CBSE Class 10 Mathematics Real Numbers MCQs Set 14

Answer: (d) \(180a^3b^4\)

Explanation:
1. Write both numbers in prime factor form: \(p = 2 \times 3^2 \times a^2b^4\) and \(q = 2^2 \times 5 \times a^3b^2\).
2. LCM takes the highest power of each prime factor: \(2^2 \times 3^2 \times 5 \times a^3 \times b^4\).
3. So LCM \((p,q) = 180a^3b^4\).

Teacher's Note:
1. For LCM, take the highest power of every prime and variable.
2. Check the powers of \(a\) and \(b\) separately; they are the easiest to mix up.

Answer: (c) rational number

Explanation:
1. \(p^2 = \frac{32}{50} = \frac{16}{25}\), so \(p = \sqrt{\frac{16}{25}} = \frac{4}{5}\).
2. \(\frac{4}{5}\) is in the form \(\frac{p}{q}\) with \(q \neq 0\), so \(p\) is a rational number.

Teacher's Note:
1. A number of the form \(\frac{p}{q}\), with integers \(p\) and \(q \neq 0\), is rational.
2. Take the square root of the numerator and the denominator separately.

Answer: (b) 4

Explanation:
1. \(196 = 2^2 \times 7^2\).
2. Sum of exponents = 2 + 2 = 4.

Teacher's Note:
1. Add the small powers, not the prime numbers themselves.
2. Writing the factorisation first makes this quick.

Answer: (a) 7, 13

Explanation:
1. \(1001 = x \times 143 \Rightarrow x = 7\).
2. \(143 = y \times 11 \Rightarrow y = 13\).

Teacher's Note:
1. In a factor tree, each node is the product of its two children.
2. Divide the parent by the known child to find the missing one.

Answer: (c) 2

Explanation:
1. A prime number has only two factors: 1 and the number itself.

Teacher's Note:
1. Remember the definition of a prime number.
2. 1 is not prime, so option (a) is wrong.

Answer: (c) 3, 420

Explanation:
1. \(12 = 2^2 \times 3\), \(21 = 3 \times 7\), \(15 = 3 \times 5\).
2. HCF = 3 (the only prime common to all three).
3. LCM = \(2^2 \times 3 \times 5 \times 7 = 420\).

Teacher's Note:
1. HCF is the product of the common primes with their lowest powers.
2. LCM is the product of all primes with their highest powers.

Answer: (c) 20

Explanation:
1. Smallest two-digit composite number = 10.
2. Smallest composite number = 4.
3. LCM \((10, 4) = 2^2 \times 5 = 20\).

Teacher's Note:
1. The smallest composite number is 4, not 1 or 2.
2. Two-digit numbers start at 10.

Answer: (d) 81

Explanation:
1. Product of two numbers = HCF × LCM.
2. \(54 \times y = 27 \times 162\).
3. \(y = \frac{27 \times 162}{54} = 81\).

Teacher's Note:
1. Use the rule: the two numbers multiply to HCF × LCM.
2. Check: 54 × 81 = 4374 = 27 × 162.

Answer: (b) 18

Explanation:
1. \(144 = 2^4 \times 3^2\), \(198 = 2 \times 3^2 \times 11\).
2. HCF \(= 2 \times 3^2 = 18\).

Teacher's Note:
1. Use the lowest power of each common prime.
2. Factorise both numbers before comparing them.

Answer: (c) \(5^2 \times 3^2\)

Explanation:
1. \(225 = 3 \times 3 \times 5 \times 5 = 3^2 \times 5^2\).

Teacher's Note:
1. Divide by the smallest prime repeatedly to build the factor list.
2. Each option can be checked by multiplying it out.

Answer: (c) an irrational number

Explanation:
1. Assume \(2\sqrt{3}\) is rational, \(2\sqrt{3} = r\), so \(\sqrt{3} = \frac{r}{2}\), which is rational.
2. But \(\sqrt{3}\) is irrational. Contradiction. So \(2\sqrt{3}\) is irrational.

Teacher's Note:
1. Multiplying an irrational number by a non-zero integer keeps it irrational.
2. Proof by contradiction is the standard method for these questions.

Answer: (a) always irrational

Explanation:
1. Example: \(\frac{3}{4} \times \sqrt{2} = \frac{3\sqrt{2}}{4}\), which is irrational.
2. In general, a non-zero rational times an irrational is irrational.

Teacher's Note:
1. Test with one example, then argue in general.
2. Zero is the only exception, and it is excluded by the question.

Answer: (b) \(xy^2\)

Explanation:
1. HCF takes the smallest power of each common prime.
2. For \(x\): \(\min(3, 1) = 1\). For \(y\): \(\min(2, 3) = 2\).
3. HCF \(= xy^2\).

Teacher's Note:
1. HCF uses the smallest power of each common prime.
2. LCM uses the largest power, so do not mix the two rules.

Answer: (c) \(a^3b^2\)

Explanation:
1. LCM takes the highest power of each prime.
2. For \(a\): \(\max(1, 3) = 3\). For \(b\): \(\max(2, 1) = 2\).
3. LCM \(= a^3b^2\).

Teacher's Note:
1. Take the larger exponent for each variable in an LCM.
2. Option (d) has the wrong power of \(b\).

Answer: (d) 2520

Explanation:
1. LCM \((1, 2, \ldots, 10) = 2^3 \times 3^2 \times 5 \times 7 = 2520\).

Teacher's Note:
1. Each prime power that appears up to 10 must be included once at its highest power.
2. 504 is too small: 504 is not divisible by 9.

Answer: (a) an even number

Explanation:
1. Take \(p_1 = 5\), \(p_2 = 3\): \(p_1^2 - p_2^2 = 25 - 9 = 16\), which is even.
2. In general, odd² − odd² is even, since both squares are odd.

Teacher's Note:
1. The difference of two odd numbers is even.
2. One example is enough to rule out options (b), (c) and (d).

Answer: (d) 2, 3 and 13

Explanation:
1. \(3^{13} - 3^{10} = 3^{10}(3^3 - 1) = 3^{10} \times 26 = 2 \times 13 \times 3^{10}\).
2. So it is divisible by 2, 3 and 13.

Teacher's Note:
1. Take out the common factor \(3^{10}\) first.
2. Then factorise the remaining number, 26.

Answer: (d) 4

Explanation:
1. LCM × HCF = product of the two numbers.
2. \(x = \frac{36 \times 2}{18} = 4\).

Teacher's Note:
1. The same product rule as in the two-number HCF and LCM questions applies here.
2. Check: LCM(4, 18) = 36 and HCF(4, 18) = 2.

Answer: (b) 2

Explanation:
1. The highest power of 3 in the LCM is \(3^2\).
2. Only \(c\) can contribute \(3^2\), so \(n = 2\).

Teacher's Note:
1. Find which number supplies the highest power of each prime.
2. Here \(c\) is the only number that can give \(3^2\).

Answer: (d) 3600

Explanation:
1. LCM \((16, 20, 24) = 2^4 \times 3 \times 5 = 240\).
2. To make it a perfect square, use \(3^2\) and \(5^2\): \(2^4 \times 3^2 \times 5^2 = 3600\).

Teacher's Note:
1. Every exponent in a perfect square must be even.
2. 240 is not a perfect square, and 1600 is not divisible by 3.

Answer: (c) A is true, but R is false.

Explanation:
1. \(\text{LCM} \times \text{HCF} = a \times b\), so \(\text{LCM} \times 5 = 150\), giving \(\text{LCM} = 30\). Assertion (A) is true.
2. The reason is wrong: the rule uses a product, \(\text{HCF} \times \text{LCM} = a \times b\), not a sum.

Teacher's Note:
1. Reason (R) can be false even when it looks like a familiar rule.
2. Check the operation carefully: product, not sum.