The chapter on "Real Numbers" is crucial for understanding the properties and relationships of numbers, forming the basis for solving a wide range of mathematical problems in various fields.
This doc provides Class 10 Maths Formulas for Real Numbers that can help you to succeed in board exams and in future competitive exams.
Example 1: What is the HCF of 36 and 48 using prime factorisation?
Solution: To find the HCF using prime factorisation, follow these steps:
1. Find the prime factors of each number.
Prime factors of 36: 2 × 2 × 3 × 3 (2^2 × 3^2)
Prime factors of 48: 2 × 2 × 2 × 2 × 3 (2^4 × 3)
2. Find the common prime factors.
Common prime factors: 2 × 2 × 3 (2^2 × 3)
3. Multiply the common prime factors to get the HCF.
HCF = 2 × 2 × 3 = 12
So, the HCF of 36 and 48 using prime factorisation is 12.
The decimal expansion of every rational number is either terminating or nonterminating repeating.
A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and the number itself.
In other words, a prime number cannot be formed by multiplying two smaller natural numbers. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, and so on.
For any two positive integers p and q, we have:
HCF (p,q) x LCM [p, q] = p x q
Example 2: There are two positive numbers such that one exceeds the other by 20 and their LCM and HCF are 504 and 4 respectively. Find the numbers.
Solution: Let the numbers be 'x' and 'x + 20'.
We know that:
HCF (p,q) x LCM (p, q) = p x q
x(x + 20) = 4 x 504
x^{2} + 20x = 2016
x^{2} + (56  36)x  2016 = 0
(x + 56)(x  36) = 0
x = 56 & x = 36
x cannot be negative.
So two numbers are: x = 36 & (x + 20) = 56
Example 3: Two positive numbers have a difference of 15. If their LCM (Least Common Multiple) is 360 and their HCF (Highest Common Factor) is 5, what are the values of these numbers?
Let's assume the smaller number is 'x'. Since one number exceeds the other by 15, the larger number would be 'x + 15'.
Using the relationship between LCM, HCF, and the product of two numbers, we have:
HCF(p, q) x LCM(p, q) = p x q
Substituting the given values, we get:
5 x 360 = x(x + 15)
1800 = x^{2}+ 15x
Rearranging the equation, we have:
x^{2} + 15x  1800 = 0
To factorize this quadratic equation, we find two numbers whose sum is 15 and product is 1800. One such pair is 45 and 40.
(x + 45)(x  40) = 0
Setting each factor to zero, we get:
x + 45 = 0 or x  40 = 0
x = 45 or x = 40
Since the numbers are specified to be positive, we discard the negative value. Thus, the smaller number 'x' is 40.
The larger number would be x + 15:
40 + 15 = 55
Therefore, the two numbers are 40 and 55, respectively.
For any three positive integers a, b and c, we have:
Example 4:
Let's apply Euclid's Division Lemma to find the quotient and remainder when 87 is divided by 9.
According to the lemma, for any positive integers a and b, there exist unique integers q and r satisfying:
a = bq + r, 0 ≤ r < b.
In our case, a = 87 and b = 9. We need to find the values of q and r.
By applying Euclid's Division Lemma, we have:
87 = 9q + r
To find the quotient and remainder, we can divide 87 by 9:
87 ÷ 9 = 9 remainder 6
So, in this case, the quotient (q) is 9, and the remainder (r) is 6.
Hence, when 87 is divided by 9, the quotient is 9, and the remainder is 6, as per Euclid's Division Lemma.
Note: It can be extended for all integers, except zero i.e., b ≠ 0.
Example 5: Find HCF of 56 and 72.
Solution:
Apply lemma to 56 and 72.
Take a bigger number and locate ‘b’ and ‘r’. 72 = 56 × 1 + 16
Since 16 ≠ 0, consider 56 as the new dividend and 16 as the new divisor. 56 = 16 × 3 + 8
Again, 8 ≠ 0, consider 16 as the new dividend and 8 as the new divisor. 16 = 8 × 2 + 0.
Since the remainder is zero, divisor 8 is HCF.
116 videos420 docs77 tests

1. What are real numbers? 
2. What is the Fundamental Theorem of Arithmetic? 
3. What are rational numbers? 
4. What is an irrational number? 
5. What is a prime number? 

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