CBSE Class 10 Mathematics Real Numbers Worksheet Set 01

Choose the correct answer from the given options:

Question. The LCM of smallest two-digit composite number and smallest composite number is:
(a) 12
(b) 4
(c) 20
(d) 44
Answer : C

Question. 325 can be expressed as a product of its primes as:
(a) 52 × 7
(b) 52 × 13
(c) 5 × 132
(d) 2 × 32 × 52
Answer : B

Question. HCF (a, b) × LCM (a, b) is equal to
(a) a + b
(b) a – b
(c) a × b
(d) a ÷ b
Answer : C

Question. The decimal expansion (without actual division) and its nature (terminating or non-terminating) of 15/1600 will be
(a) Terminating after 6 places
(b) Non-terminating but repeating
(c) Non-terminating and non-repeating
(d) Terminating after 2 places
Answer : A

Question. The decimal expansion (without actual division) and its nature (terminating or non terminating) of 17/8 will be
(a) Terminating after 2 places
(b) Non-terminating but repeating
(c) Non-terminating but non-repeating
(d) Terminating after 3 places
Answer : D

Question. If a and b are co-prime, then a2 and b2 are
(a) primes
(b) composites
(c) co-primes
(d) None of these
Answer : C

Question. When 429 is expressed as a product of its prime factors, we get
(a) 2 × 5 × 29
(b) 33 × 13 × 1
(c) 3 × 11 × 9
(d) 3 × 11 × 13
Answer : D

Question. The values of x and y in the given figure respectively are

(a) x = 84, y = 21
(b) x = 21, y = 84
(c) x = 42, y = 24
(d) x = 24, y = 42
Answer : B

Question. A rational and an irrational number lying between 0.25 and 0.32 are respectively.
(a) 0.30, 0.3010203040...
(b) 0.20, 0.2010203040...
(c) 0.33, 0.3510203040...
(d) None of these
Answer : A

Question. The HCF and LCM of 404 and 96 respectively are
(a) 2, 9696
(b) 4, 9696
(c) 8, 3636
(d) 10, 2020
Answer : B

Question. The 2ⁿ5^m (where n and m are non-negative integers) from of denominator of 3/5 and its decimal expansion respectively are
(a) 2⁰ × 5¹, 0.6
(b) 2¹ × 5⁰, 0.5
(c) 2¹ × 5¹, 0.6
(d) 2² × 5⁰, 0.8
Answer : A

Question. 3 bells ring at an interval of 4, 7 and 14 minutes. All three bells rang at 6 am. When the three bells will ring together next?
(a) 6:20 am
(b) 6:24 am
(c) 6:28 am
(d) 6:30 am
Answer : C

Question. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, the other number is
(a) 31
(b) 71
(c) 61
(d) 91
Answer : D

Question. When 156 is expressed as the product of primes, we get
(a) 22 × 3 × 13
(b) 22 × 3 × 11
(c) 2 × 32 × 13
(d) 2 × 32 × 11
Answer : C

Question. The decimal expansion (without actual division) and its nature (terminating or non-terminating) of 64/455 will be
(a) Terminating after 2 places
(b) Non-terminating but repeating
(c) Non-terminating but non-repeating
(d) Terminating after 3 places
Answer : B

Question. The 2ⁿ5^m (where n and m are non-negative integers) form of denominator of 13/80 and its decimal expansion respectively are
(a) 2³ × 5², 0.1248
(b) 2² × 5³, 0.1698
(c) 2⁴ × 5¹, 0.1625
(d) None of these
Answer : C

Question. The LCM and the HCF of 15, 18, 45 respectively are
(a) 3, 30
(b) 4, 40
(c) 5, 50
(d) 3, 90
Answer : D

Question. A rational number can be expressed as a terminating decimal if the denominator has factors
(a) 2, 3 or 5
(b) 2 or 3
(c) 3 or 5
(d) 2 or 5
Answer : D

Question. Rational number p/q, q ≠ 0, will be terminating decimal if the prime factorisation of q is of the form (m and n are non-negative integers)
(a) 2^m × 3ⁿ
(b) 2^m × 5ⁿ
(c) 3^m × 5ⁿ
(d) 3^m × 7ⁿ
Answer : B

Question. Which of the following is the decimal expansion of an irrational number?
(a) 4.561
(b) 0.12
(c) 5.010010001…
(d) 6.03
Answer : C

Question. 2.35 is
(a) an integer
(b) a rational number
(c) an irrational number
(d) a natural number
Answer : B

Question. The LCM of 150 and 200 is
(a) 320
(b) 400
(c) 550
(d) 600
Answer : D

B. Assertion-Reason Type Questions
In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose
the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): The sum or difference of a rational number and an irrational number is irrational.
Reason (R): Negative of an irrational number is rational
Answer : B

Question. Assertion (A): If m and n are odd positive integers, then m² + n² is even but not divisible by 4.
Reason (R): 3 × 5 × 7 + 7 is a composite number
Answer : B

Question. Assertion (A): 5 + √3 is an irrational number.
Reason (R): The sum or difference of a rational and an irrational number is always irrational.
Answer : A

Question. Assertion (A): The number 6ⁿ, n being a natural number, ends with the digit 5.
Reason (R): The number 9ⁿ cannot end with digit 0 for any natural number n.
Answer : D

Very Short Answer Type Questions

Question. The LCM of two numbers is 182 and their HCF is 13. If one of the number is 26, find the other.
Answer : We know that HCF (a, b) × LCM (a, b) = a × b
So, 13 × 182 = 26 × b
⇒ b = 13x182/26 = 91
Thus, the other number is 91.

Question. Given that HCF (135, 225) = 45, find the LCM (135, 225).
Answer : We know that
LCM × HCF = Product of two numbers
LCM (135, 225) = Product of 135 and 225/HCF(135, 225)
= 135×225/45
= 675

Question. After how many decimal places will the decimal representation of the rational number 229/2² × 5⁷ terminate ?
Answer : Here, 
229/2²×5⁷ = 229x2⁵/2⁷x5⁷ = 229x2⁵/(10)⁷
Hence, the given rational number will terminate after 7 decimal places.

Question. Are the smallest prime and the smallest composite numbers co-prime? Justify.
Answer : No.
We know that,
Smallest prime number is 2 and smallest composite number is 4.
HCF of (2, 4) = 2
Since, there is a common factor 2.
So, they are not co-prime.

Question. The HCF of two numbers a and b is 5 and their LCM is 200. Find the product ab.
Answer : Given, HCF (a, b) = 5
LCM (a, b) = 200
HCF × LCM = Product of the numbers
⇒ a × b = 5 × 200
⇒ ab = 1000
Hence, the product of ab is 100.

Question. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
Answer : No.
We know that:
“The HCF of any two numbers must be a factor of the LCM of those numbers.”
So, two numbers cannot have their HCF 18 and LCM 380, as 18 does not divide 380.

Question. Find a rational number between 2 and 7 .
Answer : √2 = 1.414
and √7 = 2.6
Let the rational number be x.
∴ √2 < x < √7
or 1.4 < x < 2.6
Hence, any rational number like 1.5, 2.0, 2.5, can be the answer.

Question. Write the number of zeroes in the end of a number whose prime factorization is 22 × 53 × 32 × 17.
Answer : Given, 2² × 5² × 5 × 3² × 17
= (2 × 5)² × 5 × 3² × 17
[∵ on multiplying 2 × 5 we get 10]
= (10)² × 5 × 3² × 17
The power of 10 in the given expression is 2.
Hence, the number of zeroes in the end will be = 2.

Question. If the HCF of (336, 54) = 6, find the LCM (336, 54).
Answer : The HCF of (336, 54) = 6.
We know that:
LCM × HCF = Product of two numbers
⇒ LCM = 336 x 54/6
= 336 × 9 = 3024
Hence, the LCM of the two numbers is 3024.

Question. Find a rational number betwen 2 and 3.
Answer : Rational number between √2 (1.41 approx)  and √3 (1.73 approx) can be 1.5, 1.6, 1.63 etc.
So, a required rational number may be 1.5.

Question. Write one rational and one irrational number lying between 0.25 and 0.32.
Answer : Rational number= 0.30
Irrational number = 0.3010203040…
Or any other correct rational and irrational number.

Question. Write the exponent of 3 in the prime factorization of 144.
Answer : Prime factorization of 14⁴ = 2⁴ × 3²
So, exponent of 3 = 2.

Question. Write the sum of exponents of prime factors of 98.
Answer: 1+2=3

Question. Given HCF of (16, 100) = 4. find L.C.M of (16, 100).
Answer: 400

Question. What is the maximum no. of factors of a prime number?
Answer: 2

Question. State if (√2 – √3)( √2 + √3) is rational or irrational.
Answer: rational

Question. Write a rational no. between √2 and √3 .
Answer: 2/3

Question. If x and y are two irrational numbers then tell whether x + y is always irrational or not.
Answer: Not, x + y may be rational

Question. Write if 343/28 is a terminating or non-terminating repeating decimal without doing actual division.
Answer: Non terminating & repeating decimal

Question. What is the L.C.M of x and y if y is a multiple of x?
Answer: y

Question. Express 0.03 as a rational number in the form of p/q.
Answer: 1/10

Question. Tell whether the prime factorization of 15 is 1X 3 X 5 or not.
Answer: Not, 15=3X5

Real Numbers

Key Points

• Real Numbers
  • Rational Numbers (\(Q\)): Can be represented in the form \( \frac{p}{q} \) where \( q \neq 0 \).
    • Natural Numbers (\(N\)): Also known as counting numbers. Examples: \(1, 2, 3, \ldots\)
    • Whole Numbers (\(W\)): Include all natural numbers along with zero. Examples: \(0, 1, 2, 3, \ldots\)
    • Integers (\(Z\)):
      • Negative Integers: \(\{-1, -2, -3, \ldots\}\)
      • Zero: \(\{0\}\)
      • Positive Integers: \(\{1, 2, 3, \ldots\}\) (equivalent to Natural Numbers)
  • Irrational Numbers (\(I\)): Numbers that cannot be expressed as a simple fraction. Examples: \(\sqrt{2}, \pi, \text{etc.}\)

Decimal Form of Real Numbers

• Rational Numbers:
  • Terminating Decimals: Decimals that end after a finite number of digits. Examples: \(\frac{2}{5} = 0.4\), \(\frac{3}{4} = 0.75\).
  • Non-Terminating Repeating Decimals (Recurring): Decimals that do not end but repeat a block of digits infinitely. Examples: \(\frac{1}{3} = 0.333\ldots\), \(\frac{2}{7} = 0.285714\ldots\), \(\frac{3}{11} = 0.2727\ldots\)
• Irrational Numbers:
  • Non-Terminating Non-Repeating Decimals: Decimals that continue indefinitely without repeating any pattern. Example: \(1.010010001\ldots\)

1. Euclid's Division Lemma

For any two positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) such that they satisfy the relation:
\[ a = bq + r \quad \text{where} \quad 0 \leq r < b \]
Here, \(a\) represents the dividend, \(b\) is the divisor, \(q\) is the quotient, and \(r\) is the remainder.

2. Euclid's Division Algorithm

This is a step-by-step technique to compute the Highest Common Factor (HCF) of two positive integers, say \(c\) and \(d\), where \(c > d\):

• Step I: Apply Euclid's division lemma to \(c\) and \(d\). Thus, we find integers \(q\) and \(r\) satisfying \(c = dq + r\) where \(0 \leq r < d\).
• Step II: If the remainder \(r = 0\), then the divisor \(d\) is the HCF of \(c\) and \(d\). If the remainder \(r \neq 0\), we apply the division lemma again to \(d\) and \(r\).
• Step III: We repeat this process recursively until the remainder becomes zero. The divisor at this final stage is the required HCF.

Note: If \(a\) and \(b\) are positive integers and \(a = bq + r\) where \(0 \leq r < b\), then the HCF of \(a\) and \(b\) is equal to the HCF of \(b\) and \(r\): \(\text{HCF}(a, b) = \text{HCF}(b, r)\).

3. The Fundamental Theorem of Arithmetic

Every composite number can be written as a product of prime numbers. This prime factorization is entirely unique, regardless of the order in which the prime factors are multiplied.
For example: \(24 = 2 \times 2 \times 2 \times 3 = 3 \times 2 \times 2 \times 2\).

Theorems on Rational Numbers

• Theorem 1: If \(x\) is a rational number with a terminating decimal expansion, then it can be written as \( \frac{p}{q} \), where \(p\) and \(q\) are co-prime integers. The prime factorization of the denominator \(q\) is of the form \(2^n \cdot 5^m\), where \(n\) and \(m\) are non-negative integers.
  Example: \(\frac{7}{10} = \frac{7}{2 \times 5} = 0.7\).
• Theorem 2: Let \(x = \frac{p}{q}\) be a rational number. If the prime factorization of \(q\) is not of the form \(2^n \cdot 5^m\) (where \(n, m\) are non-negative integers), then \(x\) has a decimal expansion that is non-terminating and repeating (recurring).
  Example: \(\frac{7}{6} = \frac{7}{2 \times 3} = 1.1666\ldots\)
• Theorem 3: For any two positive integers \(a\) and \(b\), the product of their HCF and LCM is equal to the product of the two numbers themselves:
  \[ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \]
  Example: For numbers 4 and 6, \(\text{HCF}(4, 6) = 2\) and \(\text{LCM}(4, 6) = 12\). Thus, \(2 \times 12 = 4 \times 6 = 24\).

Level 1 Page 9

Level-I

Question 1. If \(\frac{p}{q}\) is a rational number (\(q \neq 0\)).What is the condition on q so that the decimal representation of is \(\frac{p}{q}\) terminating?
Answer: The denominator \(q\) must have a prime factorization of the form \(2^n \cdot 5^m\), where both \(n\) and \(m\) are non-negative integers.
In simple words: For a fraction to end cleanly as a decimal, its bottom number must only have 2s and 5s as its prime building blocks.

Exam Tip: Ensure you specify that n and m are non-negative integers, not just any integers.

Question 2. Write a rational number between \(\sqrt{2}\) and \(\sqrt{3}\).
Answer: A rational number lying between these two values is \(1.5\). Since \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\), any terminating decimal like \(1.5\) (or \(\frac{3}{2}\)) fits perfectly in this interval.
In simple words: Since \(\sqrt{2}\) is about 1.41 and \(\sqrt{3}\) is about 1.73, the number 1.5 lies between them and is rational because it can be written as 3/2.

Exam Tip: You can choose any simple terminating decimal like 1.5 or 1.6 that lies between 1.414 and 1.732.

Question 3. The decimal expansion oftherationalno.43/2\(^4\)5\(^3\) will terminate after how many places of decimal?
Answer: The decimal expansion will end after \(4\) decimal places. This is determined by the highest power of 2 or 5 in the prime factorization of the denominator, which is \(4\) in \(2^4\).
In simple words: Look at the powers of 2 and 5 in the denominator. The larger exponent is 4, which means the decimal will stop after exactly 4 places.

Exam Tip: In \(\frac{p}{2^n \cdot 5^m}\), the decimal expansion always terminates after \(\max(n, m)\) decimal places.

Question 4. Find the (HCF X LCM) for the numbers 100 and 190.
Answer: The product of HCF and LCM of any two numbers is equal to the product of those two numbers themselves. Thus, \(\text{HCF}(100, 190) \times \text{LCM}(100, 190) = 100 \times 190 = 19000\).
In simple words: You do not need to calculate the HCF or LCM. Just multiply the two numbers together to find the product of their HCF and LCM.

Exam Tip: Remember the formula \(\text{HCF} \times \text{LCM} = a \times b\). It saves a lot of time on such direct questions.

Question 5. State whether the number \((\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})\) is rational or irrational justify.
Answer: The given expression is rational. Using the algebraic identity \((a - b)(a + b) = a^2 - b^2\), we can simplify the expression: \((\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1\). Since \(-1\) is an integer, it is a rational number.
In simple words: Multiply the brackets using the formula \((a-b)(a+b) = a^2-b^2\). This simplifies to -1, which is a plain rational number.

Exam Tip: Always simplify the expression completely before declaring whether a number is rational or irrational.

Question 6. Write one rational and one irrational number lying between 0.25and 0.32.
Answer: A rational number in this range is \(0.26\) (which can be written as \(\frac{13}{50}\)), and an irrational number is \(0.27010010001\ldots\) because its decimal representation is non-terminating and non-repeating.
In simple words: For a rational number, choose a simple terminating decimal like 0.26. For an irrational number, create a decimal like 0.27010010001... that goes on forever without any repeating pattern.

Exam Tip: To write an irrational number, construct a non-terminating, non-repeating pattern such as \(0.28010010001\ldots\)

Question 7. Express 107 in the form of 4q+3 for some positive integer q.
Answer: By dividing \(107\) by \(4\), we get a quotient of \(26\) and a remainder of \(3\). Thus, \(107\) can be expressed as: \(107 = 4 \times 26 + 3\), where \(q = 26\).
In simple words: Divide 107 by 4. You get 26 with 3 left over, so write it as \(4 \times 26 + 3\).

Exam Tip: Use Euclid's Division Lemma (\(a = bq + r\)) to systematically find the quotient and remainder.

Question 8. Write whether the rational number \(\frac{51}{1500}\) will have a terminating decimal expansion or a non Terminating repeating decimal expansion.
Answer: The rational number \(\frac{51}{1500}\) simplifies to \(\frac{17}{500}\) by dividing both numerator and denominator by 3. The prime factorization of the simplified denominator \(500\) is \(2^2 \times 5^3\). Since the prime factors consist only of 2 and 5, the decimal expansion is terminating.
In simple words: First, reduce the fraction to \(\frac{17}{500}\). The bottom number has only 2s and 5s in its prime factorization, which means the decimal will stop.

Exam Tip: Always simplify the fraction to its lowest terms (co-prime) before checking the prime factors of the denominator.

Question 9. Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5, where q is some integer.
Answer: According to Euclid's division lemma, any positive integer \(a\) can be written in the form \(a = 6q + r\) where \(0 \leq r < 6\) and \(q\) is a non-negative integer. The possible values for the remainder \(r\) are \(0, 1, 2, 3, 4, 5\). This gives six possible forms for \(a\):
• If \(a = 6q\), \(a = 6q + 2 = 2(3q + 1)\), or \(a = 6q + 4 = 2(3q + 2)\), the integer is divisible by 2 and is therefore even.
• If \(a = 6q + 1\), \(a = 6q + 3 = 2(3q + 1) + 1\), or \(a = 6q + 5 = 2(3q + 2) + 1\), the integer is not divisible by 2. Thus, any positive odd integer must be of the form \(6q + 1\), \(6q + 3\), or \(6q + 5\).
In simple words: When you divide any number by 6, the remainder can be 0, 1, 2, 3, 4, or 5. The remainders 0, 2, and 4 give even numbers, so odd numbers must end with remainders 1, 3, or 5.

Exam Tip: Start your proof by stating Euclid's Division Lemma and specifying all possible values for the remainder \(r\).

Question 10. Express 0.2545454…………..As a fraction in simplest form.
Answer: Let \(x = 0.2\overline{54}\). Multiplying by \(10\) gives \(10x = 2.\overline{54}\), and multiplying the original equation by \(1000\) gives \(1000x = 254.\overline{54}\). Subtracting these two equations yields \(990x = 252\), which simplifies to \(x = \frac{252}{990} = \frac{14}{55}\).
In simple words: Let \(x = 0.25454...\). By using equations to eliminate the repeating part, we find the fraction is 252/990, which simplifies to 14/55.

Exam Tip: Always reduce the fraction to its lowest terms by dividing the numerator and denominator by their greatest common divisor.

LEVEL-II

Question 1. Use Euclid’s division algorithm to find the HCF of 1288 and 575.
Answer: We apply Euclid's division algorithm to the numbers \(1288\) and \(575\):
• Step 1: \(1288 = 575 \times 2 + 138\) (since remainder \(138 \neq 0\), we apply the algorithm to \(575\) and \(138\))
• Step 2: \(575 = 138 \times 4 + 23\) (since remainder \(23 \neq 0\), we apply the algorithm to \(138\) and \(23\))
• Step 3: \(138 = 23 \times 6 + 0\)
Since the remainder is now \(0\), the divisor at this stage, which is \(23\), is the HCF.
In simple words: Keep dividing the larger number by the smaller one and using the remainder as the new divisor. When the remainder becomes 0, the last divisor (23) is the HCF.

Exam Tip: Show all division steps clearly to secure full marks in long-answer questions.

Question 2. Check whether 5 x 3 x 11+11 and 5x7+7X3 are composite number and justify.
Answer: Both numbers are composite. For the first expression, \(5 \times 3 \times 11 + 11 = 11(5 \times 3 + 1) = 11(16)\), which has factors other than 1 and itself. Similarly, for the second expression, \(5 \times 7 + 7 \times 3 = 7(5 + 3) = 7(8)\), which also has multiple prime factors. Since both expressions can be factored into a product of primes, they are composite numbers.
In simple words: Factor out 11 from the first number to get \(11 \times 16\), and factor out 7 from the second to get \(7 \times 8\). Since both have factors other than 1 and themselves, they are composite numbers.

Exam Tip: Avoid multiplying out the entire expression; instead, show they are composite by factoring out common terms.

Question 3. Check whether \(6^n\) can end with the digit 0, where n is any natural number.
Answer: For any positive integer \(6^n\) to end with the digit \(0\), its prime factorization must contain both \(2\) and \(5\). However, the prime factorization of \(6^n\) is \((2 \times 3)^n = 2^n \times 3^n\). Since \(5\) is not a prime factor in this representation, \(6^n\) can never end with \(0\) for any natural number \(n\).
In simple words: To end in 0, a number's prime factors must include both 2 and 5. Since \(6^n\) only has 2 and 3 as prime factors, it can never end in 0.

Exam Tip: Use the Fundamental Theorem of Arithmetic to justify that the prime factorization of a number is unique.

Page 10

Question 4. Given that LCM (26,169) = 338, write HCF (26,169).]
Answer: Using the relation \(\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\), we can substitute the given values:
\(\text{HCF}(26, 169) \times 338 = 26 \times 169\)
\(\text{HCF}(26, 169) = \frac{26 \times 169}{338} = \frac{4394}{338} = 13\).
In simple words: Multiply 26 by 169, then divide the result by 338 to get the HCF, which is 13.

Exam Tip: Use the HCF-LCM product relationship to directly solve for the missing value without prime factoring.

Question 5. Find the HCF and LCM of 6, 72and 120 using the prime factorization method.
Answer: Using prime factorization, we find:
• \(6 = 2^1 \times 3^1\)
• \(72 = 2^3 \times 3^2\)
• \(120 = 2^3 \times 3^1 \times 5^1\)
• The HCF is the product of the lowest powers of common prime factors: \(2^1 \times 3^1 = 6\).
• The LCM is the product of the highest powers of all prime factors involved: \(2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360\).
In simple words: Write down the prime factors for each number. The HCF uses the smallest power of the shared factors (2 and 3), giving 6. The LCM uses the largest power of all factors, giving 360.

Exam Tip: Make sure to include 5 in the LCM calculation, even though it is not a common factor for all three numbers.

Question 6. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Answer: Let \(a\) be any positive integer. By Euclid's division lemma with divisor \(b = 3\), we can express \(a\) as \(a = 3q + r\), where \(0 \leq r < 3\). The possible values of \(r\) are \(0, 1, 2\). This gives three cases for \(a\):
• Case 1: If \(a = 3q\), then \(a^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m\), where \(m = 3q^2\).
• Case 2: If \(a = 3q + 1\), then \(a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1\), where \(m = 3q^2 + 2q\).
• Case 3: If \(a = 3q + 2\), then \(a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 = 3m + 1\), where \(m = 3q^2 + 4q + 1\).
Thus, the square of any positive integer is always of the form \(3m\) or \(3m+1\).
In simple words: Any number can be written as \(3q\), \(3q+1\), or \(3q+2\). Squaring each of these forms and factoring out 3 always leaves either a remainder of 0 or 1.

Exam Tip: Clearly define what 'm' stands for in each case to make your proof mathematically rigorous.

Question 7. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8 for some integer m.
Answer: Let \(a\) be any positive integer. Applying Euclid's division lemma with divisor \(3\), we can write \(a = 3q + r\), where \(0 \leq r < 3\). The possible values for \(r\) are \(0, 1, 2\). Cubing these three forms gives:
• Case 1: If \(a = 3q\), then \(a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m\), where \(m = 3q^3\).
• Case 2: If \(a = 3q + 1\), then \(a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1\), where \(m = 3q^3 + 3q^2 + q\).
• Case 3: If \(a = 3q + 2\), then \(a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8\), where \(m = 3q^3 + 6q^2 + 4q\).
Thus, the cube of any positive integer is of the form \(9m\), \(9m+1\), or \(9m+8\).
In simple words: Cube any number written in terms of divisor 3. Factoring out 9 from the expanded cube will always leave a remainder of 0, 1, or 8.

Exam Tip: Use the divisor 3 instead of 9 for Euclid's Lemma on cubing problems; it is much shorter and completely correct since any multiple of 9 is also a multiple of 3.

LEVEL-III

Question 1. Show that √3 is an irrational number.
Answer: Let us assume, on the contrary, that \(\sqrt{3}\) is a rational number. Therefore, we can write \(\sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are co-prime integers and \(b \neq 0\).
Squaring both sides gives \(3 = \frac{a^2}{b^2} \implies a^2 = 3b^2\). This implies that \(3\) divides \(a^2\), which means \(3\) must also divide \(a\) (by theorem).
Let \(a = 3c\) for some integer \(c\). Substituting this gives \((3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies b^2 = 3c^2\). This means \(3\) divides \(b^2\), so \(3\) also divides \(b\).
Thus, both \(a\) and \(b\) share a common factor of \(3\), which contradicts our assumption that they are co-prime. Hence, \(\sqrt{3}\) must be irrational.
In simple words: Assume \(\sqrt{3}\) is a regular fraction in simplest form. By squaring and rearranging, we prove both the top and bottom numbers can be divided by 3, which is impossible for a fraction in simplest form.

Exam Tip: Always state the theorem: "If a prime p divides \(a^2\), then p divides a", as it is critical for full marks.

Question 2. Show that is an irrational number.
Answer: (Note: The question is to show that \(5 + 3\sqrt{2}\) is irrational).
Let us assume that \(5 + 3\sqrt{2}\) is a rational number, say \(r\).
We can write \(5 + 3\sqrt{2} = r \implies 3\sqrt{2} = r - 5 \implies \sqrt{2} = \frac{r - 5}{3}\).
Since \(r\) is a rational number, \(\frac{r - 5}{3}\) must also be rational. However, this means \(\sqrt{2}\) is rational, which contradicts the known fact that \(\sqrt{2}\) is irrational. Thus, our assumption is incorrect, and \(5 + 3\sqrt{2}\) is irrational.
In simple words: Assume the number is rational. Rearranging it shows that \(\sqrt{2}\) must equal a rational fraction, which is impossible because we know \(\sqrt{2}\) is irrational.

Exam Tip: Isolate the root term (like \(\sqrt{2}\)) on one side to show the contradiction between rational and irrational sides.

Question 3. Show that square of an odd positive integer is of the form 8m+1, for some integer m.
Answer: Any positive odd integer \(a\) can be expressed in the form \(a = 4q + 1\) or \(a = 4q + 3\) for some integer \(q\).
• Case 1: If \(a = 4q + 1\), then \(a^2 = (4q + 1)^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1 = 8m + 1\), where \(m = 2q^2 + q\).
• Case 2: If \(a = 4q + 3\), then \(a^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 16q^2 + 24q + 8 + 1 = 8(2q^2 + 3q + 1) + 1 = 8m + 1\), where \(m = 2q^2 + 3q + 1\).
Thus, the square of any positive odd integer is always of the form \(8m + 1\).
In simple words: Odd numbers can be written as \(4q+1\) or \(4q+3\). Squaring these forms and factoring out 8 always leaves a remainder of 1.

Exam Tip: Using the form \(4q+1\) or \(4q+3\) is much more direct for showing divisibility by 8 than using \(2q+1\).

Question 4. Find the LCM &HCF of 26 and 91 and verify that
Answer: First, we find the prime factors of both numbers:
• \(26 = 2 \times 13\)
• \(91 = 7 \times 13\)
The HCF is \(13\) and the LCM is \(2 \times 7 \times 13 = 182\).
Verification:
\(\text{HCF} \times \text{LCM} = 13 \times 182 = 2366\).
Product of the two numbers \(= 26 \times 91 = 2366\).
Since both values are equal, the relation \(\text{HCF} \times \text{LCM} = a \times b\) is verified.
In simple words: The HCF of 26 and 91 is 13, and their LCM is 182. Multiplying 13 by 182 gives 2366, which is exactly the same as multiplying 26 by 91.

Exam Tip: Always show the final calculation for both sides of the identity to prove the verification is complete.

Question 5. Prove that ∛7 is irrational.
Answer: Assume, on the contrary, that \(\sqrt[3]{7}\) is a rational number. Thus, we can express it as \(\sqrt[3]{7} = \frac{a}{b}\), where \(a\) and \(b\) are co-prime integers and \(b \neq 0\).
Cubing both sides gives \(7 = \frac{a^3}{b^3} \implies a^3 = 7b^3\). This implies that \(7\) divides \(a^3\), which means \(7\) must divide \(a\).
Let \(a = 7c\) for some integer \(c\). Substituting this gives \((7c)^3 = 7b^3 \implies 343c^3 = 7b^3 \implies b^3 = 49c^3 = 7(7c^3)\). This means \(7\) divides \(b^3\), so \(7\) also divides \(b\).
Thus, both \(a\) and \(b\) share a common factor of \(7\), contradicting their co-primality. Hence, \(\sqrt[3]{7}\) is irrational.
In simple words:

Real Numbers (Key Points)

Classification of Real Numbers

• Real Numbers: Comprise both rational and irrational numbers.
  • Rational Numbers (Q): Can be written in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \). These include:
    • Natural Numbers (N): Counting numbers starting from 1 - \( (1, 2, 3, \dots) \).
    • Whole Numbers (W): Natural numbers along with zero - \( (0, 1, 2, 3, \dots) \).
    • Integers (Z): Includes negative integers \( (\dots, -3, -2, -1) \), zero \( (0) \), and positive integers \( (1, 2, 3, \dots) \).
  • Irrational Numbers (I): Numbers that cannot be expressed as a ratio of integers.

Decimal Form of Real Numbers

• Terminating Decimal: Decimals that end after a finite number of digits (e.g., \( \frac{2}{5} \), \( \frac{3}{4} \)). These are Rational Numbers.
• Non-Terminating Repeating (Recurring) Decimal: Decimals that do not end but repeat a pattern of digits (e.g., \( \frac{1}{3} \), \( \frac{2}{7} \), \( \frac{3}{11} \)). These are Rational Numbers.
• Non-Terminating Non-Repeating Decimal: Decimals that never end and never repeat a regular pattern (e.g., \( 1.010010001\dots \)). These are Irrational Numbers.

Theorems and Lemmas

• Euclid's Division Lemma: Given positive integers \( a \) and \( b \), there exist unique integers \( q \) and \( r \) satisfying:
  \( a = bq + r \), where \( 0 \leq r < b \)
  Here, \( a, b, q, \) and \( r \) are respectively called the dividend, divisor, quotient, and remainder.
• Euclid's Division Algorithm: To obtain the HCF of two positive integers \( c \) and \( d \) with \( c > d \):
  Step I: Apply Euclid's division lemma to \( c \) and \( d \) to find whole numbers \( q \) and \( r \) such that \( c = dq + r \), where \( 0 \leq r < d \).
  Step II: If \( r = 0 \), then \( d \) is the HCF of \( c \) and \( d \). If \( r \neq 0 \), apply the division lemma to \( d \) and \( r \).
  Step III: Continue the process until the remainder is zero. The divisor at this stage will be the required HCF.
  Note: If \( a = bq + r \), where \( 0 \leq r < b \), then \( \text{HCF}(a, b) = \text{HCF}(b, r) \).
• The Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorized) as a unique product of primes, apart from the order in which the prime factors occur.
  Example: \( 24 = 2 \times 2 \times 2 \times 3 = 3 \times 2 \times 2 \times 2 \).
• Theorem on Terminating Decimals: Let \( x \) be a rational number whose decimal expansion terminates. Then \( x \) can be expressed in the form \( \frac{p}{q} \), where \( p \) and \( q \) are co-prime, and the prime factorization of \( q \) is of the form \( 2^n \cdot 5^m \), where \( n \) and \( m \) are non-negative integers.
  Example: \( \frac{7}{10} = \frac{7}{2 \times 5} = 0.7 \).
• Theorem on Non-Terminating Repeating Decimals: Let \( x = \frac{p}{q} \) be a rational number such that the prime factorization of \( q \) is not of the form \( 2^n \cdot 5^m \), where \( n \) and \( m \) are non-negative integers. Then \( x \) has a decimal expansion that is non-terminating repeating (recurring).
  Example: \( \frac{7}{6} = \frac{7}{2 \times 3} = 1.1666\dots \).
• Theorem on HCF and LCM Relation: For any two positive integers \( a \) and \( b \):
  \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \)
  Example: For 4 and 6, \( \text{HCF}(4, 6) = 2 \) and \( \text{LCM}(4, 6) = 12 \).
  \( \text{HCF} \times \text{LCM} = 2 \times 12 = 24 \), and \( a \times b = 4 \times 6 = 24 \).

LEVEL-I

Question 1. If \( \frac{p}{q} \) is a rational number (\( q \neq 0 \)). What is the condition on q so that the decimal representation of \( \frac{p}{q} \) is terminating?
Answer: The prime factorization of the denominator \( q \) must be of the form \( 2^n \cdot 5^m \), where \( n \) and \( m \) are non-negative integers.
In simple words: For a fraction to have a terminating decimal, the bottom number must only have 2 and 5 as its prime factors.

Exam Tip: Remember to specify that \( n \) and \( m \) must be non-negative integers, meaning they can also be zero.

Question 2. Write a rational number between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer: A rational number lying in the interval between \( \sqrt{2} \approx 1.414 \) and \( \sqrt{3} \approx 1.732 \) is \( 1.5 \) (or \( \frac{3}{2} \)).
In simple words: Since \( \sqrt{2} \) is about 1.41 and \( \sqrt{3} \) is about 1.73, any standard decimal like 1.5 lies between them and is rational.

Exam Tip: To find numbers between square roots, convert them to approximate decimal forms first to verify your choice.

Question 3. The decimal expansion of the rational no. \( \frac{43}{2^4 \cdot 5^3} \) will terminate after how many places of decimal?
Answer: The prime factors in the denominator have exponents of 4 and 3. The higher power is 4, so the decimal expansion stops after 4 decimal places.
In simple words: Look at the powers of 2 and 5 in the denominator. Whichever power is larger tells you the exact number of decimal places before the number terminates.

Exam Tip: The terminating decimal place corresponds to \( \max(n, m) \) in the denominator \( 2^n \cdot 5^m \).

Question 4. Find the HCF X LCM for the numbers 100 and 190.
Answer: Using the property \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \):
\( \text{HCF} \times \text{LCM} = 100 \times 190 = 19000 \)
In simple words: Multiplying the HCF and LCM of two numbers is exactly the same as multiplying the two original numbers together.

Exam Tip: This formula only holds true for two numbers, not three or more.

Question 5. State whether the number \( (\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3}) \) is rational or irrational justify.
Answer: Applying the difference of squares identity \( (a-b)(a+b) = a^2 - b^2 \):
\( (\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1 \)
Since \( -1 \) is an integer, it is a rational number.
In simple words: Multiplying these terms removes the square roots and leaves us with -1, which is a rational number.

Exam Tip: Always fully simplify the expression before writing down whether it is rational or irrational.

Question 6. Write one rational and one irrational number lying between 0.25and 0.32.
Answer:
A rational number between \( 0.25 \) and \( 0.32 \) is \( 0.26 \) (which can be written as \( \frac{26}{100} = \frac{13}{50} \)).
An irrational number between \( 0.25 \) and \( 0.32 \) is \( 0.27010010001\dots \) (a non-terminating, non-repeating decimal).
In simple words: A simple ending decimal like 0.26 is rational. An endless decimal with no repeating pattern like 0.27010010001... is irrational.

Exam Tip: For irrational numbers, create a clear pattern that does not repeat to demonstrate it is non-recurring.

Question 7. Express 107 in the form of 4q+3 for some positive integer q.
Answer: Dividing 107 by 4 gives a quotient of 26 and a remainder of 3. Thus:
\( 107 = 4 \times 26 + 3 \)
Here, the integer \( q = 26 \).
In simple words: If we divide 107 by 4, it goes in 26 times with a leftover of 3.

Exam Tip: This is a direct application of Euclid's Division Lemma where \( a = bq + r \).

Question 8. Write whether the rational number \( \frac{51}{1500} \) will have a terminating decimal expansion or a non Terminating repeating decimal expansion.
Answer: First, simplify the rational number to its lowest terms:
\( \frac{51}{1500} = \frac{17}{500} \)
Now, find the prime factorization of the simplified denominator \( 500 \):
\( 500 = 2^2 \times 5^3 \)
Since the prime factors of the denominator consist solely of 2 and 5, it has a terminating decimal expansion.
In simple words: Simplify the fraction first! Once simplified, the bottom number only has 2 and 5 as its prime factors, meaning it will terminate.

Exam Tip: Never analyze the denominator before simplifying the fraction, as common factors can lead to incorrect conclusions.

Question 9. Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5, where q is some integer.
Answer: Let \( a \) be any positive odd integer and let \( b = 6 \). Applying Euclid's division lemma:
\( a = 6q + r \), where \( 0 \leq r < 6 \)
The possible remainders are \( r = 0, 1, 2, 3, 4, 5 \).
This gives the possible forms for \( a \):
\( 6q, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5 \)
Since \( a \) is odd, we exclude the even forms \( 6q = 2(3q) \), \( 6q+2 = 2(3q+1) \), and \( 6q+4 = 2(3q+2) \).
Thus, any positive odd integer must be of the form \( 6q+1 \), \( 6q+3 \), or \( 6q+5 \).
In simple words: When dividing by 6, the remainders can be 0 to 5. Even remainders produce even numbers, so odd integers must have odd remainders (1, 3, or 5).

Exam Tip: State Euclid's lemma clearly at the beginning and explicitly show why the even cases are divisible by 2.

Question 10. Express 0.2545454…………..As a fraction in simplest form.
Answer: Let \( x = 0.2\overline{54} \)
Multiplying by 10 to shift the non-repeating part:
\( 10x = 2.\overline{54} \) — (Equation 1)
Multiplying Equation 1 by 100 since there are two repeating digits:
\( 1000x = 254.\overline{54} \) — (Equation 2)
Subtracting Equation 1 from Equation 2:
\( 1000x - 10x = 254.\overline{54} - 2.\overline{54} \)
\( 990x = 252 \)
\( x = \frac{252}{990} \)
Simplifying by dividing both the numerator and denominator by 18:
\( x = \frac{14}{55} \)
In simple words: We multiply by 10 and 1000 to align the repeating decimal parts. Subtracting these equations cancels out the infinite repeating digits, leaving a simple fraction that simplifies to \( \frac{14}{55} \).

Exam Tip: Double-check your final fraction by performing long division to make sure it matches the original repeating decimal.

LEVEL-II

Question 1. Use Euclid’s division algorithm to find the HCF of 1288 and 575.
Answer: We apply Euclid's division algorithm to 1288 and 575:
\( 1288 = 575 \times 2 + 138 \)
Since the remainder \( 138 \neq 0 \), we apply the division lemma to 575 and 138:
\( 575 = 138 \times 4 + 23 \)
Since the remainder \( 23 \neq 0 \), we apply the division lemma to 138 and 23:
\( 138 = 23 \times 6 + 0 \)
Since the remainder has reached zero, the divisor at this step is the HCF.
\( \text{HCF}(1288, 575) = 23 \)
In simple words: Keep dividing the previous divisor by the remainder until you get a remainder of zero. The last divisor you used is the HCF.

Exam Tip: Ensure you show every single step of the division algorithm to secure full marks.

Question 2. Check whether \( 5 \times 3 \times 11 + 11 \) and \( 5 \times 7 + 7 \times 3 \) are composite number and justify.
Answer:
1) For \( 5 \times 3 \times 11 + 11 \):
We can factor out the common term 11:
\( 11 \times (5 \times 3 + 1) = 11 \times (15 + 1) = 11 \times 16 \)
Since this number can be expressed as a product of prime factors \( 11 \times 2^4 \), it is a composite number.

2) For \( 5 \times 7 + 7 \times 3 \):
We can factor out the common term 7:
\( 7 \times (5 + 3) = 7 \times 8 \)
Since this number can be expressed as a product of prime factors \( 7 \times 2^3 \), it is also a composite number.
In simple words: Any number with factors other than 1 and itself is composite. Factoring out the common terms shows that both values are composite.

Exam Tip: Factoring out the common term is quicker and less error-prone than calculating the final product and then finding its factors.

Question 3. Check whether \( 6^n \) can end with the digit 0, where n is any natural number.
Answer: For any positive integer to end with the digit 0, its prime factorization must contain both 2 and 5.
The prime factorization of \( 6^n \) is:
\( 6^n = (2 \times 3)^n = 2^n \times 3^n \)
The prime factors of \( 6^n \) are only 2 and 3. Since 5 is not present in its prime factorization, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).
In simple words: To end in a zero, a number must be a multiple of 10, meaning it needs 2 and 5 as prime factors. Since \( 6^n \) only contains 2 and 3, it can never end in zero.

Exam Tip: Cite the Fundamental Theorem of Arithmetic to justify why the prime factorization of \( 6^n \) is unique and cannot contain 5.

Question 4. Given that LCM (26,169) = 338, write HCF (26,169).
Answer: We use the mathematical identity:
\( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \)
Given \( a = 26 \), \( b = 169 \), and \( \text{LCM} = 338 \):
\( \text{HCF}(26, 169) = \frac{26 \times 169}{338} \)
\( \text{HCF}(26, 169) = \frac{4394}{338} = 13 \)
In simple words: Multiply the two numbers and divide the result by their LCM to find their HCF.

Exam Tip: Simplify the calculation by recognizing that \( 338 = 2 \times 169 \), which leaves \( \frac{26}{2} = 13 \).

Question 5. Find the HCF and LCM of 6, 72and 120 using the prime factorization method.
Answer: First, find the prime factorization of each of the numbers:
\( 6 = 2^1 \times 3^1 \)
\( 72 = 2^3 \times 3^2 \)
\( 120 = 2^3 \times 3^1 \times 5^1 \)

To find the HCF, take the product of the lowest power of each common prime factor:
\( \text{HCF} = 2^1 \times 3^1 = 6 \)

To find the LCM, take the product of the highest power of all prime factors involved:
\( \text{LCM} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \)
In simple words: HCF uses the smallest shared powers of the common factors, while LCM takes the largest powers of all the prime factors found in the numbers.

Exam Tip: Expressing the factorizations with exponents helps prevent errors when selecting the correct powers for HCF and LCM.

Question 6. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Answer: Let \( a \) be any positive integer and \( b = 3 \). By Euclid's division lemma:
\( a = 3q + r \), where \( q \geq 0 \) and \( 0 \leq r < 3 \) (meaning \( r = 0, 1, 2 \)).
Now, let's square \( a \) for each possible remainder:

Case 1: If \( r = 0 \), then \( a = 3q \).
\( a^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m \), where \( m = 3q^2 \) is an integer.

Case 2: If \( r = 1 \), then \( a = 3q + 1 \).
\( a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1 \), where \( m = 3q^2 + 2q \) is an integer.

Case 3: If \( r = 2 \), then \( a = 3q + 2 \).
\( a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1 \), where \( m = 3q^2 + 4q + 1 \) is an integer.

Thus, the square of any positive integer is always of the form \( 3m \) or \( 3m+1 \) for some integer \( m \).
In simple words: Any positive integer can be written relative to 3. Squaring these options always allows us to group terms to show a multiple of 3, or a multiple of 3 plus 1.

Exam Tip: In Case 3, remember to split 4 into \( 3 + 1 \) so you can factor out the 3 from the rest of the terms.

Question 7. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8 for some integer m.
Answer: Let \( a \) be any positive integer and \( b = 3 \). By Euclid's division lemma:
\( a = 3q + r \), where \( 0 \leq r < 3 \) (meaning \( r = 0, 1, 2 \)).
Let's examine the cube of \( a \) for each possible remainder:

Case 1: If \( r = 0 \), then \( a = 3q \).
\( a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m \), where \( m = 3q^3 \) is an integer.

Case 2: If \( r = 1 \), then \( a = 3q + 1 \).
\( a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1 \), where \( m = 3q^3 + 3q^2 + q \) is an integer.

Case 3: If \( r = 2 \), then \( a = 3q + 2 \).
\( a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8 \), where \( m = 3q^3 + 6q^2 + 4q \) is an integer.

Thus, the cube of any positive integer is of the form \( 9m \), \( 9m + 1 \), or \( 9m + 8 \) for some integer \( m \).
In simple words: Any positive integer can be written relative to 3. Cubing these forms and factoring out 9 leaves us with remainders of only 0, 1, or 8.

Exam Tip: Using \( b = 3 \) instead of \( b = 9 \) is much faster and completely correct, since any multiple of 9 is also a multiple of 3.

LEVEL-III

Question 1. Show that √3 is an irrational number.
Answer: Assume, on the contrary, that \( \sqrt{3} \) is a rational number. Let:
\( \sqrt{3} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).
Squaring both sides:
\( 3 = \frac{a^2}{b^2} \)
\( \implies a^2 = 3b^2 \) — (Equation 1)
Since 3 divides \( a^2 \), it must also divide \( a \). Let \( a = 3c \) for some integer \( c \).
Substituting into Equation 1:
\( (3c)^2 = 3b^2 \)
\( \implies 9c^2 = 3b^2 \)
\( \implies b^2 = 3c^2 \)
This means 3 divides \( b^2 \), and consequently, 3 must also divide \( b \).
Since 3 divides both \( a \) and \( b \), they have a common factor of 3, which contradicts our assumption that \( a \) and \( b \) are co-prime.
Thus, \( \sqrt{3} \) is irrational.
In simple words: We assume the root is a fully simplified fraction. We then prove that both the numerator and denominator can still be divided by 3, creating a contradiction that proves the assumption wrong.

Exam Tip: Be sure to write the statement "if a prime divides \( a^2 \), it must divide \( a \)" to make the proof mathematically complete.

Question 2. Show that 5 + 3√2 is an irrational number.
Answer: Let us assume that \( 5 + 3\sqrt{2} \) is a rational number. Then:
\( 5 + 3\sqrt{2} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).
Rearranging the terms:
\( 3\sqrt{2} = \frac{a}{b} - 5 \)
\( \implies 3\sqrt{2} = \frac{a - 5b}{b} \)
\( \implies \sqrt{2} = \frac{a - 5b}{3b} \)
Since \( a \), \( b \), 5, and 3 are integers, \( \frac{a - 5b}{3b} \) is rational. This implies \( \sqrt{2} \) must be rational.
However, this contradicts the established fact that \( \sqrt{2} \) is irrational.
Thus, our assumption is incorrect, and \( 5 + 3\sqrt{2} \) is irrational.
In simple words: We isolate the square root term. If the starting number were rational, then \( \sqrt{2} \) would have to equal a rational fraction, which is impossible.

Exam Tip: In questions like this, you can directly assume \( \sqrt{2} \) is irrational unless the question explicitly asks you to prove it first.

Question 3. Show that square of an odd positive integer is of the form 8m+1, for some integer m.
Answer: Any positive odd integer can be written in the form \( 4q + 1 \) or \( 4q + 3 \) for some integer \( q \). Let's square both forms:

Case 1: For \( (4q + 1) \):
\( (4q + 1)^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1 = 8m + 1 \), where \( m = 2q^2 + q \) is an integer.

Case 2: For \( (4q + 3) \):
\( (4q + 3)^2 = 16q^2 + 24q + 9 = 16q^2 + 24q + 8 + 1 = 8(2q^2 + 3q + 1) + 1 = 8m + 1 \), where \( m = 2q^2 + 3q + 1 \) is an integer.

Thus, the square of any odd positive integer is of the form \( 8m + 1 \) for some integer \( m \).
In simple words: Representing an odd number relative to 4 and squaring it always lets us group terms into a multiple of 8 plus 1.

Exam Tip: Expressing the odd integer as \( 4q+1 \) and \( 4q+3 \) is much better than \( 2q+1 \), as squaring \( 2q+1 \) only directly gives a factor of 4, making the factor of 8 harder to show.

Question 4. Find the LCM &HCF of 26 and 91 and verify that [HCF X LCM = product of two numbers].
Answer: Prime factorization of the numbers:
\( 26 = 2 \times 13 \)
\( 91 = 7 \times 13 \)

\( \text{HCF}(26, 91) = 13 \)
\( \text{LCM}(26, 91) = 2 \times 7 \times 13 = 182 \)

Verification:
\( \text{HCF} \times \text{LCM} = 13 \times 182 = 2366 \)
\( \text{Product of the numbers} = 26 \times 91 = 2366 \)
Since \( \text{HCF} \times \text{LCM} = \text{Product of the numbers} \), the relationship is verified.
In simple words: We find the prime factors, get the HCF and LCM, and then multiply them. The result is 2366, which is exactly equal to the product of 26 and 91.

Exam Tip: State the final values of both sides clearly and write "Hence verified" to conclude the proof.

Question 5. Prove that ∛7 is irrational.
Answer: Assume, on the contrary, that \( \sqrt[3]{7} \) is rational. Let:
\( \sqrt[3]{7} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).
Cubing both sides:
\( 7 = \frac{a^3}{b^3} \)
\( \implies a^3 = 7b^3 \) — (Equation 1)
This implies 7 divides \( a^3 \). Since 7 is a prime number, 7 must also divide \( a \). Let \( a = 7c \) for some integer \( c \).
Substituting this into Equation 1:
\( (7c)^3 = 7b^3 \)
\( \implies 343c^3 = 7b^3 \)
\( \implies b^3 = 49c^3 = 7(7c^3) \)
This implies 7 divides \( b^3 \), which means 7 also divides \( b \).
Since 7 divides both \( a \) and \( b \), they share a common factor of 7, which contradicts that they are co-prime.
Thus, \( \sqrt[3]{7} \) is irrational.
In simple words: This proof is just like the square root proof, except we cube both sides. We show that both parts of the fraction must be divisible by 7, which means it cannot be a rational fraction.

Exam Tip: Make sure you state that because 7 is a prime, if it divides \( a^3 \), it must divide \( a \).

Question 6. Show that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer.
Answer: Any positive integer \( n \) can be expressed in one of the forms: \( 3q \), \( 3q+1 \), or \( 3q+2 \).

Case 1: If \( n = 3q \):
- \( n = 3q \) is divisible by 3.
- \( n+2 = 3q+2 \) is not divisible by 3.
- \( n+4 = 3q+4 = 3(q+1) + 1 \) is not divisible by 3.
Only \( n \) is divisible by 3.

Case 2: If \( n = 3q+1 \):
- \( n = 3q+1 \) is not divisible by 3.
- \( n+2 = 3q+3 = 3(q+1) \) is divisible by 3.
- \( n+4 = 3q+5 = 3(q+1) + 2 \) is not divisible by 3.
Only \( n+2 \) is divisible by 3.

Case 3: If \( n = 3q+2 \):
- \( n = 3q+2 \) is not divisible by 3.
- \( n+2 = 3q+4 = 3(q+1) + 1 \) is not divisible by 3.
- \( n+4 = 3q+6 = 3(q+2) \) is divisible by 3.
Only \( n+4 \) is divisible by 3.

Therefore, for any positive integer \( n \), one and only one of \( n \), \( n+2 \), or \( n+4 \) is divisible by 3.
In simple words: Any starting number will be a multiple of 3, a multiple of 3 plus 1, or a multiple of 3 plus 2. Testing each case shows that only one of the three numbers will divide by 3.

Exam Tip: Clearly list out the three distinct cases and evaluate all three expressions (\( n, n+2, n+4 \)) under each case.

Question 7. Find the HCF of 65 & 117 and express it in the form of 65m + 117n.
Answer: Using Euclid's division algorithm to find HCF:
\( 117 = 65 \times 1 + 52 \) — (Equation 1)
\( 65 = 52 \times 1 + 13 \) — (Equation 2)
\( 52 = 13 \times 4 + 0 \)
Thus, the HCF is 13.

To express HCF in the form \( 65m + 117n \), we work backwards from Equation 2:
\( 13 = 65 - 52 \times 1 \)
From Equation 1, substitute \( 52 = 117 - 65 \times 1 \):
\( 13 = 65 - (117 - 65 \times 1) \)
\( 13 = 65 - 117 + 65 \)
\( 13 = 65(2) + 117(-1) \)
Comparing with \( 65m + 117n \), we get:
\( m = 2 \) and \( n = -1 \).
In simple words: First, find the HCF using division, which is 13. Then, substitute back through the division equations to express 13 as \( 65(2) + 117(-1) \).

Exam Tip: Be very careful with positive and negative signs when substituting and regrouping the terms.

(PROBLEMS FOR SELF EVALUATION/HOTS)

Question 1. State the fundamental theorem of Arithmetic.
Answer: The Fundamental Theorem of Arithmetic states that every composite number can be uniquely factorized as a product of prime numbers, regardless of the order in which these prime factors are written.
In simple words: Any composite number can be broken down into a unique set of prime multiplication blocks.

Exam Tip: Remember to use the keyword "unique" to describe the prime factorization to secure full marks.

Question 2. Express 2658 as a product of its prime factors.
Answer: We factorize 2658 step-by-step:
\( 2658 = 2 \times 1329 \)
\( \implies 2658 = 2 \times 3 \times 443 \)
Since 443 has no divisors up to its square root, it is a prime number.
Thus, the prime factorization is \( 2 \times 3 \times 443 \).
In simple words: Break the number down by dividing by prime numbers until you are left with only prime numbers: 2, 3, and 443.

Exam Tip: To verify if a number like 443 is prime, check for divisibility by prime numbers up to its approximate square root, which is around 21.

Question 3. Find the LCM and HCF of 17, 23 and 29.
Answer: The numbers 17, 23, and 29 are all prime numbers.
Since they share no common factors other than 1:
\( \text{HCF} = 1 \)
\( \text{LCM} = 17 \times 23 \times 29 = 11339 \)
In simple words: Since all three numbers are prime, their HCF is 1, and their LCM is simply the product of the three numbers.

Exam Tip: For any set of prime numbers, the HCF is always 1 and the LCM is always their product.

Question 4. Prove that √2 is not a rational number.
Answer: Assume, on the contrary, that \( \sqrt{2} \) is rational. Let:
\( \sqrt{2} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).
Squaring both sides:
\( 2 = \frac{a^2}{b^2} \)
\( \implies a^2 = 2b^2 \) — (Equation 1)
Since 2 divides \( a^2 \), it must divide \( a \). Let \( a = 2c \) for some integer \( c \).
Substituting this into Equation 1:
\( (2c)^2 = 2b^2 \)
\( \implies 4c^2 = 2b^2 \)
\( \implies b^2 = 2c^2 \)
This means 2 divides \( b^2 \), so 2 must also divide \( b \).
Since 2 divides both \( a \) and \( b \), they share a common factor of 2, contradicting that they are co-prime.
Thus, \( \sqrt{2} \) is not rational.
In simple words: If \( \sqrt{2} \) were rational, we could write it as a simplified fraction. But we prove both top and bottom can be divided by 2, which contradicts that the fraction was fully simplified.

Exam Tip: This is a fundamental proof in real numbers; practice writing it out step-by-step as it is frequently asked in exams.

Question 5. Find the largest positive integer that will divide 122, 150 and 115 leaving remainder 5,7 and 11 respectively.
Answer: The required integer is the Highest Common Factor (HCF) of the numbers after subtracting their respective remainders:
\( 122 - 5 = 117 \)
\( 150 - 7 = 143 \)
\( 115 - 11 = 104 \)
Now, find the HCF of 117, 143, and 104 using prime factorization:
\( 117 = 3^2 \times 13 \)
\( 143 = 11 \times 13 \)
\( 104 = 2^3 \times 13 \)
The greatest common factor is 13.
Thus, the largest positive integer is 13.
In simple words: Subtract the remainders from each of the numbers first, then find the HCF of the resulting numbers, which is 13.

Exam Tip: Subtract the remainders first before calculating any prime factorizations or HCF.

Question 6. Show that there is no positive integer n for which √𝑛 − 1 + √𝑛 + 1 is rational.
Answer: Assume there exists a positive integer \( n \) such that:
\( \sqrt{n-1} + \sqrt{n+1} = p \) is rational (where \( p > 0 \)).
Taking the reciprocal:
\( \frac{1}{\sqrt{n-1} + \sqrt{n+1}} = \frac{1}{p} \)
Multiplying by the conjugate \( \sqrt{n+1} - \sqrt{n-1} \):
\( \frac{\sqrt{n+1} - \sqrt{n-1}}{(n+1) - (n-1)} = \frac{1}{p} \)
\( \implies \sqrt{n+1} - \sqrt{n-1} = \frac{2}{p} \)
Adding our two equations:
\( 2\sqrt{n+1} = p + \frac{2}{p} \implies \sqrt{n+1} = \frac{p^2 + 2}{2p} \) (rational)
Subtracting them:
\( 2\sqrt{n-1} = p - \frac{2}{p} \implies \sqrt{n-1} = \frac{p^2 - 2}{2p} \) (rational)
This implies both \( n+1 \) and \( n-1 \) must be perfect squares of integers.
The difference between these perfect squares is:
\( (n+1) - (n-1) = 2 \)
However, the minimum difference between two distinct positive perfect squares of integers is at least 3 (since \( 2^2 - 1^2 = 3 \)).
This is a contradiction, meaning no such positive integer \( n \) exists.
In simple words: If this sum were rational, both square roots would have to be rational numbers, making \( n-1 \) and \( n+1 \) perfect squares of integers. But there are no integer perfect squares that have a difference of only 2.

Exam Tip: Rationalizing the reciprocal of the expression is the key step to finding a system of equations to solve for the individual roots.

Question 7. Using prime factorization method, find the HCF and LCM of 72, 126 and 168. Also show that HCF X LCM ≠ product of three numbers.
Answer: Prime factorization of the three numbers:
\( 72 = 2^3 \times 3^2 \)
\( 126 = 2^1 \times 3^2 \times 7^1 \)
\( 168 = 2^3 \times 3^1 \times 7^1 \)

To find HCF:
\( \text{HCF} = 2^1 \times 3^1 = 6 \)

To find LCM:
\( \text{LCM} = 2^3 \times 3^2 \times 7^1 = 8 \times 9 \times 7 = 504 \)

Now verify the product rule:
\( \text{HCF} \times \text{LCM} = 6 \times 504 = 3024 \)
\( \text{Product of the three numbers} = 72 \times 126 \times 168 = 1,524,096 \)
Since \( 3024 \neq 1,524,096 \), it is shown that \( \text{HCF} \times \text{LCM} \neq \text{product of the three numbers} \).
In simple words: We find the HCF is 6 and the LCM is 504. Multiplying them gives 3024, which is not equal to the product of the three original numbers, proving the product rule only applies to two numbers.

Exam Tip: Explicitly state both products at the end to make your proof clear and complete.

Question 8. Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. How many stacks will be there?
Answer: To find the maximum number of books per stack (so that all stacks have the same height), we calculate the HCF of 336, 240, and 96:
\( 336 = 2^4 \times 3 \times 7 \)
\( 240 = 2^4 \times 3 \times 5 \)
\( 96 = 2^5 \times 3 \)
\( \text{HCF} = 2^4 \times 3 = 16 \times 3 = 48 \) books per stack.

Now, we calculate the number of stacks for each subject:
- English stacks: \( \frac{336}{48} = 7 \)
- Mathematics stacks: \( \frac{240}{48} = 5 \)
- Science stacks: \( \frac{96}{48} = 2 \)
Total number of stacks = \( 7 + 5 + 2 = 14 \).
In simple words: Find the HCF of the three book counts to get the size of each stack, which is 48. Then, divide the book counts of each subject by 48 and add them together to get 14 total stacks.

Exam Tip: Be careful not to stop after calculating the HCF (48), as the question asks for the total number of stacks, not the number of books in each stack.

Value Based Questions

Question 1. A person wanted to distribute 96 apples and 112 oranges among poor children in an orphanage. He packed all the fruits in boxes in such a way that each box contains fruits of the same variety, and also every box contains an equal number of fruits.
(i) Find the maximum number of boxes in which all the fruits can be packed.
(ii) Which concept have you used to find it?
(iii) Which values of this person have been reflected in above situation?
Answer:
(i) To determine the distribution of fruits, we calculate the Highest Common Factor (HCF) of 96 and 112:
\( 96 = 2^5 \times 3 \)
\( 112 = 2^4 \times 7 \)
\( \text{HCF} = 2^4 = 16 \)
Thus, the maximum number of boxes is 16.

(ii) We have used the concept of prime factorization and HCF (Highest Common Factor) from the number system.

(iii) The actions of the person reflect the values of kindness, social responsibility, generosity, and an attitude of helping the underprivileged.
In simple words: Finding the HCF of 96 and 112 gives us 16, which represents the boxes. This charitable work highlights the person's helpful and caring nature.

Exam Tip: Value-based questions often require both a mathematical calculation and a brief, thoughtful moral reflection.

Question 2. A teacher draws the factor tree given in figure and ask the students to find the value of x without finding the value of y and z. Shaurya gives the answer x=136.
a) Is his answer correct?
b) Give reason for your answer.
c) Which value is depicted in this?

Answer:
a) Yes, Shaurya's answer is correct.

b) According to the Fundamental Theorem of Arithmetic, the value of \( x \) is equal to the product of all its prime factors shown at the ends of the branches:
\( x = 2 \times 2 \times 2 \times 17 = 136 \)
Alternatively, we can calculate bottom-up:
\( z = 2 \times 17 = 34 \)
\( y = 2 \times z = 2 \times 34 = 68 \)
\( x = 2 \times y = 2 \times 68 = 136 \)

c) This depicts the value of logical reasoning, critical thinking, and the application of prime factorization concepts.
In simple words: The top number in a factor tree is the product of all the prime numbers at the tips of the branches. Multiplying \( 2 \times 2 \times 2 \times 17 \) gives 136 directly.

Exam Tip: You do not need to calculate every single intermediate variable in a factor tree; multiplying the final leaf nodes directly yields the root number.