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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
Solution :
Given: Two integers, $a = 26$ and $b = 91$.
To Find: The Highest Common Factor (HCF) and Least Common Multiple (LCM) of 26 and 91, and to verify the relationship: $LCM \times HCF = a \times b$.
Step 1: Prime Factorization of the given numbers
To find the HCF and LCM, we first express each number as a product of its prime factors.
For 26:
$26 = 2 \times 13$
[Since 26 is an even number, we divide by the smallest prime 2, leaving 13, which is prime.]
For 91:
$91 = 7 \times 13$
[Since 91 is not divisible by 2, 3, or 5, we test 7: $91 \div 7 = 13$. Both 7 and 13 are prime numbers.]
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: $13^1$
$HCF(26, 91) = 13$
LCM (Least Common Multiple): The LCM is the product of the greatest power of each prime factor involved in the numbers.
Prime factors involved: $2^1, 7^1, 13^1$
$LCM(26, 91) = 2 \times 7 \times 13$
$LCM(26, 91) = 14 \times 13$
$LCM(26, 91) = 182$
Step 3: Verification of the relationship $LCM \times HCF = a \times b$
Left Hand Side (LHS):
$LCM \times HCF = 182 \times 13$
Calculation:
$182 \times 10 = 1820$
$182 \times 3 = 546$
$1820 + 546 = 2366$
$LHS = 2366$
Right Hand Side (RHS):
$Product of numbers = 26 \times 91$
Calculation:
$26 \times 90 = 2340$
$26 \times 1 = 26$
$2340 + 26 = 2366$
$RHS = 2366$
Conclusion:
Since $LHS = RHS$ ($2366 = 2366$), the relationship is verified.
Final Answer: The HCF is 13, the LCM is 182, and the relationship $LCM \times HCF = \text{product of the two numbers}$ is verified.
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(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
- 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
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