Chapter 1: Real Number - Case Base Questions with solutions (CBSE) | Class 10 Mathematics : | Toppers Hub

 REAL NUMBERS- CASE STUDY

Class 10 Maths Case Study Questions with Solutions – Chapter 1 Real Numbers (Case Study 1 & 2)

The Central Board of Secondary Education (CBSE) has introduced case study-based questions in Class 10 Mathematics to strengthen problem-solving and application skills. Below are Case Study 1, Case Study 2 and Case Study 3 from Chapter 1 – Real Numbers along with step-by-step professional solutions.

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 Case Study 1: Class Library (Sections A & B)

To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections — Section A and Section B of grade X. There are 32 students in Section A and 36 students in Section B.

1. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?
   a) 144  b) 128  c) 288  d) 272

2. If the product of two positive integers is equal to the product of their HCF and LCM, then the HCF (32, 36) is
   a) 2  b) 4  c) 6  d) 8

3. 36 can be expressed as a product of its primes as
   (options provided)

4. 7 is a
   a) Prime number  b) Composite number  c) Neither prime nor composite  d) None of the above

5. If $p$ and $q$ are positive integers such that $p=a^m$ and $q=b^n$, where $a, b$ are prime numbers, then the LCM$(p, q)$ is
   (options provided)

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✅ Solutions – Case Study 1

Q1. Minimum number of books
We need the LCM of 32 and 36.

• $32 = 2^5$

• $36 = 2^2 \times 3^2$
  LCM = $2^5 \times 3^2 = 288$.
  Answer: 288 (option c).

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Q2. HCF (32, 36)
Common factor: $2^2 = 4$.
Answer: 4 (option b).

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Q3. Prime factorisation of 36
$36 = 2^2 \times 3^2$.
Answer: $2^2 \cdot 3^2$.

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Q4. Classification of 7
7 has only two factors (1 and 7).
Answer: Prime number (option a).

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Q5. LCM of $p=a^m, q=b^n$
If $a$ and $b$ are distinct primes, LCM = $a^m \cdot b^n$.
Answer: $a^m b^n$.

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 Case Study 2: Seminar Participants

A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.

1. In each room the same number of participants are to be seated and all of them being in the same subject. Hence, the maximum number of participants that can be accommodated in each room is:
   a) 14  b) 12  c) 16  d) 18

2. What is the minimum number of rooms required during the event?
   a) 11  b) 31  c) 41  d) 21

3. The LCM of 60, 84 and 108 is
   a) 3780  b) 3680  c) 4780  d) 4680

4. The product of HCF and LCM of 60, 84 and 108 is
   a) 55360  b) 35360  c) 45500  d) 45360

5. 108 can be expressed as a product of its primes as
   (options provided)

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✅ Solutions – Case Study 2

Q1. Maximum participants per room
HCF(60, 84, 108) = $2^2 \times 3 = 12$.
Answer: 12 (option b).

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Q2. Minimum number of rooms required
Total participants = 60 + 84 + 108 = 252.
Rooms = $252 \div 12 = 21$.
Answer: 21 (option d).

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Q3. LCM of 60, 84 and 108
LCM = $2^2 \times 3^3 \times 5 \times 7 = 3780$.
Answer: 3780 (option a).

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Q4. Product of HCF and LCM
= $12 \times 3780 = 45360$.
Answer: 45360 (option d).

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Q5. Prime factorisation of 108
$108 = 2^2 \times 3^3$.
Answer: $2^2 \cdot 3^3$.

Case Study 3: Factor Tree & Fundamental Theorem of Arithmetic

A factor tree shows a number $x$ broken into factors: $x = 5 \times 2783$. The number 2783 is further split into $23 \times 121$. The number 121 is then broken into $11 \times 11$. Some values in the tree are marked as $y$ and $z$.

Answer the following questions:

1. What will be the value of $x$?
   a) 15005  b) 13915  c) 56920  d) 17429

2. What will be the value of $y$?
   a) 23  b) 22  c) 11  d) 19

3. What will be the value of $z$?
   a) 22  b) 23  c) 17  d) 19

4. According to Fundamental Theorem of Arithmetic, 13915 is a
   a) Composite number  b) Prime number  c) Neither prime nor composite  d) Even number

5. The prime factorisation of 13915 is
   (options provided in PDF)

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✅ Solutions – Case Study 3

Q1. Value of $x$
From the tree:

$$x = 5 \times 2783,\quad 2783 = 23 \times 121,\quad 121 = 11 \times 11$$

Thus,

$$x = 5 \times 23 \times 11 \times 11 = 13915$$

Answer: 13915 (option b).

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Q2. Value of $y$
From the tree, $y$ corresponds to one of the 11’s from 121.
Answer: 11 (option c).

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Q3. Value of $z$
From the tree, $z = 23$.
Answer: 23 (option b).

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Q4. Classification of 13915
Since 13915 can be expressed as a product of primes, it is a composite number.
Answer: Composite number (option a).

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Q5. Prime factorisation of 13915

$$13915 = 5 \times 11^2 \times 23$$

Answer: $5 \cdot 11^2 \cdot 23$.

 Summary

• Case Study 1 (Library): Introduced the use of HCF and LCM in distributing books and prime factorisation.
• Case Study 2 (Seminar): Applied HCF and LCM to real-life situations involving groups and rooms.
• Case Study 3: is a direct application of the Fundamental Theorem of Arithmetic, a foundational idea in number theory

These cases highlight the importance of the Fundamental Theorem of Arithmetic, prime factorisation, and the relationship between HCF and LCM.

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