Real Numbers | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

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10th

10th - Mathematics

Real Numbers

Chapter Overview: Real Numbers

This chapter introduces the concept of Real Numbers, which includes rational and irrational numbers. Students will learn about the fundamental theorem of arithmetic, Euclid's division algorithm, and the properties of real numbers through various theorems and examples.

1. Introduction to Real Numbers

Real Numbers: The collection of all rational and irrational numbers forms the set of real numbers.

Real numbers can be represented on the number line and include integers, fractions, and non-terminating non-repeating decimals.

2. Euclid's Division Algorithm

Euclid's Division Lemma: Given two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

This lemma is the basis for Euclid's division algorithm, which is used to find the HCF (Highest Common Factor) of two numbers.

3. The Fundamental Theorem of Arithmetic

Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order of the prime factors.

This theorem helps in understanding the prime factorization of numbers and their properties.

4. Irrational Numbers

Irrational Numbers: A number is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.

Examples include √2, √3, and π. Proofs of irrationality for numbers like √2 are discussed in detail.

5. Rational Numbers and Their Decimal Expansions

Rational Numbers: A number is rational if it can be expressed in the form p/q, where p and q are integers and q ≠ 0.

The decimal expansion of a rational number is either terminating or non-terminating repeating.

6. Operations on Real Numbers

Real numbers follow the commutative, associative, and distributive laws under addition and multiplication. The chapter also covers the method of rationalizing denominators.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

What is the HCF of 12 and 18?

Answer:

Question 2:

Express 0.375 as a fraction in simplest form.

Answer:

3/8

Question 3:

Is √2 a rational number?

Answer:

Question 4:

Find the LCM of 8 and 12.

Answer:

Question 5:

Write the decimal expansion of 3/8.

Answer:

0.375

Question 6:

What is the Euclid's division lemma for 17 and 5?

Answer:

17 = 5 × 3 + 2

Question 7:

Is 1 a prime number?

Answer:

Question 8:

Find the HCF of 15 and 20 using prime factorization.

Answer:

Question 9:

What is the fundamental theorem of arithmetic?

Answer:

Every integer >1 has a unique prime factorization.

Question 10:

Express 5/6 as a recurring decimal.

Answer:

0.83̅

Question 11:

Find the irrational number between 1 and 2.

Answer:

√2 (≈1.414)

Question 12:

What is the HCF of two consecutive integers?

Answer:

Question 13:

What is the Euclid's division lemma?

Answer:

Euclid's division lemma states that for any two positive integers a and b, there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b.

Question 14:

Find the HCF of 56 and 72 using the prime factorization method.

Answer:

Prime factors of 56: 2 × 2 × 2 × 7 = 2³ × 7
Prime factors of 72: 2 × 2 × 2 × 3 × 3 = 2³ × 3²
HCF is the product of the lowest powers of common prime factors: 2³ = 8

Question 15:

State whether the number 3.24636363... is rational or irrational.

Answer:

The number 3.24636363... is rational because it has a repeating decimal pattern (63), which means it can be expressed as a fraction.

Question 16:

Express 0.375 as a fraction in its simplest form.

Answer:

0.375 = 375/1000
Simplify by dividing numerator and denominator by 125: 3/8

Question 17:

Find the LCM of 12 and 18 using the division method.

Answer:

Divide by common prime factors:
12, 18 ÷ 2 → 6, 9
6, 9 ÷ 3 → 2, 3
2, 3 ÷ 2 → 1, 3
1, 3 ÷ 3 → 1, 1
LCM = 2 × 3 × 2 × 3 = 36

Question 18:

Prove that √5 is an irrational number.

Answer:

Assume √5 is rational, so √5 = a/b where a and b are co-prime.
Squaring both sides: 5 = a²/b² → 5b² = a²
Thus, a² is divisible by 5, so a is divisible by 5.
Let a = 5k, then 5b² = (5k)² → 5b² = 25k² → b² = 5k².
This implies b² is divisible by 5, so b is divisible by 5.
But this contradicts the assumption that a and b are co-prime. Hence, √5 is irrational.

Question 19:

What is the HCF of two consecutive integers?

Answer:

The HCF of two consecutive integers is always 1 because they have no common prime factors other than 1.

Question 20:

Write the decimal expansion of 1/7 and identify its type.

Answer:

1/7 = 0.̅142857̅ (repeating decimal)
It is a non-terminating repeating decimal, hence a rational number.

Question 21:

If the LCM of two numbers is 120 and their HCF is 4, and one number is 20, find the other number.

Answer:

Using the formula: LCM × HCF = Product of the two numbers
120 × 4 = 20 × other number
480 = 20 × other number
Other number = 480 ÷ 20 = 24

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

State the Euclid's Division Lemma for positive integers.

Answer:

For any two positive integers a and b, there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b.

Question 2:

What is the HCF of the smallest prime number and the smallest composite number?

Answer:

The smallest prime number is 2 and the smallest composite number is 4.
HCF of 2 and 4 is 2.

Question 3:

Express 0.375 as a fraction in simplest form.

Answer:

0.375 = 375/1000
Simplify by dividing numerator and denominator by 125:
= 3/8.

Question 4:

If the HCF of two numbers is 12 and their LCM is 180, and one number is 36, find the other number.

Answer:

Using the formula:
HCF × LCM = Product of two numbers
12 × 180 = 36 × Other number
Other number = (12 × 180)/36 = 60.

Question 5:

Find the HCF of 96 and 404 using the Euclid's Division Algorithm.

Answer:

Using Euclid's Division Algorithm:
404 = 96 × 4 + 20
96 = 20 × 4 + 16
20 = 16 × 1 + 4
16 = 4 × 4 + 0
HCF is the last non-zero remainder, which is 4.

Question 6:

Write the decimal expansion of 13/6250 without actual division.

Answer:

13/6250 = 13/(2 × 5⁵) = (13 × 2⁴)/(2⁵ × 5⁵) = 208/100000 = 0.00208.

Question 7:

Check whether 6ⁿ can end with the digit 0 for any natural number n.

Answer:

For a number to end with 0, it must have both 2 and 5 as prime factors.
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ.
Since 5 is not a prime factor of 6ⁿ, it can never end with 0.

Question 8:

Find the LCM of 12, 15, and 21 using the prime factorization method.

Answer:

Prime factors:
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
LCM = 2² × 3 × 5 × 7 = 420.

Question 9:

Explain why 7 × 11 × 13 + 13 is a composite number.

Answer:

7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78.
Since it has factors other than 1 and itself (13 and 78), it is a composite number.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

Explain why every composite number can be expressed as a product of prime numbers, and give an example.

Answer:

Every composite number can be broken down into a product of prime numbers because of the Fundamental Theorem of Arithmetic, which states that every number greater than 1 is either a prime or can be uniquely factorized into primes.

For example, take the number 24:
24 = 2 × 12
12 = 2 × 6
6 = 2 × 3
So, 24 = 2 × 2 × 2 × 3 (or 2³ × 3).

This factorization is unique, regardless of the order of the primes.

Question 2:

Using the Euclidean algorithm, find the HCF of 56 and 72.

Answer:

To find the HCF of 56 and 72 using the Euclidean algorithm, follow these steps:

Step 1: Divide 72 by 56.
72 = 56 × 1 + 16 (remainder is 16).

Step 2: Now, divide 56 by the remainder 16.
56 = 16 × 3 + 8 (remainder is 8).

Step 3: Next, divide 16 by the remainder 8.
16 = 8 × 2 + 0 (remainder is 0).

Since the remainder is now 0, the HCF is the last non-zero remainder, which is 8.

Question 3:

Explain how the decimal expansion of a rational number is either terminating or non-terminating repeating.

Answer:

A rational number is of the form p/q, where p and q are integers and q ≠ 0.

The decimal expansion depends on the denominator q after simplifying the fraction:

If the prime factorization of q contains only 2 and/or 5, the decimal expansion is terminating (e.g., 3/8 = 0.375, since 8 = 2³).
If q has any prime factor other than 2 or 5, the decimal expansion is non-terminating repeating (e.g., 1/3 = 0.333..., since 3 is a prime other than 2 or 5).

This is because the decimal system is based on powers of 10 (2 × 5), so only these primes ensure exact division.

Question 4:

Find the LCM and HCF of 12 and 18 using the prime factorization method.

Answer:

First, find the prime factors of both numbers:

12 = 2 × 2 × 3 = 2² × 3¹
18 = 2 × 3 × 3 = 2¹ × 3²

HCF is the product of the lowest power of common primes:
HCF = 2¹ × 3¹ = 6

LCM is the product of the highest power of all primes:
LCM = 2² × 3² = 4 × 9 = 36

Thus, HCF = 6 and LCM = 36.

Question 5:

Explain why every composite number can be expressed as a product of primes, with an example.

Answer:

The Fundamental Theorem of Arithmetic states that every composite number can be uniquely expressed as a product of prime numbers, regardless of the order.

For example, take the number 12:
12 = 2 × 2 × 3
Here, 2 and 3 are primes, and their product gives the composite number 12.

This property is crucial for simplifying fractions, finding LCM, and HCF.

Question 6:

Using the Euclid's Division Lemma, find the HCF of 56 and 72.

Answer:

Euclid's Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that:
a = bq + r, where 0 ≤ r < b.

Applying this to 72 and 56:
72 = 56 × 1 + 16
56 = 16 × 3 + 8
16 = 8 × 2 + 0

Since the remainder is now 0, the HCF is the last non-zero remainder, which is 8.

Question 7:

Prove that √5 is an irrational number.

Answer:

Assume √5 is rational. Then, it can be written as √5 = a/b, where a and b are co-prime integers.

Squaring both sides:
5 = a²/b² ⇒ a² = 5b²
This means a² is divisible by 5, so a must also be divisible by 5.

Let a = 5k. Substituting:
(5k)² = 5b² ⇒ 25k² = 5b² ⇒ b² = 5k²
Thus, b² is divisible by 5, so b must also be divisible by 5.

But this contradicts our assumption that a and b are co-prime. Hence, √5 is irrational.

Question 8:

Find the LCM and HCF of 12 and 18 using the prime factorization method.

Answer:

Question 9:

Explain why the decimal expansion of a rational number is either terminating or non-terminating repeating.

Answer:

A rational number can be written as p/q, where q ≠ 0 and p, q are integers. The decimal expansion depends on the prime factors of q after simplifying:

If q has only 2 and/or 5 as prime factors, the decimal expansion is terminating (e.g., 1/2 = 0.5).
If q has other primes, the expansion is non-terminating repeating (e.g., 1/3 = 0.333...).

This is because in base 10, only denominators with factors 2 or 5 divide evenly, while others lead to repeating remainders.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

Prove that √5 is an irrational number using the method of contradiction. Explain each step clearly.

Answer:

Introduction

We studied that irrational numbers cannot be expressed as p/q where p and q are integers. Here, we prove √5 is irrational.

Argument 1

Assume √5 is rational, so √5 = a/b (a, b are co-prime).
Squaring both sides: 5 = a²/b² → 5b² = a².

Argument 2

Thus, a² is divisible by 5 → a is divisible by 5 (Let a = 5c).
Substituting: 5b² = 25c² → b² = 5c² → b is divisible by 5.

Conclusion

This contradicts our assumption that a and b are co-prime. Hence, √5 is irrational.

Question 2:

A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to stack them in the least number of boxes such that each box has the same number of sweets. Find the maximum number of sweets per box and the total boxes needed.

Answer:

Introduction

Our textbook shows that such problems use HCF to find the maximum equal distribution. Here, we find HCF of 420 and 130.

Argument 1

Using Euclid’s algorithm: 420 = 130 × 3 + 30.
130 = 30 × 4 + 10 → 30 = 10 × 3 + 0.

Argument 2

HCF is 10, so max sweets per box = 10.
Total boxes = (420 + 130)/10 = 55.

Conclusion

Thus, 10 sweets per box and 55 boxes are needed. This ensures minimal wastage.

Question 3:

Prove that √5 is an irrational number using the method of contradiction. Refer to NCERT Class 10 Example 10.

Answer:

Introduction

We studied that irrational numbers cannot be expressed as p/q where p and q are integers. Our textbook shows a proof for √2, and we apply the same logic here.

Argument 1

Assume √5 is rational, so √5 = a/b (a, b co-prime).
Squaring both sides: 5 = a²/b² → 5b² = a².

Argument 2

Thus, a² is divisible by 5 → a is divisible by 5.
Let a = 5k, then 5b² = 25k² → b² = 5k² → b is divisible by 5.

Conclusion

This contradicts our assumption that a and b are co-prime. Hence, √5 is irrational.

Question 4:

A sweet seller has 420 rasgullas and 130 barfis. He wants to stack them in equal rows with no sweets left over. Find the maximum number of rows possible using the Euclid's division algorithm (NCERT Class 10 Example 2).

Answer:

Introduction

We need to find the HCF of 420 and 130 to determine the maximum rows. Our textbook uses Euclid's algorithm for such problems.

Argument 1

Apply Euclid’s lemma: 420 = 130 × 3 + 30.
Now, 130 = 30 × 4 + 10.

Argument 2

Next, 30 = 10 × 3 + 0.
Since remainder is 0, HCF is 10.

Conclusion

The seller can make a maximum of 10 rows, with each row having 42 rasgullas and 13 barfis.

Question 5:

Prove that √5 is an irrational number using the method of contradiction. Explain each step clearly.

Answer:

Introduction

We studied that irrational numbers cannot be expressed as fractions. Here, we prove √5 is irrational.

Argument 1

Assume √5 is rational, so √5 = a/b (a, b co-prime).
Squaring both sides: 5 = a²/b² → 5b² = a².

Argument 2

Thus, a² is divisible by 5 → a is divisible by 5 (Let a = 5c).
Substituting: 5b² = 25c² → b² = 5c² → b is divisible by 5.

Conclusion

This contradicts our assumption that a and b are co-prime. Hence, √5 is irrational.

Question 6:

A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to stack them in identical boxes so that each box has equal numbers of both sweets. What is the maximum number of boxes she can use? Explain using the Euclid's division algorithm.

Answer:

Introduction

Our textbook shows that Euclid's algorithm helps find HCF, which solves such problems.

Argument 1

Find HCF of 420 and 130 using Euclid's algorithm.
420 = 130 × 3 + 30 → 130 = 30 × 4 + 10 → 30 = 10 × 3 + 0.

Argument 2

HCF is the last non-zero remainder, which is 10.
Thus, maximum boxes = HCF = 10.

Conclusion

She can use 10 boxes, each containing 42 kaju and 13 badam barfis.

Question 7:

A sweet seller has 420 kaju barfis and 130 badam barfis. He wants to stack them in the least number of boxes such that each box has the same number of sweets. Find the maximum number of sweets per box and the total boxes needed.

Answer:

Introduction

Our textbook shows that such problems use HCF to find the maximum equal distribution. Here, we find HCF of 420 and 130.

Argument 1

Using Euclid’s algorithm: 420 = 130 × 3 + 30.
130 = 30 × 4 + 10 → 30 = 10 × 3 + 0.

Argument 2

HCF = 10. So, 10 sweets per box.
Total boxes = (420 + 130)/10 = 55.

Conclusion

The seller needs 55 boxes with 10 sweets each, ensuring no wastage.

Question 8:

A sweet seller has 420 gulab jamuns and 130 rasgullas. He wants to stack them in identical trays with no sweets left over. Find the maximum number of trays he can use and the number of each sweet per tray.

Answer:

Introduction

Our textbook shows that such problems require finding the HCF of given numbers to determine the maximum identical groups.

Argument 1

We find HCF of 420 and 130 using Euclid's algorithm:
420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 + 0.

Argument 2

The HCF is 10. Thus, the seller can use 10 trays. Each tray will have 420/10 = 42 gulab jamuns and 130/10 = 13 rasgullas.

Conclusion

The maximum number of trays is 10, with 42 and 13 sweets per tray respectively.

Question 9:

Prove that √5 is an irrational number using the method of contradiction. Refer to NCERT Example 5.

Answer:

Introduction

We studied that a number is irrational if it cannot be expressed as a fraction p/q where p and q are co-prime integers. Here, we prove √5 is irrational.

Argument 1

Assume √5 is rational, so √5 = p/q (p, q co-prime).
Squaring both sides: 5 = p²/q² ⇒ p² = 5q².

Argument 2

This implies p² is divisible by 5, so p is divisible by 5.
Let p = 5k, then (5k)² = 5q² ⇒ q² = 5k², making q also divisible by 5.

Conclusion

This contradicts our assumption that p and q are co-prime. Hence, √5 is irrational.

Question 10:

Answer:

Introduction

Our textbook shows that such problems require finding the HCF of given numbers. Here, we find HCF of 420 and 130.

Argument 1

Using Euclid’s algorithm: 420 = 130 × 3 + 30.
130 = 30 × 4 + 10, then 30 = 10 × 3 + 0.

Argument 2

The last non-zero remainder is 10, so HCF = 10.
Thus, maximum boxes = 10, each with 42 kaju and 13 badam barfis.

Conclusion

The seller can use 10 boxes, ensuring equal distribution as per the problem.

Question 11:

Prove that √2 is an irrational number using the method of contradiction.

Answer:

To prove that √2 is an irrational number, we assume the opposite, i.e., √2 is a rational number. This means it can be expressed as a fraction in its simplest form a/b, where a and b are coprime integers (having no common factors other than 1) and b ≠ 0.

Step 1: Assume √2 = a/b.
Step 2: Square both sides: 2 = a²/b².
Step 3: Rearrange to get a² = 2b².
Step 4: This implies a² is even, so a must also be even (since the square of an odd number is odd).
Step 5: Let a = 2k, where k is an integer.
Step 6: Substitute into a² = 2b²: (2k)² = 2b² → 4k² = 2b² → b² = 2k².
Step 7: Thus, b² is even, so b must also be even.
Step 8: But if both a and b are even, they share a common factor of 2, contradicting our assumption that they are coprime.
Step 9: Hence, our initial assumption is false, and √2 must be irrational.

This proof is a classic example of the method of contradiction and is fundamental in understanding the nature of real numbers.

Question 12:

Explain the Fundamental Theorem of Arithmetic with an example. How is it used to find the LCM and HCF of two numbers?

Answer:

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a unique product of prime numbers, regardless of the order.
Example: Take the number 60.
Its prime factorization is 60 = 2 × 2 × 3 × 5 = 2² × 3¹ × 5¹.
No other combination of primes will give 60, proving uniqueness.

To find HCF and LCM of two numbers (e.g., 24 and 36):
Step 1: Prime factorize both numbers.
24 = 2³ × 3¹
36 = 2² × 3²
Step 2: For HCF, take the lowest power of common primes.
HCF = 2² × 3¹ = 12
Step 3: For LCM, take the highest power of all primes.
LCM = 2³ × 3² = 72
This method ensures accuracy and relies on the uniqueness guaranteed by the theorem.

Question 13:

Prove that the square of any positive integer is of the form 3m or 3m + 1 for some integer m. Use Euclid's division lemma to justify your answer.

Answer:

To prove that the square of any positive integer is of the form 3m or 3m + 1, we will use Euclid's division lemma.

According to Euclid's division lemma, any positive integer a can be expressed as:

a = 3q + r, where q is an integer and r can be 0, 1, or 2.

Now, let's consider the square of a:

a² = (3q + r)² = 9q² + 6qr + r²

We will analyze the possible values of r:

Case 1: If r = 0, then a² = 9q² + 0 + 0 = 3(3q²), which is of the form 3m.
Case 2: If r = 1, then a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, which is of the form 3m + 1.
Case 3: If r = 2, then a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1, which is again of the form 3m + 1.

Thus, the square of any positive integer is either of the form 3m or 3m + 1.

Question 14:

Explain why (17 × 11 × 2) + (17 × 11 × 5) is a composite number. Justify your answer using the Fundamental Theorem of Arithmetic.

Answer:

To determine why (17 × 11 × 2) + (17 × 11 × 5) is a composite number, we will simplify the expression and apply the Fundamental Theorem of Arithmetic.

First, let's factor out the common terms:

(17 × 11 × 2) + (17 × 11 × 5) = 17 × 11 × (2 + 5) = 17 × 11 × 7

Now, the expression simplifies to 17 × 11 × 7, which is a product of prime numbers.

According to the Fundamental Theorem of Arithmetic, every composite number can be expressed as a unique product of primes. Here, the number 17 × 11 × 7 = 1309 is clearly a product of primes, making it a composite number.

Additionally, since the original expression was rewritten as a product of primes, it confirms that the number is not prime but composite.

Question 15:

Prove that √5 is an irrational number using the method of contradiction.

Answer:

To prove that √5 is irrational, we assume the opposite, i.e., √5 is rational.
Let √5 = a/b, where a and b are coprime integers (no common factors other than 1) and b ≠ 0.
Squaring both sides, we get 5 = a²/b² ⇒ a² = 5b².
This implies a² is divisible by 5, so a must also be divisible by 5 (since 5 is a prime number).
Let a = 5k for some integer k.
Substituting back, (5k)² = 5b² ⇒ 25k² = 5b² ⇒ b² = 5k².
This means b² is divisible by 5, so b must also be divisible by 5.
But this contradicts our assumption that a and b are coprime (both divisible by 5).
Hence, √5 cannot be expressed as a fraction, proving it is irrational.

Question 16:

Prove that the square of any positive integer is of the form 3m or 3m + 1 for some integer m. Use Euclid's Division Lemma to justify your answer.

Answer:

To prove that the square of any positive integer is of the form 3m or 3m + 1, we will use Euclid's Division Lemma, which states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Let us consider any positive integer a. According to Euclid's Division Lemma, when a is divided by 3, the possible remainders (r) are 0, 1, or 2. Therefore, a can be expressed in one of the following forms:

a = 3q (when r = 0)
a = 3q + 1 (when r = 1)
a = 3q + 2 (when r = 2)

Now, let's find the square of a in each case:

Case 1: a = 3q
Square of a = (3q)² = 9q² = 3(3q²), which is of the form 3m, where m = 3q².

Case 2: a = 3q + 1
Square of a = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, which is of the form 3m + 1, where m = 3q² + 2q.

Case 3: a = 3q + 2
Square of a = (3q + 2)² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1, which is again of the form 3m + 1, where m = 3q² + 4q + 1.

Thus, the square of any positive integer is always of the form 3m or 3m + 1.

Question 17:

Explain why 7 × 11 × 13 + 13 is a composite number. Justify your answer using the Fundamental Theorem of Arithmetic.

Answer:

To determine whether 7 × 11 × 13 + 13 is a composite number, we will use the Fundamental Theorem of Arithmetic, which states that every composite number can be expressed as a unique product of prime numbers.

First, let's simplify the expression:

7 × 11 × 13 + 13 = 13(7 × 11 + 1)
= 13(77 + 1)
= 13 × 78

Now, let's factorize 78 further:
78 = 2 × 3 × 13

Thus, the expression becomes:
13 × 2 × 3 × 13 = 2 × 3 × 13²

Since the expression can be written as a product of prime factors (2, 3, 13), it satisfies the definition of a composite number. Additionally, the presence of multiple prime factors confirms that 7 × 11 × 13 + 13 is not a prime number but a composite number.

Therefore, 7 × 11 × 13 + 13 is a composite number as it has more than two distinct prime factors.

Question 18:

Prove that √5 is an irrational number using the method of contradiction. Explain each step clearly.

Answer:

To prove that √5 is irrational, we assume the opposite—that it is rational. This means it can be expressed as a fraction in its simplest form a/b, where a and b are coprime integers (i.e., their HCF is 1).

Step 1: Assume √5 = a/b.
Step 2: Squaring both sides, we get 5 = a²/b² ⇒ 5b² = a².
Step 3: This implies a² is divisible by 5, so a must also be divisible by 5 (since 5 is a prime number).
Step 4: Let a = 5k for some integer k. Substituting, we get 5b² = (5k)² ⇒ 5b² = 25k² ⇒ b² = 5k².
Step 5: Thus, b² is divisible by 5, meaning b is also divisible by 5.

This contradicts our assumption that a and b are coprime (as both are divisible by 5). Hence, √5 cannot be rational and must be irrational.

Question 19:

Using the Euclid’s Division Algorithm, find the HCF of 420 and 130. Show all steps systematically and verify your answer.

Answer:

To find the HCF of 420 and 130 using Euclid’s Division Algorithm, follow these steps:

Step 1: Apply the division lemma to 420 and 130:
420 = 130 × 3 + 30 (Here, quotient = 3, remainder = 30).
Step 2: Now, divide 130 by the remainder 30:
130 = 30 × 4 + 10 (Quotient = 4, remainder = 10).
Step 3: Next, divide 30 by the new remainder 10:
30 = 10 × 3 + 0 (Quotient = 3, remainder = 0).

Since the remainder is now 0, the HCF is the last non-zero remainder, which is 10.

Verification:
The factors of 420: 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 420.
The factors of 130: 1, 2, 5, 10, 13, 26, 65, 130.
The common factors are 1, 2, 5, 10. The highest is 10, confirming our answer.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A school library has 96 Mathematics books and 72 Science books. The librarian wants to arrange them in shelves with equal numbers of books, without mixing subjects. What is the maximum number of books per shelf? (4 marks)

Answer:

Problem Interpretation
We need to find the HCF of 96 and 72 to determine the maximum equal arrangement.

Mathematical Modeling
Prime factorize 96: 2⁵ × 3
Prime factorize 72: 2³ × 3²

Solution
HCF = 2³ × 3 = 24. Thus, 24 books per shelf (2m comprehension + 2m application).

Question 2:

A traffic light cycles every 45 seconds, and another every 60 seconds. If they turn green together at 8:00 AM, when will they synchronize again? (4 marks)

Answer:

Problem Interpretation
We studied LCM to find the next common time interval.

Mathematical Modeling
Prime factors: 45 = 3² × 5, 60 = 2² × 3 × 5

Solution
LCM = 2² × 3² × 5 = 180 seconds. They synchronize at 8:03 AM (2m comprehension + 2m application).

Question 3:

Prove that 3 + √5 is an irrational number using contradiction. (4 marks)

Answer:

Problem Interpretation
Our textbook shows irrationality proofs by assuming the opposite.

Mathematical Modeling
Assume 3 + √5 = a/b (a, b co-prime)
Rearrange: √5 = (a − 3b)/b

Solution
Since √5 is irrational, RHS must be irrational, contradicting a/b being rational (2m comprehension + 2m application).

Question 4:

A school auditorium has 420 seats and 130 chairs. The principal wants to arrange them in rows such that each row has equal number of seats and chairs. Using Euclid's division lemma, find the maximum number of rows possible and the number of seats and chairs in each row.

Answer:

Problem Interpretation
We need to find the HCF of 420 and 130 to determine the maximum rows with equal seats and chairs.
Mathematical Modeling
Using Euclid's division lemma: 420 = 130 × 3 + 30, 130 = 30 × 4 + 10, 30 = 10 × 3 + 0.
Solution
The HCF is 10. Thus, 10 rows are possible with 42 seats and 13 chairs per row.

Question 5:

A bakery sells cookies in packs of 12 and muffins in packs of 18. A customer wants to buy the same number of cookies and muffins. What is the smallest number of packs of each they must buy? Use the Fundamental Theorem of Arithmetic.

Answer:

Problem Interpretation
We need the LCM of 12 and 18 to find the smallest equal quantity of cookies and muffins.
Mathematical Modeling
Prime factors: 12 = 2² × 3, 18 = 2 × 3². LCM = 2² × 3² = 36.
Solution
The customer must buy 3 packs of cookies (12×3=36) and 2 packs of muffins (18×2=36).

Question 6:

A school auditorium has 156, 208, and 260 students in three different grades. The principal wants to divide them into equal groups for a competition with no students left out. Find the maximum number of students that can be placed in each group.

Answer:

Problem Interpretation
We need to find the largest number that divides 156, 208, and 260 exactly, which is their HCF.
Mathematical Modeling
Using the Euclidean algorithm, we find HCF step-by-step:
HCF(156, 208) = 52
HCF(52, 260) = 52
Solution
The maximum number of students per group is 52, ensuring no one is left out.

Question 7:

A bakery sells cookies in packs of 12, 18, and 24. A customer wants to buy the smallest number of cookies that can be equally divided into any of these pack sizes. How many cookies should they purchase?

Answer:

Problem Interpretation
We need the smallest number divisible by 12, 18, and 24, which is their LCM.
Mathematical Modeling
Prime factorization gives:
12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3
Solution
LCM = 2³ × 3² = 72. The customer should buy 72 cookies for equal distribution.

Question 8:

A school auditorium has 156 seats arranged in rows. The number of seats per row is equal to the number of rows. Using the Euclid's division lemma, find the exact number of rows and seats per row if 12 extra seats are added to the auditorium.

Answer:

Problem Interpretation
We need to find the original number of rows (and seats per row) in the auditorium, then adjust for 12 extra seats.
Mathematical Modeling
Let original rows = seats/row = x. Total seats = x². After adding 12 seats: x² + 12 = 156 → x² = 144.
Solution
Using Euclid's lemma, we verify 144 is perfect square. √144 = 12. Thus, original rows = 12, seats/row = 12.

Question 9:

A bakery packs 84 almond cookies and 108 raisin cookies in identical boxes. Using the fundamental theorem of arithmetic, determine the minimum number of boxes required if each box must contain equal quantities of both cookie types with no leftovers.

Answer:

Problem Interpretation
We must find HCF of 84 and 108 to determine maximum cookies per box, then calculate minimum boxes.
Mathematical Modeling
Prime factorization: 84 = 2²×3×7, 108 = 2²×3³. HCF = product of common primes = 2²×3 = 12.
Solution
Cookies/box = 12. Almond boxes = 84÷12 = 7, raisin boxes = 108÷12 = 9. Total boxes = 7 + 9 = 16.

Question 10:

A school librarian arranges books in stacks of 12, 15, and 20. She notices that each arrangement leaves no books remaining. Using Euclid’s Division Lemma, find the minimum number of books she could have.

Answer:

Problem Interpretation
We need to find the smallest number divisible by 12, 15, and 20, which is their LCM.
Mathematical Modeling
Prime factorize: 12 = 2²×3, 15 = 3×5, 20 = 2²×5
Solution
LCM = 2²×3×5 = 60. Thus, the librarian has 60 books.

Question 11:

A tailor cuts ribbons of lengths 18 cm and 24 cm without waste. Prove that the longest tape to measure both exactly is their HCF.

Answer:

Problem Interpretation
We must find the largest length dividing 18 cm and 24 cm perfectly.
Mathematical Modeling
Apply Euclid’s algorithm: 24 = 18×1 + 6, 18 = 6×3 + 0
Solution
HCF is 6 cm. Our textbook shows this method ensures exact measurement.

Question 12:

A school is organizing a sports event and needs to distribute 120 footballs and 90 cricket balls equally among the maximum number of teams such that each team gets the same number of footballs and cricket balls. Based on this situation, answer the following:
What is the maximum number of teams that can be formed?
How many footballs and cricket balls will each team get?

Answer:

To find the maximum number of teams, we need to calculate the Highest Common Factor (HCF) of 120 and 90.

Step 1: Prime factorize 120 and 90.
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
90 = 2 × 3 × 3 × 5 = 2 × 3² × 5

Step 2: Identify the common prime factors with the lowest exponents.
Common factors: 2, 3, 5
HCF = 2 × 3 × 5 = 30

Conclusion:
The maximum number of teams that can be formed is 30.
Each team will get:
Footballs = 120 ÷ 30 = 4
Cricket balls = 90 ÷ 30 = 3

Question 13:

Two friends, Riya and Priya, are practicing for a marathon. Riya takes 18 minutes to complete one round of the park, while Priya takes 12 minutes. If they start together at 6:00 AM, after how many minutes will they meet again at the starting point?

Answer:

To find the time when Riya and Priya meet again, we need to calculate the Least Common Multiple (LCM) of their individual round times (18 and 12 minutes).

Step 1: Prime factorize 18 and 12.
18 = 2 × 3 × 3 = 2 × 3²
12 = 2 × 2 × 3 = 2² × 3

Step 2: Identify the highest exponents of all prime factors.
LCM = 2² × 3² = 4 × 9 = 36 minutes

Conclusion:
Riya and Priya will meet again at the starting point after 36 minutes, i.e., at 6:36 AM.

Question 14:

A school is organizing a sports event and needs to distribute 120 footballs, 180 basketballs, and 240 cricket balls equally among the maximum number of teams such that each team gets the same number of each type of ball. Based on this situation, answer the following:
Find the maximum number of teams that can be formed.
How many footballs, basketballs, and cricket balls will each team receive?

Answer:

To solve this problem, we need to find the Highest Common Factor (HCF) of the given numbers of balls.

Step 1: List the numbers to find HCF: 120 (footballs), 180 (basketballs), 240 (cricket balls).
Step 2: Find the prime factorization of each number:
120 = 2³ × 3 × 5
180 = 2² × 3² × 5
240 = 2⁴ × 3 × 5
Step 3: Identify the lowest power of each common prime factor:
HCF = 2² × 3 × 5 = 4 × 3 × 5 = 60.
Conclusion: The maximum number of teams that can be formed is 60.

Now, calculate the number of balls per team:
Footballs per team = 120 ÷ 60 = 2
Basketballs per team = 180 ÷ 60 = 3
Cricket balls per team = 240 ÷ 60 = 4

Question 15:

Two friends, Rahul and Priya, are practicing LCM and HCF concepts. Rahul says the HCF of two numbers is 12, and their LCM is 180. Priya claims one of the numbers is 36. Verify Priya's claim and find the other number.

Answer:

We know the relationship between two numbers, their HCF, and LCM:
Product of two numbers = HCF × LCM.

Given: HCF = 12, LCM = 180, one number = 36.
Step 1: Let the other number be x.
Step 2: Apply the formula: 36 × x = 12 × 180
36x = 2160
x = 2160 ÷ 36 = 60.
Verification:
HCF of 36 and 60:
36 = 2² × 3²
60 = 2² × 3 × 5
HCF = 2² × 3 = 12 (matches given HCF).
LCM of 36 and 60:
LCM = 2² × 3² × 5 = 180 (matches given LCM).
Conclusion: Priya's claim is correct, and the other number is 60.

Question 16:

A school is organizing a sports event and needs to distribute identical medals to the winners. The medals are to be arranged in stacks of equal height, either in stacks of 12 or stacks of 18. The organizers want to ensure that no medal is left out. Based on this situation, answer the following:
What is the minimum number of medals the organizers must have to arrange them in stacks of 12 or 18 without any leftovers?
Explain the mathematical concept used to solve this problem.

Answer:

The minimum number of medals required is the Least Common Multiple (LCM) of 12 and 18.

To find the LCM of 12 and 18:

Step 1: Prime factorize both numbers.

12 = 2² × 3¹

18 = 2¹ × 3²

Step 2: Take the highest power of each prime factor.

LCM = 2² × 3² = 4 × 9 = 36

Thus, the organizers must have at least 36 medals to arrange them in stacks of 12 or 18 without leftovers.

The concept used here is the LCM, which ensures that the number is divisible by both given numbers. This is a fundamental application of Real Numbers in practical scenarios.

Question 17:

Riya and Priya are practicing for a math quiz. Riya claims that the decimal expansion of 13/125 will terminate after 3 decimal places, while Priya argues it will terminate after 5 decimal places. Based on this, answer:
Who is correct and why?
Explain the condition for a rational number to have a terminating decimal expansion.

Answer:

Riya is correct. The decimal expansion of 13/125 terminates after 3 decimal places.

Step 1: Simplify the denominator into its prime factors.

125 = 5³

Step 2: The denominator must be of the form 2^m × 5ⁿ for the decimal to terminate.

Here, 125 = 5³ × 2⁰, so it satisfies the condition.

Step 3: The number of decimal places is determined by the higher power of 2 or 5 in the denominator.

Since the highest power is 3 (for 5), the decimal terminates after 3 places.

13/125 = 0.104 (3 decimal places).

Terminating decimal condition: A rational number p/q (in simplest form) has a terminating decimal expansion if the prime factorization of q consists only of the primes 2 and/or 5. This is a key concept in Real Numbers.

Question 18:

A school is organizing a sports event and needs to distribute 120 footballs and 90 cricket balls equally among the maximum number of teams such that each team gets the same number of footballs and cricket balls. Based on this scenario, answer the following:
Find the maximum number of teams that can be formed.
How many footballs and cricket balls will each team get?

Answer:

To solve this problem, we need to find the HCF (Highest Common Factor) of 120 and 90, as it represents the maximum number of teams that can be formed with equal distribution.

Step 1: Find the prime factorization of 120 and 90.
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
90 = 2 × 3 × 3 × 5 = 2 × 3² × 5

Step 2: Identify the common prime factors with the lowest exponents.
Common factors: 2, 3, and 5 (minimum exponents: 2¹, 3¹, 5¹).

Step 3: Multiply the common factors to get the HCF.
HCF = 2 × 3 × 5 = 30.

Thus, the maximum number of teams that can be formed is 30.

Step 4: Calculate the number of footballs and cricket balls per team.
Footballs per team = 120 ÷ 30 = 4.
Cricket balls per team = 90 ÷ 30 = 3.

Each team will receive 4 footballs and 3 cricket balls.

Question 19:

Two friends, Riya and Priya, are practicing for a marathon. Riya takes 12 minutes to complete one round of the track, while Priya takes 18 minutes. If they start running together at the same time, after how many minutes will they meet again at the starting point?

Answer:

To determine when Riya and Priya will meet again at the starting point, we need to find the LCM (Least Common Multiple) of their individual round times.

Step 1: Find the prime factorization of 12 and 18.
12 = 2 × 2 × 3 = 2² × 3
18 = 2 × 3 × 3 = 2 × 3²

Step 2: Identify the highest exponents of all prime factors.
For 2: highest exponent is 2².
For 3: highest exponent is 3².

Step 3: Multiply the factors with the highest exponents to get the LCM.
LCM = 2² × 3² = 4 × 9 = 36.

Riya and Priya will meet again at the starting point after 36 minutes.

Note: LCM gives the smallest time interval where both runners complete whole numbers of rounds, ensuring they meet at the starting point.

Question 20:

A school is organizing a sports event and needs to distribute 120 footballs and 84 basketballs equally among the maximum number of teams such that each team gets the same number of footballs and the same number of basketballs.
(i) Find the maximum number of teams that can be formed.

(ii) How many footballs and basketballs will each team get?

Answer:

To solve this problem, we need to find the Highest Common Factor (HCF) of the given numbers of footballs and basketballs.

Step 1: Identify the numbers.
Number of footballs = 120
Number of basketballs = 84

Step 2: Find the HCF of 120 and 84 using the Euclid's division algorithm.
120 = 84 × 1 + 36
84 = 36 × 2 + 12
36 = 12 × 3 + 0

Since the remainder is now 0, the HCF is the last non-zero remainder, which is 12.

(i) The maximum number of teams that can be formed is 12.

(ii) Number of footballs per team = 120 ÷ 12 = 10
Number of basketballs per team = 84 ÷ 12 = 7

Thus, each team will get 10 footballs and 7 basketballs.

Question 21:

A bakery sells cookies in packs of 15 and muffins in packs of 18. A customer wants to buy the same number of cookies and muffins.
(i) What is the smallest number of each item the customer must buy?

(ii) How many packs of cookies and muffins will the customer need to purchase?

Answer:

This problem involves finding the Least Common Multiple (LCM) of the pack sizes to determine the smallest number where the quantities match.

Step 1: Identify the pack sizes.
Cookies per pack = 15
Muffins per pack = 18

Step 2: Find the LCM of 15 and 18 using prime factorization.
15 = 3 × 5
18 = 2 × 3²

The LCM is the product of the highest powers of all prime factors:
LCM = 2 × 3² × 5 = 2 × 9 × 5 = 90

(i) The smallest number of each item the customer must buy is 90.

(ii) Packs of cookies needed = 90 ÷ 15 = 6
Packs of muffins needed = 90 ÷ 18 = 5

Therefore, the customer needs to buy 6 packs of cookies and 5 packs of muffins to have an equal number of both items.

Real Numbers – CBSE NCERT Study Resources

Chapter Overview: Real Numbers

1. Introduction to Real Numbers

2. Euclid's Division Algorithm

3. The Fundamental Theorem of Arithmetic

4. Irrational Numbers

5. Rational Numbers and Their Decimal Expansions

6. Operations on Real Numbers

All Question Types with Solutions – CBSE Exam Pattern

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)