IMO Maths Olympiad Class 10 Chapter 01 Real Numbers MCQs with Answers

Answer: (b) 2

Explanation:
1. Write n = 9q + 7, where q is the quotient.
2. Then 3n = 27q + 21, so 3n − 1 = 27q + 20 = 9(3q + 2) + 2.
3. So the remainder is 2.

Teacher's Note:
1. Express the number using the division algorithm before operating on it.
2. Rewrite the result as 9 × (something) + remainder with 0 ≤ remainder < 9.

Answer: (c) \( \dfrac{551}{450} \)

Explanation:
1. Let \( x = 1.22\overline{4} \) = 1.22444… …(i)
2. Multiplying (i) by 10: \( 10x = 12.2\overline{4} \) = 12.2444… …(ii)
3. Subtracting (i) from (ii): 9x = 11.02, so \( x = \dfrac{1102}{900} = \dfrac{551}{450} \).

Teacher's Note:
1. Multiply by a power of 10 that lines up the repeating parts.
2. Subtracting then removes the endless tail.

Answer: (c) 1183914

Explanation:
1. 306 = 2 × 3² × 17, 954 = 2 × 3² × 53 and 1314 = 2 × 3² × 73.
2. The HCF is 2 × 3² = 18, and the LCM takes every prime factor with its highest power.
3. LCM = 2 × 3² × 17 × 53 × 73 = 18 × 65773 = 1183914.

Teacher's Note:
1. For three numbers, HCF × LCM is not always equal to the product, so use prime factorisation.
2. Take the highest power of each prime for the LCM.

Answer: (c) \( \sqrt[3]{2},\ \sqrt[6]{3},\ \sqrt[9]{4} \)

Explanation:
1. L.C.M. of 3, 6 and 9 is 18, so write each surd as an 18th root.
2. \( \sqrt[3]{2} = 2^{6/18} = (64)^{1/18} \), \( \sqrt[6]{3} = 3^{3/18} = (27)^{1/18} \), \( \sqrt[9]{4} = 4^{2/18} = (16)^{1/18} \).
3. Since 64 > 27 > 16, the descending order is \( \sqrt[3]{2} > \sqrt[6]{3} > \sqrt[9]{4} \).

Teacher's Note:
1. Surds of different orders are compared by bringing them to the same order.
2. Use the L.C.M. of the orders.

Answer: (c) 2

Explanation:
1. Rationalising, \( \dfrac{1}{\sqrt{k}+\sqrt{k+1}} = \sqrt{k+1}-\sqrt{k} \).
2. The sum becomes (√2 − 1) + (√3 − √2) + (√4 − √3) + … + (√9 − √8).
3. All middle terms cancel, leaving √9 − 1 = 3 − 1 = 2.

Teacher's Note:
1. This is a telescoping sum: each term cancels part of the next.
2. Only the first and last parts survive.

Answer: (b) \( 10^{10^{8}} \)

Explanation:
1. The 100th root means the power \( \dfrac{1}{100} \): \( \left(10^{10^{10}}\right)^{\frac{1}{100}} = 10^{10^{10} \times 10^{-2}} \).
2. \( 10^{10} \times 10^{-2} = 10^{8} \), so the root is \( 10^{10^{8}} \).

Teacher's Note:
1. (a^m)ⁿ = a^mn; here the exponent itself is a power of 10.
2. Dividing 10¹⁰ by 100 gives 10⁸.

Answer: (a) \( \sqrt[7]{z^5y^2x^4} \)

Explanation:
1. We need a factor that makes every power inside the 7th root a multiple of 7.
2. \( \sqrt[7]{x^3y^5z^2} \times \sqrt[7]{x^4y^2z^5} = \sqrt[7]{x^7y^7z^7} = xyz \), which is rational.
3. So the rationalising factor is \( \sqrt[7]{x^4y^2z^5} \), written in option (a) as \( \sqrt[7]{z^5y^2x^4} \).

Teacher's Note:
1. For an nth root, raise each power to the next multiple of n.
2. The order of the factors inside the root does not matter.

Answer: (d) Both (b) and (c)

Explanation:
1. Cancelling common factors, the number is \( \dfrac{1}{3^3 \times 5^3 \times 7} \).
2. The denominator has prime factors 3 and 7 besides 5, so the expansion does not terminate.
3. Being rational, it is non-terminating and repeating, so the answer is both (b) and (c).

Teacher's Note:
1. Simplify before testing the denominator.
2. A rational number never has a non-terminating, non-repeating expansion.

Answer: (c) p divides b

Explanation:
1. The prime factors of b² are exactly the prime factors of b, each taken twice.
2. So if a prime p divides b², p is one of the prime factors of b.
3. Hence p divides b.

Teacher's Note:
1. This result is used to prove that √2, √3 and √5 are irrational.
2. It holds because p is prime.

Answer: (d) 1120

Explanation:
1. The distance must be a common multiple of 35, 32 and 40, and the least one is wanted.
2. 35 = 5 × 7, 32 = 2⁵, 40 = 2³ × 5, so L.C.M. = 2⁵ × 5 × 7 = 1120.
3. Each should walk 1120 cm.

Teacher's Note:
1. ‘Minimum common distance’ points to the LCM.
2. Take the highest power of each prime.

Answer: (d) 136

Explanation:
1. Let the number be x. Then \( \dfrac{x}{8/17} - \dfrac{8x}{17} = 225 \).
2. \( \dfrac{17x}{8} - \dfrac{8x}{17} = \dfrac{289x - 64x}{136} = \dfrac{225x}{136} \).
3. \( \dfrac{225x}{136} = 225 \Rightarrow x = 136 \).

Teacher's Note:
1. Dividing by 8/17 is the same as multiplying by 17/8.
2. Form one equation from the difference given.

Answer: (d) 9 a.m.

Explanation:
1. In minutes the intervals are 30, 60, 90 and 105.
2. L.C.M. (30, 60, 90, 105) = 2² × 3² × 5 × 7 = 1260 minutes = 21 hours.
3. 21 hours after 12 noon is 9 a.m. the next day.

Teacher's Note:
1. Convert all intervals to the same unit first.
2. The devices beep together again after the LCM of the intervals.

Answer: (d) P: Real, Q: Rational, R: Irrational, S: Rational

Explanation:
1. Every point on the number line represents a real number, and every real number is either rational or irrational.
2. Rational numbers are exactly the numbers whose decimal expansions terminate or recur.
3. So P = Real, Q = Rational, R = Irrational and S = Rational.

Teacher's Note:
1. Real numbers = rational numbers + irrational numbers.
2. Irrational numbers are non-terminating and non-repeating.

Answer: (c) (P) → (iii); (Q) → (i); (R) → (iv); (S) → (ii)

Explanation:
1. (P) x = 0.3232…; 100x − x = 32, so \( x = \dfrac{32}{99} \) → (iii).
2. (Q) x = 0.25454…; 100x − x = 25.2, so \( x = \dfrac{252}{990} = \dfrac{14}{55} \) → (i).
3. (R) x = 0.1222…; 10x − x = 1.1, so \( x = \dfrac{11}{90} \) → (iv).
4. (S) x = 0.2444…; 10x − x = 2.2, so \( x = \dfrac{22}{90} = \dfrac{11}{45} \) → (ii).

Teacher's Note:
1. Multiply by 10 for a one-digit repeat and by 100 for a two-digit repeat.
2. Always simplify the final fraction.

Answer: (b) (i) T, (ii) F, (iii) T, (iv) T

Explanation:
1. (i) 25 = 5², so 501/25 = 20.04 terminates: T.
2. (ii) 125 = 5³, so 621/125 = 4.968 terminates: the statement is F.
3. (iii) 6805 = 5 × 1361 is not divisible by 3, so the factor 27 stays in the denominator and the decimal does not terminate: T.
4. (iv) 7105 = 5 × 7² × 29, so the 7 cancels: \( \dfrac{7105}{7 \times 5^2} = \dfrac{1015}{25} = 40.6 \), which terminates: T.

Teacher's Note:
1. Cancel common factors before deciding.
2. A 7 in the denominator can disappear if the numerator is a multiple of 7.