RBSE Solutions Class 10 Maths Chapter 2 Real Numbers Important Questions

Objective Type Questions

Question 1. Prime factor of 4050 is
(A) \( 2 \times 3^5 \times 5 \)
(B) \( 2 \times 3^4 \times 5 \)
(C) \( 2 \times 3^4 \times 5^2 \)
(D) \( 2 \times 3^4 \times 5^3 \)
Answer: (C) \( 2 \times 3^4 \times 5^2 \)
In simple words: To find the prime factors of 4050, you divide it by the smallest prime numbers until you cannot divide anymore. The factors are 2, 3, and 5. The number 3 appears four times and 5 appears twice.

🎯 Exam Tip: Always make sure to divide completely until all factors are prime numbers and express them with their correct powers.

Question 2. For any integer m, the form of each even integer is
(A) m
(B) m + 1
(C) 2m
(D) 2m + 1.
Answer: (C) 2m
In simple words: Any even number can be written as "2 multiplied by an integer". This means that when you multiply any whole number (m) by 2, you always get an even number.

🎯 Exam Tip: Remember that an odd integer is represented as \( 2m + 1 \) (or \( 2m - 1 \)), while an even integer is always \( 2m \), where m is an integer.

Question 4. If p is a prime number then LCM of p, \( p^2 \), and \( p^3 \) 
(A) p
(B) \( p^3 \)
(C) \( p^2 \)
(D) \( p^6 \)
Answer: (B) \( p^3 \)
In simple words: The Least Common Multiple (LCM) of powers of the same prime number is simply the highest power of that prime number. Here, the highest power of p is \( p^3 \).

🎯 Exam Tip: For numbers that are powers of a prime number, the LCM is the highest power, and the HCF is the lowest power.

Question 5. If \( p^2 \) is an even number then p also a/an
(A) odd number
(B) even number
(C) even and odd number
(D) none of there.
Answer: (B) even number
In simple words: If the square of a number is even, then the number itself must also be even. For example, if \( p^2 = 16 \) (an even number), then \( p = 4 \) (also an even number).

🎯 Exam Tip: Remember the property that if a prime number divides \( a^2 \), it also divides a. Similarly, if \( a^2 \) is even, then 'a' must be even.

Question 6. If number \( n^2 - 1 \), divides by 8 then n is an 
(A) integer
(B) natural number
(C) odd number
(D) even number
Answer: (C) odd number
In simple words: If \( n^2 - 1 \) can be perfectly divided by 8, then the number 'n' must always be an odd number. Try with an odd number like 3: \( 3^2 - 1 = 9 - 1 = 8 \), which divides by 8.

🎯 Exam Tip: To solve this, test values. If n is even, let \( n = 2k \), then \( n^2 - 1 = 4k^2 - 1 \), which is not always divisible by 8. If n is odd, let \( n = 2k+1 \), then \( n^2 - 1 = (2k+1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k(k+1) \). Since either k or k+1 is even, \( k(k+1) \) is even, so \( k(k+1) = 2m \). Thus, \( n^2 - 1 = 4(2m) = 8m \), which is divisible by 8.

Question 8. The product of a non-zero rational number and an irrational number is 
(A) always an irrational number
(B) always a rational number
(C) a rational or irrational number
(D) one
Answer: (A) always an irrational number
In simple words: When you multiply a rational number (like 2 or 1/2) that is not zero by an irrational number (like \( \sqrt{2} \) or \( \pi \)), the answer will always be an irrational number. For example, \( 2 \times \sqrt{2} = 2\sqrt{2} \), which is irrational.

🎯 Exam Tip: Remember that the product of two irrational numbers can be rational (e.g., \( \sqrt{2} \times \sqrt{2} = 2 \)), but this is not the case when one number is a non-zero rational and the other is irrational.

Question 9. If p is a positive prime number then \( \sqrt{p} \) is a/an
(A) irrational number
(B) rational number
(C) integer number
Answer: (A) irrational number
In simple words: If 'p' is a prime number, its square root \( \sqrt{p} \) will always be an irrational number. For example, \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \) are all irrational because their decimal representation is non-terminating and non-repeating.

🎯 Exam Tip: The square root of any prime number (or any non-perfect square integer) is always irrational. This is a fundamental concept in number theory.

Question 10. If L.C.M and H.C.F of two rational numbers are the same, the numbers will be
(A) prime
(B) co-prime
(C) composite
(D) equal
Answer: (D) equal
In simple words: If two numbers have the same Least Common Multiple (LCM) and Highest Common Factor (HCF), it means the numbers must be identical. For example, if LCM(a,b) = HCF(a,b) = 5, then a and b must both be 5.

🎯 Exam Tip: This property arises because the product of two numbers is equal to the product of their HCF and LCM. If HCF = LCM, then the numbers themselves must be identical.

Question 11. If two positive integers p and q are expressible in the form \( p = a^2b^3 \) and \( q = ab \), where a and b are co-prime, then LCM (p, q) will be
(A) ab
(B) \( a^2b^2 \)
(C) \( a^2b^3 \)
(D) \( a^3b^3 \)
Answer: (C) \( a^2b^3 \)
In simple words: To find the LCM of two numbers given by their prime factors, you take the highest power of each prime factor present in either number. Here, the highest power of 'a' is \( a^2 \) and the highest power of 'b' is \( b^3 \), so the LCM is \( a^2b^3 \).

🎯 Exam Tip: For LCM, consider all unique prime factors from both numbers and select the highest power for each. For HCF, consider only common prime factors and select the lowest power for each.

Question 12. If HCF and LCM of two numbers are 6 and 60 respectively, then those numbers are
(A) 10 and 36
(B) 24 and 15
(C) 12 and 30
(D) 18 and 20.
Answer: (C) 12 and 30
In simple words: The product of two numbers is always equal to the product of their HCF and LCM. Here, HCF \( \times \) LCM = \( 6 \times 60 = 360 \). From the options, only \( 12 \times 30 = 360 \).

🎯 Exam Tip: Remember the fundamental relationship: Product of two numbers = HCF \( \times \) LCM. Test each option to see which pair satisfies both the HCF and LCM conditions.

Question 13. Find the H.C.F. of 17, 23 and 29.
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (A) 1
In simple words: The numbers 17, 23, and 29 are all prime numbers. Prime numbers only have two factors: 1 and themselves. Since there are no common factors other than 1, their Highest Common Factor (HCF) is 1.

🎯 Exam Tip: The HCF of any set of prime numbers is always 1, as 1 is the only common factor they share.

Question 14. If a rational number \( x = \frac{p}{q} \) such as q is not the form of \( 2^n \times 5^m \) where n, m is a non-zero integer, then the decimal expansion of x is
(A) terminating decimal expansion
(B) non-terminating repeating decimal
(C) choice (A) and (B) both correct
(D) none of there
Answer: (B) non-terminating repeating decimal
In simple words: If the bottom part (denominator 'q') of a fraction, when simplified, has prime factors other than just 2 or 5, then its decimal form will go on forever and have a repeating pattern. This is what we call a non-terminating repeating decimal.

🎯 Exam Tip: To determine if a rational number has a terminating or non-terminating repeating decimal expansion, simplify the fraction and then check the prime factors of the denominator. If only 2s and 5s are present, it terminates; otherwise, it repeats.

Question 15. First Greek mathematician was
(A) AL-knwarijmi
(B) Aryabhata
(C) Einstein
(D) Euclid
Answer: (D) Euclid
In simple words: Euclid was a very famous ancient Greek mathematician, often called the "Father of Geometry." He wrote a book called "Elements" which is one of the most important works in the history of mathematics.

🎯 Exam Tip: Knowing key historical figures in mathematics can help contextualize concepts, especially in topics like geometry or number theory.

Very Short/Short Answer Type Questions

Question 2. Find the HCF of numbers 44 and 99 
Answer: First, we find the prime factors for each number. For 44, the prime factors are \( 2 \times 2 \times 11 \) or \( 2^2 \times 11 \). For 99, the prime factors are \( 3 \times 3 \times 11 \) or \( 3^2 \times 11 \). The common prime factor is 11. Therefore, the Highest Common Factor (HCF) of 44 and 99 is 11. Knowing prime factorization helps identify shared factors easily.
In simple words: To find the HCF of 44 and 99, list out their prime factors. Both numbers share the prime factor 11. So, 11 is their HCF.

🎯 Exam Tip: To find the HCF, always identify the common prime factors and take the lowest power of each common factor.

Question 3. Is the HCF of two numbers 15 and LCM 175 possible? Give reason. 
Answer: No, it is not possible for the HCF of two numbers to be 15 and their LCM to be 175. This is because a fundamental property of numbers states that the HCF of any two numbers must always divide their LCM completely. In this case, 175 divided by 15 gives \( 11.66... \), which is not a whole number. Since 15 does not perfectly divide 175, this combination of HCF and LCM is not possible. The LCM must always be a multiple of the HCF.
In simple words: No, this is not possible. The HCF must always be a factor of the LCM. Since 15 does not divide 175 evenly, you cannot have these HCF and LCM values for two numbers.

🎯 Exam Tip: A quick check for the validity of given HCF and LCM values is to ensure that LCM is perfectly divisible by HCF.

Question 4. Whether the decimal expansion of number \( \frac{3}{625} \) is terminating or non-terminating repeating? write it in decimal form? 
Answer: To determine the type of decimal expansion, we first look at the prime factors of the denominator. The denominator is 625. When we prime factorize 625, we get \( 5 \times 5 \times 5 \times 5 \), which is \( 5^4 \). Since the prime factors of the denominator only include 5 (which can be written as \( 2^0 \times 5^4 \)), the decimal expansion will be terminating. To write it in decimal form, we can multiply the numerator and denominator by \( 2^4 \) to make the denominator a power of 10.
\[ \frac{3}{625} = \frac{3}{5^4} = \frac{3 \times 2^4}{5^4 \times 2^4} = \frac{3 \times 16}{(5 \times 2)^4} = \frac{48}{10^4} = \frac{48}{10000} = 0.0048 \]
The decimal expansion is 0.0048, which is a terminating decimal. This method simplifies conversion to decimal form.
In simple words: The number \( \frac{3}{625} \) has a denominator whose prime factors are only 5s. This means its decimal form will stop (terminate). To find the decimal, we can make the bottom number 10,000, which gives us 0.0048.

🎯 Exam Tip: A rational number \( \frac{p}{q} \) (in simplest form) has a terminating decimal expansion if and only if the prime factorization of its denominator q is of the form \( 2^n \times 5^m \), where n and m are non-negative integers.

Question 6. Prove that \( 3 - \sqrt{5} \) is an irrational number? 
Answer: Let us assume, for the sake of contradiction, that \( 3 - \sqrt{5} \) is a rational number. If it is rational, then we can write it as \( \frac{a}{b} \), where 'a' and 'b' are co-prime integers and \( b \neq 0 \).
So, \( 3 - \sqrt{5} = \frac{a}{b} \)
Now, we can rearrange the equation to isolate \( \sqrt{5} \):
\( 3 - \frac{a}{b} = \sqrt{5} \)
\( \frac{3b - a}{b} = \sqrt{5} \)
Since 'a' and 'b' are integers, \( 3b - a \) is an integer, and 'b' is a non-zero integer. Therefore, \( \frac{3b - a}{b} \) must be a rational number. This implies that \( \sqrt{5} \) is also a rational number. However, we know that \( \sqrt{5} \) is an irrational number. This creates a contradiction with our initial assumption. Thus, our assumption that \( 3 - \sqrt{5} \) is rational must be false. Therefore, \( 3 - \sqrt{5} \) is an irrational number, which is an important concept in number theory.
In simple words: We pretend that \( 3 - \sqrt{5} \) is a rational number (a fraction). When we move things around, it makes \( \sqrt{5} \) look like a rational number too. But we know \( \sqrt{5} \) is not a rational number. This contradiction means our first guess was wrong, so \( 3 - \sqrt{5} \) must be irrational.

🎯 Exam Tip: When proving a number is irrational, always use the method of contradiction: assume it's rational, show this leads to a known false statement (like \( \sqrt{5} \) being rational), and conclude the initial assumption was false.

Question 7. Prove that \( 7 + \sqrt{5} \) is an irrational number.
Answer: Let us assume, contrary to our belief, that \( 7 + \sqrt{5} \) is a rational number. If it is rational, then it can be written in the form \( \frac{a}{b} \), where 'a' and 'b' are co-prime integers and \( b \neq 0 \).
So, \( 7 + \sqrt{5} = \frac{a}{b} \)
Now, we isolate \( \sqrt{5} \) by moving 7 to the other side:
\( \sqrt{5} = \frac{a}{b} - 7 \)
\( \sqrt{5} = \frac{a - 7b}{b} \)
Since 'a' and 'b' are integers, \( a - 7b \) is also an integer, and 'b' is a non-zero integer. This means that \( \frac{a - 7b}{b} \) is a rational number. Therefore, this implies that \( \sqrt{5} \) is also a rational number. However, we already know that \( \sqrt{5} \) is an irrational number. This contradiction arises because our initial assumption that \( 7 + \sqrt{5} \) is rational was incorrect. Thus, we conclude that \( 7 + \sqrt{5} \) is an irrational number. This illustrates how properties of rational and irrational numbers interact.
In simple words: We assume \( 7 + \sqrt{5} \) is a fraction. By moving 7 to the other side, we make \( \sqrt{5} \) look like a fraction. But we know \( \sqrt{5} \) is not a fraction. So our first guess was wrong, meaning \( 7 + \sqrt{5} \) is irrational.

🎯 Exam Tip: When dealing with sums or differences involving rational and irrational numbers, always remember that the result will be irrational. This method of proof by contradiction is standard for such problems.

Question 8. Find the HCF and LCM of two numbers 3364 and 54.
Answer: To find the HCF and LCM, we first perform the prime factorization for each number:
For 3364:
\( 3364 = 2 \times 2 \times 29 \times 29 = 2^2 \times 29^2 \)
For 54:
\( 54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 \)
Now, we find the HCF (Highest Common Factor) by taking the lowest power of the common prime factors. The only common prime factor is 2, with the lowest power being \( 2^1 \). So, H.C.F. = 2.
Next, we find the LCM (Least Common Multiple) by taking the highest power of all prime factors present in either number.
The prime factors are 2, 3, and 29.
Highest power of 2 is \( 2^2 \).
Highest power of 3 is \( 3^3 \).
Highest power of 29 is \( 29^2 \).
So, L.C.M. \( = 2^2 \times 3^3 \times 29^2 = 4 \times 27 \times 841 = 90,828 \). This comprehensive approach ensures all factors are accounted for.
In simple words: Break down both 3364 and 54 into their prime numbers. The HCF is the common prime number with the smallest power, which is 2. The LCM is all the prime numbers from both, using their largest powers, which is \( 2^2 \times 3^3 \times 29^2 \), making 90,828.

🎯 Exam Tip: Always list out the prime factors with their powers clearly for both numbers before calculating HCF (common factors, lowest powers) and LCM (all factors, highest powers).

Long Answer Type Questions

Question 1. Prove that only one out of three consecutive positive integers is divisible by 3. 
Answer: Let the three consecutive positive integers be n, \( n+1 \), and \( n+2 \). We can use the Euclidean Division Lemma to show that any integer n can be written in one of these forms: \( 3q \), \( 3q+1 \), or \( 3q+2 \) for some integer q. We will examine each case:
Case 1: If \( n = 3q \)
In this case, n is clearly divisible by 3.
Then \( n+1 = 3q+1 \), which is not divisible by 3.
And \( n+2 = 3q+2 \), which is also not divisible by 3.
So, in this case, only n is divisible by 3.
Case 2: If \( n = 3q+1 \)
In this case, n is not divisible by 3.
Then \( n+1 = (3q+1)+1 = 3q+2 \), which is not divisible by 3.
And \( n+2 = (3q+1)+2 = 3q+3 = 3(q+1) \). This means \( n+2 \) is divisible by 3.
So, in this case, only \( n+2 \) is divisible by 3.
Case 3: If \( n = 3q+2 \)
In this case, n is not divisible by 3.
Then \( n+1 = (3q+2)+1 = 3q+3 = 3(q+1) \). This means \( n+1 \) is divisible by 3.
And \( n+2 = (3q+2)+2 = 3q+4 = 3q+3+1 = 3(q+1)+1 \), which is not divisible by 3.
So, in this case, only \( n+1 \) is divisible by 3.
In all three possible cases, exactly one of the three consecutive positive integers is divisible by 3. This proves the statement, showing the cyclical nature of divisibility by 3.
In simple words: Imagine any three numbers that come one after the other. If the first number can be divided by 3, the next two cannot. If the second number can be divided by 3, the first and third cannot. If the third number can be divided by 3, the first two cannot. So, only one of them will always be a multiple of 3.

🎯 Exam Tip: For problems involving divisibility by an integer 'k', use the Euclidean Division Lemma to express numbers in the form \( kq \), \( kq+1 \), ..., \( kq+(k-1) \) and analyze each case.

Question 2. Show that cube of any positive integer is either of the form \( 4m \), \( 4m + 1 \) or \( 4m + 3 \) for some integer m. 
Answer: Let x be any positive integer. According to the Euclidean Division Lemma, any positive integer x can be expressed in the form \( 4q \), \( 4q+1 \), \( 4q+2 \), or \( 4q+3 \) for some integer q. We need to find the cube of x for each case.
Case 1: If \( x = 4q \)
\( x^3 = (4q)^3 = 64q^3 = 4(16q^3) \)
Let \( m = 16q^3 \). Then \( x^3 = 4m \).
Case 2: If \( x = 4q+1 \)
\( x^3 = (4q+1)^3 \)
Using the identity \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \):
\( x^3 = (4q)^3 + 3(4q)^2(1) + 3(4q)(1)^2 + (1)^3 \)
\( x^3 = 64q^3 + 48q^2 + 12q + 1 \)
\( x^3 = 4(16q^3 + 12q^2 + 3q) + 1 \)
Let \( m = 16q^3 + 12q^2 + 3q \). Then \( x^3 = 4m + 1 \).
Case 3: If \( x = 4q+2 \)
\( x^3 = (4q+2)^3 \)
\( x^3 = (2(2q+1))^3 = 8(2q+1)^3 \)
\( x^3 = 8(8q^3 + 12q^2 + 6q + 1) \)
\( x^3 = 64q^3 + 96q^2 + 48q + 8 \)
\( x^3 = 4(16q^3 + 24q^2 + 12q + 2) \)
Let \( m = 16q^3 + 24q^2 + 12q + 2 \). Then \( x^3 = 4m \).
Case 4: If \( x = 4q+3 \)
\( x^3 = (4q+3)^3 \)
\( x^3 = (4q)^3 + 3(4q)^2(3) + 3(4q)(3)^2 + (3)^3 \)
\( x^3 = 64q^3 + 144q^2 + 108q + 27 \)
We can rewrite 27 as \( 24+3 \):
\( x^3 = 64q^3 + 144q^2 + 108q + 24 + 3 \)
\( x^3 = 4(16q^3 + 36q^2 + 27q + 6) + 3 \)
Let \( m = 16q^3 + 36q^2 + 27q + 6 \). Then \( x^3 = 4m + 3 \).
From all cases, the cube of any positive integer is either of the form \( 4m \), \( 4m+1 \), or \( 4m+3 \). This demonstrates the pattern of cubes under modulo 4.
In simple words: We take any whole number and group it based on what's left over when divided by 4. Then we cube these numbers. We find that the result, when divided by 4, will always have a remainder of 0, 1, or 3. It will never have a remainder of 2.

🎯 Exam Tip: When using Euclidean Division Lemma for powers, expand carefully and group terms to show the desired form. Remember that \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).

Question 3. Find the LCM and HCF of the given pair of integers and check product of two numbers = H.C.F x L.C.M
(i) 26 and 91,
(ii) 510 and 92,
Answer: We will find the prime factorization for each pair of numbers to determine their HCF and LCM, and then verify the product rule.
(i) For 26 and 91:
Prime factorization of 26: \( 26 = 2 \times 13 \)
Prime factorization of 91: \( 91 = 7 \times 13 \)
H.C.F. (26, 91): The common prime factor is 13, so H.C.F. = 13.
L.C.M. (26, 91): Take all prime factors with their highest powers: \( 2 \times 7 \times 13 = 182 \).
Check: Product of the two numbers \( = 26 \times 91 = 2366 \).
Product of H.C.F. and L.C.M. \( = 13 \times 182 = 2366 \).
Since \( 2366 = 2366 \), the product of the two numbers equals the product of their HCF and LCM.
(ii) For 510 and 92:
Prime factorization of 510: \( 510 = 2 \times 3 \times 5 \times 17 \)
Prime factorization of 92: \( 92 = 2 \times 2 \times 23 = 2^2 \times 23 \)
H.C.F. (510, 92): The common prime factor is 2, with the lowest power being \( 2^1 \). So, H.C.F. = 2.
L.C.M. (510, 92): Take all prime factors with their highest powers: \( 2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = 23460 \).
Check: Product of the two numbers \( = 510 \times 92 = 46920 \).
Product of H.C.F. and L.C.M. \( = 2 \times 23460 = 46920 \).
Since \( 46920 = 46920 \), the product of the two numbers equals the product of their HCF and LCM. This important property holds true for all pairs of positive integers.
In simple words: For each pair of numbers, first break them down into their prime factors. Then, find the HCF (smallest common prime factors) and LCM (all prime factors with their largest powers). Finally, multiply the original two numbers and compare it with the product of their HCF and LCM. They should always be equal.

🎯 Exam Tip: Always remember the fundamental theorem of arithmetic: "Product of two numbers = HCF \( \times \) LCM". This is a crucial identity for verifying your calculations and solving related problems.

Objective Type Questions

Question 1. Prime factor of 4050 is
(a) \( 2 \times 3^5 \times 5 \)
(b) \( 2 \times 3^4 \times 5 \)
(c) \( 2 \times 3^4 \times 5^2 \)
(d) \( 2 \times 3^4 \times 5^3 \)
Answer: (c) \( 2 \times 3^4 \times 5^2 \)
In simple words: To find the prime factors of 4050, we divide it by the smallest prime numbers until we can no longer divide. This process gives us the factors that multiply together to make 4050, showing that two 5s are involved, making it \( 5^2 \).

🎯 Exam Tip: Always perform prime factorization systematically, starting with the smallest prime numbers (2, 3, 5, etc.) to ensure no factors are missed.

Question 2. For any integer m, the form of each even integer is
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1.
Answer: (c) 2m
In simple words: An even number is any number that can be divided by 2 without a remainder. If you multiply any whole number (integer) m by 2, the result will always be an even number.

🎯 Exam Tip: Remember that "m" can be any integer, so "m" alone could be odd or even. "m+1" would be odd if m is even, and even if m is odd. Only "2m" always produces an even number.

Question 4. If p is a prime number then LCM of p, \( p^2 \), and \( p^3 \) is
(a) p
(b) \( p^3 \)
(c) \( p^2 \)
(d) \( p^6 \)
Answer: (b) \( p^3 \)
In simple words: When finding the Least Common Multiple (LCM) of powers of the same prime number, you always pick the highest power among them. Here, \( p^3 \) is the highest power.

🎯 Exam Tip: For LCM of powers of the same base, the result is the highest power. For HCF, it's the lowest power.

Question 5. If \( p^2 \) is an even number then p also a/an
(a) odd number
(b) even number
(c) even and odd number
(d) none of there.
Answer: (b) even number
In simple words: If the square of a number is even, then the number itself must also be even. For example, if \( p^2 = 4 \) (which is even), then p must be 2 (which is also even). If p were odd (like 3), then \( p^2 \) would be 9 (which is odd).

🎯 Exam Tip: This is a fundamental property in number theory: the square of an odd number is always odd, and the square of an even number is always even.

Question 6. If number \( n^2 – 1 \), divides by 8 then n is an (NCERT Exemplar Problem)
(a) integer
(b) natural number
(c) odd number
(d) even number
Answer: (c) odd number
In simple words: If you take a number, square it, and subtract 1, and the result can be divided by 8, then the original number must be odd. For example, if n=3, \( n^2 - 1 = 9 - 1 = 8 \), which is divisible by 8. If n=2, \( n^2 - 1 = 4 - 1 = 3 \), not divisible by 8.

🎯 Exam Tip: Test simple odd and even values (like 1, 2, 3, 4) to quickly confirm the property for \( n^2-1 \). This often reveals the pattern.

Question 8. The product of a non-zero rational number and an irrational number is (NCERT Exemplar Problem)
(a) always an irrational number
(b) always a rational number
(c) a rational or irrational number
(d) one
Answer: (a) always an irrational number
In simple words: If you multiply a normal fraction-type number (rational) that isn't zero by a number that cannot be written as a simple fraction (irrational), the answer will always be irrational. For instance, \( 2 \times \sqrt{2} = 2\sqrt{2} \), which is irrational.

🎯 Exam Tip: Remember that a non-zero rational number is crucial. If the rational number is zero, the product would be zero, which is rational.

Question 9. If p is a positive prime number then \( \sqrt{p} \) is a/an
(a) irrational number
(b) rational number
(c) integer number
Answer: (a) irrational number
In simple words: If you take the square root of any prime number (like 2, 3, 5, etc.), the result will always be a number that cannot be written as a simple fraction. This means it is an irrational number.

🎯 Exam Tip: The square root of any non-square integer is irrational. Since prime numbers are not perfect squares, their square roots must be irrational.

Question 10. If L.C.M and H.C.F of two rational numbers are the same, the number will be
(a) prime
(b) co-prime
(c) composite
(d) equal
Answer: (d) equal
In simple words: If the Least Common Multiple (LCM) and the Highest Common Factor (HCF) of two numbers are exactly the same, it means the two numbers themselves must be identical. For example, the HCF and LCM of 5 and 5 are both 5.

🎯 Exam Tip: This property arises because LCM and HCF are related to the prime factors. If they are equal, all prime factors and their powers must be identical, leading to equal numbers.

Question 11. If two positive integers p and q are expressible in the form \( p = a^2b^3 \) and \( q = ab \), where a and b are co-prime, then LCM (p, q) will be
(a) ab
(b) \( a^2b^2 \)
(c) \( a^2b^3 \)
(d) \( a^3b^3 \)
Answer: (c) \( a^2b^3 \)
In simple words: To find the Least Common Multiple (LCM) of two numbers expressed in terms of their prime factors, you take the highest power for each common prime factor. Here, the highest power of 'a' is \( a^2 \) and the highest power of 'b' is \( b^3 \).

🎯 Exam Tip: When finding LCM from prime factorizations, select the highest power of each prime factor present in either number. For HCF, select the lowest power of common prime factors.

Question 12. If HCF and LCM of two numbers are 6 and 60 respectively, then those numbers are
(a) 10 and 36
(b) 24 and 15
(c) 12 and 30
(d) 18 and 20.
Answer: (c) 12 and 30
In simple words: The product of two numbers is always equal to the product of their HCF and LCM. Here, HCF \( \times \) LCM = \( 6 \times 60 = 360 \). Only the pair 12 and 30 has a product of 360, and also HCF(12,30) = 6 and LCM(12,30) = 60.

🎯 Exam Tip: A quick way to check is to multiply the HCF and LCM (6 x 60 = 360). Then, multiply the numbers in each option. The pair whose product is 360 is a strong candidate, then verify HCF and LCM directly.

Question 13. Find the H.C.F. of 17, 23, and 29.
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
In simple words: 17, 23, and 29 are all prime numbers, meaning they can only be divided by 1 and themselves. When numbers are all prime, the only common factor they share is 1, so their Highest Common Factor (HCF) is 1.

🎯 Exam Tip: The HCF of any set of prime numbers is always 1, as 1 is their only common factor.

Question 14. If a rational number \( x = \frac { p }{ q } \) such that q is not of the form of \( 2^n \times 5^m \) where n, m is a non-zero integer, then the decimal expansion of x is
(a) terminating decimal expansion
(b) non-terminating repeating decimal
(c) choice (A) and (B) both correct
(d) none of there
Answer: (b) non-terminating repeating decimal
In simple words: If the denominator of a fraction, once fully simplified, has prime factors other than just 2s and 5s, then when you divide it out, the decimal will go on forever and have a repeating pattern.

🎯 Exam Tip: A rational number has a terminating decimal expansion only if its denominator (in simplest form) has prime factors of only 2 and/or 5. Any other prime factor in the denominator leads to a non-terminating, repeating decimal.

Question 15. First Greek mathematician was
(a) AL-knwarijmi
(b) Aryabhata
(c) Einstein
(d) Euclid
Answer: (d) Euclid
In simple words: Euclid is widely known as the "Father of Geometry" and was a very important Greek mathematician. He lived a long time ago and wrote a famous book called "Elements."

🎯 Exam Tip: For general knowledge questions like this, associating key figures with their primary fields or nationalities can help recall the correct answer.

Very Short/Short Answer Type Questions

Question 2. Find the HCF of numbers 44 and 99 
Answer: First, we find the prime factors for each number:
For 44: \( 44 = 2 \times 2 \times 11 = 2^2 \times 11 \)
For 99: \( 99 = 3 \times 3 \times 11 = 3^2 \times 11 \)
The common prime factor is 11. The lowest power of this common factor is \( 11^1 \).
So, the H.C.F. is 11. Finding common factors is useful for simplifying fractions.
In simple words: Break down 44 into \( 2 \times 2 \times 11 \). Break down 99 into \( 3 \times 3 \times 11 \). The only number that appears in both lists is 11. So, 11 is their HCF.

🎯 Exam Tip: Always list all prime factors for each number before identifying the common ones. The HCF is the product of the lowest powers of these common prime factors.

Question 3. Is the HCF of two numbers 15 and LCM 175 possible? Give reason. 
Answer: No, it is not possible for the HCF of two numbers to be 15 and their LCM to be 175. This is because the Least Common Multiple (LCM) of any two numbers must always be perfectly divisible by their Highest Common Factor (HCF). Here, we check if 175 can be divided by 15.
When we divide 175 by 15, we get \( 175 \div 15 = 11 \) with a remainder of 10. Since 175 is not exactly divisible by 15, these values for HCF and LCM are not possible. The HCF must always be a factor of the LCM.
In simple words: No, it's not possible. The big number (LCM) must always be a multiple of the small number (HCF). 175 cannot be perfectly divided by 15, so these numbers can't be the HCF and LCM.

🎯 Exam Tip: The golden rule for HCF and LCM is that LCM is always a multiple of HCF. If LCM % HCF is not zero, the given HCF and LCM values are incorrect for any pair of numbers.

Question 4. Whether the decimal expansion of number \( \frac {3}{625} \) is terminating or non-terminating repeating? write it in decimal form? 
Answer: To determine if the decimal expansion is terminating, we look at the prime factors of the denominator. We write the denominator, 625, in its prime factor form:
\( 625 = 5 \times 5 \times 5 \times 5 = 5^4 \)
The denominator is of the form \( 2^m \times 5^n \) (here, \( m=0 \) and \( n=4 \)). Because the denominator only has prime factors of 5 (and implicitly 2 to the power of zero), its decimal expansion will terminate. Any fraction whose denominator only contains prime factors of 2 or 5 will result in a terminating decimal.
To represent it in decimal form:
\( \frac { 3 }{ 625 } = \frac { 3 \times 2^4 }{ 5^4 \times 2^4 } = \frac { 3 \times 16 }{ (2 \times 5)^4 } = \frac { 48 }{ 10^4 } = 0.0048 \)
In simple words: The bottom part of the fraction, 625, can be written as \( 5 \times 5 \times 5 \times 5 \). Since it only has 5s as factors, the decimal will stop (terminate). To find the decimal, we can multiply the top and bottom by 16 (which is \( 2^4 \)) to get 10,000 at the bottom, making it 0.0048.

🎯 Exam Tip: Always fully simplify the fraction before checking the denominator's prime factors. If the denominator has only 2s or 5s (or both), the decimal terminates. Otherwise, it is non-terminating repeating.

Question 5. Find the minimum number of books required in a library if it has two sections with 48 and 60 students respectively.
Answer: To find the minimum number of books required so that students from both sections (48 students and 60 students) can be given an equal number of books, we need to find the Least Common Multiple (LCM) of 48 and 60. The LCM is the smallest number that is a multiple of both 48 and 60. We can find this using prime factorization:
For 48: \( 48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3^1 \)
For 60: \( 60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3^1 \times 5^1 \)
To find the LCM, we take the highest power of all prime factors present in either number:
\( LCM(48, 60) = 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 = 240 \)
So, the minimum number of required books is 240. This ensures enough books for both groups without any leftover.
In simple words: We need to find the smallest number that can be divided perfectly by both 48 and 60. This is called the LCM. By breaking down 48 and 60 into their prime factors, we find the LCM is 240. So, 240 books are needed.

🎯 Exam Tip: "Minimum" or "least" often hints that you need to find the LCM, especially when dealing with quantities that need to be grouped or distributed evenly.

Question 6. Prove that \( 3 - \sqrt{5} \) is an irrational number? 
Answer: To prove that \( 3 - \sqrt{5} \) is an irrational number, we will use the method of contradiction. We start by assuming the opposite.
Let's assume that \( 3 - \sqrt{5} \) is a rational number.
If \( 3 - \sqrt{5} \) is rational, then we can write it in the form \( \frac{a}{b} \), where a and b are co-prime integers and \( b \neq 0 \).
So, \( 3 - \sqrt{5} = \frac{a}{b} \)
Now, we rearrange the equation to isolate \( \sqrt{5} \):
\( 3 - \frac{a}{b} = \sqrt{5} \)
\( \frac{3b - a}{b} = \sqrt{5} \)
Since a and b are integers, \( (3b - a) \) is also an integer, and b is a non-zero integer. Therefore, the expression \( \frac{3b - a}{b} \) must be a rational number. This means that \( \sqrt{5} \) would also be a rational number.
However, we already know that \( \sqrt{5} \) is an irrational number. This creates a contradiction. Our initial assumption that \( 3 - \sqrt{5} \) is rational must be false. Hence, \( 3 - \sqrt{5} \) is an irrational number. This type of proof is common in number theory.
In simple words: We pretend \( 3 - \sqrt{5} \) is a normal fraction. If it were, we could move numbers around to show \( \sqrt{5} \) as a fraction too. But we know \( \sqrt{5} \) cannot be a fraction, it's an irrational number. This means our first guess was wrong, so \( 3 - \sqrt{5} \) must be irrational.

🎯 Exam Tip: For proofs involving irrational numbers, always start by assuming the number is rational, express it as \( \frac{a}{b} \), isolate the known irrational part, and show that this leads to a contradiction.

Question 7. Prove that \( 7 + \sqrt{5} \) is an irrational number.
Answer: To prove that \( 7 + \sqrt{5} \) is an irrational number, we will use the method of contradiction. We begin by assuming that \( 7 + \sqrt{5} \) is a rational number.
If \( 7 + \sqrt{5} \) is rational, then it can be written in the form \( \frac{a}{b} \), where a and b are co-prime integers and \( b \neq 0 \).
So, \( 7 + \sqrt{5} = \frac{a}{b} \)
Now, we rearrange the equation to isolate \( \sqrt{5} \):
\( \sqrt{5} = \frac{a}{b} - 7 \)
\( \sqrt{5} = \frac{a - 7b}{b} \)
Since a and b are integers, \( (a - 7b) \) is also an integer, and b is a non-zero integer. Therefore, the expression \( \frac{a - 7b}{b} \) must be a rational number. This means that \( \sqrt{5} \) would also be a rational number.
However, we know that \( \sqrt{5} \) is an irrational number. This is a contradiction, which means our initial assumption was wrong. Thus, \( 7 + \sqrt{5} \) must be an irrational number. This demonstrates that adding a rational number to an irrational number results in an irrational number.
In simple words: Let's pretend \( 7 + \sqrt{5} \) is a simple fraction. If it is, then we can rearrange the equation to show that \( \sqrt{5} \) is also a simple fraction. But we know \( \sqrt{5} \) is not a simple fraction; it's irrational. This clash means our first idea was wrong, so \( 7 + \sqrt{5} \) must be irrational.

🎯 Exam Tip: Always make sure to clearly state your initial assumption, show the algebraic manipulation to isolate the known irrational part, and explicitly mention the contradiction to complete the proof.

Question 8. Find the HCF and LCM of two numbers 3364 and 54.
Answer: To find the HCF (Highest Common Factor) and LCM (Least Common Multiple) of 3364 and 54, we first perform their prime factorization:
For 3364:
\( 3364 = 2 \times 2 \times 29 \times 29 = 2^2 \times 29^2 \)
For 54:
\( 54 = 2 \times 3 \times 3 \times 3 = 2^1 \times 3^3 \)

To find the HCF, we take the lowest power of the common prime factors:
The only common prime factor is 2. The lowest power of 2 is \( 2^1 \).
So, H.C.F. (3364, 54) = 2.

To find the LCM, we take the highest power of all prime factors (common and uncommon):
Highest power of 2 is \( 2^2 \).
Highest power of 3 is \( 3^3 \).
Highest power of 29 is \( 29^2 \).
So, L.C.M. (3364, 54) = \( 2^2 \times 3^3 \times 29^2 = 4 \times 27 \times 841 = 90,828 \).
The HCF helps simplify fractions, while the LCM is useful for finding common multiples.
In simple words: We break down 3364 and 54 into their smallest prime numbers. For HCF, we find what prime numbers they share and take the smallest power. They only share one '2'. So HCF is 2. For LCM, we take all prime numbers from both lists and use their biggest powers. This gives us 90,828.

🎯 Exam Tip: Carefully list all prime factors and their powers for each number. HCF uses the minimum power of common factors, while LCM uses the maximum power of all factors.

Long Answer Type Questions

Question 1. Prove that only one out of three consecutive positive integers is divisible by 3. 
Answer: Let the three consecutive positive integers be n, n+1, and n+2. We use Euclid's Division Lemma to show that any positive integer n can be written in one of these three forms: \( 3q \), \( 3q+1 \), or \( 3q+2 \) for some integer q.

**Case 1: n = 3q**
If \( n = 3q \), then n is directly divisible by 3.
For \( n+1 = 3q+1 \), it leaves a remainder of 1 when divided by 3, so it is not divisible by 3.
For \( n+2 = 3q+2 \), it leaves a remainder of 2 when divided by 3, so it is not divisible by 3.
In this case, only n is divisible by 3.

**Case 2: n = 3q+1**
If \( n = 3q+1 \), then n is not divisible by 3.
For \( n+1 = (3q+1)+1 = 3q+2 \), it leaves a remainder of 2 when divided by 3, so it is not divisible by 3.
For \( n+2 = (3q+1)+2 = 3q+3 = 3(q+1) \). This is clearly divisible by 3.
In this case, only n+2 is divisible by 3.

**Case 3: n = 3q+2**
If \( n = 3q+2 \), then n is not divisible by 3.
For \( n+1 = (3q+2)+1 = 3q+3 = 3(q+1) \). This is clearly divisible by 3.
For \( n+2 = (3q+2)+2 = 3q+4 = 3q+3+1 = 3(q+1)+1 \), which leaves a remainder of 1 when divided by 3, so it is not divisible by 3.
In this case, only n+1 is divisible by 3.

In all possible cases, exactly one out of the three consecutive positive integers is divisible by 3. This mathematical principle is very fundamental. Hence proved.
In simple words: Take any three numbers in a row. One of them will always divide evenly by 3. We show this by checking all the ways a number can be written when divided by 3 (like 3 times something, or 3 times something plus 1, or 3 times something plus 2). In each case, only one of the three numbers ends up being a perfect multiple of 3.

🎯 Exam Tip: When proving divisibility properties for consecutive integers, Euclid's Division Lemma is often the most effective tool. Systematically cover all possible remainders when divided by the specified number (e.g., 0, 1, 2 for division by 3).

Question 2. Show that cube of any positive integer is either of the form 4m, 4m + 1 or 4m + 3 for some integer m. 
Answer: Let x be any positive integer. According to Euclid's Division Lemma, x can be expressed in one of the following forms when divided by 4: \( 4q \), \( 4q+1 \), \( 4q+2 \), or \( 4q+3 \). We will examine the cube of x for each of these cases.

**Case 1: When \( x = 4q \)**
Cube both sides:
\( x^3 = (4q)^3 \)
\( x^3 = 64q^3 \)
We can write this as \( x^3 = 4(16q^3) \).
Let \( m = 16q^3 \). Then \( x^3 = 4m \).

**Case 2: When \( x = 4q+1 \)**
Cube both sides (using the formula \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \)):
\( x^3 = (4q+1)^3 \)
\( x^3 = (4q)^3 + 3(4q)^2(1) + 3(4q)(1)^2 + (1)^3 \)
\( x^3 = 64q^3 + 48q^2 + 12q + 1 \)
We can factor out 4 from the first three terms:
\( x^3 = 4(16q^3 + 12q^2 + 3q) + 1 \)
Let \( m = 16q^3 + 12q^2 + 3q \). Then \( x^3 = 4m + 1 \).

**Case 3: When \( x = 4q+2 \)**
Cube both sides:
\( x^3 = (4q+2)^3 \)
\( x^3 = (4q)^3 + 3(4q)^2(2) + 3(4q)(2)^2 + (2)^3 \)
\( x^3 = 64q^3 + 96q^2 + 48q + 8 \)
We can factor out 4 from all terms:
\( x^3 = 4(16q^3 + 24q^2 + 12q + 2) \)
Let \( m = 16q^3 + 24q^2 + 12q + 2 \). Then \( x^3 = 4m \).

**Case 4: When \( x = 4q+3 \)**
Cube both sides:
\( x^3 = (4q+3)^3 \)
\( x^3 = (4q)^3 + 3(4q)^2(3) + 3(4q)(3)^2 + (3)^3 \)
\( x^3 = 64q^3 + 144q^2 + 108q + 27 \)
We can write 27 as \( 24 + 3 \) to factor out 4:
\( x^3 = 64q^3 + 144q^2 + 108q + 24 + 3 \)
\( x^3 = 4(16q^3 + 36q^2 + 27q + 6) + 3 \)
Let \( m = 16q^3 + 36q^2 + 27q + 6 \). Then \( x^3 = 4m + 3 \).

From all these cases, we see that the cube of any positive integer can only be of the form \( 4m \), \( 4m+1 \), or \( 4m+3 \) for some integer m. This proves the statement. This property helps understand number patterns.
In simple words: We take any whole number and imagine dividing it by 4. It can either have a remainder of 0, 1, 2, or 3. Then we cube (multiply by itself three times) each of these possibilities. After cubing, we see that the result always looks like "4 times some number" (4m), "4 times some number plus 1" (4m+1), or "4 times some number plus 3" (4m+3). This means the cube of any positive integer will always fit one of these three forms.

🎯 Exam Tip: For problems involving cubes or squares of integers and divisibility, use Euclid's Division Lemma. Clearly list all possible remainder cases and apply the relevant algebraic expansion (like \( (a+b)^3 \)) to each case. Remember to factor out the divisor (here, 4) to show the required form.

Question 3. Find the LCM and HCF of the given pair of integers and check product of two numbers = H.C.F x L.C.M
(i) 26 and 91,
(ii) 510 and 92,
Answer: We will find the HCF and LCM for each pair of numbers using prime factorization and then verify the relationship.

**(i) For numbers 26 and 91:**
First, find the prime factors of each number:
\( 26 = 2 \times 13 \)
\( 91 = 7 \times 13 \)

To find the HCF (Highest Common Factor), we take the lowest power of the common prime factors:
The only common prime factor is 13, with the lowest power \( 13^1 \).
So, H.C.F. (26, 91) = 13.

To find the LCM (Least Common Multiple), we take the highest power of all prime factors (common and uncommon):
\( LCM (26, 91) = 2^1 \times 7^1 \times 13^1 = 2 \times 7 \times 13 = 182 \).

Now, let's check the product relationship:
Product of the two numbers = \( 26 \times 91 = 2366 \).
Product of HCF and LCM = \( 13 \times 182 = 2366 \).
Since Product of numbers = HCF \( \times \) LCM, the relationship is verified.

**(ii) For numbers 510 and 92:**
First, find the prime factors of each number:
For 510:
\( 510 = 2 \times 3 \times 5 \times 17 \)
For 92:
\( 92 = 2 \times 2 \times 23 = 2^2 \times 23 \)

To find the HCF, we take the lowest power of the common prime factors:
The only common prime factor is 2. The lowest power of 2 is \( 2^1 \).
So, H.C.F. (510, 92) = 2.

To find the LCM, we take the highest power of all prime factors:
\( LCM (510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1 = 4 \times 3 \times 5 \times 17 \times 23 = 23460 \).

Now, let's check the product relationship:
Product of the two numbers = \( 510 \times 92 = 46920 \).
Product of HCF and LCM = \( 2 \times 23460 = 46920 \).
Since Product of numbers = HCF \( \times \) LCM, the relationship is verified. This relationship is a very important theorem in number theory.
In simple words: For both pairs of numbers, we first find their prime factors. Then, for HCF, we pick the common prime factors with the smallest powers. For LCM, we pick all prime factors with the biggest powers. After calculating both, we multiply the original numbers together and separately multiply the HCF and LCM. We see that both products are always the same, proving the rule.

🎯 Exam Tip: The fundamental theorem of arithmetic states that for any two positive integers 'a' and 'b', \( a \times b = HCF(a,b) \times LCM(a,b) \). Always use prime factorization for HCF and LCM, and then perform the multiplication to verify the identity.