Real Numbers – Class 10 Mathematics (Lecture 5)
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HCF of even prime number and 10 is:
A) 2 B) 1 C) 3 D) 4
Solution:
Even prime number = 2 (only even prime number)
10 = 2 × 5
HCF(2, 10) = 2
Answer: A) 2
HCF of smallest prime and smallest composite number is:
A) 1 B) 2 C) 3 D) 4
Solution:
Smallest prime number = 2
Smallest composite number = 4
HCF(2, 4) = 2
Answer: B) 2
If \(p\) and \(q\) are two prime numbers, what is their LCM?
A) 1 B) 0 C) Cannot be determined D) \(p \times q\)
Solution:
If \(p\) and \(q\) are distinct prime numbers, they have no common factors other than 1.
LCM of two numbers with no common factors = product of the numbers
LCM(\(p, q\)) = \(p \times q\)
Note: If \(p = q\) (same prime number), then LCM = \(p\) (or \(q\)), but in most questions, distinct primes are assumed.
Answer: D) \(p \times q\)
If \(p\) and \(q\) are two prime numbers, what is their HCF?
A) 1 B) 0 C) \(p \times q\) D) Cannot be determined
Solution:
If \(p\) and \(q\) are distinct prime numbers, they have no common factors other than 1.
HCF of two numbers with no common factors = 1
HCF(\(p, q\)) = 1
Note: If \(p = q\) (same prime number), then HCF = \(p\) (or \(q\)), but for distinct primes, HCF = 1.
Answer: A) 1
The sum of the exponents of the prime factors in the prime factorisation of 196 is:
A) 1 B) 2 C) 4 D) 6
Solution:
Prime factorisation of 196:
\(196 = 2^2 \times 7^2\)
Exponents: 2 (from \(2^2\)) and 2 (from \(7^2\))
Sum of exponents = 2 + 2 = 4
Answer: C) 4
If two positive numbers \(a\) and \(b\) are expressible as \(a = pq^2\) and \(b = p^3 q\), where \(p\) and \(q\) are prime numbers, then LCM(\(a, b\)) is:
A) \(pq\) B) \(p^3 q^3\) C) \(p^3 q^2\) D) \(p^2 q^2\)
Solution:
Given: \(a = p^1 q^2\) (explicitly: \(p^1 \times q^2\))
\(b = p^3 q^1\)
To find LCM, take the highest power of each prime:
For prime \(p\): highest power = \(p^3\)
For prime \(q\): highest power = \(q^2\)
LCM = \(p^3 \times q^2 = p^3 q^2\)
Answer: C) \(p^3 q^2\)
If the LCM of \(a\) and 18 is 36 and the HCF of \(a\) and 18 is 2, then \(a =\)
A) 2 B) 3 C) 4 D) 1
Solution:
Using the formula: \(a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)\)
Here, \(b = 18\), HCF = 2, LCM = 36
\(a \times 18 = 2 \times 36\)
\(a \times 18 = 72\)
\(a = \frac{72}{18} = 4\)
Answer: C) 4
Check: HCF(4, 18) = 2, LCM(4, 18) = 36 ✓
If HCF(26, 169) = 13, then LCM(26, 169) is:
A) 26 B) 52 C) 338 D) 13
Solution:
Using the formula: \(a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)\)
Here, \(a = 26\), \(b = 169\), HCF = 13
\(26 \times 169 = 13 \times \text{LCM}\)
\(\text{LCM} = \frac{26 \times 169}{13} = 2 \times 169 = 338\)
Answer: C) 338
Check:
26 = 2 × 13
169 = 13 × 13
LCM = 2 × 13 × 13 = 338 ✓