Board Exam 2025
1 Mark
Q7. The total number of factors of the square
of a prime number is :
Let the prime number be \(p\). Its square is \(p^2\).
The factors of \(p^2\) are \(1, p, p^2\).
Thus, there are exactly 3 factors.
Answer: (C)
The factors of \(p^2\) are \(1, p, p^2\).
Thus, there are exactly 3 factors.
Answer: (C)
2 Marks
Q22. Show that \(14^n\) cannot end with the
digit 0 or 5 for any natural number n.
We know that any number ending with the digit 0 or 5 must be divisible by 5.
Therefore, its prime factorization must contain the prime number 5.
Now, \(14^n = (2 \times 7)^n = 2^n \times 7^n\).
The only primes in the factorization of \(14^n\) are 2 and 7.
There is no other prime in the factorization of \(14^n\) (by the uniqueness of the Fundamental Theorem of Arithmetic).
Since the prime factor 5 is clearly missing from the prime factorization of \(14^n\), the number \(14^n\) can never end with the digit 0 or 5 for any natural number \(n\).
Therefore, its prime factorization must contain the prime number 5.
Now, \(14^n = (2 \times 7)^n = 2^n \times 7^n\).
The only primes in the factorization of \(14^n\) are 2 and 7.
There is no other prime in the factorization of \(14^n\) (by the uniqueness of the Fundamental Theorem of Arithmetic).
Since the prime factor 5 is clearly missing from the prime factorization of \(14^n\), the number \(14^n\) can never end with the digit 0 or 5 for any natural number \(n\).
3 Marks
Q30. Ranjita, Neha and Salma start weaving
sweaters at the same time for the children of an orphan home. They need 15, 18 and 20 days,
respectively, to complete a sweater. After how many days will all of them start making a new sweater
again? By that time how many sweaters will have been competed by them?
They will start together again after a time equal to the LCM of their individual times.
\(15 = 3 \times 5\)
\(18 = 2 \times 3^2\)
\(20 = 2^2 \times 5\)
LCM(15, 18, 20) = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\) days.
Number of sweaters:
Ranjita: \(180/15 = 12\)
Neha: \(180/18 = 10\)
Salma: \(180/20 = 9\)
Total sweaters = \(12 + 10 + 9 = 31\).
\(15 = 3 \times 5\)
\(18 = 2 \times 3^2\)
\(20 = 2^2 \times 5\)
LCM(15, 18, 20) = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\) days.
Number of sweaters:
Ranjita: \(180/15 = 12\)
Neha: \(180/18 = 10\)
Salma: \(180/20 = 9\)
Total sweaters = \(12 + 10 + 9 = 31\).
1 Mark
Q9. If \(HCF(98, 28) = m\) and \(LCM(98, 28)
= n\), then the value of \(n - 7m\) is:
\(98 = 2 \times 7^2, 28 = 2^2 \times 7\).
\(HCF(m) = 2 \times 7 = 14\).
\(LCM(n) = 2^2 \times 7^2 = 4 \times 49 = 196\).
\(n - 7m = 196 - 7(14) = 196 - 98 = 98\).
Answer: (C)
\(HCF(m) = 2 \times 7 = 14\).
\(LCM(n) = 2^2 \times 7^2 = 4 \times 49 = 196\).
\(n - 7m = 196 - 7(14) = 196 - 98 = 98\).
Answer: (C)
1 Mark
Q10. Which of the following is a rational
number between \(\sqrt{3}\) and \(\sqrt{5}\)?
\(\sqrt{3} \approx 1.732\) and \(\sqrt{5} \approx 2.236\).
Option (D) \(1.857142\) is a terminating decimal, thus rational, and lies between 1.732 and 2.236.
Answer: (D)
Option (D) \(1.857142\) is a terminating decimal, thus rational, and lies between 1.732 and 2.236.
Answer: (D)
1 Mark
Q15. If \((-1)^n + (-1)^8 = 0\), then \(n\)
is:
\((-1)^n + 1 = 0 \Rightarrow (-1)^n = -1\).
\((-1)\) raised to a power is \(-1\) only if the power is an odd integer.
Answer: (C)
\((-1)\) raised to a power is \(-1\) only if the power is an odd integer.
Answer: (C)
2 Marks
Q26. Prove that \(\sqrt{5}\) is an irrational
number.
Let us assume, to the contrary, that \(\sqrt{5}\) is a rational number.
Then, \(\sqrt{5} = \frac{a}{b}\), where \(a\) and \(b\) are co-prime integers and \(b \neq 0\).
Squaring both sides:
\(5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2\) ...(i)
This means \(a^2\) is divisible by 5. Hence, \(a\) is also divisible by 5.
Let \(a = 5c\) for some integer \(c\).
Substituting \(a = 5c\) in (i):
\((5c)^2 = 5b^2 \Rightarrow 25c^2 = 5b^2 \Rightarrow b^2 = 5c^2\).
This means \(b^2\) is divisible by 5. Hence, \(b\) is also divisible by 5.
Thus, \(a\) and \(b\) have at least 5 as a common factor.
But this contradicts the fact that \(a\) and \(b\) are co-prime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{5}\) is rational.
Hence, \(\sqrt{5}\) is an irrational number.
Then, \(\sqrt{5} = \frac{a}{b}\), where \(a\) and \(b\) are co-prime integers and \(b \neq 0\).
Squaring both sides:
\(5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2\) ...(i)
This means \(a^2\) is divisible by 5. Hence, \(a\) is also divisible by 5.
Let \(a = 5c\) for some integer \(c\).
Substituting \(a = 5c\) in (i):
\((5c)^2 = 5b^2 \Rightarrow 25c^2 = 5b^2 \Rightarrow b^2 = 5c^2\).
This means \(b^2\) is divisible by 5. Hence, \(b\) is also divisible by 5.
Thus, \(a\) and \(b\) have at least 5 as a common factor.
But this contradicts the fact that \(a\) and \(b\) are co-prime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{5}\) is rational.
Hence, \(\sqrt{5}\) is an irrational number.
1 Mark
Q4. If \(a^b = 32\), where 'a' and 'b' are
positive integers, then the value of \(b^{ab}\) is:
\(32 = 2^5\). Since \(a, b\) are integers, \(a=2, b=5\).
\(b^{ab} = 5^{(2)(5)} = 5^{10}\).
Answer: (B)
\(b^{ab} = 5^{(2)(5)} = 5^{10}\).
Answer: (B)
1 Mark
Q7. The sum of the exponents of prime factors
in the prime factorisation of 4004 is :
\(4004 = 4 \times 1001 = 2^2 \times 7 \times 11 \times 13\).
Prime factorization: \(2^2 \times 7^1 \times 11^1 \times 13^1\).
Sum of exponents = \(2 + 1 + 1 + 1 = 5\).
Answer: (A)
Prime factorization: \(2^2 \times 7^1 \times 11^1 \times 13^1\).
Sum of exponents = \(2 + 1 + 1 + 1 = 5\).
Answer: (A)
1 Mark
Q16. The least number which is a perfect
square and is divisible by each of 16, 20 and 50, is :
LCM of 16, 20, 50:
\(16 = 2^4\)
\(20 = 2^2 \times 5\)
\(50 = 2 \times 5^2\)
LCM = \(2^4 \times 5^2 = 16 \times 25 = 400\).
400 is a perfect square (\(20^2\)). However, 400 is not in the options.
Check options for divisibility and perfect square property:
(A) 1200: Not a square.
(B) 100: Not divisible by 16.
(C) 3600: \(60^2\). Divisible by 16 (\(3600/16 = 225\)), 20 (\(180\)), 50 (\(72\)).
(D) 2400: Not a square.
Answer: (C)
\(16 = 2^4\)
\(20 = 2^2 \times 5\)
\(50 = 2 \times 5^2\)
LCM = \(2^4 \times 5^2 = 16 \times 25 = 400\).
400 is a perfect square (\(20^2\)). However, 400 is not in the options.
Check options for divisibility and perfect square property:
(A) 1200: Not a square.
(B) 100: Not divisible by 16.
(C) 3600: \(60^2\). Divisible by 16 (\(3600/16 = 225\)), 20 (\(180\)), 50 (\(72\)).
(D) 2400: Not a square.
Answer: (C)
2 Marks
Q23. Prove that \(2 - \sqrt{3}\) is an
Let us assume, to the contrary, that \(2 - \sqrt{3}\) is a rational number.
That is, we can find coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that
\(2 - \sqrt{3} = \frac{a}{b}\)
Rearranging the terms:
\(\sqrt{3} = 2 - \frac{a}{b} = \frac{2b - a}{b}\)
Since \(a\) and \(b\) are integers, we get \(2b - a\) is an integer, and so \(\frac{2b - a}{b}\) is rational.
Thus, \(\sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is irrational.
This contradiction has arisen because of our incorrect assumption that \(2 - \sqrt{3}\) is rational.
So, we conclude that \(2 - \sqrt{3}\) is irrational.
That is, we can find coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that
\(2 - \sqrt{3} = \frac{a}{b}\)
Rearranging the terms:
\(\sqrt{3} = 2 - \frac{a}{b} = \frac{2b - a}{b}\)
Since \(a\) and \(b\) are integers, we get \(2b - a\) is an integer, and so \(\frac{2b - a}{b}\) is rational.
Thus, \(\sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is irrational.
This contradiction has arisen because of our incorrect assumption that \(2 - \sqrt{3}\) is rational.
So, we conclude that \(2 - \sqrt{3}\) is irrational.
3 Marks
Q31. Prove that \(\sqrt{3}\) is an
irrational
number.
Let us assume, to the contrary, that \(\sqrt{3}\) is rational.
That is, we can find integers \(a\) and \(b\) (\(\neq 0\)) such that \(\sqrt{3} = \frac{a}{b}\).
Suppose \(a\) and \(b\) have a common factor other than 1, then we can divide by the common factor, and assume that \(a\) and \(b\) are coprime.
So, \(b\sqrt{3} = a\).
Squaring on both sides, and rearranging, we get \(3b^2 = a^2\).
Therefore, \(a^2\) is divisible by 3, and from the Fundamental Theorem of Arithmetic, it follows that \(a\) is also divisible by 3.
So, we can write \(a = 3c\) for some integer \(c\).
Substituting for \(a\), we get \(3b^2 = (3c)^2 = 9c^2\), that is, \(b^2 = 3c^2\).
This means that \(b^2\) is divisible by 3, and so \(b\) is also divisible by 3.
Therefore, \(a\) and \(b\) have at least 3 as a common factor.
But this contradicts the fact that \(a\) and \(b\) are coprime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{3}\) is rational.
So, we conclude that \(\sqrt{3}\) is irrational.
That is, we can find integers \(a\) and \(b\) (\(\neq 0\)) such that \(\sqrt{3} = \frac{a}{b}\).
Suppose \(a\) and \(b\) have a common factor other than 1, then we can divide by the common factor, and assume that \(a\) and \(b\) are coprime.
So, \(b\sqrt{3} = a\).
Squaring on both sides, and rearranging, we get \(3b^2 = a^2\).
Therefore, \(a^2\) is divisible by 3, and from the Fundamental Theorem of Arithmetic, it follows that \(a\) is also divisible by 3.
So, we can write \(a = 3c\) for some integer \(c\).
Substituting for \(a\), we get \(3b^2 = (3c)^2 = 9c^2\), that is, \(b^2 = 3c^2\).
This means that \(b^2\) is divisible by 3, and so \(b\) is also divisible by 3.
Therefore, \(a\) and \(b\) have at least 3 as a common factor.
But this contradicts the fact that \(a\) and \(b\) are coprime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{3}\) is rational.
So, we conclude that \(\sqrt{3}\) is irrational.
1 Mark
Q16. If x is the LCM of 4, 6, 8 and y is
the
LCM of 3, 5, 7 and p is the LCM of x and y, then which of the following is true?
\(x = LCM(4, 6, 8) = 24\).
\(y = LCM(3, 5, 7) = 105\).
\(p = LCM(24, 105)\).
\(24 = 2^3 \times 3\), \(105 = 3 \times 5 \times 7\).
\(p = 2^3 \times 3 \times 5 \times 7 = 840\).
Check (A): \(35x = 35 \times 24 = 840\). Correct.
Answer: (A)
\(y = LCM(3, 5, 7) = 105\).
\(p = LCM(24, 105)\).
\(24 = 2^3 \times 3\), \(105 = 3 \times 5 \times 7\).
\(p = 2^3 \times 3 \times 5 \times 7 = 840\).
Check (A): \(35x = 35 \times 24 = 840\). Correct.
Answer: (A)
2 Marks
Q25 (a). Find the smallest number which
is
divisible by both 644 and 462.
OR
Q25 (b). Two numbers are in the ratio 4:5
and
their HCF is 11. Find the LCM of these numbers.
(a): Smallest number = LCM(644, 462).
\(644 = 2^2 \times 7 \times 23\).
\(462 = 2 \times 3 \times 7 \times 11\).
LCM \(= 2^2 \times 3 \times 7 \times 11 \times 23 = 21252\).
(b): Numbers \(4x, 5x\). HCF \(= x = 11\).
Numbers: 44, 55.
LCM \(= 4 \times 5 \times 11 = 220\).
\(644 = 2^2 \times 7 \times 23\).
\(462 = 2 \times 3 \times 7 \times 11\).
LCM \(= 2^2 \times 3 \times 7 \times 11 \times 23 = 21252\).
(b): Numbers \(4x, 5x\). HCF \(= x = 11\).
Numbers: 44, 55.
LCM \(= 4 \times 5 \times 11 = 220\).
3 Marks
Q29. Prove that \((5\sqrt{3} +
\frac{2}{3})\)
is an irrational number, given that \(\sqrt{3}\) is an irrational number.
Let us assume, to the contrary, that \(5\sqrt{3} + \frac{2}{3}\) is a rational number.
Let \(5\sqrt{3} + \frac{2}{3} = r\), where \(r\) is a rational number.
Rearranging the terms:
\(5\sqrt{3} = r - \frac{2}{3} = \frac{3r - 2}{3}\)
\(\Rightarrow \sqrt{3} = \frac{3r - 2}{15}\)
Since \(r\) is rational, \(3r - 2\) and \(15\) are integers (\(15 \neq 0\)).
Thus, \(\frac{3r - 2}{15}\) is a rational number.
This implies that \(\sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is an irrational number.
So, our assumption is incorrect.
Hence, \(5\sqrt{3} + \frac{2}{3}\) is an irrational number.
Let \(5\sqrt{3} + \frac{2}{3} = r\), where \(r\) is a rational number.
Rearranging the terms:
\(5\sqrt{3} = r - \frac{2}{3} = \frac{3r - 2}{3}\)
\(\Rightarrow \sqrt{3} = \frac{3r - 2}{15}\)
Since \(r\) is rational, \(3r - 2\) and \(15\) are integers (\(15 \neq 0\)).
Thus, \(\frac{3r - 2}{15}\) is a rational number.
This implies that \(\sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is an irrational number.
So, our assumption is incorrect.
Hence, \(5\sqrt{3} + \frac{2}{3}\) is an irrational number.
1 Mark
Q10. If \(x = ab^3\) and \(y = a^3b\),
where
a and b are prime numbers, then \([HCF(x, y) - LCM(x, y)]\) is equal to:
\(x = a^1b^3\), \(y = a^3b^1\).
HCF = \(a^1b^1 = ab\). LCM = \(a^3b^3\).
Value = \(ab - a^3b^3 = ab(1 - a^2b^2)\).
Using \(x^2 - y^2\): \(ab(1 - ab)(1 + ab)\).
Answer: (d)
HCF = \(a^1b^1 = ab\). LCM = \(a^3b^3\).
Value = \(ab - a^3b^3 = ab(1 - a^2b^2)\).
Using \(x^2 - y^2\): \(ab(1 - ab)(1 + ab)\).
Answer: (d)
1 Mark
Q11. \((1 + \sqrt{3})^2 - (1 -
\sqrt{3})^2\)
is:
Using \((a+b)^2 - (a-b)^2 = 4ab\):
Here \(a=1, b=\sqrt{3}\).
Expression = \(4(1)(\sqrt{3}) = 4\sqrt{3}\).
This is a positive irrational number.
Answer: (c)
Here \(a=1, b=\sqrt{3}\).
Expression = \(4(1)(\sqrt{3}) = 4\sqrt{3}\).
This is a positive irrational number.
Answer: (c)
1 Mark
Q20. Assertion (A): Unit
digit of \(3^n\) cannot be an even number for any natural number n.
Reason (R): 2 is not a prime factor of \(3^n\) for any natural number n.
Reason (R): 2 is not a prime factor of \(3^n\) for any natural number n.
Powers of 3 end in 3, 9, 7, 1. None are even, so A is true.
For a number to be even, it must have 2 as a prime factor. \(3^n\) only has prime factor 3. Thus R is true and explains A.
Answer: (a)
For a number to be even, it must have 2 as a prime factor. \(3^n\) only has prime factor 3. Thus R is true and explains A.
Answer: (a)
3 Marks
Q30 (A). Prove that \(\sqrt{2}\) is an
irrational number.
OR
Q30 (B). Let x and y be two distinct
prime
numbers and \(p = x^2y^3, q = xy^4, r = x^5y^2\). Find the HCF and LCM of p, q and r. Further
check
if \(HCF(p, q, r) \times LCM(p, q, r) = p \times q \times r\) or not.
(A): Let us assume technically that \(\sqrt{2}\) is a rational number.
\(\therefore \sqrt{2} = \frac{p}{q}\), where \(p\) and \(q\) are integers, \(q \neq 0\), and \(p, q\) are co-prime (have no common factor other than 1).
Squaring both sides:
\(2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2\) ...(i)
\(\Rightarrow p^2\) is divisible by 2.
\(\Rightarrow p\) is divisible by 2. (If a prime \(p\) divides \(a^2\), then it divides \(a\)).
Let \(p = 2m\) for some integer \(m\).
Substituting in (i):
\((2m)^2 = 2q^2 \Rightarrow 4m^2 = 2q^2 \Rightarrow q^2 = 2m^2\).
\(\Rightarrow q^2\) is divisible by 2.
\(\Rightarrow q\) is divisible by 2.
Thus, both \(p\) and \(q\) integers have a common factor 2.
This contradicts our assumption that \(p\) and \(q\) are co-prime.
\(\therefore\) Our assumption was wrong.
\(\therefore \sqrt{2}\) is an irrational number.
(B): \(p=x^2y^3, q=xy^4, r=x^5y^2\).
HCF = \(xy^2\). LCM = \(x^5y^4\).
HCF \(\times\) LCM = \(xy^2 \cdot x^5y^4 = x^6y^6\).
Product \(pqr = (x^2y^3)(xy^4)(x^5y^2) = x^{2+1+5}y^{3+4+2} = x^8y^9\).
Thus, HCF \(\times\) LCM \(\neq pqr\).
\(\therefore \sqrt{2} = \frac{p}{q}\), where \(p\) and \(q\) are integers, \(q \neq 0\), and \(p, q\) are co-prime (have no common factor other than 1).
Squaring both sides:
\(2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2\) ...(i)
\(\Rightarrow p^2\) is divisible by 2.
\(\Rightarrow p\) is divisible by 2. (If a prime \(p\) divides \(a^2\), then it divides \(a\)).
Let \(p = 2m\) for some integer \(m\).
Substituting in (i):
\((2m)^2 = 2q^2 \Rightarrow 4m^2 = 2q^2 \Rightarrow q^2 = 2m^2\).
\(\Rightarrow q^2\) is divisible by 2.
\(\Rightarrow q\) is divisible by 2.
Thus, both \(p\) and \(q\) integers have a common factor 2.
This contradicts our assumption that \(p\) and \(q\) are co-prime.
\(\therefore\) Our assumption was wrong.
\(\therefore \sqrt{2}\) is an irrational number.
(B): \(p=x^2y^3, q=xy^4, r=x^5y^2\).
HCF = \(xy^2\). LCM = \(x^5y^4\).
HCF \(\times\) LCM = \(xy^2 \cdot x^5y^4 = x^6y^6\).
Product \(pqr = (x^2y^3)(xy^4)(x^5y^2) = x^{2+1+5}y^{3+4+2} = x^8y^9\).
Thus, HCF \(\times\) LCM \(\neq pqr\).
1 Mark
Q10. \((\sqrt{3} + 2)^2 + (\sqrt{3} -
2)^2\)
is a/an:
Using \((a + b)^2 + (a - b)^2 = 2(a^2 + b^2)\):
Here \(a = \sqrt{3}, b = 2\).
Expression = \(2((\sqrt{3})^2 + 2^2) = 2(3 + 4) = 14\).
14 is a positive rational number.
Correct Option: (A)
Here \(a = \sqrt{3}, b = 2\).
Expression = \(2((\sqrt{3})^2 + 2^2) = 2(3 + 4) = 14\).
14 is a positive rational number.
Correct Option: (A)
1 Mark
Q11. Let \( p = x^2 y^3 z^n \) and \( q =
x^3
y^m z^2 \), where \( x, y, z \) are prime numbers. If LCM(p, q) = \( x^3 y^4 z^3 \), then the
value
of \( (2m + 3n) \) is
LCM takes the highest power of each prime.
For \(y\): \(\max(3, m) = 4 \Rightarrow m = 4\).
For \(z\): \(\max(n, 2) = 3 \Rightarrow n = 3\).
\(2m + 3n = 2(4) + 3(3) = 17\).
Correct Option: (B)
For \(y\): \(\max(3, m) = 4 \Rightarrow m = 4\).
For \(z\): \(\max(n, 2) = 3 \Rightarrow n = 3\).
\(2m + 3n = 2(4) + 3(3) = 17\).
Correct Option: (B)
1 Mark
Q12. For any prime number p, if p divides
\(
a^2 \), where a is any positive integer, then p also divides
By Fundamental Theorem of Arithmetic, if prime \(p\) divides \(a^2\), then
\(p\)
divides \(a\).
Correct Option: (A)
Correct Option: (A)
3 Marks
Q30 (a). Prove that \(\sqrt{3}\) is an
irrational number.
OR
Q30 (b). State true or false for each of
the
following statements and justify in each case :
(i) \(2 \times 3 \times 5 \times 7 + 7\) is a composite number.
(ii) \(2 \times 3 \times 5 \times 7 + 1\) is a composite number.
(i) \(2 \times 3 \times 5 \times 7 + 7\) is a composite number.
(ii) \(2 \times 3 \times 5 \times 7 + 1\) is a composite number.
(a): Assume \(\sqrt{3} = a/b\) (coprime). \(a^2 = 3b^2\). So
\(a\) is div by 3. Let \(a = 3c\). Then \(b^2 = 3c^2\). Contradiction. Thus
irrational.
(b)(i): \(7(31) = 217\). Composite. True.
(b)(ii): \(211\) is prime. False.
(b)(i): \(7(31) = 217\). Composite. True.
(b)(ii): \(211\) is prime. False.
1 Mark
Q10. \(\sqrt{0.4}\) is a/an
We have \(\sqrt{0.4} = \sqrt{\frac{4}{10}} = \frac{\sqrt{4}}{\sqrt{10}} =
\frac{2}{\sqrt{10}}\).
Since \(\sqrt{10}\) is an irrational number and the quotient of a non-zero rational number (2) and an irrational number is irrational, \(\sqrt{0.4}\) is an irrational number.
Correct Option: (D)
Since \(\sqrt{10}\) is an irrational number and the quotient of a non-zero rational number (2) and an irrational number is irrational, \(\sqrt{0.4}\) is an irrational number.
Correct Option: (D)
1 Mark
Q11. Which of the following cannot be the
unit digit of \(8^n\), where \(n\) is a natural number?
The powers of 8 repeat in a cycle of 4:
\(8^1 = 8\) (ends in 8)
\(8^2 = 64\) (ends in 4)
\(8^3 = 512\) (ends in 2)
\(8^4 = 4096\) (ends in 6)
The cycle of unit digits is {8, 4, 2, 6}. Zero (0) is never a unit digit for \(8^n\).
Correct Option: (C)
\(8^1 = 8\) (ends in 8)
\(8^2 = 64\) (ends in 4)
\(8^3 = 512\) (ends in 2)
\(8^4 = 4096\) (ends in 6)
The cycle of unit digits is {8, 4, 2, 6}. Zero (0) is never a unit digit for \(8^n\).
Correct Option: (C)
1 Mark
Q20. Assertion (A): For
two
odd prime numbers \(x\) and \(y\), \((x \neq y)\), \(LCM(2x, 4y) = 4xy\)
Reason (R): \(LCM(x, y)\) is a multiple of \(HCF(x, y)\).
Reason (R): \(LCM(x, y)\) is a multiple of \(HCF(x, y)\).
Assertion Check:
Given numbers: \(2x\) and \(4y\). Prime factorizations are \(2^1 \cdot x\) and \(2^2 \cdot y\) (since \(x, y\) are distinct odd primes).
LCM = Highest power of each prime factor = \(2^2 \cdot x \cdot y = 4xy\). Assertion is True.
Reason Check: The LCM of any two numbers is always a multiple of their HCF. Reason is True.
Explanation: The reason (general property of LCM/HCF) does not explain the specific calculation for \(2x\) and \(4y\). The assertion is true based on the fundamental theorem of arithmetic/prime factorization method.
Correct Option: (B)
Given numbers: \(2x\) and \(4y\). Prime factorizations are \(2^1 \cdot x\) and \(2^2 \cdot y\) (since \(x, y\) are distinct odd primes).
LCM = Highest power of each prime factor = \(2^2 \cdot x \cdot y = 4xy\). Assertion is True.
Reason Check: The LCM of any two numbers is always a multiple of their HCF. Reason is True.
Explanation: The reason (general property of LCM/HCF) does not explain the specific calculation for \(2x\) and \(4y\). The assertion is true based on the fundamental theorem of arithmetic/prime factorization method.
Correct Option: (B)
3 Marks
Q30 (a). Prove that \(\sqrt{5}\) is an
irrational number.
OR
Q30 (b). Let \(p, q\) and \(r\) be three
distinct prime numbers. Check whether \(p \cdot q \cdot r + q\) is a composite number or not.
Further, give an example for 3 distinct primes \(p, q, r\) such that:
(i) \(p \cdot q \cdot r + 1\) is a composite number.
(ii) \(p \cdot q \cdot r + 1\) is a prime number.
(i) \(p \cdot q \cdot r + 1\) is a composite number.
(ii) \(p \cdot q \cdot r + 1\) is a prime number.
Solution Q30 (a):
Let us assume, to the contrary, that \(\sqrt{5}\) is rational. So, we can find coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that \(\sqrt{5} = \frac{a}{b}\).
Squaring: \(5b^2 = a^2\). Therefore, 5 divides \(a^2\), so 5 divides \(a\). Let \(a = 5c\).
Then \(5b^2 = 25c^2 \Rightarrow b^2 = 5c^2\). So 5 divides \(b\).
This contradicts the fact that \(a\) and \(b\) are coprime. Hence, \(\sqrt{5}\) is irrational.
Solution Q30 (b):
\(p \cdot q \cdot r + q = q(p \cdot r + 1)\). Since \(q > 1\) and \(pr + 1 > 1\), it is a composite number.
Examples:
(i) For composite: \(p=3, q=5, r=7 \Rightarrow 3 \cdot 5 \cdot 7 + 1 = 105 + 1 = 106\) (even, composite).
(ii) For prime: \(p=2, q=3, r=5 \Rightarrow 2 \cdot 3 \cdot 5 + 1 = 30 + 1 = 31\) (prime).
Let us assume, to the contrary, that \(\sqrt{5}\) is rational. So, we can find coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that \(\sqrt{5} = \frac{a}{b}\).
Squaring: \(5b^2 = a^2\). Therefore, 5 divides \(a^2\), so 5 divides \(a\). Let \(a = 5c\).
Then \(5b^2 = 25c^2 \Rightarrow b^2 = 5c^2\). So 5 divides \(b\).
This contradicts the fact that \(a\) and \(b\) are coprime. Hence, \(\sqrt{5}\) is irrational.
Solution Q30 (b):
\(p \cdot q \cdot r + q = q(p \cdot r + 1)\). Since \(q > 1\) and \(pr + 1 > 1\), it is a composite number.
Examples:
(i) For composite: \(p=3, q=5, r=7 \Rightarrow 3 \cdot 5 \cdot 7 + 1 = 105 + 1 = 106\) (even, composite).
(ii) For prime: \(p=2, q=3, r=5 \Rightarrow 2 \cdot 3 \cdot 5 + 1 = 30 + 1 = 31\) (prime).
Board Exam 2024
2 Marks
Find the LCM and HCF of 6 and 20 by the prime factorization method.
\( 6 = 2 \times 3 \)
\( 20 = 2^2 \times 5 \)
HCF = \( 2 \) (lowest power of common factors)
LCM = \( 2^2 \times 3 \times 5 = 60 \)
\( 20 = 2^2 \times 5 \)
HCF = \( 2 \) (lowest power of common factors)
LCM = \( 2^2 \times 3 \times 5 = 60 \)
Board Exam 2023
3 Marks
Prove that \( 5 - \sqrt{3} \) is irrational, given that \( \sqrt{3} \) is
irrational.
Assume \( 5 - \sqrt{3} \) is rational, say \( = r \).
Then \( \sqrt{3} = 5 - r \), which is rational (difference of two rationals).
Contradiction: \( \sqrt{3} \) is given to be irrational.
? \( 5 - \sqrt{3} \) is irrational. ?
Then \( \sqrt{3} = 5 - r \), which is rational (difference of two rationals).
Contradiction: \( \sqrt{3} \) is given to be irrational.
? \( 5 - \sqrt{3} \) is irrational. ?
Board Exam 2022
2 Marks
Explain why \( 7 \times 11 \times 13 + 13 \) is a composite number.
\( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 \)
Since it has factors other than 1 and itself, it is composite.
Since it has factors other than 1 and itself, it is composite.