Real Numbers — Previous Year Questions (Class 10 Mathematics)
Real Numbers covers Euclid's division lemma, HCF and LCM, and the fundamental theorem of arithmetic. This chapter forms the foundation for algebraic proble
TL;DR: Real Numbers covers Euclid's division lemma, HCF and LCM, and the fundamental theorem of arithmetic. This chapter forms the foundation for algebraic p…
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Real Numbers covers Euclid's division lemma, HCF and LCM, and the fundamental theorem of arithmetic. This chapter forms the foundation for algebraic proble
Real Numbers — Previous Year Questions with Solutions
Q (2023, 3 marks): Using Euclid's division lemma, find the HCF of 135 and 225.
Answer: Apply Euclid's division lemma: a = bq + r
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
Since remainder is 0, HCF(135, 225) = 45.
Q (2022, 3 marks): Find LCM and HCF of 26 and 91. Verify that HCF × LCM = 26 × 91.
Answer: Using prime factorization:
26 = 2 × 13
91 = 7 × 13
HCF = 13 (common factor)
LCM = 2 × 7 × 13 = 182
Verification: 13 × 182 = 2366 and 26 × 91 = 2366
HCF × LCM = Product of numbers ✓
Q (2021, 2 marks): Prove that 3 + 2√5 is irrational.
Answer: Assume 3 + 2√5 is rational = p/q (p, q coprime integers)
Then 2√5 = p/q - 3 = (p - 3q)/q
√5 = (p - 3q)/(2q)
Since p, q are integers, RHS is rational, but √5 is irrational.
This is a contradiction. Therefore, 3 + 2√5 is irrational.
Q (2023, 2 marks): Express 0.342342342... as a fraction in lowest terms.
Answer: Let x = 0.342342342... (period = 3)
1000x = 342.342342...
1000x - x = 342
999x = 342
x = 342/999
Find HCF(342, 999): 342 = 2 × 171 = 2 × 9 × 19, 999 = 27 × 37 = 3^3 × 37
Wait, recalculate: 999 = 3 × 333 = 3 × 3 × 111 = 9 × 111 = 9 × 3 × 37 = 27 × 37
342 = 2 × 171 = 2 × 9 × 19... Let me check: 342/9 = 38, so 342 = 9 × 38
HCF(342, 999) = 9
342/999 = 38/111
Simplify further: 38 = 2 × 19, 111 = 3 × 37, so HCF = 1
Answer: 38/111
Q (2022, 3 marks): Show that the square of an odd integer leaves remainder 1 when divided by 8.
Answer: Let odd integer = 2k + 1 (k is integer)
(2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1
Note: k(k + 1) is always even (product of consecutive integers)
Let k(k + 1) = 2m for some integer m
(2k + 1)^2 = 4(2m) + 1 = 8m + 1
Therefore, square of odd integer leaves remainder 1 when divided by 8.
Q (2023, 1 mark): Check whether 6^n can end with 0 for any positive integer n.
Answer: For a number to end with 0, it must be divisible by 10 = 2 × 5
Prime factorization of 6^n = (2 × 3)^n = 2^n × 3^n
For divisibility by 10, we need both 2 and 5 as factors.
6^n contains 2^n and 3^n but no factor of 5.
Therefore, 6^n can never end with 0 for any positive integer n.
Frequently Asked Questions
What is Euclid's division lemma?
For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Here q is the quotient and r is the remainder.
Why is the relationship HCF × LCM = product of two numbers important?
This relationship allows us to find LCM easily if HCF is known, or vice versa. It's useful in solving problems related to common multiples and divisors.
More Class 10 Mathematics PYQs
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