Euclid's lemma, fundamental theorem, irrationals, HCF/LCM MCQs
Practice Euclid's lemma, fundamental theorem, irrationals, HCF/LCM multiple-choice questions from Real Numbers (Class 10 Maths) - tap an answer for instant feedback and a step-by-step solution. Practice the full set free on the RankByte app.
Euclid's lemma, fundamental theorem, irrationals, HCF/LCMQuiz - Solve & Score
Q1. How many positive integers from 1 to 1000 are divisible by 2 or 5?
- A.600
- B.500
- C.700
- D.550
Answer: A. 600
Spot-the-setup - a typical math numerical. The data on the table: 1, 1000, 2, 5. We are after How many positive integers from 1 to 1000 are divisible by 2 or 5?. Tool of choice - Inclusion-exclusion: ⌊1000/2⌋ + ⌊1000/5⌋ − ⌊1000/10⌋ = 500 + 200 − 100 = 600. Rearrange it for the unknown before substituting. Numbers in: Inclusion-exclusion: ⌊1000/2⌋ + ⌊1000/5⌋ − ⌊1000/10⌋ = 500 + 200 − 100 = 600. Lock in option A) 600.
Q2. The decimal expansion of 17/3125 terminates after how many places?
- A.5
- B.4
- C.3
- D.Does not terminate
Answer: A. 5
Given: 17; 3125. To find: how many places?. The relation that links these is - 3125 = 5⁵, of the form 2⁰·5⁵, so it terminates after max(0,5)=5 places. This is the equation that links the given quantities to the unknown (math, chapter 'Real Numbers'). Substituting: 3125 = 5⁵, of the form 2⁰·5⁵, so it terminates after max(0,5)=5 places → 17/3125 = 17·32/10⁵ = 544/100000 = 0.00544. Thus the answer is A) 5.
Q3. Which of the following has a non-terminating recurring decimal?
- A.23/(2³·5²·3)
- B.13/3125
- C.7/8
- D.23/200
Answer: A. 23/(2³·5²·3)
Key chapter idea: A reduced rational p/q has terminating decimal iff q is of the form 2^a·5^b. 23/(2³·5²·3) has factor 3 in the denominator, so it recurs. So the correct option is A) 23/(2³·5²·3).
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