Real Numbers CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Real Numbers CBQs

1. Question 1MCQ1 mark
   There are five bells placed at different swings in a park, which toll at intervals of 2, 3, 5, 6, and 10 minutes, respectively. They all toll together when the park is open for visitors at 10:00 AM. How many more times do they all toll together till the park is closed at 8:00 PM?
   • A. 10
   • B. 20
   • C. 30
   • D. 60
   Show step-wise solution

   1. [1 mark] LCM(2, 3, 5, 6, 10) = 30, so the bells toll together every 30 minutes. From 10:00 AM to 8:00 PM is 10 hours = 600 minutes; 600 ÷ 30 = 20 more times. Answer: B.

   Answer: B. 20

2. Question 2MCQ1 mark
   Let p be a prime number and k be a positive integer. If p divides k², then which of these is DEFINITELY divisible by p? k/2 | k | 7k | k³
   • only k
   • only k and 7k
   • only k, 7k and k³
   • all - k/2, k, 7k and k³
   Show step-wise solution

   1. [1 mark] Correct option: (3) only k, 7k and k³. Since p is prime and p divides k², p must itself divide k. Then 7k and k³ are multiples of k, so p divides both of them as well. k/2 need not even be an integer, so it is not definitely divisible by p.

   Answer: only k, 7k and k³

3. Question 3Assertion-Reason1 mark
   In a laboratory, the number of cells in a culture after n hours is 6ⁿ, where n is a natural number. Assertion (A): The number of cells in the culture can never end with the digit 0. Reason (R): The prime factorisation of 6ⁿ is 2ⁿ × 3ⁿ, which does not contain the prime 5.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): A number ending in 0 is divisible by 10 = 2 × 5, so its prime factorisation must contain 5; by the Fundamental Theorem of Arithmetic 6ⁿ = 2ⁿ × 3ⁿ uniquely, with no factor 5. So A and R are true and R explains A.

   Answer: (A)

4. Question 4Short answer2 marks
   LED light arrangements are made in a marriage function. Yellow lights will flicker every 3 seconds, red lights will flicker every 4 seconds, and green lights will flicker every 5 seconds. How many times will all the three lights flicker together in 30 minutes?
   Show step-wise solution

   1. [1 mark] LCM(3, 4, 5) = 60, so all three lights flicker together every 60 seconds = 1 minute.
   2. [1 mark] In 30 minutes the lights flicker together 30 ÷ 1 = 30 times.

   Answer: 30 times

5. Question 5Case study4 marks
   A sweet shop in Kolkata receives a festival order of 420 rasgullas, 588 sandesh and 252 cham-chams. The shopkeeper packs them in boxes so that each box contains only one kind of sweet, every box contains the same number of sweets, and the number of boxes is as small as possible. Based on the above information, answer the following questions: (i) Write 588 as a product of its prime factors. (ii) How many sweets should be packed in each box? (iii) Find the number of boxes of each kind and the total number of boxes needed.
   Show step-wise solution

   1. [1 mark] (i) 588 = 2² × 3 × 7².
   2. [1 mark] (ii) 420 = 2² × 3 × 5 × 7 and 252 = 2² × 3² × 7, so sweets per box = HCF(420, 588, 252) = 2² × 3 × 7 = 84.
   3. [1 mark] (iii) Boxes: rasgullas = 420 ÷ 84 = 5, sandesh = 588 ÷ 84 = 7, cham-chams = 252 ÷ 84 = 3.
   4. [1 mark] (iii) Total boxes = 5 + 7 + 3 = 15.

   Answer: (i) 588 = 2² × 3 × 7² (ii) 84 sweets per box (iii) 5, 7, 3 boxes; 15 in all

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