Kerala Syllabus Class 5 Maths Chapter 11 Solutions Within Numbers

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:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 4
This shows 91 = 13 × 7
That is, 91 is a multiple of 7.

ii) 9, 127
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 5
This shows 127 = (14 × 9) + 1
That is, 91 is not a multiple of 7.

iii) 12, 136
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 6
This shows
136 = (11 × 12) + 4
That is, 136 is not a multiple of 12.

iv) 15, 225
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 7
This shows
225 = 15 × 15
That is, 225 is a multiple of 15.

Kerala Syllabus Class 5 Maths Chapter 11 Solutions Within Numbers

Question 2.
Now write the relation between each pair of numbers below like this, as multiplication-multiple and division-factor.
i) 9,72
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 9

ii) 12, 156
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 10

iii) 13, 169
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 11

iv) 25, 375
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 12

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

Kerala Syllabus Class 5 Maths Chapter 11 Solutions Within Numbers

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,
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 2
This shows
215 = 43 × 5
Thus, 215 is indeed a multiple of 5.

(ii) Let’s divide 168 by 5,
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 3
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:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 14

ii) 15, 120
Answer:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 15

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.
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 1
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.
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 8

Kerala Syllabus Class 5 Maths Chapter 11 Solutions Within Numbers

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:
Kerala Syllabus Class 7 Maths Chapter 11 Solutions Within Numbers 13
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.

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