NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 2(ii)

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 2(ii)

Question:

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: 510 and 92.

Given:

The two numbers are 510 and 92.

To Find:

1. The Highest Common Factor (HCF) of 510 and 92.
2. The Least Common Multiple (LCM) of 510 and 92.
3. Verify that $$\text{LCM} \times \text{HCF} = 510 \times 92$$.

Formula:

$$ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b $$

Solution:

Step 1: Find the prime factorization of 510.

$$ 510 \div 2 = 255 \\ 255 \div 3 = 85 \\ 85 \div 5 = 17 \\ 17 \text{ is prime} $$

So, $$510 = 2 \times 3 \times 5 \times 17$$

Step 2: Find the prime factorization of 92.

$$ 92 \div 2 = 46 \\ 46 \div 2 = 23 \\ 23 \text{ is prime} $$

So, $$92 = 2^2 \times 23$$

Step 3: Find the HCF (Highest Common Factor).

List the prime factors:

  • 510: \(2^1 \times 3^1 \times 5^1 \times 17^1\)
  • 92: \(2^2 \times 23^1\)

Common prime factor: 2
The lowest power of 2 is 1.

$$ \text{HCF}(510, 92) = 2^1 = 2 $$

Step 4: Find the LCM (Least Common Multiple).

Take all prime factors with the highest exponent:

  • 2: highest is 2 (\(2^2\))
  • 3: highest is 1 (\(3^1\))
  • 5: highest is 1 (\(5^1\))
  • 17: highest is 1 (\(17^1\))
  • 23: highest is 1 (\(23^1\))
$$ \text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 $$

Step 5: Calculate the value of the LCM.

$$ 2^2 = 4 \\ 4 \times 3 = 12 \\ 12 \times 5 = 60 \\ 60 \times 17 = 1020 \\ 1020 \times 23 = 23,460 $$

So, $$\text{LCM}(510, 92) = 23,460$$

Step 6: Verify the relationship $$\text{LCM} \times \text{HCF} = 510 \times 92$$.

$$ \text{LCM} \times \text{HCF} = 23,460 \times 2 = 46,920 $$ $$ 510 \times 92 = 46,920 $$

Both values are equal.

Result:

  • HCF(510, 92) = 2
  • LCM(510, 92) = 23,460
  • Verification: $$23,460 \times 2 = 46,920 = 510 \times 92$$ ✓
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