CBSE Class 9 Maths Advanced Number System MCQs Set 03

Answer: (d) \( q^{\frac{1}{q-1}} \)

Explanation:
1. (x^p)^q = x^pq, so p^q = pq.
2. For p ≠ 0, divide by p: p^q−1 = q.
3. So p = \( q^{\frac{1}{q-1}} \).

Teacher's Note:
1. Equate the exponents of x.
2. Divide by p (taking p non-zero).

Answer: (b) \( 1 \)

Explanation:
1. As in the earlier question, a^xyz = a, so xyz = 1.
2. Then x²y²z² = (xyz)² = 1.

Teacher's Note:
1. First find xyz.
2. Squaring 1 gives 1.

Answer: (c) 7

Explanation:
1. 28 = 2² × 7, 35 = 5 × 7, 91 = 7 × 13.
2. The only common prime factor is 7, so the H.C.F. is 7.

Teacher's Note:
1. Factorise all three numbers.
2. The H.C.F. uses only the common primes.

Answer: (a) 15 minute

Explanation:
1. 1 hour = 60 minutes.
2. 1/4 of 60 minutes = 15 minutes.

Teacher's Note:
1. The question needs 'of an hour' to make sense.
2. Quarter of 60 is 15.

Answer: (d) \( \sqrt[3]{9} \)

Explanation:
1. 2 = ∛8 and 2.5 = ∛15.625, so a cube root of a number between 8 and 15.625 works.
2. ∛9 lies in that range. Also √11 ≈ 3.32, √8 ≈ 2.83 and ∛7 ≈ 1.91 are outside (2, 2.5).
3. The answer is \( \sqrt[3]{9} \) (≈ 2.08).

Teacher's Note:
1. Write 2 and 2.5 as cube roots.
2. Compare decimal values of the options.

Answer: (c) \( (17)^{105} \)

Explanation:
1. Write each as a power with the same exponent 105: 3²¹⁰ = 9¹⁰⁵; 7¹⁴⁰ = (7^4/3)¹⁰⁵ ≈ 13.1¹⁰⁵; 31⁸⁴ = (31^4/5)¹⁰⁵ ≈ 15.6¹⁰⁵.
2. 17¹⁰⁵ has the largest base, so (17)¹⁰⁵ is largest.
3. Alternatively, 17¹⁰⁵ > 16¹⁰⁵ = 2⁴²⁰ while 3²¹⁰ < 4²¹⁰, 7¹⁴⁰ < 8¹⁴⁰ and 31⁸⁴ < 32⁸⁴ are all below 2⁴²⁰.

Teacher's Note:
1. Compare with powers of 2: 4²¹⁰, 8¹⁴⁰, 32⁸⁴ are all 2⁴²⁰.
2. Only 17¹⁰⁵ exceeds 16¹⁰⁵ = 2⁴²⁰.

Answer: (c) \( \sqrt[n]{\frac{a^{n-1}}{b^{n-1}}} \)

Explanation:
1. The R.F. of (a/b)^1/n is (a/b)^{1 − 1/n}, because their product is (a/b)¹, which is rational.
2. (a/b)^(n−1)/n = ⁿ√((a/b)ⁿ⁻¹) = \( \sqrt[n]{\frac{a^{n-1}}{b^{n-1}}} \).

Teacher's Note:
1. The factor must raise the total exponent to 1.
2. 1 − 1/n = (n − 1)/n.

Answer: (d) 3

Explanation:
1. The number is n = 6k + 3.
2. n² = 36k² + 36k + 9 = 6(6k² + 6k + 1) + 3.
3. The remainder is 3.

Teacher's Note:
1. Write the number as 6k + 3 and square it.
2. Check with n = 3: 9 = 6 + 3.

Answer: (c) 34

Explanation:
1. 1/x = 1/(3 + √8) = 3 − √8, so x + 1/x = 6.
2. x² + 1/x² = (x + 1/x)² − 2 = 36 − 2 = 34.

Teacher's Note:
1. Find x + 1/x first.
2. Then use (a + b)² − 2ab.

Answer: (c) 2

Explanation:
1. Write (b/a)^x−3 as (a/b)^3−x.
2. Then x − 1 = 3 − x, so 2x = 4 and x = 2.

Teacher's Note:
1. Make both sides use the base a/b.
2. Equal bases mean equal exponents.

Answer: (b) \( 3^{1/3} \)

Explanation:
1. ³√72 = ³√(2³ × 3²) = 2 × 3^2/3.
2. Multiplying by 3^1/3 gives 2 × 3 = 6, a rational number.
3. The simplest factor is 3^1/3.

Teacher's Note:
1. Take out the perfect cube 8.
2. Complete 3^{2/3} to 3^1.

Answer: (c) 1

Explanation:
1. The expression is x^(b−c)/(bc) × x^(c−a)/(ca) × x^(a−b)/(ab).
2. The exponents are 1/c − 1/b, 1/a − 1/c and 1/b − 1/a, which add to 0.
3. So the value is x⁰ = 1.

Teacher's Note:
1. Split each exponent into two fractions.
2. The three differences cancel in pairs.

Answer: (a) \( \sqrt2 \)

Explanation:
1. y = 1/x = 2 − √3. Note x = ((√3 + 1)/√2)² and y = ((√3 − 1)/√2)², so √x = (√3 + 1)/√2 and √y = (√3 − 1)/√2.
2. √2 + √x = (2 + √3 + 1)/√2 = (3 + √3)/√2 and √2 − √y = (2 − √3 + 1)/√2 = (3 − √3)/√2.
3. x/(√2 + √x) + y/(√2 − √y) = √2 [(2 + √3)/(3 + √3) + (2 − √3)/(3 − √3)] = √2 × [(2 + √3)(3 − √3) + (2 − √3)(3 + √3)]/6 = √2 × 6/6 = \( \sqrt2 \).

Teacher's Note:
1. Show that x and y are perfect squares of surds.
2. Check numerically: the value is about 1.4142.

Answer: (b) \( 18 \)

Explanation:
1. Common denominator (√5)² − 2² = 1.
2. Numerator: (√5 − 2)² + (√5 + 2)² = 2(5 + 4) = 18.
3. The value is 18.

Teacher's Note:
1. The two fractions are reciprocals of each other.
2. Cross terms cancel when you add the squares.

Answer: (d) 60

Explanation:
1. Product of the two numbers = L.C.M. × H.C.F. = 2520 × 12.
2. Other number = 2520 × 12 ÷ 504 = 60.

Teacher's Note:
1. Use the relation N₁ × N₂ = L.C.M. × H.C.F.
2. Check: H.C.F. of 504 and 60 is 12.

Answer: (b) \( 1\frac35 \)

Explanation:
1. The opposite (additive inverse) of −5/8 is 5/8.
2. The inverse (reciprocal) of 5/8 is 8/5 = \( 1\frac35 \).

Teacher's Note:
1. Do the 'opposite' first, then the 'inverse'.
2. Opposite changes the sign, inverse flips the fraction.

Answer: (a) 7

Explanation:
1. ⁴√16 = 2 and ⁴√625 = 5.
2. x = 2 + 5 = 7.

Teacher's Note:
1. 16 = 2⁴ and 625 = 5⁴.
2. Add the two roots.

Answer: (d) \( \frac{1}{\sqrt2} \)

Explanation:
1. Let a = √2 − √5 and b = √3. The two denominators are a + b and a − b.
2. Sum = (a − b + a + b)/(a² − b²) = 2a/(a² − b²).
3. a² − b² = 2 + 5 − 2√10 − 3 = 4 − 2√10 = 2(2 − √10), so the sum = (√2 − √5)/(2 − √10).
4. Since 2 − √10 = √2(√2 − √5), the sum is \( \frac{1}{\sqrt2} \).

Teacher's Note:
1. Group √2 − √5 together as one term.
2. Numerically the value is about 0.7071.

Answer: (b) 492

Explanation:
1. 1 to 9: 9 numbers × 1 digit = 9 key presses.
2. 10 to 99: 90 numbers × 2 digits = 180.
3. 100 to 199: 100 numbers × 3 digits = 300, and 200 needs 3 more.
4. Total = 9 + 180 + 300 + 3 = 492.

Teacher's Note:
1. Count by number of digits.
2. Do not forget the number 200 itself.

Answer: (c) \( \sqrt[3]{12} \)

Explanation:
1. L.C.M. of 6, 3, 2 is 6.
2. ⁶√100, ³√12 = ⁶√144 and √3 = ⁶√27.
3. 144 is the largest radicand, so \( \sqrt[3]{12} \) is greatest.

Teacher's Note:
1. Bring all three to the sixth root.
2. Compare radicands 100, 144 and 27.

Answer: (b) \( 3^{200} \)

Explanation:
1. Write each as a 50th power: 2²⁵⁰ = 32⁵⁰, 3²⁰⁰ = 81⁵⁰, 4¹⁵⁰ = 64⁵⁰, 5¹⁰⁰ = 25⁵⁰.
2. The greatest base is 81, so 3²⁰⁰ is greatest.

Teacher's Note:
1. Find the H.C.F. of the exponents (50) and take 50th powers.
2. Then compare the bases 32, 81, 64, 25.

Answer: (d) \( \frac{1}{\sqrt2} \)

Explanation:
1. Take a = √2 − √5 and b = √3: the sum is 1/(a + b) + 1/(a − b) = 2a/(a² − b²).
2. a² − b² = (7 − 2√10) − 3 = 2(2 − √10) = 2√2(√2 − √5).
3. So the sum = 2(√2 − √5)/(2√2(√2 − √5)) = \( \frac{1}{\sqrt2} \).

Teacher's Note:
1. This is the same expression as an earlier question; the answer is unchanged.
2. The value is about 0.707.

Answer: (b) \( 1-\frac17 \)

Explanation:
1. 1 − 127/343 = 216/343.
2. The cube root of 216/343 is 6/7 = 1 − 1/7.
3. The answer is \( 1-\frac17 \).

Teacher's Note:
1. Subtract first: 343 − 127 = 216.
2. 216 = 6³ and 343 = 7³.

Answer: (a) \( 0.\overline6 \)

Explanation:
1. 2.666... = 8/3 and 1.999... = 2.
2. The difference is 8/3 − 2 = 2/3 = \( 0.\overline6 \).

Teacher's Note:
1. Note that 1.999... equals exactly 2.
2. Convert the other decimal to a fraction.

Answer: (a) \( \frac{20\sqrt5-35}{31} \)

Explanation:
1. Multiply by (7 − 4√5): numerator 5(7 − 4√5) = 35 − 20√5; denominator 49 − 80 = −31.
2. The value is (35 − 20√5)/(−31) = \( \frac{20\sqrt5-35}{31} \).

Teacher's Note:
1. Denominator: 7² − (4√5)² = 49 − 80.
2. Moving the minus sign flips the numerator.

Answer: (a) 16, 24, 5, 80

Explanation:
1. Let the common value be k. The numbers are k − 4, k + 4, k/4 and 4k.
2. Sum: k − 4 + k + 4 + k/4 + 4k = 6.25k = 125, so k = 20.
3. The integers are 16, 24, 5, 80, which is option (a).

Teacher's Note:
1. Introduce one unknown for the final equal value.
2. Check: 16 + 4 = 24 − 4 = 5 × 4 = 80 ÷ 4 = 20.

Answer: (c) 39

Explanation:
1. a² = 11 + 2√30, b² = 11 − 2√30 and ab = 6 − 5 = 1.
2. 2a² + 2b² = 2(22) = 44, and 5ab = 5.
3. The value is 44 − 5 = 39.

Teacher's Note:
1. ab = 1 simplifies things.
2. The √30 terms cancel in a² + b².

Answer: (b) 3

Explanation:
1. √(m⁴n⁴) = m²n² = (mn)².
2. ⁶√(m²n²) = (mn)^1/3 and ³√(m²n²) = (mn)^2/3.
3. Product = (mn)^{2 + 1/3 + 2/3} = (mn)³, so k = 3.

Teacher's Note:
1. Write every root as a power of mn.
2. Add the exponents.