Real Number
Previous Year Solved Questions 2019
Q1: If HCF(336, 54) = 6, find LCM(336, 54). (CBSE 2019)
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Ans: 3024
Using HCF × LCM = Product of numbers
6 × LCM = 336 × 54
LCM = (336 × 54) ÷ 6 = 336 × 9 = 3024
Using HCF × LCM = Product of numbers
6 × LCM = 336 × 54
LCM = (336 × 54) ÷ 6 = 336 × 9 = 3024
Q2: The HCF of two numbers a and b is 5 and their LCM is 200. Find the product of ab. (CBSE 2019)
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Ans: 1000
Product = HCF × LCM = 5 × 200 = 1000
Product = HCF × LCM = 5 × 200 = 1000
Q3: If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n. (CBSE 2019)
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Ans: 2
HCF(65,117) = 13
Given: 65n – 117 = 13
65n = 13 + 117 = 130
n = 130 ÷ 65 = 2
HCF(65,117) = 13
Given: 65n – 117 = 13
65n = 13 + 117 = 130
n = 130 ÷ 65 = 2
Q4: Find the HCF of 612 and 1314 using prime factorization. (CBSE 2019)
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Ans: 18
612 = 2² × 3² × 17
1314 = 2 × 3² × 73
Common factors: 2 × 3² = 2 × 9 = 18
612 = 2² × 3² × 17
1314 = 2 × 3² × 73
Common factors: 2 × 3² = 2 × 9 = 18
Q5: Prove that √5 is an irrational number. (CBSE 2019)
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Proof:
Assume √5 is rational: √5 = a/b (a,b coprime, b≠0)
Squaring: 5 = a²/b² ⇒ a² = 5b²
Thus 5 divides a² ⇒ 5 divides a
Let a = 5k
Then: (5k)² = 5b² ⇒ 25k² = 5b² ⇒ b² = 5k²
Thus 5 divides b² ⇒ 5 divides b
Both a and b divisible by 5, contradicting coprime assumption
Hence √5 is irrational.
Assume √5 is rational: √5 = a/b (a,b coprime, b≠0)
Squaring: 5 = a²/b² ⇒ a² = 5b²
Thus 5 divides a² ⇒ 5 divides a
Let a = 5k
Then: (5k)² = 5b² ⇒ 25k² = 5b² ⇒ b² = 5k²
Thus 5 divides b² ⇒ 5 divides b
Both a and b divisible by 5, contradicting coprime assumption
Hence √5 is irrational.
Q6: Prove that √2 is an irrational number. (CBSE 2019)
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Proof:
Assume √2 is rational: √2 = p/q (p,q coprime, q≠0)
Squaring: 2 = p²/q² ⇒ p² = 2q²
Thus p² is even ⇒ p is even
Let p = 2k
Then: (2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k²
Thus q² is even ⇒ q is even
Both p and q even, contradicting coprime assumption
Hence √2 is irrational.
Assume √2 is rational: √2 = p/q (p,q coprime, q≠0)
Squaring: 2 = p²/q² ⇒ p² = 2q²
Thus p² is even ⇒ p is even
Let p = 2k
Then: (2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k²
Thus q² is even ⇒ q is even
Both p and q even, contradicting coprime assumption
Hence √2 is irrational.
Q7: Prove that 2 + 5√3 is an irrational number given that √3 is irrational. (CBSE 2019)
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Proof:
Assume 2 + 5√3 is rational = a/b
Then 5√3 = a/b – 2
√3 = (a – 2b)/(5b)
RHS is rational, but LHS (√3) is irrational → Contradiction
Hence 2 + 5√3 is irrational.
Assume 2 + 5√3 is rational = a/b
Then 5√3 = a/b – 2
√3 = (a – 2b)/(5b)
RHS is rational, but LHS (√3) is irrational → Contradiction
Hence 2 + 5√3 is irrational.
Q8: Write the smallest number which is divisible by both 306 and 657. (CBSE 2019)
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Ans: 22338
306 = 2 × 3² × 17
657 = 3² × 73
LCM = 2 × 3² × 17 × 73 = 2 × 9 × 17 × 73 = 18 × 1241 = 22338
306 = 2 × 3² × 17
657 = 3² × 73
LCM = 2 × 3² × 17 × 73 = 2 × 9 × 17 × 73 = 18 × 1241 = 22338