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Q2(ii):
Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (ii) 510 and 92
Solution :
Given: Two integers, $a = 510$ and $b = 92$.
To Find: The Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the given integers, and to verify the relationship: $LCM \times HCF = \text{Product of the two numbers}$.
Step 1: Prime Factorization of the given numbers
We express each number as a product of its prime factors.
For $510$:
$510 = 2 \times 255$
$255 = 3 \times 85$
$85 = 5 \times 17$
$17 = 17 \times 1$
So, $510 = 2^1 \times 3^1 \times 5^1 \times 17^1$
For $92$:
$92 = 2 \times 46$
$46 = 2 \times 23$
$23 = 23 \times 1$
So, $92 = 2^2 \times 23^1$
Step 2: Determining HCF and LCM
HCF (Highest Common Factor): The HCF is the product of the smallest power of each common prime factor in the numbers.
Common prime factor is $2$. The smallest power of $2$ is $2^1$.
$HCF(510, 92) = 2^1 = 2$.
LCM (Least Common Multiple): The LCM is the product of the greatest power of each prime factor involved in the numbers.
Prime factors involved are $2, 3, 5, 17, 23$.
Greatest powers are $2^2, 3^1, 5^1, 17^1, 23^1$.
$LCM(510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1$
$LCM(510, 92) = 4 \times 3 \times 5 \times 17 \times 23$
$LCM(510, 92) = 12 \times 5 \times 17 \times 23$
$LCM(510, 92) = 60 \times 17 \times 23$
$LCM(510, 92) = 1020 \times 23$
$LCM(510, 92) = 23460$.
Step 3: Verification of the relationship
We need to verify if $LCM \times HCF = a \times b$.
Left Hand Side (LHS):
$LCM \times HCF = 23460 \times 2$
$LCM \times HCF = 46920$.
Right Hand Side (RHS):
Product of numbers $= 510 \times 92$
$510 \times 92 = 510 \times (90 + 2)$
$= 45900 + 1020$
$= 46920$.
Since $LHS = RHS$, the relationship is verified.
Final Answer: The HCF is $2$, the LCM is $23460$, and the relationship $LCM \times HCF = \text{Product of the two numbers}$ is verified as $46920 = 46920$.
More Questions from Class 10 Mathematics Real numbers EXERCISE 1.1
- Q1(i): Express each number as a product of its prime factors: (i) 140
- Q1(ii): Express each number as a product of its prime factors: (ii) 156
- Q1(iii): Express each number as a product of its prime factors: (iii) 3825
- Q1(iv): Express each number as a product of its prime factors: (iv) 5005
- Q1(v): Express each number as a product of its prime factors: (v) 7429
- Q2(i): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (i) 26 and 91
- Q2(iii): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (iii) 336 and 54
- Q3(i): Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21
- Q3(ii): Find the LCM and HCF of the following integers by applying the prime factorisation method. (ii) 17, 23 and 29
- Q3(iii): Find the LCM and HCF of the following integers by applying the prime factorisation method. (iii) 8, 9 and 25
- Q4: Given that HCF (306, 657) = 9, find LCM (306, 657).
- Q5: Check whether $6^n$ can end with the digit 0 for any natural number $n$.
- Q6: Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.
- Q7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
CBSE Solutions for Class 10 Mathematics Real numbers
Chapters in CBSE - Class 10 Mathematics
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