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.