CBSE Class 10 Mathematics Real Numbers MCQs Set 28

Answer: (d) xy > yz

Explanation:
1. x and y are positive, so xy is positive.
2. z is negative and y is positive, so yz is negative.
3. A positive number is greater than a negative number, so xy > yz.
4. Check (a): multiplying x < y by negative z reverses the sign, so xz > yz. Options (b) and (c) are also false for the same reason (try x = 1, y = 2, z = -1).

Teacher's Note:
1. Multiplying or dividing an inequality by a negative number reverses it.
2. Testing with small numbers quickly eliminates wrong options.

Answer: (b) (18, 25)

Explanation:
1. Co-primes have HCF 1.
2. HCF(14, 35) = 7, HCF(18, 25) = 1, HCF(31, 93) = 31, HCF(32, 62) = 2.
3. Only (18, 25) has HCF 1.

Teacher's Note:
1. Two numbers are co-prime when they share no common prime factor.
2. Co-prime numbers need not be prime themselves, as 18 and 25 show.

Answer: (c) 60

Explanation:
1. Take each common prime with its smallest power.
2. Power of 2: smallest is \( 2^2 \). Power of 3: smallest is \( 3^1 \). Power of 5: smallest is \( 5^1 \). Prime 7 is not common.
3. HCF = 4 × 3 × 5 = 60.

Teacher's Note:
1. HCF uses the smallest power of every common prime; LCM uses the greatest power of every prime.
2. Leave out any prime that is missing in even one number.

Answer: (b) 36, 60

Explanation:
1. Let the numbers be 12a and 12b with a and b co-prime. Then 144ab = 2160, so ab = 15.
2. Co-prime pairs with product 15 are (1, 15) and (3, 5).
3. (1, 15) gives 12 and 180, and 180 is not a two-digit number. (3, 5) gives 36 and 60.
4. The numbers are 36 and 60.

Teacher's Note:
1. Write the numbers as HCF times co-prime factors.
2. Always check the extra condition (two-digit) to reject extra pairs.

Answer: (c) 80

Explanation:
1. Let HCF = h, so LCM = 14h and h + 14h = 600, giving h = 40 and LCM = 560.
2. Product of the two numbers = HCF × LCM = 40 × 560 = 22400.
3. Other number = 22400 ÷ 280 = 80.

Teacher's Note:
1. Use HCF × LCM = product of the two numbers.
2. Solve for the HCF first from the given sum.

Answer: (d) 48

Explanation:
1. The numbers are 3 × 4 = 12 and 4 × 4 = 16.
2. LCM(12, 16) = 48.

Teacher's Note:
1. When the HCF is h and the ratio is a : b (co-prime), the LCM is h × a × b.
2. Here the LCM is 4 × 3 × 4 = 48.

Answer: (a) -2

Explanation:
1. The LCM of two different primes is their product, so xy = 161 = 7 × 23.
2. As x > y, x = 23 and y = 7.
3. 3y - x = 21 - 23 = -2.

Teacher's Note:
1. The LCM of two distinct primes is simply their product.
2. Factorise the product to recover the primes.

Answer: (b) (2ab - 3)

Explanation:
1. \( 4a^2b^3 - 9b = b(4a^2b^2 - 9) = b(2ab - 3)(2ab + 3) \).
2. \( 2a^2b^2 - ab - 3 = (2ab - 3)(ab + 1) \).
3. The common factor is (2ab - 3).

Teacher's Note:
1. Factorise each expression completely before looking for the HCF.
2. The HCF of algebraic expressions is the product of their common factors.

Answer: (c) 27

Explanation:
1. Subtract the remainder: 305 - 8 = 297 and 629 - 8 = 621.
2. 297 = 3³ × 11 and 621 = 3³ × 23.
3. HCF = 27, which is greater than the remainder 8.

Teacher's Note:
1. Subtract the remainder from each number, then take the HCF.
2. The divisor must be greater than the remainder.

Answer: (c) x is not divisible by n

Explanation:
1. The numerator is \( 21n^2 + 5n + x \), so the expression equals \( 21n + 5 + \frac{x}{n} \).
2. \( 21n + 5 \) is always an integer, so the whole expression is an integer exactly when n divides x.
3. It is not an integer exactly when x is not divisible by n, whatever the sign of n.

Teacher's Note:
1. Split the fraction and see which part can fail to be an integer.
2. Whether n is positive has no effect on divisibility.

Answer: (c) \( \frac{13}{325} \)

Explanation:
1. A fraction in lowest terms terminates only if its denominator has no prime factor other than 2 and 5.
2. 455 = 5 × 7 × 13 and 64/455 is in lowest terms, so it does not terminate. 343 = 7³, so 29/343 does not terminate.
3. 13/325 = 1/25 and 25 = 5², so it terminates (0.04). 308 = 2² × 7 × 11 has the factors 7 and 11, so 1/308 does not terminate.

Teacher's Note:
1. Reduce the fraction to lowest terms before checking the denominator.
2. 13/325 looks complicated but simplifies to 1/25.

Answer: (c) Addition, subtraction and multiplication

Explanation:
1. Sum, difference and product of two even integers are always even.
2. Division can give an odd number or a non-integer, for example 2 ÷ 4 = 0.5 and 6 ÷ 2 = 3.
3. So the set is closed under addition, subtraction and multiplication only.

Teacher's Note:
1. A set is closed under an operation if the result always stays in the set.
2. One counter-example is enough to show that closure fails.

Answer: (c) a natural number or an irrational number

Explanation:
1. If n is a perfect square, \( \sqrt{n} \) is a natural number (for example \( \sqrt{9} = 3 \)).
2. Otherwise \( \sqrt{n} \) is irrational (for example \( \sqrt{2} \)).
3. It is never a non-integer rational number.

Teacher's Note:
1. The square root of a natural number is either a natural number or irrational.
2. Give one example of each to check the claim.

Answer: (b) (1), (2) and (3) only

Explanation:
1. \( a^2 \) has factors 1, a, \( a^2 \): 3 factors, so (1) is true.
2. ab has factors 1, a, b, ab: 4 factors, so (2) is true.
3. \( a^3 \) has factors 1, a, \( a^2 \), \( a^3 \): 4 factors, so (3) is true.
4. \( a^2b^2 \) has (2 + 1)(2 + 1) = 9 factors, so (4) is false.

Teacher's Note:
1. If \( n = p^x q^y \), the number of factors is (x + 1)(y + 1).
2. Count the exponents plus one for each prime.

Answer: (d) (1), (2), (3) and (4)

Explanation:
1. (1) a = bk and b = cm give a = c(km), so it is true.
2. (2) and (3): if b = ap and c = aq then b ± c = a(p ± q), so both are true.
3. (4) A common factor of a and c would also divide b, and then it would be a common factor of b and c, which is only 1. So it is true.
4. All four statements are true.

Teacher's Note:
1. Divisibility is preserved under sums, differences and chains.
2. For (4), reason with a common prime factor.

Answer: (c) rational and equal to 6

Explanation:
1. \( 14 + 6\sqrt{5} = (3 + \sqrt{5})^2 \) because \( 9 + 5 + 6\sqrt{5} = 14 + 6\sqrt{5} \).
2. Similarly \( 14 - 6\sqrt{5} = (3 - \sqrt{5})^2 \), and \( 3 - \sqrt{5} > 0 \).
3. The sum is \( (3 + \sqrt{5}) + (3 - \sqrt{5}) = 6 \).

Teacher's Note:
1. Try to write the expression under the root as a perfect square \( (a + b\sqrt{5})^2 \).
2. The irrational parts cancel in the sum.

Answer: (a) A-Q, B-S, C-P, D-R

Explanation:
1. -10 is a negative integer, so it is an integer but not a natural number (Q).
2. \( \pi \) is irrational (S).
3. 3 is a natural number (P).
4. 5/2 is rational but not an integer (R).

Teacher's Note:
1. Natural numbers start from 1; integers include negatives and zero.
2. A fraction that does not simplify to a whole number is rational but not an integer.

Answer: (c) a rational number or an irrational number

Explanation:
1. \( \sqrt{2} + \sqrt{3} \) is irrational.
2. \( \sqrt{2} + (-\sqrt{2}) = 0 \) is rational.
3. So the sum can be either.

Teacher's Note:
1. The sum of two irrationals is not always irrational.
2. One example of each type settles the question.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The assertion is Euclid's division lemma, which is true.
2. The reason is also true: dividing by 3 leaves remainder 0, 1 or 2, and remainder 0 is excluded.
3. The reason is an application of the lemma with b = 3 but it does not explain why q and r are unique, so it is not the explanation of the assertion.

Teacher's Note:
1. The division lemma holds for every non-zero divisor.
2. The reason is a special case, not an explanation.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 9/4608 is a ratio of two integers with a non-zero denominator, so it is rational.
2. The reason is true: a rational number has a terminating or non-terminating recurring decimal expansion, and so does every such decimal represent a rational number.
3. The assertion follows from the reason, so R explains A.

Teacher's Note:
1. Rational numbers are exactly those with terminating or recurring decimal expansions.
2. Irrational numbers have non-terminating non-recurring expansions.