CBSE Class 10 Mathematics Worksheet
Chapter 1: Real Numbers – Important Practice Questions
General Instructions:
- All questions are compulsory
- Questions 1-5 carry 1 mark each
- Questions 6-8 carry 2 marks each
- Questions 9-10 carry 3 marks each
- Question 11 carries 5 marks
The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
96 = 25 × 3
404 = 22 × 101
HCF = Product of smallest power of common prime factors = 22 = 4
17/8 = 17/23
Since denominator is of form 2n, it has terminating decimal expansion.
156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
A rational number p/q has terminating decimal expansion if q can be expressed in the form 2n × 5m where n, m are non-negative integers.
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
HCF = 45
6n = (2 × 3)n = 2n × 3n
For a number to end with 0, its prime factors must include both 2 and 5.
Since there’s no factor of 5, 6n cannot end with digit 0.
7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 13 × 2 × 3 × 13
Since it has more than two distinct prime factors (2, 3, and 13), it is a composite number.
Assume √5 is rational. Then √5 = a/b where a, b are coprime integers.
Squaring: 5 = a²/b² ⇒ 5b² = a² ⇒ a² is divisible by 5 ⇒ a is divisible by 5.
Let a = 5k. Then 5b² = (5k)² ⇒ 5b² = 25k² ⇒ b² = 5k² ⇒ b is divisible by 5.
This contradicts that a, b are coprime. Hence, √5 is irrational.
12 = 22 × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3 (only common prime factor)
LCM = 22 × 3 × 5 × 7 = 420
(a) Let a be any positive integer and b = 3. By Euclid’s division lemma:
a = 3q + r where r = 0, 1, or 2
Case 1: r = 0 ⇒ a = 3q ⇒ a² = 9q² = 3(3q²) = 3m
Case 2: r = 1 ⇒ a = 3q + 1 ⇒ a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1
Case 3: r = 2 ⇒ a = 3q + 2 ⇒ a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1
Hence proved.
(b) Time until they meet = LCM(18, 12)
18 = 2 × 32
12 = 22 × 3
LCM = 22 × 32 = 36 minutes