test by ruchi

Basic Concepts

• Fundamental theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique apart from the order in which the prime factors occur.
• If \( x \) is a positive prime, then \( \sqrt{x} \) is an irrational number.
  For example, 7 is a positive prime
  \( \implies \) \( \sqrt{7} \) is an irrational number.

Some Important Facts/Tips:

(i) If ‘p’ is a prime and \( p \) divides \( a^2 \), then \( p \) divides ‘a’ also, where \( a \) is positive integer.
For example: 3 divides 36 i.e., \( 6^2 \)
\( \implies \) 3 divides 6.

(ii) For any two positive integers \( a \) and \( b \); \( HCF (a, b) \times LCM (a, b) = a \times b \)
For example: \( a = 6, b = 4 \)
\( LCM of (6, 4) = 12 \)
\( HCF of (6, 4) = 2 \)
\( LCM (6, 4) \times HCF (6, 4) = 6 \times 4 \)
\( 12 \times 2 = 6 \times 4 = 24 \)

(iii) The sum or difference of a rational and an irrational number is irrational.

(iv) The product and quotient of a non-zero rational number and an irrational number is irrational.

Selected NCERT Questions

Question. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer: For the maximum number of columns, we have to find the HCF of 616 and 32.
\( 616 = 2 \times 2 \times 2 \times 7 \times 11 \)
\( = 2^3 \times 7 \times 11 \)
\( 32 = 2 \times 2 \times 2 \times 2 \times 2 \)
\( = 2^3 \times 2^2 \)
HCF of 616, 32 \( = 2^3 = 8 \)
Hence, maximum number of columns is 8.

Question. Check whether \( 6^n \) can end with the digit 0 for any natural number \( n \).
Answer: If the number \( 6^n \), for any \( n \), were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of \( 6^n \) would contain the prime 5. But \( 6^n = (2 \times 3)^n = 2^n \times 3^n \) so the primes in factorisation of \( 6^n \) are 2 and 3. So the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes except 2 and 3 in the factorisation of \( 6^n \). So there is no natural number \( n \) for which \( 6^n \) ends with digit zero.

Question. Explain why \( 7 \times 11 \times 13 + 13 \) and \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) are composite numbers?
Answer: We have, \( 7 \times 11 \times 13 + 13 = 1001 + 13 = 1014 \)
\( 1014 = 2 \times 3 \times 13 \times 13 \)
So, it is the product of more than two prime numbers. 2, 3 and 13.
Hence, it is a composite number.
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5040 + 5 = 5045 \)
\( \implies \) \( 5045 = 5 \times 1009 \)
It is the product of prime factor 5 and 1009.
Hence, it is a composite number.

Question. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start from the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Answer: To find the time after which they meet again at the starting point, we have to find LCM of 18 and 12 minutes. We have
\( 18 = 2 \times 3^2 \)
and \( 12 = 2^2 \times 3 \)
Therefore, LCM of 18 and 12 \( = 2^2 \times 3^2 = 36 \)
So, they will meet again at the starting point after 36 minutes.

Question. Show that \( 5 - \sqrt{3} \) is irrational.
Answer: Let us assume, to the contrary, that \( 5 - \sqrt{3} \) is rational.
That is, we can find coprime \( a \) and \( b \) (\( b \neq 0 \)) such that \( 5 - \sqrt{3} = \frac{a}{b} \).
Therefore, \( 5 - \frac{a}{b} = \sqrt{3} \).
Rearranging this equation, we get \( \sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b} \).
Since \( a \) and \( b \) are integers, we get \( \frac{5b - a}{b} \) is rational, and so \( \sqrt{3} \) is rational.
But this contradicts the fact that \( \sqrt{3} \) is irrational.
This contradiction has arisen because of our incorrect assumption that \( 5 - \sqrt{3} \) is rational.
So, we conclude that \( 5 - \sqrt{3} \) is irrational.

Question. Prove that \( 3 + 2\sqrt{5} \) is an irrational number.
Answer: Let \( 3 + 2\sqrt{5} \) is a rational number.
\( \implies \) \( 3 + 2\sqrt{5} = \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \)
\( \implies \) \( 2\sqrt{5} = \frac{p}{q} - 3 \)
\( \implies \) \( 2\sqrt{5} = \frac{p - 3q}{q} \)
\( \implies \) \( \sqrt{5} = \frac{p - 3q}{2q} \) ...(i)
Since, \( p, q, 2 \) and \( -3 \) are integers, \( p, -3q, 2q \) are also integers.
Also, \( 2 \neq 0, q \neq 0 \implies 2q \neq 0 \)
[\(\because\) Product of two non-zero numbers can never be zero]
Therefore, RHS of (i) is rational number and LHS \( = \sqrt{5} \) is an irrational number.
But this is not possible.
So, our assumption is wrong.
Hence, \( 3 + 2\sqrt{5} \) is irrational number.

Question. Prove that \( 7\sqrt{5} \) is an irrational number.
Answer: Let \( 7\sqrt{5} \) be a rational number.
\( \implies \) \( 7\sqrt{5} = \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \)
\( \implies \) \( \sqrt{5} = \frac{p}{7q} \) ...(i)
\(\because\) \( p, 7, q \) are integers \( \implies \) \( p, 7q \) are integers
Also \( 7 \neq 0, q \neq 0, \implies 7q \neq 0 \)
Therefore RHS of (i) is rational number but LHS \( = \sqrt{5} \) is irrational, which is contradiction.
Hence, \( 7\sqrt{5} \) is an irrational number.

Question. Prove that \( 6 + \sqrt{2} \) is an irrational number.
Answer: Let \( 6 + \sqrt{2} \) be a rational number.
\( \implies \) \( 6 + \sqrt{2} = \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \).
\( \implies \) \( \sqrt{2} = \frac{p}{q} - 6 \)
\( \implies \) \( \sqrt{2} = \frac{p - 6q}{q} \) ...(i)
\(\because\) \( p, q, -6 \) are integers \( \implies \) \( p - 6q, q \) are integers.
Also, \( q \neq 0 \)
Therefore, RHS of (i) is rational number but LHS \( = \sqrt{2} \) is irrational, which is contradiction.
Hence, \( 6 + \sqrt{2} \) is irrational.

Multiple Choice Questions

Question. \( n^2 - 1 \) is divisible by 8 if \( n \) is [NCERT Exemplar]
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Answer: (c) an odd integer

Question. The product of three consecutive integers is divisible by
(a) 5
(b) 6
(c) 7
(d) None of the options
Answer: (b) 6

Question. The largest number which divides 615 and 963 leaving remainder 6 in each case is
(a) 82
(b) 95
(c) 87
(d) 93
Answer: (c) 87

Question. The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively is [NCERT Exemplar]
(a) 13
(b) 65
(c) 875
(d) 1750
Answer: (a) 13

Question. If two positive integers \( a \) and \( b \) are written as \( a = x^3 y^2 \) and \( b = x y^3 \); \( x, y \) are prime numbers, then LCM (\( a, b \)) is [NCERT Exemplar]
(a) \( xy \)
(b) \( xy^2 \)
(c) \( x^3 y^3 \)
(d) \( x^2 y^2 \)
Answer: (c) \( x^3 y^3 \)

Question. If HCF (26, 169) = 13 then LCM (26, 169) is
(a) 26
(b) 52
(c) 338
(d) 13
Answer: (c) 338

Question. The HCF and the LCM of 12, 21, 15 respectively are [CBSE 2020 (30/1/1)]
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer: (c) 3, 420

Question. The product of two irrational numbers is
(a) always irrational
(b) always rational
(c) rational or irrational
(d) one
Answer: (c) rational or irrational

Question. \( 3.\overline{27} \) is
(a) an integer
(b) a rational number
(c) a natural number
(d) an irrational number
Answer: (b) a rational number

Question. If 3 is the least prime factor of number \( a \) and 7 is the least prime factor of number \( b \), then the least prime factor of (\( a + b \)) is [Competency Based Question]
(a) 2
(b) 3
(c) 5
(d) 10
Answer: (a) 2

Question. Which of these is a RATIONAL number? [CBSE Question Bank]
(a) \( 3\pi \)
(b) \( 5\sqrt{5} \)
(c) 0.346666...
(d) 0.345210651372849...
Answer: (c) 0.346666...

Question. Which of these numbers can be expressed as a product of two or more prime numbers?
(i) 15
(ii) 34568
(iii) (\( 15 \times 13 \))
[CBSE Question Bank]
(a) only (ii)
(b) only (iii)
(c) only (i) and (ii)
(d) all-(i), (ii) and (iii)
Answer: (d) all-(i), (ii) and (iii)

Question. A number of the form \( 8^n \), where \( n \) is a natural number greater than 1, cannot be divisible by [CBSE Question Bank]
(a) 1
(b) 40
(c) 64
(d) \( 2^{2n} \)
Answer: (b) 40

Question. 1245 is a factor of the numbers \( p \) and \( q \). Which of the following will always have 1245 as a factor?
(i) \( p + q \)
(ii) \( p \times q \)
(iii) \( p \div q \)
[Competency Based Question]
(a) only (ii)
(b) only (i) and (ii)
(c) only (ii) and (iii)
(d) all-(i), (ii) and (iii)
Answer: (b) only (i) and (ii)

Very Short Answer Questions

Question. What is the HCF of the smallest composite number and the smallest prime number? [CBSE 2018]
Answer: Smallest prime \( = 2 \)
Smallest composite \( = 4 \)
HCF (2, 4) \( = 2 \)
The HCF of the smallest prime and smallest composite is 2.