Class 10 Maths Chapter 1 Real Numbers Case Study with Answers

This set of case-study questions for Chapter 1: Real Numbers Class 10 aligned to CBSE and NCERT provides short, exam-style scenarios that build reasoning and calculation skills in topics like Euclid’s division algorithm, HCF–LCM, rationality and irrationality, and the Fundamental Theorem of Arithmetic; each question includes step-by-step solutions, common-trap warnings, and a downloadable PDF with answer keys and timed practice sets for quick revision, classroom worksheets, and focused pre-exam practice.

Solved Real Numbers Case Study Questions and Answers

CBSE's case-based questions usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Case Study 1: The Class Library

Setting up a shared reading corner

To improve reading habits, you and two classmates have been asked to set up a small library for the two Class 10 sections. Section A has 32 students and Section B has 36. The school wants identical sets of books bought in bulk, enough that either section, working alone, could split the full stock into equal stacks with nothing left over.

Q1: What is the minimum number of books needed so they divide equally between the two sections?

(a) 144  (b) 128

(c) 288  (d) 272

Q2: Using product = HCF × LCM, the HCF(32, 36) is:

(a) 2  (b) 4

(c) 6  (d) 8

Q3: 36 expressed as a product of its primes is:

(a) 2² × 3²   (b) 2¹ × 3³

(c) 2³ × 3¹   (d) 2⁰ × 3⁰

Q4: 7 is a:

(a) Prime number

(b) Composite number

(c) Neither prime nor composite

(d) None of these

Q5: If p = ab² and q = a²b, where a and b are prime numbers, then LCM(p, q) is:

(a) ab     (b) a²b²

(c) a³b²  (d) a³b³

Solution:

Q1: 32 = 2⁵ and 36 = 2² × 3². The smallest number divisible by both is the LCM. Take the highest power of every prime involved: 2⁵ × 3² = 32 × 9 = 288.

Q2: 32 × 36 = 1152. Since HCF × LCM = 1152 and we already found LCM = 288, HCF = 1152 ÷ 288 = 4.
You can also read it straight off the factorisations: the only shared prime is 2, and its smallest shared power is 2² = 4.

Q3: 36 → 18 → 9 → 3, dividing by 2 twice and then 3 twice: 36 = 2² × 3².

Q4: 7 has exactly two factors, 1 and itself, so it's prime.

Q5: p has a to the power 1 and b to the power 2; q has a to the power 2 and b to the power 1. LCM takes the higher power of each: a²b². 

Case Study 2: The Seminar Hall

Allocating rooms for three workshops

A college fest is running three workshops at the same time: Hindi with 60 participants, English with 84, and Mathematics with 108. The organisers want every room to hold the same number of attendees, with everyone in a room from the same workshop, and they want each room as full as possible.

Q1: The largest number of participants a room can hold, keeping every room subject-pure, is:

(a) 6    (b) 12

(c)18   (d)24

Q2: How many rooms are needed in total?

Q3: LCM(60, 84, 108) equals:

(a) 1260  (b)2520

(c) 3780  (d)4200

Q4: Does HCF × LCM = product of the numbers still hold for these three numbers? Check it.

Solution: 

Q1: 60 = 2²×3×5, 84 = 2²×3×7, 108 = 2²×3³. The only primes common to all three are 2 and 3, at their smallest shared powers: HCF = 2²×3 = 12.

Q2: Divide each workshop's headcount by 12: 60÷12 = 5 rooms, 84÷12 = 7 rooms, 108÷12 = 9 rooms. Total = 5 + 7 + 9 = 21 rooms.

Q3: Take the highest power of every prime appearing anywhere: 2²×3³×5×7 = 4 × 27 × 5 × 7 = 3780.

Q4: 60 × 84 × 108 = 544,320. But HCF × LCM = 12 × 3780 = 45,360. The rule fails for three numbers. It's only guaranteed for exactly two positive integers; for three or more, you have to calculate HCF and LCM directly from the prime factorisations.

Case Study 3: The Irrationality Showdown

A whiteboard argument at debate club

At the school debate club, Meher claims that √5 can be written as a simple fraction. Kabir disagrees, and offers to prove her wrong on the whiteboard using a method called proof by contradiction.

Q1: Kabir's proof will lean most directly on which earlier result?

(a) Euclid's Division Lemma

(b) If a prime divides a², it divides a

(c) BODMAS rule

(d) Pythagoras theorem

Q2: Kabir assumes √5 = p/q with p, q coprime. Squaring gives 5q² = ?, which means 5 divides ?.

Q3: True or False: once Kabir shows 5 also divides q, the proof is complete because it confirms the original assumption.

(a) True

(b) False

Q4: Which of these could be proved irrational using exactly the same method?

(a) √9   (b)√16

(c) √7   (c)√25

Q5: Is 5 − √3 rational or irrational? Justify in one line.

Solution

Q1: Every irrationality proof in this chapter is built on Theorem 1.2: if a prime p divides a², then p divides a. Thus ‘5 divides p squared’ turns into ‘5 divides p’ itself.

Q2: 5q² = p², so 5 divides p², which by Theorem 1.2 means 5 divides p as well.

Q3: False. The proof is complete because it does the opposite of confirming anything. It shows p and q share a factor of 5, which directly contradicts the starting assumption that they were coprime. Based on contradiction, we prove that √5 to be irrational.

Q4: √7, because 7 is prime, the same contradiction argument applies directly. √9, √16 and √25 are 3, 4 and 5 respectively. They are already whole numbers, so there's no irrationality to prove.

Q5: Irrational. If 5 − √3 were rational, rearranging would express √3 as a difference of two rational numbers, which would make √3 rational too, contradicting what's already proven about √3.

Quick Overview of Real Numbers

 

Click below to download your free Case Study Questions PDF with worked-out examples for Class 10 Chapter 1: Real Numbers, perfect for last-minute CBSE exam revision.

Class 10 Chapter 1: Real Numbers Case Study PDF