Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 9 Number Relations Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 7 Maths Chapter 9 Solutions Number Relations
Class 7 Maths Chapter 9 Number Relations Questions and Answers Kerala State Syllabus
Number Relations Class 7 Questions and Answers Kerala Syllabus
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Now try these problems:
Question 1.
Find the number of factors of each number below:
i) 40
ii) 54
iii) 60
iv) 100
v) 210
Answer:
i) The factors of 40, expressed as powers of prime numbers, are:,
40 = 23 × 5 = 8 × 5
Therefore, number of factors = (3 + 1)(1 + 1) = 4 × 2 = 8
ii) The factors of 54, expressed as powers of prime numbers, are:
54 = 33 × 2 ,
Therefore, number of factors = (3 + 1)(1 + 1) = 4 × 2 = 8
iii) The factors of 60, expressed as powers of prime numbers, are:
60 = 22 × 3 × 5
Therefore, number of factors = (2 + 1) (1 + 1) (1 + 1) = 3 × 2 × 2 = 12
iv) The factors of 100, expressed as powers of prime numbers, are:
100 = 52 × 22
Therefore, number of factors = (2 + 1)(2 + 1) = 3 × 3 = 9
v) The factors of 210, expressed as powers of prime numbers, are:
210 = 7 × 5 × 3 × 2
Therefore, number of factors = (1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 × 2 = 16
Question 2.
From the number of factors of a number, we can deduce some peculiarities of the number. The table below lists these for number of factors up to 5. Extend it to number of factors up to 10
Answer:
| 6 | Fifth power of a prime | p5(p, a prime) |
| Product of two primes | pq2 or p2q(p, q primes) | |
| 7 | Sixth power of a prime | p6(p, a prime) |
| 8 | Seventh power of a prime | p7 (p, a prime) |
| Product of three primes | pqr (p, q, r primes) | |
| Product of two primes | p3q or pq3 (p, q primes) | |
| 9 | Eight power of a prime | p8 (p, a prime) |
| Product of two primes | p2q2 (p, q primes) | |
| 10 | Ninth power of a prime | p9(p, a prime) |
| Product of two primes | pq4 or p4q (p,q primes) |
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Question 1.
For each pair of numbers given below, find the largest common factor and all other common factors:
i) 45, 75
ii) 225, 275
iii) 360, 300
iv) 210, 504
v) 336, 588
Answer:
i) Factors of 45 = 32 × 5
Factors of 75 = 3 × 52
Common primes: 3 and 5
Take the lowest powers: 31 × 51 = 15
Largest common factor: 15
Other common factors: 1,3,5
ii) Factors of 225 = 32 × 52
Factors of 275 = 52 × 11
Common prime: 5
Take the lowest power: 2 = 25
Largest common factor: 25
Other common factors: 1,5
iii) Factors of 360 = 23 × 32 × 5
Factors of 300 = 22 × 3 × 52
Common primes: 2, 3, and 5
Take the lowest powers: 22, 31, 51
Largest common factor = 22 × 31 × 51 = 4 × 3 × 5 = 60
Other common factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
iv) Factors of 210 = 2 × 3 × 5 × 7
Factors of 504 = 23 × 32 × 7
Common primes: 2, 3, and 5
Take the lowest powers: 21, 31, 51
Largest common factor = 21 × 31 × 51 = 2 × 3 × 5 = 30
Other common factors: 1, 2, 3, 5, 6, 10, 15, 30
v) Factors of 336 = 24 × 3 × 7
Factors of 558 = 22 × 3 × 72
Common primes: 2, 3, and 7
Take the lowest powers: 22, 31, 71
Largest common factor = 22 × 31 × 71 = 4 × 3 × 7 = 84
Other common factors: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Question 2.
i) What is the largest common factor of two different prime numbers?
ii) Can the largest common factor of two composite numbers be 1?
iii) If two numbers are divided by their largest common factor, what would be the largest common factor of the quotients?
Answer:
i) The largest common factor of two different prime numbers is 1, since prime numbers have no common factors other than 1.
ii) Yes, the largest common factor of two composite numbers can be 1 if they are co prime, meaning they have no common factors other than 1.
iii) If two numbers are divided by their largest common factor, the largest common factor of the quotients will be I. This is because the largest common factor is the greatest number that divides both original numbers, and dividing by it eliminates any common factors from the quotients.
Class 7 Maths Chapter 9 Kerala Syllabus Number Relations Questions and Answers
Question 1.
Find the number of factors of each number below:
i) 36
ii) 84
iii) 144
Answer:
i) The factors of 36, expressed as powers of prime numbers, are:
36 = 22 × 32
Therefore, number of factors = (2 + 1) (2 +1) = 3 × 3 = 9
ii) The factors of 84, expressed as powers of prime numbers, are:
84 = 22 × 31 × 71
Therefore, the number of factors = (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12
iii) The factors of 144, expressed as powers of prime numbers, are:
144 = 24 × 32
Therefore, the number of factors = (4 + 1)(2 + 1) = 5 × 3 = 15
Question 2.
For each pair of numbers given below, find the largest common factor:
i) 48, 180
ii) 90, 150
iii) 84, 126
Answer:
i) Prime factors of 48 = 24 × 31
Prime factors of 180 = 22 × 32 × 51
Common primes: 2 and 3
Take the lowest powers: 22, 31
Largest Common Factor = 22 × 31 = 4 × 3 = 12
ii) Prime factors of 90 = 21 × 32 × 51
Prime factors of 150 = 21 × 31 × 52
Common primes: 2,3, and 5
Take the lowest powers: 21, 31 , 51
Largest Common Factor = 21 × 31 × 51 = 2 × 3 × 5 = 30
iii) Prime factors of 84 = 22 × 31 × 71
Prime factors of 126 = 21 × 32 × 71
Common primes: 2, 3, and 7
Take the lowest powers: 21, 31, 71
Largest Common Factor = 21 × 31 × 71 = 2 × 3 × 7 = 42
Class 7 Maths Chapter 9 Notes Kerala Syllabus Number Relations
In this chapter, we will explore important concepts related to numbers and their factors. You’ll learn how to determine the number of factors for prime numbers and how to identify common factors between two numbers.
In short, we can explain as,
- To find the number of factors of the product of powers of two prime numbers, we have to add one to each exponent and multiply these numbers.
- To find the common factors of two numbers, first write the factors of each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.
- These fundamental ideas are essential for understanding the relationships between numbers and will help you solve problems with greater confidence.
Number Of Factors
Prime numbers are natural numbers that can only be divided by 1 and the number itself. Examples include 2, 3, 5, 7, and 11.
Each of these numbers has exactly two factors.,
Now, let’s examine how the number of factors changes as the power of a prime number varies.
In general,
The number of factors of a power of any prime number is one more than the exponent.
Algebraically we can say,
If p is a prime number and n is a natural number, then the number of factors of pn is n + 1.
Now, let’s explore how the number of factors changes when a prime number is multiplied by another number.
For 3 × 5 = 15
Factors of 3 = 1, 3
Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Thus, the number of factors = 4
We can tabulate it like this:
| 1 | 3 | (Factors of 3) |
| 5 | 15 | (Factors of 3 multiplied by 5) |
Thus, the number of factors = 4
For 32 × 5 = 45
Factors of 32 as the power of prime numbers = 1, 3, 32 that is 1, 3, 9
Factors when each of them multiplied by 5 are:
Thus, the number of factors = 3 + 3 = 3 × 2 = 6
For 32 × 52 = 225
Thus, the number of factors = 3 × 3 = 9
For 33 × 53
The factors are:
Thus, the number of factors = 4 × 4 = 16
In general,
The number of factors of the product of powers of two prime numbers is got by adding one to each exponent and multiplying these numbers.
Algebraically we can say,
If p and q are two different primes and m, n are any two natural numbers, then the number of factors of pm qn is (m + 1)(n + 1).
Now, let’s consider the situation with three different prime numbers.
Examine the expression 33 × 52 × 11
First, tabulate the factors of 33 × 52
Here, the number of factors for 33 × 52 = 4 × 3 = 12
Then, tabulate the factors of 33 × 52 × 11
Thus, all together 33 × 52 × 11 = 12 × 2 = 24
In general,
To compute the number of factors of a number, we write it a product of powers of different prime numbers, and find the product of the numbers got by adding one to each exponent.
Common Factors
Let’s understand what a common factor of two numbers means:
For that consider the two numbers 180 and 270.
Factors of 180 = 22 × 32 × 5
Factors of 270 = 2 × 33 × 5
Here, the prime common for both numbers are 2, 3, 5
The smaller power of these primes in the factorizations is 2, 32, 5
Here, the common factors are the factors of 2 × 32 × 5
We can tabulate it as follows:
In this case, the largest common factor of 180 and 270 is 90.
In short,
To find the common factors of two numbers, first write each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.
- The number of factors of a power of any prime number is one more than the exponent.
- If p is a prime number and n is a natural number, then the number of factors of pn is n + 1.
- The number of factors of the product of powers of two prime numbers is got by adding one to each exponent and multiplying these numbers.
- If p and q are two different primes and m, n are any two natural numbers, then the number of factors of pmqn is (m + 1)(n + 1).
- To find the common factors of two numbers, first write each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.