CBSE Class 9 Maths Advanced Number System MCQs Set 01

Answer: (a) Terminating

Explanation:
1. 400 = 2⁴ × 5², so \( \frac{27}{400} \) = \( \frac{27}{2^4\times5^2} \).
2. The denominator has only the prime factors 2 and 5, so the decimal stops: 27/400 = 0.0675.
3. The representation is terminating.

Teacher's Note:
1. Factorise the denominator of a fraction in lowest terms.
2. Only 2s and 5s in the denominator give a terminating decimal.

Answer: (c) Non-terminating and repeating

Explanation:
1. The bar shows that the digits 34 repeat for ever, so the decimal never ends.
2. A decimal that never ends but repeats a block is non-terminating and repeating (it is rational).
3. The answer is non-terminating and repeating.

Teacher's Note:
1. A bar over digits means they repeat.
2. Repeating decimals always represent rational numbers.

Answer: (d) Infinite

Explanation:
1. The mean of two rational numbers is rational and lies between them.
2. Taking means of the new pairs again and again never stops.
3. So there are infinitely many rational numbers between any two.

Teacher's Note:
1. There is no 'next' rational number after a given one.
2. Repeated averaging is the quickest argument.

Answer: (c) \( \frac{27161}{9900} \)

Explanation:
1. Let x = 2.74353535... Two digits repeat after two non-repeating digits.
2. 10000x = 27435.3535... and 100x = 274.3535...; subtract: 9900x = 27161.
3. x = \( \frac{27161}{9900} \).

Teacher's Note:
1. Direct method: numerator 27435 − 274 = 27161; denominator 9900 (two 9s, then two 0s).
2. Check: 27161 ÷ 9900 = 2.74353535...

Answer: (c) \( \sqrt5+3 \)

Explanation:
1. √16 − 4 = 4 − 4 = 0, rational.
2. (3 − √3)(3 + √3) = 9 − 3 = 6, rational.
3. √5 + 3 is a rational plus an irrational, so it is irrational.
4. −√25 = −5, rational.
5. The irrational number is \( \sqrt5+3 \).

Teacher's Note:
1. Simplify each option before judging it.
2. Rational ± irrational is always irrational.

Answer: (d) Non-terminating non recurring

Explanation:
1. √2 is irrational.
2. The decimal form of an irrational number neither ends nor repeats.
3. So it is non-terminating non-recurring (1.41421356...).

Teacher's Note:
1. 1.4121 is only a rough value of √2, not its expansion.
2. Irrational means non-terminating and non-repeating.

Answer: (a) Irrational number

Explanation:
1. If r (non-zero rational) × s were rational, then s = (r × s) ÷ r would be rational, which is false.
2. So the product is an irrational number.

Teacher's Note:
1. Example: 2 × √2 = 2√2 is irrational.
2. The rational number must be non-zero, as 0 × √2 = 0.

Answer: (b) irrational

Explanation:
1. Multiply top and bottom by 3 − √3: \( \frac{(3-\sqrt3)^2}{9-3} = \frac{12-6\sqrt3}{6} = 2-\sqrt3 \).
2. 2 − √3 is a rational number minus an irrational number, so it is irrational.

Teacher's Note:
1. Rationalise first, then examine the result.
2. A number cannot be both rational and irrational.

Answer: (b) A real number

Explanation:
1. Every real number has exactly one point on the number line and every point corresponds to a real number.
2. Natural, rational or irrational numbers cover only some points.
3. So the answer is a real number.

Teacher's Note:
1. The number line is also called the real line.
2. Rational and irrational numbers together make up the reals.

Answer: (c) \( 1.6 \)

Explanation:
1. √2 ≈ 1.414 and √3 ≈ 1.732.
2. (√2 + √3)/2 and √6 lie in between but are irrational. 1.9 is greater than √3.
3. 1.6 is rational and 1.414 < 1.6 < 1.732, so the answer is 1.6.

Teacher's Note:
1. Look for the option that is both rational and in the interval.
2. Use decimal values of the roots.

Answer: (b) 3.146

Explanation:
1. Rationalise: \( \frac{1}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2} = \sqrt3+\sqrt2 \).
2. Substitute: 1.732 + 1.414 = 3.146.

Teacher's Note:
1. The denominator becomes 3 − 2 = 1 after rationalising.
2. Do not subtract the decimals in the denominator directly.

Answer: (c) \( 4 \)

Explanation:
1. 1/x = 1/(2 + √3) = (2 − √3)/(4 − 3) = 2 − √3.
2. x + 1/x = (2 + √3) + (2 − √3) = 4.

Teacher's Note:
1. The irrational parts cancel.
2. x and 1/x are conjugates here.

Answer: (c) \( 6^{1/4} \)

Explanation:
1. √2 × √3 = √6 = 6^1/2.
2. The outer square root gives (6^1/2)^1/2 = 6^1/4.

Teacher's Note:
1. Square root means power 1/2.
2. Powers of powers multiply: 1/2 × 1/2 = 1/4.

Answer: (b) \( \sqrt2 \)

Explanation:
1. √50 = 5√2.
2. Multiplying by √2 gives 5 × 2 = 10, a rational number.
3. So the simplest factor is \( \sqrt2 \).

Teacher's Note:
1. Simplify the surd first.
2. The simplest factor has no unnecessary square factors.

Answer: (b) \( 34 \)

Explanation:
1. xy = 9 − 8 = 1, so 1/x = y and 1/y = x.
2. 1/x² + 1/y² = y² + x² = (3 − √8)² + (3 + √8)² = 2(9 + 8) = 34.

Teacher's Note:
1. Notice that x and y are reciprocals.
2. The cross terms with √8 cancel.

Answer: (d) \( (8, 3) \)

Explanation:
1. Multiply by 3 + √7: numerator (3 + √7)² = 16 + 6√7, denominator 9 − 7 = 2.
2. The value is 8 + 3√7, so a = 8 and b = 3.
3. (a, b) = (8, 3).

Teacher's Note:
1. Rationalise, then compare rational and irrational parts.
2. The denominator 3² − (√7)² = 2.

Answer: (c) \( \sqrt[3]{4} \)

Explanation:
1. Take the L.C.M. of the orders (2 and 3) = 6.
2. √2 = 2^3/6 = ⁶√8; ∛3 = ⁶√9; ∛4 = ⁶√16; ∛2 = ⁶√4.
3. The greatest radicand is 16, so \( \sqrt[3]{4} \) is greatest.

Teacher's Note:
1. Bring all surds to a common order before comparing.
2. Larger radicand means larger value when the order is the same.

Answer: (a) 1/8

Explanation:
1. 32 = 2⁵, so (32)⁻³ = 2⁻¹⁵.
2. The fifth root gives 2⁻³ = 1/8.

Teacher's Note:
1. Write 32 as a power of 2.
2. Multiply the exponents: −15 × 1/5 = −3.

Answer: (a) 99

Explanation:
1. x = 5 − 2√6 and y = 5 + 2√6, so xy = 25 − 24 = 1.
2. x² + y² = (x + y)² − 2xy = 10² − 2 = 98.
3. x² + xy + y² = 98 + 1 = 99.

Teacher's Note:
1. Rationalising shows that x and y are reciprocals.
2. Use (x + y)² − 2xy for x² + y².

Answer: (b) 0

Explanation:
1. 2/(√5 + √3) = 2(√5 − √3)/2 = √5 − √3.
2. 1/(√3 + √2) = √3 − √2.
3. 3/(√5 + √2) = 3(√5 − √2)/3 = √5 − √2.
4. Sum: (√5 − √3) + (√3 − √2) − (√5 − √2) = 0.

Teacher's Note:
1. Rationalise each fraction separately.
2. Most terms cancel in the end.

Answer: (b) \( \sqrt[5]{4} \)

Explanation:
1. Use L.C.M. of 4, 5, 2, 2 = 20.
2. ⁴√5 = ²⁰√(5⁵) = ²⁰√3125; ⁵√4 = ²⁰√(4⁴) = ²⁰√256; √4 = ²⁰√(4¹⁰); √3 = ²⁰√(3¹⁰) = ²⁰√59049.
3. 256 is the smallest radicand, so \( \sqrt[5]{4} \) is smallest.

Teacher's Note:
1. Convert all four surds to the same order 20.
2. √4 is just 2.

Answer: (b) \( \sqrt[6]{675} \)

Explanation:
1. 3^1/2 × 5^1/3 = (3³)^1/6 × (5²)^1/6.
2. = (27 × 25)^1/6 = \( \sqrt[6]{675} \).

Teacher's Note:
1. Make the order 6, the L.C.M. of 2 and 3.
2. 27 × 25 = 675.

Answer: (a) 10

Explanation:
1. √20 × √5 = √(20 × 5) = √100 = 10.

Teacher's Note:
1. Multiply under one root, then take the root.
2. 100 is a perfect square.

Answer: (b) \( \frac{3}{5} \)

Explanation:
1. 54/250 = 27/125.
2. The cube root of 27/125 is \( \frac{3}{5} \).

Teacher's Note:
1. Reduce the fraction first (divide by 2).
2. 27 = 3³ and 125 = 5³.

Answer: (d) 5

Explanation:
1. 225 = 25 × 9 = 5² × 3².
2. Compare exponents: x − 3 = 2 and 2x − 8 = 2, both give x = 5.

Teacher's Note:
1. Prime factorise 225 and compare the exponents of 5 and 3.
2. Check both equations give the same x.

Answer: (b) x = −3

Explanation:
1. 81² = 3⁸ and 6561 = 3⁸.
2. The left side is 3^{5x + 8 + 8 − 2x} = 3^{3x + 16}.
3. Equate to 3⁷: 3x + 16 = 7, so x = −3.

Teacher's Note:
1. Write 81 and 6561 as powers of 3.
2. Subtract the exponent in the denominator.

Answer: (b) 27

Explanation:
1. Write everything with base 3: 3ⁿ × 3²ⁿ × 3^{3(1 − n)} = 3^{n + 2n + 3 − 3n}.
2. The exponent is 3, so the value is 3³ = 27.

Teacher's Note:
1. Convert 9 and 27 to powers of 3.
2. The n terms cancel completely.

Answer: (d) \( None \)

Explanation:
1. 3⁻¹ + 5⁻¹ + 2⁻¹ = 1/3 + 1/5 + 1/2 = (10 + 6 + 15)/30 = 31/30.
2. The power −1 gives the reciprocal: (31/30)⁻¹ = 30/31.
3. 30/31 is not among options (a), (b), (c), so the answer is None.

Teacher's Note:
1. The outer −1 power turns the sum into its reciprocal.
2. Do not stop at 31/30; that is only the sum inside the bracket.

Answer: (d) 6 : 3 : 2

Explanation:
1. Write all as powers of 2: 2^x = 2^2y = 2^3z, so x = 2y = 3z.
2. Let x = 6k: y = 3k and z = 2k.
3. So x : y : z = 6 : 3 : 2.

Teacher's Note:
1. Equal powers of the same base have equal exponents.
2. Choose a common multiple such as 6 for x.