 4 = (300 + 4) ÷ 4
. = (300 ÷ 4) + (4 ÷ 4)
= 75 + 1
= 76

v) 165 ÷ 15 = (150 + 15) ÷ 15
= (150 ÷ 15) + (15 ÷ 15)
= 10 + 1
= 11

vi) 105 ÷ 5 = (100 + 5) ÷ 5
= (100 ÷ 5) + (5 ÷ 5)
= 20 + 1
= 21

vii) 44 ÷ 2 = (50 — 6) ÷ 2
= (50 ÷ 2) – (6 ÷ 2)
= 25 – 3
= 22

Question 3.
Complete the following table:

Answer:

Class 5 Maths Chapter 9 Notes Kerala Syllabus Number Relations

Mathematics is all about working with numbers, and there are four basic operations that help us do just that: addition, subtraction, multiplication, and division. Each operation has its own purpose and rules, and together, they form the foundation of all math concepts. By mastering these basic operations, you’ll build a strong foundation for more advanced math concepts in the future. There are certain relationship between these operations which help us to make calculations easier. Some of these relations are:

• The product of a sum is the sum of the products.
• The product of a difference is the difference of the product.
• In division without remainder, the quotient of a sum is the sum of quotient.
• The quotient of difference is the difference of quotient.

Divisibility rule:
The divisibility rule is a guideline that tells us if a number can be evenly divided by another number. If a number can be divided without leaving a remainder, we say it is divisible by that number. Divisibility rules help us quickly determine whether one number can be divided by another without doing long division. They make it easier to understand factors and multiples and can save time when solving math problems.
In this chapter we also study the divisibility by certain numbers.

Addition, Subtraction And Multiplication
To multiply the sum of two numbers by a number, we need to multiply each number in the sum and add them together.
In other words, the product of a sum is the sum of the products.

Addition, Subtraction And Division
If two numbers can be divided without remainder by a number, then their sum can also be divided without remainder, and the quotient is the sum of quotients.
In other words, in division without remainder, the quotient of a sum is the sum of the quotients.

The remainder on dividing any number by 10 is the last digit of the number.
Also, numbers ending in 0 can be divided by 10 without remainder. In other words, numbers ending in 0 are divisible by 10.
Consider a number 37 where,
37 is 3 tens and 7 ones.
∴ 37 = (3 × 10) + 7
Last digit 7 is the remainder on dividing 37 by 10.

If the last digit of a number is less than 5, the remainder on dividing the number by 5 is the last digit itself. If the last digit is greater than or equal to 5 the remainder is 5 subtracted from the last digit. Also, the numbers ending in 0 or 5 are divisible by 5.
34 ÷ 5 = (6 × 5) + 4
37 ÷ 5 = (7 × 5) + 2
35 ÷ 5 = (7 × 5) + 0
30 ÷ 5 = (6 × 5) + 0

If the last digit of a number is divisible by 2, then the number itself is divisible by 2; if the last digit is not divisible by 2, the remainder is 1 on division by 2.
48 ÷ 2 = (24 × 2) + 0
Here the last digit 8 is divisible by 2. So, the remainder is 0.
51 ÷ 2 = (25 × 2) + 1
Here the last digit 1 is not divisible by 2. So, the remainder is 1.

For any number with three or more digits, the remainder on dividing it by 100 is the number formed by the last two digits.
329 ÷ 100 = (3 × 100) + 29
Here the remainder is the number formed by the last two digits.
7654 ÷ 100 = (76 × 100) + 54
Here also the remainder is the number formed by the last two digits.

For any number with two or more digits, the remainder on dividing it by 4 is the remainder on dividing the last two digits of the number by 4.
In other words, for any number with two or more digits, if the number formed by the last two digits is divisible by 4, the number itself is divisible by 4.
329 ÷ 4 = (82 × 4) + 1
Here the number formed by the last two digits is 29, which is not divisible by 4. So, the remainder is the remainder on dividing 29 by 4, which is 1.
424 ÷ 4 = (106 × 4) + 0
Here the number formed by the last two digits is 24, which is divisible by 4. So, the number itself is divisible by 4.

• The product of a sum is the sum of the products.
• The product of a difference is the difference of the products.
• In division without remainder, the quotient of a sum is the sum of the quotients.
• The remainder on dividing any number by 10 is the last digit of the number. Also, numbers ending in 0 can be divided by 10 without remainder. In other words, numbers ending in 0 are divisible by 10.
• If the last digit of a number is less than 5, the remainder on dividing the number by 5 is the last digit itself. If the last digit is greater than or equal to 5 the remainder is 5 subtracted from the last digit. Also, the numbers ending in 0 or 5 are divisible by 5.
• If the last digit of a number is divisible by 2, then the number itself is divisible by 2; if the last digit is not divisible by 2, the remainder is 1 on division by 2.
• For any number with three or more digits, the remainder on dividing it by 100 is the number formed by the last two digits.
• For any number with two or more digits, the remainder on dividing it by 4 is the remainder on dividing the last two digits of the number by 4.
• In other words, for-any number with two or more digits, if the number formed by the last two digits is divisible by 4, the number itself is divisible by 4.

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 8 Fractions Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 8 Solutions Fractions

Class 5 Maths Chapter 8 Fractions Questions and Answers Kerala State Syllabus

Fractions Class 5 Questions and Answers Kerala Syllabus

Question 1.
i) If 3 cakes are divided equally among 4 persons, what fraction of a cake would each one get?
ii) If a 3 metre long ribbon is cut into four equal pieces, what would be the length of a piece in metres?
iii) If 3 litres of milk is divided equally among 4 persons, how much milk would each get, in litres?
Answer:
i) If one cake is divided equally among 4 persons, each will get \(\frac{1}{4}\) of the cake.
We have 3 cakes, each will get \(\frac{1}{4}\) from each cake.

So, altogether each person gets \(\frac{3}{4}\) of the whole cake.

ii) If one-meter long ribbon cut into four equal pieces, length of each piece is \(\frac{1}{4}\) metre.
Consider 3 metre long ribbon as 3 one metre long ribbons.

So, the length of a piece will be \(\frac{3}{4}\) metres.

iii) If one litre of milk is divided equally among 4 persons, each will get \(\frac{1}{4}\) litre.
Consider 3 litres of milk as 3 one litre of milk.
So, each get \(\frac{3}{4}\) litre.

Question 2.
If a 2 metre long rope is cut into 5 equal pieces, what would be the length of each piece in metres?
Answer:
If one metre long rope is cut into 5 equal pieces, each piece will be \(\frac{1}{5}\) metre.
Consider 2 metre long rope as two 1 metre long ropes.
So, the length of each piece will be \(\frac{2}{5}\) metre.

Question 3.
4 kilograms of sugar is divided into 5 packets of the same weight. How much sugar, in kilograms, is there in each packet?
Answer:
If one kilogram sugar is divided into 5 packets of the same weight each packet will have \(\frac{1}{5}\) kilograms.
Consider 4 kilograms as four 1 kilogram sugar.
So, each packet will have \(\frac{4}{5}\) kilogram.

For each of the problems below, write the answer as a single fraction and as a whole number and a fraction.

Question 4.
If an 8 metre long string is cut into 5 equal parts, what would be the length of each piece?
Answer:
Total length of the string = 8 metres
It is divided into 5 equal parts.
Giving one metre to each , it left with 3 metres.
Dividing 3 metre to 5 equal pieces each gets
Length of each piece = \(\frac{8}{5}\) = 1 \(\frac{3}{5}\) metres

Question 5.
15 litres of kerosene is divided equally into 4 cans, How much does each can contain, in litres?
Answer:
Total amount of kerosene =15 litres
Number of cans = 4
Giving 3 litres to each can, it left with 3 litres.
Distributing this 3 litres equally among 4 cans , each get \(\frac{3}{4}\) litre.
∴ Amount of kerosene in each can = \(\frac{15}{4}\) = 3 \(\frac{3}{4}\) litres

Question 6.
If 25 kilograms of sugar is divided equally among 8 persons, how much would each get, in kilograms?
Answer:
Total amount of sugar = 25 kilogram
Number of person = 8
Giving 3 kilogram of sugar to each , it left with 1 kilogram.
Distributing this 1 kilogram equally among 8 persons, each get \(\frac{1}{8}\) kilogram.
∴ Amount of sugar each would get = 25 ÷ 8 = \(\frac{25}{8}\) = 3\(\frac{1}{8}\) kilogram

For each of the problems, write the answer either as a w hole number and fraction or as quotient and remainder, according to the situation.

Question 7.
If 15 metres of cloth is divided equally among 4 persons, how many metres of cloth would each one get?
Answer:
Total length of cloth =15 metres
Number of persons = 4
Each get.= 15 ÷ 4 = \(\frac{15}{4}\) = 3 \(\frac{3}{4}\) metres

Question 8.
If 250 rupees is equally shared by 8 persons, how much money would each one get? Write this in rupees and paise.
Answer:
Total amount = 250 rupees
Number of persons = 8
Amount of money each gets = 250 ÷ 8 = \(\frac{250}{8}\) = 31 \(\frac{1}{4}\) rupees.
That is 31 rupees and 25 paise.

Question 9.
How many boxes are required to pack 100 eggs with 12 in each box?
Answer:
Total number of eggs = 100
Number of eggs in each box =12
No of boxes required = 100 ÷ 12 = 8 \(\frac{4}{12}\) = 8 \(\frac{1}{3}\)
So, 8 boxes are required to pack 100 eggs with 12 in each box and it left with 4 eggs.

Question 10.
If 15 litres of milk is divided equally among 4 persons, how much would each get, in litres?
Answer:
Total amount of milk =15 litres
Number of persons = 4
Amount of milk each get = 15 ÷ 4 = \(\frac{15}{4}\) = 3\(\frac{3}{4}\) litres

Question 11.
Among how many persons can 15 litres of milk be shared, if each is to get 4 litres?
Answer:
Total amount of milk =15 litres
Amount of milk each get = 4 litres
No of people can have milk = 15 ÷ 4 = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)
So, 3 persons can share 15 litres of milk such that each get 4 litres and left with 3 litres.

Intext Questions And Answers

Question 1.
If a three metre long ribbon is cut into two equal pieces, what would be the length of each piece?
Answer:
3 metre means 2 metre and 1 metre
To cut 3 metre into two equal pieces , then half of three metre is half of 2 metre and half of 1 metre together.

That is one and a half meter

Question 2.
If 2 litre milk is divided equally among 3 persons, how much would each get?
Answer:
Each would get 2 ÷ 3 = \(\frac{2}{3}\) litre

Question 3.
If seven litres of milk is divided equally among two kids, how much would each get, in litres?
Answer:
If we give three litres to each then it left with one litre of milk. The one litre left can also be shared by giving half a litre to each.
i. e, Each would get 7 ÷ 2, which gives quotient 3 and remainder 1.
7 ÷ 2 = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)litres.

Question 4.
If 7 pens are divided between two persons, how many pens would each get?
Answer:
Total no of pens = 7
No of pen would each get = \(\frac{7}{2}\) = 3 \(\frac{1}{2}\)

Class 5 Maths Chapter 8 Kerala Syllabus Fractions Questions and Answers

Question 1.
Ali divided one fruit cake equally among six persons. What part of the cake he gave to each person?
Answer:
Number of fruits = 1
Number of persons = 6
Part of cake each person get = \(\frac{1}{6}\) part

Question 2.
Sidharth has a cake. He cuts it into 10 equal parts. He gave 2 parts to Naman, 3 parts to Nidhi, 1 part to Seema and the remaining four parts he kept for himself.Find:
i) What fraction of cake he gave to Naman?
ii) What fraction of cake he gave to Nidhi?
iii) What fraction of cake,he gave to Seema?
Answer:
Number of cakes = 1
Number of parts = 10
Each part is \(\frac{1}{10}\)

i) Sidharth gave 2 parts to Naman.
∴ Fraction of cake with Naman = 2 × \(\frac{1}{10}\) = \(\frac{2}{10}\)

ii) Sidharth gave 3 parts to Nidhi.
∴ Fraction of cake with Nidhi = 3 × \(\frac{1}{10}\) = \(\frac{3}{10}\)

iii) Sidharth gave 1 part to Seema.
∴ Fraction of cake with Seema = \(\frac{1}{10}\)

Question 3.
If 6500 rupees is equally shared by 16 persons, how much money would each one get? Write this in rupees and paise.
Answer:
Total amount = 6500 rupees
Number of persons =16
Amount each person get = 6500 ÷ 16 = \(\frac{6500}{16}\)
= 406 \(\frac{1}{4}\) rupees
= 406 rupees and 25

Question 4.
Nine litres of milk is equally shared by four kids. How much does each get?
What if it were shared by three?
Answer:
Amount of milk = 9 litres
Number of kids = 4
Amount of milk each kid get = \(\frac{9}{4}\) = 2 \(\frac{1}{2}\) litres.
If it is shared by three, amount of milk each kid get = \(\frac{9}{3}\) = 3 litres.

Question 5.
Six kilogram of rice is packed into four identical bags. How much rice is in each bag?
Answer:
Amount of rice = 6 kilogram
Number of bags = 4
∴ Amount of rice in each bag = 6 ÷ 4 = \(\frac{6}{4}\)
= 1 \(\frac{2}{4}\)
= 1 \(\frac{1}{2}\) kilograms

Question 6.
Among how many persons can 23 litres of milk be shared, if each is to get 5 litres?
Answer:
Total amount of milk = 23 litres
Amount of milk each get = 5 litres
23 ÷ 5 = \(\frac{23}{5}\) = 4\(\frac{3}{5}\)
Distributing 5 litres of milk each to 4 persons, it left with 3 litres of milk.
So, 4 persons can share 23 litres of milk such that each get 5 litres and left with 3 litres.

Question 7.
4 metre long ribbon equally divided into 3 equal parts. What is the length of each piece ?
Answer:
Total length of the ribbon = 4 metres
Number of equal parts = 3
Length of each piece = 4 ÷ 3 = \(\frac{4}{3}\)
= 1 \(\frac{1}{3}\) metres

Class 5 Maths Chapter 8 Notes Kerala Syllabus Fractions

What is equal sharing?
Imagine you have 12 cookies and you want to share them equally among your 4 friends. How many cookies will each friend get? 3 cookies each, right? Equal sharing means dividing something (like cookies, toys, or money) into equal parts so everyone gets the same amount.
In this chapter we deal with equal sharing. Portions that are exactly the same size are referred to as equal parts. Equal sharing is mathematically called as division and division is denoted by the symbol “÷”. This can also represented by fractions, i.e. fractional share.

Division rule :

• Dividend = (Divisor × Quotient) + Remainder
• The dividened is the number that is to be divided.
• The divisor is the number to be divided with.
• The Quotient is the result.

Fractional Share
Suppose we have a cake and we have to divide it equally among two persons. Then each will get half of the original cake.

What if we divide equally among three persons?

That is each will get \(\frac{1}{3}\) of the original cake.
Now, if we cut other cake and give equally to three persons, again each get another \(\frac{1}{3}\) part of the cake.
So, each person gets \(\frac{2}{3}\) of the whole cake.

Now look at another problem:

We have to divide 2 metre long ribbon into three equal parts.

Next fold each 1 metre part into 3 and mark the folds.

Now, if we cut off pairs of these parts, we get three equal parts.

Dividing one into three equal parts and taking two of them, and dividing two into three equal 2 parts and taking one of them, both are \(\frac{2}{3}\).

New Fractions
Suppose we have to divide 3 cake among two persons. First, we can give one whole cake to each.

Thus, each one gets one and a half cake.

• Portions that are exactly the same size is equal parts.
• Equal sharing is mathematically called as division, which is denoted by the symbol “÷”
• Division Rule:
  Dividend = (Divisor × Quotient) + Remainder
• The dividend is the number that is to be divided.
• The divisor is the number to be divided with
• Quotient is the result.

SCERT Class 5 Maths Solutions Pdf

Kerala Syllabus 5th Standard Maths Textbook Solutions Part 1

• Class 5 Maths Chapter 1 Kerala Syllabus – Lines and Circles
• Class 5 Maths Chapter 2 Kerala Syllabus – Number World
• Class 5 Maths Chapter 3 Kerala Syllabus – Multiplication Methods
• Class 5 Maths Chapter 4 Kerala Syllabus – Division Methods
• Class 5 Maths Chapter 5 Kerala Syllabus – Part Number
• Class 5 Maths Chapter 6 Kerala Syllabus – With the Times

Class 5 Maths SCERT Solutions Part 2

• Class 5 Maths Chapter 7 Kerala Syllabus – Measure Math
• Class 5 Maths Chapter 8 Kerala Syllabus – Fractions
• Class 5 Maths Chapter 9 Kerala Syllabus – Number Relations
• Class 5 Maths Chapter 10 Kerala Syllabus – Rectangle Math
• Class 5 Maths Chapter 11 Kerala Syllabus – Within Numbers
• Class 5 Maths Chapter 12 Kerala Syllabus – Tables and Graphs

SCERT Class 5 Maths Solutions Malayalam Medium

5th Std Maths Textbook Solutions Pdf Malayalam Medium Part 1

• Class 5 Maths Chapter 1 Malayalam Medium – വരകളും വട്ടങ്ങളും
• Class 5 Maths Chapter 2 Malayalam Medium – സംഖ്യാലോകം
• Class 5 Maths Chapter 3 Malayalam Medium – ഗുണനരീതികൾ
• Class 5 Maths Chapter 4 Malayalam Medium – ഹരണരീതികൾ
• Class 5 Maths Chapter 5 Malayalam Medium – ഭാഗങ്ങളുടെ സംഖ്യകൾ
• Class 5 Maths Chapter 6 Malayalam Medium – സമയത്തോടൊപ്പം

5th Class Maths Textbook Solutions Pdf Malayalam Medium Part 2

• Class 5 Maths Chapter 7 Malayalam Medium – അളവുകണക്ക്
• Class 5 Maths Chapter 8 Malayalam Medium – ഭിന്നസംഖ്യകൾ
• Class 5 Maths Chapter 9 Malayalam Medium – സംഖ്യാബന്ധങ്ങൾ
• Class 5 Maths Chapter 10 Malayalam Medium – ചതുരക്കണക്കുകൾ
• Class 5 Maths Chapter 11 Malayalam Medium – സംഖ്യകൾക്കുള്ളിൽ
• Class 5 Maths Chapter 12 Malayalam Medium – പട്ടികകളും ചിത്രങ്ങളും

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 7 Measure Math Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 7 Solutions Measure Math

Class 5 Maths Chapter 7 Measure Math Questions and Answers Kerala State Syllabus

Measure Math Class 5 Questions and Answers Kerala Syllabus

Question 1.
How do we write 1.75 metres in fractional form? How many metres and centimetres is this? How much is it in centimetre alone?
Answer:
In fractional form,
1.75metres = 1 \(\frac{75}{100}\) metres.
It is 1 metre and 75 centimetres.
1 metre = 100 centimetres
So, 1.75 metres is 1 metre and 75 centimetres, which is 175 centimetres.

Question 2.
How do we write 0.38 metres in fractional form? How many centimetres is this?
Answer:
In fractional form,
0.38 metres = \(\frac{38}{100}\) metres 100
1 metre =100 centimetres
0.38 metres = \(\frac{38}{100}\) metres = 38 centimetres

Question 3.
i) How do we write 3.275 kilograms in fractional form?
ii) How do we write this in kilograms and grams?
iii) How do we write this in grams alone?
Answer:
i) 3.275 kilograms = 3 \(\frac{275}{1000}\) kilograms
ii) 3 kilograms and 275 grams
iii) 1 kilogram = 1000 grams
So, 3 kilograms = 3000 grams
∴ 3.275 kilograms = 3275 grams.

Question 4.
Complete the table below:

Answer:

Decimal form	Using fractions	Without fractions
7.3 centimetres	7\(\frac{3}{10}\) centimetres	7 centimetres, 3 millimetres = 73 millimetres
3.5 centimetres	3 \(\frac{5}{10}\) centimetres	3 centimetres, 5 millimetres = 35 millimetres
0.7 centimetres	\(\frac{7}{10}\)centimetres	0 centimetres, 7 millimetres = 7 millimetres
5.42 metres	5 \(\frac{42}{100}\) metres	5 metres, 42 centimetres =542 centimetres
10.27 metres	10 \(\frac{27}{100}\) metres	10 metres, 27 centimetres = 1027 centimetres
6.875 kilograms	6\(\frac{875}{1000}\) kilograms	6 kilograms, 875 grams = 6875 grams
0.432 kilograms	\(\frac{432}{1000}\) kilograms	0 kilograms, 432 grams = 432 grams
3.425 litres	3\(\frac{425}{1000}\) litres	3 litres, 425 millilitres = 3425 millilitres
5.763 litres	5\(\frac{763}{1000}\) litres	5 litres, 763 millilitres = 5763 millilitres

Intext Questions And Answers

Question 1.
Consider the following picture :

Answer:
7 centimetres and 3 millimetres.
That is, the length of the pencil = 7\(\frac{3}{10}\) centimetres or 7.3 centimetres.

Question 2.
Complete the following table:

Answer:

If the length of an object is just 8 centimetres, it has only whole number part and has no fractional part. So, we write it in decimal form as 8.0 centimetres.
If the length of an object is less than a centimetre, for example 6 millimetre, which means centimetres. It has no whole number part and is just a fraction. So, we can write it in decimal form as 0.6 centimetres.

Question 3.
Write the measurements in the following table in fractional form and decimal form.

Answer:

Question 4.
Convert 1.25 metres into centimetres.
Answer:
1.25 metres = 1 \(\frac{25}{100}\) metres 100
1 metre = 100 centimetres
1 centimetre = \(\frac{1}{100}\) metres
So, \(\frac{25}{100}\) metres = 25 centimetre
Hence, 1.25 metres = 125 centimetres.

Question 5.
Convert 7.125 kilogram into grams.
Answer:
7.125 kilograms = 7\(\frac{125}{1000}\) kilograms
1 kilogram = 1000 grams
∴ 7 kilograms = 7000 grams
1 gram = \(\frac{1}{1000}\) kilograms
So, kilograms =125 grams
Hence, 7.125 kilograms = 7125 grams.

Question 6.
Convert 2.25 litres into millilitres.
Answer:
1 litre = 1000 millilitres
So, 2 litres = 2000 millilitres
1 millilitre = \(\frac{1}{1000}\) litres
∴ \(\frac{25}{100}=\frac{250}{1000}\) litres = 250 millilitres
Hence, 2.25 litres = 2250 millilitres

Class 5 Maths Chapter 7 Kerala Syllabus Measure Math Questions and Answers

Question 1.
Fill in the blanks:

• 200 grams = ________ Kilogram
• 0.850 kilograms = ________ Grams
• 4063 millilitres = ________ litres
• 5006 millilitres = ________ Litres
• 5006 metres = ________ Kilometres
• 3000 centimetres = ________ metres
• 888 grams = ________ kilograms

Answer:

• 200 grams = 0.2 Kilogram
• 0.850 kilograms = 850 Grams
• 4063 millilitres = \(\frac{4063}{1000}\) = 4.063 litres
• 5006 millilitres = \(\frac{5006}{1000}\) = 5.006 Litres
• 5006 metres = \(\frac{5006}{1000}\) = 5.006 Kilometres
• 3000 centimetres = \(\frac{3000}{1000}\) = 3 metres
• 888 grams = \(\frac{888}{1000}\) = 0.888 kilograms

Question 2.
Tick (✓) the correct ones:
(i) 3 litres is equal to –
a) 1000 millilitres
b) 2000 millilitres
c) 3000 millilitres
d) 6000 millilitres
Answer:
3000 millilitres

(ii) 3 litres 250 millilitres is equal to –
a) 8050 millilitres
b) 3250 millilitres
c) 3000 millilitres
d) 1250 millilitres
Answer:
3250 millilitres

(iii) 7000 grams is equal to –
a) 7 kilograms
b) 8 kilograms
c) 3 kilograms
d) 2 kilograms
Answer:
7 kilograms

(iv) 8 metres is equal to –
a) 200 centimetres
b) 600 centimetres
c) 700 centimetres
d) 800 centimetres
Answer:
800 centimetres

Question 3.
State ‘T’ for True and ‘F’ for False:
i) 33 metres 50 centimetres is equal to 3350 centimetres.
Answer:
True

ii) 5 kilograms 250 grams is equal to 2300 grams.
Answer:
False

iii) 8150 millilitres is equal to 8050 millilitres.
Answer:
False

iv) 4000 millilitres is equal to 4 litres.
Answer:
True

v) 7 kilograms is equal to 8000 grams.
Answer:
False

Question 4.
Complete the following table.

Measure	Fractional form	Decimal form
25 kilometres 836 metres	_____ kilometres	kilometres
9 litres 375 millilitres	______ litres	litres
13 metres 38 centimetres	_______ metres	metres
2 kilograms 800 grams	_____ kilograms	kilograms
12 centimetres 3 millimetres	_____ centimetres	_____ centimetres
525 millilitres	______ litres	_______ litres
250 grams	_______ kilograms	________kilograms

Answer:

Measure	Fractional form	Decimal form
25 kilometres 836 metres	2 \(\frac{836}{1000}\) kilometres	25.836 kilometres
9 litres 375 millilitres	9 \(\frac{375}{1000}\) litres 1000	9.375 litres
13 metres 38 centimetres	13 \(\frac{38}{100}\)metres 100	13.38 metres
2 kilograms 800 grams	2 \(\frac{800}{1000}\)kilograms	2.8 kilograms
12 centimetres 3 millimetres	12\(\frac{3}{10}\) centimeters 10	12.3 centimeters
525 millilitres	\(\frac{525}{1000}\) litres	0.525 liters
250 grams	\(\frac{250}{1000}\) kilograms	0.25 kilograms

Class 5 Maths Chapter 7 Notes Kerala Syllabus Measure Math

What is Measurement?
Understanding measurement helps us make sense of the world and is essential for everyday activities. Measurement in math involves determining the length, weight, and volume of an object.

Length:
The length of an object can be measured using a scale or a tape, depending on the size of the object. Units of measurement of length are millimetre, centimetre, meter, kilometre, etc. Also, we can convert the length from one unit to another.

Weight:
The weight of an object can be measured using scales and balances. Units of measurement of weight are gram, kilogram, etc. Also, we can convert the weight from one unit to another.

Volume:
The volume of an object can be measured using measuring cups and containers. Units of measurement of volume are millilitre, litre, etc. Also, we can convert the volume from one unit to another.

Measurement is everywhere, and by the end of this chapter, you’ll be equipped with the skills to measure confidently in all sorts of situations. Measurement is not only fundamental in math but also in daily life.

By mastering these concepts, you’ll be better prepared to tackle practical challenges and understand the world around you.

Fractional Lengths
We know that,
1 centimetre = 10 millimetres
i. e, a millimetre is one part of a centimetre divided into 10 equal parts.
In other words, 1 millimetre is \(\frac{1}{10}\) of 1 centimetre.
1 millimetre = \(\frac{1}{10}\) centimetre.

Different Measures
To measure large lengths such as length of a bench, room etc. it is more convenient to use a metre scale or tape.
1 metre = 100 centimetre
So, 1 centimetre = \(\frac{1}{100}\) metres
Also, 1 metre = 1000 millimetres

Suppose, the length of a table is found to be 1 metre, 25 centimetres and 4 millimetres.
25 centimetres means 250 millimetres. So, 25 centimetres and 4 millimetres is equal to 254 millimetres, which is \(\frac{254}{1000}\) metres or in decimal form as 1.254 metres.

This can be done for another measures also.

• 1 litre = 1000 millilitre
• 1 millilitre = \(\frac{1}{1000}\) litre

So, 2 litre and 125 millilitre can be written in fractional form as 2 \(\frac{125}{1000}\) litre and in decimal form as 2.125 litres.

Similarly,

• 1 Kilogram = 1000 gram
• 1 gram = \(\frac{1}{1000}\) kilogram

So, 5 kilogram and 375 gram can be written as 5\(\frac{375}{1000}\) kilogram and in decimal form it can be written as 5.375 kilogram.

• 1 centimetre =10 millimetres
  1 millimetre = \(\frac{1}{10}\) centimetre.
• 1 metre =100 centimetre
  1 centimetre = \(\frac{1}{100}\) metres
• 1 metre = 1000 millimetres
  1 millimetre = \(\frac{1}{1000}\) metres
• 1 litre = 1000 millilitre
  1 millilitre = \(\frac{1}{1000}\) litre
• 1 Kilogram = 1000 gram 1 gram = \(\frac{1}{1000}\) kilogram

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 6 With the Times Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 6 Solutions With the Times

Class 5 Maths Chapter 6 With the Times Questions and Answers Kerala State Syllabus

With the Times Class 5 Questions and Answers Kerala Syllabus

Question 1.
Draw clocks showing the times given below.
(i) Quarter past 2
(ii) Quarter to 10
(iii) Half past 11
Answer:
(i) Quarter past 2

(ii) Quarter to 10

(iii) Half past 11

Question 2.
Complete the table below:

Answer:

120 minutes	2 hours
150 minutes	2 \(\frac{1}{2}\) hours
3 \(\frac{1}{2}\) hours	210 minutes
3 minutes	180 seconds
\(\frac{1}{2}\) hours and 45 minutes	1 hour
5 hours 59 minutes and 60 seconds	6 hours

(i) 120 minutes = 2 hours
The remaining 30 minutes is half an hour.
So, 150 minutes = 2\(\frac{1}{2}\) hours.

(ii) 60 minutes = 1 hour
So, 180 minutes = 3 hours
The remaining 30 minutes is half an hour.
So, 210 minutes = 3 – hours.

(iii) 1 minute = 60 seconds
So, 3 minutes = 180 seconds

(iv) 1 hour = 60 minutes
So, \(\frac{1}{4}\) hours = 15 minutes
\(\frac{1}{4}\) hours and 45 minutes = 15 minutes and 45 minutes = 60 minutes = 1 hour.

(v) 60 seconds = 1 minute
So, 5 hours 59 minutes and 60 seconds = 5 hours and 60 minutes = 6 hours

Question 3.
School starts at 10 in the morning and continues to 4 in the afternoon. There are intervals from 11:20 to 11:30 in the morning, 12:50 to 1:45 in the afternoon and 3:10 to 3:20 in the evening. How much time do students get for their studies?
Answer:
10 in the morning to 4 in the afternoon, total 6 hours.
Interval from 11:20 to 11:30, 10 minutes
Interval from 12:50 to 1:45, 55 minutes
Interval from 3:10 to3:20, 10 minutes
So, total interval time is 75 minutes = 1 hour and 15 minutes
Time get for studies = 6 hours – 1 hour and 15 minutes = 4 hours and 45 minutes

Question 4.
Look at the calendar for this year. How many days does February have?
Answer:
29 days

Question 5.
Which is the next leap year?
Answer:
2028

Question 6.
Which months of this year have 5 Mondays?
Answer:
January, April, July, September, December.

Question 7.
In a certain year, there were 5 Wednesdays in October, but only 4 Thursdays. What day of the week was the first of October that year?
Answer:
There were 5 Wednesdays in October, but only 4 Thursdays, which means 31 st October is Wednesday.
24, 17, 10, 3 are Wednesday.
So, 1 st of October is Monday.

Question 8.
The sum of the dates of two Sundays in a month is 11. What day of the week is the first of that month?
Answer:
The sum of the dates of two Sundays in that month is 11.
So, first Sunday is 11 – 7 = 4.
first of that month is Thursday

Question 9.
How many times each day of the week occurs this year? Which day occurs the most number of times? ‘
Answer:
This year, 2024, is a leap year. So, there are 366 days, which is 52 weeks and 2 days.
Days in which January 1 and 2 comes occur 53 times and all other days 52 times.
So, Monday and Tuesday occur 53 times and Wednesday, Thursday, Friday, Saturday, Sunday occur 52 times.

Intext Questions And Answers

Question 1.
Hands of a clock

Look at the picture of the clock.

(i) How many hands does it have?
Answer:
3

(ii) What does each hand represented?
Answer:
The hand points towards 12 represents the minute hand (longer and thicker)
The hand points towards 3 represents the hour hand (shorter and thicker)
The hand points towards 7 represents the second hand (longer and thinner)

(iii) Which hand turns the fastest?
Answer:
Second hand

(iv) How much time does it take to complete one rotation?
Answer:
60 seconds or 1 minute

(v) How much time does the minute hand take to complete one rotation?
Answer:
60 minutes or 1 hour

(vi) What about the hour hand?
Answer:
12 hours

(vii) How much time does the hour hand take to start from 12 and return to 12?
Answer:
12 hours

(a) By this time, how many rotations would have the minute hand made?
Answer:
12 rotations

(b) What about the second hand?
Answer:
60 rotations in 1 hour.
So, 12 × 60 = 720 rotations

(c) How many rotations does the hour hand complete in a day?

Answer:
24 rotations
1 Day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

(viii) How many minutes are there in a day?
Answer:
24 hours in one day and 60 minutes in one hour.
So, 24 × 60 = 1440 minutes in a day.

(ix) How many seconds?
Answer:
60 seconds in 1 minute.
So, 60 × 1440 = 86400 seconds in a day.

Question 2.
What’s the time shown in each clock?

Answer:

i. The clock shows 30 minutes past 1. So, the time is half past one. (1:30)
ii. The clock shows 15 minutes past 5. So, the time is quarter past five. (5:15)
iii. The clock shows 45 minutes past 3. So, the time is three quarter past 3 or a quarter to 4. (3:45)

Question 3.
Complete the table:

Answer:

7 in the morning	7 am
7 in the evening	7 pm
9 in the night	9 pm
5 in the evening	5 pm
11:3 0 in the morning	11:30 am
6 in the evening	6 pm
4 in the morning	4 am
11:30 in the night	11:30 pm

Question 4.
A person started from Thiruvananthapuram at 8 am on January 7th and reached Hyderabad at 3 pm on January 8th. How long did the journey take?
Answer:
Time from 8 am on January 7 th to 8 am on January 8 th = 24 hours
Time from 8 am on January 8 th to 3 pm on January 8 th = 7 hours
So, total time taken = 24 + 7 =31 hours.

Question 5.
An office works from 9 am to 5 pm. One of the employees went out from 11 am to 1:30 pm. How long was the person in the office?
Answer:
Total working hours of the office (9 am to 5 pm) = 8 hours
Time for which employee went out (11 am to 1:30 pm)
= 2 hours 30 minutes = 2 – hours
So, the time for which person was in the office = 8 hours – 2\(\frac{1}{2}\) hours
= 5\(\frac{1}{2}\) hours= 5 hours 30 minutes

Question 6.
Complete the table:

Answer:

24 hour clock	12 hour clock
23:30	11:30 pm
8:00	8:00 am
13:30	1:30 pm
16:25	4:25 pm
10:00	10 am
21:20	9:20 pm
15:00	3:00 pm

Question 7.
A bus started from Kozhikode at 15:30 and reached Thiruvananthapuram at 2:30. How much time did the journey take?
Answer:
15:30 means 3:30 pm and 2:30 means 2:30 am.
So, the bus started at 3:30 pm and reached the next day at 2:30 am.
Total time taken = 11 hours.

Question 8.
A flight started from Kozhikode at 18:40 and reached Delhi via Mumbai at 00:10. How much time did the journey take?
Answer:
18:40 means 6:40 pm and 00:10 means 12:10 am.
So, flight started at 6:40 pm and reached next day 12:10. am.
Total time taken = 5 hours and 30 minutes.

Leap Year Usually, Earth takes 365 days and 6 hours to complete one orbit around the sun. If we think of a year as having 365 days, then there will be a change of one day every 4 years. If this goes on for a few years, the seasons will not be calculated correctly from then on. As an alternative, every four years an extra day is added to make the year 366 days long. Leap years are these kinds of years. In a leap year, February has 29 days because of the extra day.

A leap year is a calendar year that contains an additional day compared to a common year. So, a leap year has 366 days, which is 52 weeks and 2 days.

Question 9.
What are the different systems shown in calendar sheets?
Answer:

• Gregorian calendar
• (sometimes called English calendar in our state).
• Jewish calendar
• Islamic calendar
• Persian calendar
• Julian calendar
• Hindu calendar
• Chinese calendar

Class 5 Maths Chapter 6 Notes Kerala Syllabus With the Times

Imagine a world without time. No clocks ticking away, no schedules to follow, and no birthdays to celebrate. Sounds strange, right? Time controls our days, seasons, and lives, so it’s everywhere.

Time:
Time is the measurement of the duration between events. It flows continuously, moving forward* from the past, through the present, and into the future. We usually use clocks to measure time. We divide time into smaller parts, like hours, minutes, and seconds.

‘am’ and ‘pm’
‘am’ denote time in the morning and ‘pm’ means times in the afternoon.
For example, 4:30 am means 4:30 in the morning and 5:00 pm means 5 in the evening.

Clock:
A clock has three hands. Second hand, minute hand and hour hand.
Second hand takes 60 second or 1 minute to complete a rotation.
Minute hand takes 60 minutes or 1 hour to complete a rotation.
Hour hand takes 12 hours to complete one rotation.
Second hand turns the fastest and hour hand, the slowest.

• 1 Day = 24 hours
• 1 hour = 60 minutes
• 1 minute = 60 seconds

In 24 hour clock, a day begins at 00:00 and ends at 23:59.

Calendar

• A calendar usually divides time into units such as days, weeks, months, and years.A year usually has 365 days, which is 52 weeks and one day.
• A leap year has 366 days, which is 52 weeks and 2 days.

Study Trip:
The class returned after a one-day study trip. They started at 6 in the morning. It was 9 at night when they came back. Complete the following:

Before Noon And After Noon
You might’ve seen am or pm together with time in many places, ‘am’ denote time in the morning and ‘pm’ means times in the afternoon.

24 Hour Clock:
Railways and some other agencies use a 24-hour clock. 24-hour clock, time convention that begins the day at 00:00 and ends at 23:59.
In the 24-hour clock format, for times between 12 to 1 in the night, hours are denoted as 0. Thus 00:30 means 12:30 am.

Through The Calendar

• A calendar usually divides time into units such as days, weeks, months, and years.
• A year usually has:
  • 365 days, which is 52 weeks and one day.
  • 12 months

A clock has three hands. Second hand, minute hand and hour hand. Second hand takes 60 second or 1 minute to complete a rotation, minute hand takes 60 minutes or 1 hour to complete a rotation and hour hand takes 12 hours to complete one rotation.Second hand turns the fastest and hour hand, the slowest.

1 Day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

• ‘am’ denote time in the morning and ‘pm’ means times in the afternoon.
• In 24 hour clock, a day begins at 00:00 and ends at 23:59.
• A year usually has 365 days, which is 52 weeks and one day.
• A leap year has 366 days, which is 52 weeks and 2 days.

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 5 Part Number Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 5 Solutions Part Number

Class 5 Maths Chapter 5 Part Number Questions and Answers Kerala State Syllabus

Part Number Class 5 Questions and Answers Kerala Syllabus

Now let’s try these on your own:

Question 1.
Draw a square with each side 3 centimetres. Mark off 1 centimeter from the left on the top and bottom sides, and join them.

i. What part of the square is the smaller rectangle?
Answer:
\(\frac{1}{3}\)

ii. And the larger?
Answer:
\(\frac{2}{3}\)

iii. Colour the \(\frac{1}{3}\) part with black and the \(\frac{2}{3}\) part with white.
Answer:

iv. Is there another way of dividing the square into \(\frac{1}{3}\) and \(\frac{2}{3}\) parts?
Answer:

Question 2.
Draw a rectangle like this:

Colour \(\frac{2}{3}\) of it with black and, \(\frac{1}{3}\) with white.
Answer:

Question 3.
The pictures below show a circle divided into eight equal parts and coloured two at a time: Describe the coloured part of each in two different ways and write both of them as fractions in the box below each picture.

Answer:

Question 4.
The pictures below show circles divided into twelve equal parts with some of the parts coloured:

Describe the coloured part of each in two different ways and write both of them as fractions in the box below each picture.
Answer:

Question 5.
Draw a square and divide it as shown below:

Colour \(\frac{1}{8}\) of the square with red, \(\frac{1}{4}\) with green and \(\frac{1}{2}\) with blue
i. Is there any part left uncoloured?
ii. What part of the whole square is it?
iii. Write it as a fraction.
Answer:

i. 1
ii. One eighth
iii. \(\frac{1}{8}\)

Question 6.
In each of the pictures below, find how much is coloured and write it as a fraction in the box on the right.

Answer:

Intext Questions And Answers

Question 1.
Any other way to split a square into halves?
Answer:

Question 2.
Let’s make a table of the different forms of ‘half ’ we’ve seen:

Answer:

Class 5 Maths Chapter 5 Notes Kerala Syllabus Part Number

If we filled one litre of milk in three bottles of the Same size, then how much would each the bottle contain? In this chapter we will discuss about it. Thus, each bottle Contain one third of a litre. The fractions used here can be stated mathematically as – litre. We can fill Six bottles of the Same Size with one litre of milk and take two bottles, or fill three bottles of the same size with one litre of milk and take one bottle. In either case, we get one third of a litre that is, \(\frac{2}{6}=\frac{1}{3}\)
Let’s examine it in more detail

Half Means
Half means one of two equal parts. We write it as \(\frac{1}{2}\)
\(\frac{1}{2}\) of a circle

Here, \(\frac{1}{2}\) of the circle is coloured as black and \(\frac{1}{2}\) as white.
Half of a meter is \(\frac{1}{2}\) meter.

Half a litre is \(\frac{1}{2}\) litre.
\(\frac{1}{2}\) of a square.

Here, are the two different ways of marking \(\frac{1}{2}\) of a square.

1. When a circle is divided into four equal parts.

Each part can be called as one-fourth of the circle and can be written as \(\frac{1}{4}\)

If we join two parts of the circle.
We can call it as two fourth of the circle, and can be written as \(\frac{2}{4}\) .

Here, we take two of four equal parts or one of two equal parts
That is, two fourth is the same as half.
In the language of maths,
\(\frac{2}{4}=\frac{1}{2}\)

2. If one metre long ribbon is cut into six equal pieces, each part is \(\frac{1}{6}\) metre long.

If we join three parts together.
We can call it a three sixth part of one metre, and it can be written as \(\frac{3}{6}\).

Thus, three sixth is also half.
In the language of maths,
\(\frac{3}{6}=\frac{1}{2}\)

Three Parts
1. If a one metre long ribbon is cut into three equal pieces, the length of each part is – metre.

When we put together two of the three equal pieces together, we call it two-third of one metre, and it can be written as \(\frac{2}{3}\).

Other Parts
1. When we divide one-metre ribbon into four equal parts and different numbers of pieces put together we get as follows:

Here,
\(\frac{1}{4}\) metre is called quarter metre
\(\frac{1}{2}\) metre is called half metre
\(\frac{3}{4}\) metre is called three quarters metre
One fourth is usually called a quarter and three fourths, three quarters.

2. Now divide the circle into six equal parts and three equal parts and put different numbers of parts together where we can see that;

Thus, two sixths is the same as one third i.e;

Thus, four sixths is the same as two third i.e;
The numbers like \(\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}\)……….. used to denote parts of a whole are called fractions.

Whole And Part
1. When we join one metre long ribbon and half of another one metre long ribbon end to end we get a one and a half metre long ribbon, which can be write as 1 metre.

Similarly, if we take two litres of milk and quarter of a litre together, which makes two and a quarter litres, and can be written as 2\(\frac{1}{4}\) litres.

1. We say half, one third, two third and so on in common language when we divide something into equal parts and take only some of the pieces.

2. Fractions are used to talk about these parts in the language of math
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)

3. Fraction in different forms may be the same part:
\(\frac{2}{4}\) = \(\frac{1}{2}\)
\(\frac{2}{6}\) = \(\frac{1}{3}\)
\(\frac{4}{6}\) = \(\frac{2}{3}\)

4. One fourth is usually called a quarter and three fourths, three quarters.

5. The numbers like \(\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}\), ………….. used to denote parts of a whole are called fractions.

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 4 Division Methods Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 4 Solutions Division Methods

Class 5 Maths Chapter 4 Division Methods Questions and Answers Kerala State Syllabus

Division Methods Class 5 Questions and Answers Kerala Syllabus

Question 1.
Total price of 7 pens of equal rate is 98 rupees. What’s the price of a pen?
Answer:

Thus, price of a pen =14 rupees

Question 2.
If 168 rupees is shared equally among 8 persons, how much will each get?
Answer:

Thus, each get 21 rupees.

Question 3.
The school store needs 1825 notebooks. How many packs of 25 notebooks, are to be bought?
Answer:

Thus, total number of packet is 73.

Question 4.
What are the numbers which leave remainder 0 on division by 2? And those that leave remainder 1?
Answer:
2 = (2 × 1) + 0
4 = (2 × 2) + 0
6 = (2 × 3) + 0
8 = (2 × 4) + 0
Thus, the numbers are 2, 4, 6, 8, 10
That is, the multiple of 2 which leave remainder 0 on division by 2.
3 = (2 × 1) + 1
5 = (2 × 2) + 1
7 = (2 × 3) + 1
9 = (2 × 4) + 1
Thus, the numbers are 3, 5, 7, 9

Question 5.
What are the possible remainders on dividing a number by 3? Write the pattern of numbers leaving the same remainder for each of these.
Answer:
3 = (3 × 1) + 0
4 = (3 × 1) + 1
5 = (3 × 1) + 2
6 = (3 × 2) + 0
7 = (3 × 2) + 1
8 = (3 × 2) + 2
Thus, the remainders on dividing a number by 3 are 0, 1 or 2.

Question 6.
See the way numbers are arranged below: 0 12 3 4

i. Is there any relation between the quotients on dividing the numbers in any one row by 5? And the remainders?
ii. What if numbers in any one column are divided by 5?
iii. What are the first and last numbers of the 10th row?
iv. What is the 4th number in the 12th row?
v. Which column and row does the number 123 belong to?
Answer:
i. On dividing any one of the row by 5 we get the quotient which leaves a remainder and the remainder are 0, 1, 2, 3 and 4.

ii. On dividing any one of the column by 5 we get the quotient with some reminder and the remainder will be same for a column.
That is,
For the first column the reminder will be 0 For the second column the reminder will be 1 For the third column the reminder will be 2 For the fourth column the reminder will be 3 For the fifth column the reminder will be 4

iii. First number of the 1st row is: 5 × 0 = 0
First number of the 2nd row is: 5 × 1 = 5
First number of the 3rd row is: 5 × 2 = 10
First number of the 4th row is: 5 × 3 = 15
Thus,
First number of the 10th row is: 5 × 9 = 45
Last number of the 1st row is: (5 × 0) + 4 = 4
Last number of the 2nd row is: (5 × 1) + 4 = 9
Last number of the 3rd row is: (5 × 2) + 4 = 14
Last number of the 4th row is: (5 × 3) + 4 = 19
Thus,
Last number of the 10th row is: (5 × 9) + 4 = 49

iv. Fourth number in every row have a reminder 3 Thus, the 4th number in the 12th row is:
(5 × 11) + 3 = 58

Here, the quotient is 24 and the remainder is 3.
Thus, 123 belongs to 4th column and 25th row.

Question 7.
Write each of the numbers as certain times 9 and a remainder
(a) 11 (b) 111 (c) 1111
i. Can you guess how 1111 is written like this? Check your guess by actual division.
ii. Write out this number pattern.
Answer:
The number can be written in certain times of 9 and a reminder as:
a) 11 = (9 × 1) + 2
b) 111 = (9 × 12) + 3
c) 1111 =(9 × 123) + 4
i.

Thus, we get the quotient 123 and the reminder 4

ii. (9 × 1) + 2= 11
(9 × 12) +3 = 111
(9 × 123)+ 4= 1111
(9 × 1234) + 5 = 11111
(9 × 12345) + 6= 111111
(9 × 123456) + 7 = 1111111

Question 8.
Write each of the numbers as certain times 8 and a remainder
(a) 9 (b) 98 (c) 987
i. Can you guess how 9876 is written like this? Check your guess by actual division.
ii. Write out this number pattern.
Answer:
The number can be written in certain times of 8 and a reminder as:
a) 9 = (8 × 1) + 1
b) 98 = (8 × 12) + 2
c) 987 = (8 × 123) + 3
i.

That is, 9876 divided by 8 gives quotient 1234, remainder 4. Thus, we can write 9876 as:
9876 = (8 × 1234) + 4
ii. (8 × 1) + 1 = 9
(8 × 12) + 2 = 98
(8 × 123) + 3 = 987
(8 × 1234) + 4 = 9876
(8 × 12345) + 5 = 98765

Intext Questions And Answers

Question 1.
How do we distribute 20 pieces of candy equally among 5 kids?
Answer:
Let’s give 5 candies to each one of them, then

• 1 each, then the total is 5 One more is given, then
• 2 each, then the total is two times of 5
  i.e, 5 × 2 = 10
  There are some more pieces left so
• 3 each, then the total is three times of 5
  i.e, 5 × 3 = 15
  Still, there are some left so
• 4 each, then the total is four times of 5
  i.e, 5 × 4 = 20

Question 2.
If 20 is split into 5 equal, parts, how much would be each part?
Answer:
4
In math language, we can shorten this as:
The number obtained by dividing 20 by 5 is 4
Using math symbols, we can write it as:
20 ÷ 5 = 4

Question 3.
So, the questions on parts and division above can be rewritten as questions on times and multiplication.

Answer:

Parts and Division	Times and Multiplication
If 20 is split into 5 equal parts, how much would each part be?	How many times of 5 makes 20?
What number do we get on dividing 20 by 5?	By what number should 5 be multiplied to get 20?
20 ÷ 5 = 4	5 × 4 = 20

Question 4.
40 litres of water was filled in 8 bottles of the same size. How many litres of water will be there in each bottle?

Answer:

Parts and Division	Times and Multiplication
If 40 is split into 8 equal parts, how much would each part be?	How many times of 8 makes 40?
What number do we get on dividing 40 by 8?	By what number should 8 be multiplied to get 40?
40 ÷ 8 = 5	8 × 5 = 40

Question 5.
6 persons equally shared 48 kilograms of rice. How many kilograms did each get?

Answer:

Parts and Division	Times and Multiplication
If 48 is split into 6 equal parts, how much would be each part?	How many times of 6 makes 48?
What number do we get on dividing 48 by 6?	By what number should 6 be multiplied to get 48?
48 ÷ 6 = 8	6 × 8 = 48

Question 6.
If a rectangle is to be made with 30 dots where 5 dots in each row.

(i) When one rows are filled, how many dots are left?
Answer:
30 – 5 = 25

(ii) How many dots were used to fill four rows?
Answer:
20

(iii) How many more rows can be made?
Answer:
6
Here we compute as;
30 is the 6 times of 5.
That is, 5 × 6 = 30 In terms of parts,
If 30 is divided into parts of 5, there would be 6 parts.

This can also be stated as follows:
If 30 is divided into 5 equal parts, each part would be 6.

In terms of division:
75 ÷ 5 = 15
Without dots, we can compute this using numbers as:

Thus, the total number of rows is 6.

Question 7.
A rectangle is to be made with 96 dots, 6 dots in each row. How many rows can be made?
Answer:
Using boxes, we can calculate this as:

Using a box, we can do the calculations one by one as:

At the end, if you don’t remember 36 = 6 × 6, you can do like this:

Question 8.
If 3000 rupees is shared equally among 12 persons, how much will each get?
Answer:

Thus, each gets 250 rupees.

Question 9.
How can 21 pens be packed into packets of 5 ?
Answer:
We know,
20 ÷ 5 = 4
That is, there will be 4 packets and one pen left over.
In the language of math, we can say this as:
21 divided by 5 gives quotient 4 and remainder 1.
Thus,
Quotient shows the number of parts, into which we can split 21
Remainder shows the number that is left over.
In terms of multiplication, we can write it as:
21 = (5 × 4) + 1
Similarly,
22 = (5 × 4) + 2
23 = (5 × 4) + 3
24 = (5 × 4) + 4
In the case of 20,
20 = (5 × 4) + 0
This means, 20 divided by 5 gives quotient 4 and remainder 0 In the case of 4,
4 = (5 × 0) + 4
That is, 4 divided by 5 gives quotient 0, remainder 4 itself.
Thus, any number can be written as so many times 5 and a remainder; and the remainder in all cases will be one of the numbers 0, 1, 2, 3 or 4.
Thus, we can conclude this as:
Any number can be written as so many times a number other than zero, and a remainder; and the remainder will be less than the second number.

Class 5 Maths Chapter 4 Notes Kerala Syllabus Division Methods

Division is one of the basic arithmetic operations, it is necessary for many mathematical computations as well as daily tasks. It represents the method of dividing a group of things into equal parts. You were involved in the division if you have ever split a bill with friends or shared a pizza. The division can be represented by the symbol.
In this chapter, let’s study the division in more detail.

1. Parts and Division
If 12 is split into 3 equal parts, how much would each part be?
What number do we get on dividing 12 by 3?
ie, 12 ÷ 3 = 4

2. When 21 divided by 5 gives quotient 4 and remainder 1.
That is,
Quotient shows the number of parts, into which we can split 21
Remainder shows the number that is left over.

3. Any number can be written as so many times a number other than zero, and a remainder; and the remainder will be less than the second number.

Students often refer to Kerala State Syllabus SCERT Class 5 Maths Solutions and Class 5 Maths Chapter 3 Multiplication Methods Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 5 Maths Chapter 3 Solutions Multiplication Methods

Class 5 Maths Chapter 3 Multiplication Methods Questions and Answers Kerala State Syllabus

Multiplication Methods Class 5 Questions and Answers Kerala Syllabus

Question 1.
Compute the products below:
(i) 12 × 34
Answer:

(ii) 23 × 45
Answer:

(iii) 75 × 75
Answer:

(iv) 123 × 45
Answer:

(v) 320 × 78
Answer:

Question 2.
Given that 36 × 15 = 540; compute the following products in your head:
(i) 36 × 16
Answer:
36 × (15 + 1)
= (36 × 15) + 36
= 540 + 36
= 576

(ii) 37 × 15
Answer:
(36 + 1) × 15
= (36 × 15) + (1 × 15)
= 540 + 15
= 555

(iii) 36 × 14
Answer:
36 × (15 – 1)
= (36 × 15) – 36
= 540 -36
= 504

(iv) 35 × 15
Answer:
(36 – 1) × 15
= (36 × 15) – 15
= 540 – 15
= 525

Question 3.
A number multiplied by 16 gave 1360.
(i) What will be the product, if the next number is multiplied by 16?
(ii) What will be the product if the number just before this is multiplied by 16?
Answer:
(i) If we look at the number just after our original number, we can find the new product by simply adding 16 to 1360
1360 + 16 = 1376
So, when the ne×t number is multiplied by 16, we get 1376

(ii) Now, let’s look at the number just before our original number. To find this new product, we subtract 16 from 1360
1360 – 16 = 1344
So, when the previous number is multiplied by 16, we get 1344

Question 4.
Rewrite each of the products below as the product of a number by itself:
(i) 9 × 16
(ii) 16 × 36
(iii) 36 × 49
(iv) 49 × 64
(v) 81 × 25
Answer:
(i) 9 × 16 = (3 × 4) × (3 × 4)
(ii) 16 × 36 = (4 × 6) × (4 × 6 )
(iii) 36 × 49 = (6 × 7) × (6 × 7 )
(iv) 49 × 64 = (7 × 8) × (7 × 8)
(v) 81 × 25 = (9 × 5) × (9 × 5 )

Question 5.
Calculate each of the products below in your head:
(i) 25 × 4
(ii) 25 × 16
(iii) 25 × 36
(iv) 25 × 64
Answer:
(i) 25 × 4 = (5 × 2)(5 × 2) =100
(ii) 25 × 16 = (5 × 4) × (5 × 4) = 400
(iii) 25 × 36 = (5 × 6) × (5 × 6) = 900
(iv) 25 × 64 = (5 × 8) × (5 × 8) = 1600

Question 6.
(i) Calculate 11 × 11 and 111 × 111
(ii) Can you guess what 1111 × 1111 would be? Check whether your guess is correct.
(iii) Write the pattern of such products.
Answer:
(i) 11 × 11 = 121
111 × 111 = 12321

(ii) 1111 × 1111= 1234321
Here the digits are count up to the middle and then count down.

(iii) 11111 × 11111 = 123454321
111111 × 111111 = 12345654321
1111111 × 1111111 = 1234567654321
11111111 × 11111111 = 123456787654321

Question 7.
Look at the following calculations:
1 + 3 = 4
4 + 5 = 9
9 + 7 = 16
Continue this to compute the squares up to 100.
Answer:
16 + 9 = 25
25 + 11 = 36
36 + 13 = 49
49 + 15 = 64
64 + 17 = 81
81 + 19 = 100

Question 8.
(i) How many odd numbers, like 1, 3, and 5, are to be added to get 400?
(ii) Which is the last odd number added to this ?
Answer:
(i) The sum of 20 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 = 400
(ii) 39 is the last odd number added to this.

Question 9.
(i) Which is the fiftieth odd number?
(ii) What is the sum of the odd numbers from 1 to this number?
Answer:
(i) The fiftieth odd number can be found by multiplying 50 by 2 and then subtracting 1,
50th odd number = (50 × 2) – 1 = 100 – 1 = 99
(ii) The sum of odd numbers from 1 to 99 = 50 × 50 = 2500

Question 10.
Look at these pictures.
(i) How do we write 25 as a sum like this?
(ii) How about 36?
(iii) Can you write 100 as such a sum.
Answer:
(i) 25 = 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1
(ii) 36 = 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1
(iii) 100 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

Intext Questions And Answers

Starting with a question
Question 1.
Which sum is larger (15 + 8) or (18 + 5)?
Answer:
15 + 8 = (10 + 5) + 8= 10 + (5 + 8)
18 + 5 = (10 +8)+ 5 = 10 +(8+ 5)
Here, (5 + 8) and (8 + 5) both equals to 13.
Therefore, their sum is equal to each other

Question 2.
What about their product?
Answer:
15 × 8 = (10 + 5) × 8 = (10 × 8) + (5 × 8)
18 × 5 = (10 + 8) × 5 = (10 × 5) + (8 × 5)
Here, (5 × 8) and (8 × 5) both equal to 40.
But, (10 × 8) is greater than (10 × 5)
Therefore, (15 × 8) is a larger product than (18×5), and their difference is (10 × 8) – (10 × 5) = 80 – 50 = 30.

Question 3.
Without multiplying, can we determine if 16 × 9 or 19 × 6 is larger?
Answer:
16 × 9 = (10 + 6) × 9 = (10 × 9) + (6 × 9)
19 × 6 = (10 + 9) × 6 = (10 × 6) + (9 × 6)
Here, (6 × 9) and (9 × 6) both equal to 54.
But, (10 × 9) is greater than (10 × 6)
Therefore, (16 × 9) is larger than (19 × 6), and their difference is
(10 × 9) – (10 × 6) = 90 – 60 = 30.

Question 4.
Now in each of the pairs of products given below, can you figure out in your head which is larger and how much more?
(1) 12 × 8; 18 × 2
(2) 17 × 6; 16 × 7
(3) 13 × 9; 19 × 3
(4) 25 × 6; 26 × 5
Answer:
(1) (12 × 8) = (10 + 2) × 8 = (10 × 8) + (2 × 8)
(18 × 2) = (10 + 8) × 2 = (10 × 2) + (8 × 2)
Clearly, we can say that (10 × 8) is greater than (10 × 2)
Therefore, (12 × 8) is greater than (18 × 2), and their difference is
(10 × 8) – (10 × 2) = 80 – 20 = 60.

(2) (17 × 6) = (10 + 7) × 6 = (10 × 6) + (7 × 6)
(16 × 7) = (10 + 6) × 7 = (10 × 7) + (6 × 7)
Clearly, we can say that (10 × 7) is greater than (10 × 6)
Therefore, (16 × 7) is greater than (17 × 6) and their difference is (10 × 7) – (10 × 6) = 70 – 60 = 10.

(3) (13 × 9) = (10 + 3) × 9 = (10 × 9) + (3 × 9)
(19 × 3) = (10 + 9) × 3 = (10 × 3) + (9 × 3)
Clearly, we can say that (10 × 9) is greater than (10 × 3)
Therefore, (13 × 9) is greater than (19 × 3) and their difference is (10 × 9) – (10 × 3) = 90 – 30 = 60.

(4) (25 × 6) = (20 + 5) × 6 = (20 × 6) + (5 × 6)
(26 × 5) = (20 + 6) × 5 = (20 × 5) + (6 × 5)
Clearly, we can say that (20 × 6) is greater than (20 × 5)
Therefore, (25 × 6) is greater than (26 × 5) and their difference is (20 × 6) – (20 × 5) = 120 – 100 = 20.

Question 5.
How can we multiply 15 × 13?
Answer:
We can write 15 and 13 as the sides of rectangle as shown in the figure below

⇒ 15 × 13 = (10 × 10) + (10 × 5) + (10 × 3) + (3 × 5)
= 100 + 50 + 30+ 15
= 195

Question 6.
Compute 16 × 17
Answer:

⇒ 100 + 60 + 70 + 42 = 272

Question 7.
How to compute the following questions
(i) 18 × 19
Answer:

(ii) 14 × 18
Answer:

(iii) 15 × 15
Answer:

(iv) 345 × 26
Answer:

Note: by computing 18 × 12 we get

But, these calculations can be done in the head and the whole multiplication can be written in this shortened form

Question 8.
How can 36 dots be represented in a rectangle?
Answer:

We saw that we can arrange 36 dots in many ways to make rectangles.
But when we make a 6 × 6 square, it’s special because both sides are the same length.

When we arrange a number of dots into a grid where the rows and columns are equal,
we create a square shape. We call such numbers ‘square numbers’

Question 9.
Can you verify if the product of 25 and 16 is a square number?
Answer:
25 = 5 × 5
16 = 4 × 4
25 × 16 = (5 × 5) (4×4)
= (5 × 4)(5 × 4)
= 20 × 20 = 400
Therefore, the product of 25 and 16 is a perfect square.
When you multiply two perfect square numbers, the product will always be another square number

Class 5 Maths Chapter 3 Kerala Syllabus Multiplication Methods Questions and Answers

Question 1.
Look at the following calculations:
4 = 1 + 2 + 1
9 = 1 + 2 + 3 + 2 + 1
16 = 1 + 2 + 3 + 4 + 2 + 1
(i) How do we write 49 as a sum like this?
(ii) How about 81?
(iii) Can you write 121 as such a sum?
Answer:
(i) 49 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 6 + 5 + 4 + 3 + 2 + 1
(ii) 81 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
(iii) 121 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

Question 2.
Look at the following calculations:
1 + 3 = 4
4 + 5 = 9
9 + 7 = 16
Continue this to compute the squares up to 144.
Answer:
1 + 3 = 4
4 + 5 = 9
9 + 7 = 16
16 + 9 = 25
25 + 11 = 36
36+ 13 = 49
49+ 15 = 64
64 + 17 = 81
81 + 19 = 100
100 + 21 = 121
121 + 23 = 144

Question 3.
Calculate each of the products below in your head:
(i) 9 × 25
(ii) 4 × 36
(iii) 4 × 25
Answer:
(i) ( 3 × 5) × (3 × 5) = 15 × 15 = 225
(ii) (2 × 6) × (2 × 6) = 12 × 12 = 144
(iii) (2 × 5) × (2 × 5) = 10 × 10 = 100

Question 4.
(i) How many odd numbers like 1, 3, 5, are to be added to get 123?
(ii) Which is the last odd number added in this?
Answer:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 123
12 odd numbers are needed to get 123 .
(ii) 23 is the last odd number added in this.

Question 5.
A number multiplied by 24 gave 1560.
(i) What will be the product, if the ne×t number is multiplied by 24 ?
(ii) What will be the product if the number just before this is multiplied by 24 ?
Answer:
(i) If we look at the number just after our original number, we can find the new product by simply adding 24 to 1560 .
1560 + 24 = 1584
So, when the ne×t number is multiplied by 24, we get 1584

(ii) Now, let’s look at the number just before our original number. To find this new product, we subtract 24 from 1560
1560- 24 = 1536
So. when the previous number is multiplied by 24. we get 1536

Class 5 Maths Chapter 3 Notes Kerala Syllabus Multiplication Methods

Welcome to the world of multiplication methods. In this chapter, we will learn some amazing techniques to make multiplication easier and more fun. Let’s take a quick look at what we will explore:

• Product Difference: Learn how to compare numbers without actually multiplying them, so you can quickly tell which product is bigger or smaller.
• Rectangular Multiplication: Discover a visual way to multiply using rectangles, making multiplication easier to understand.
• Square Numbers: Explore the special numbers that result from multiplying a number by itself, and see their cool patterns.
• Product of Square Numbers: Find out what happens when we multiply two square numbers together and uncover some interesting facts.

Multiplication with a two-digit number and a one-digit number doesn’t give the same result as when multiplication is done by interchanging the digits in the ones place

More Examples of square numbers are
1 × 1 =1
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
Now look at the pictures given below

Here, 4 can be written as sum of first 2 odd numbers.

Here, 9 can be written as sum of first 3 odd numbers.

Here, 16 can be written as sum of first 4 odd numbers.

Here, 25 can be written as sum of first 5 odd numbers.
Note :
1 = 1 × 1 = 1
1 + 3 = 2 × 2 = 4
1 + 3 + 5 = 3 × 3 = 9
1 + 3 + 5 + 7 = 4 × 4 = 16
1 + 3 + 5 + 7 + 9 = 5 × 5 = 25
When we add up odd numbers starting from 1, like 1, 3, 5, 7, and so on, the total is always a square number. Each time we add more odd numbers, we get a bigger square number as the result.

• Multiplication with a two-digit number and a one-digit number doesn’t give the same result as when a multiplication is done by interchange the digits in the ones place.
• When we arrange a number of dots into a grid where the rows and columns are equal, we create a square shape. We call such numbers ‘square numbers’
• When we add up odd numbers starting from 1, like 1, 3, 5, 7, and so on, the total is always a square number. Each time we add more odd numbers, we get a bigger square number as the result.
• When you multiply two perfect square numbers, the product will always be another square number.