CBSE Class 9 Maths Advanced Number System MCQs Set 02

Answer: (a) 1

Explanation:
1. 6/7 = \( 0.\overline{857142} \); six digits repeat.
2. 100 = 6 × 16 + 4, so the 100th digit is the 4th digit of the block 857142.
3. The 4th digit is 1.

Teacher's Note:
1. Find the length of the repeating block, then use the remainder on division.
2. Remainder 4 means the 4th digit of the block.

Answer: (c) 99

Explanation:
1. A two-digit block xy repeating gives 0.\(\overline{xy}\) = xy/99.
2. So xy/ab = xy/99 and ab = 99.

Teacher's Note:
1. Use the direct method: two repeating digits mean denominator 99.
2. Compare denominators.

Answer: (d) 16

Explanation:
1. 0.4545... = 45/99 = 5/11 in lowest terms.
2. The sum 5 + 11 = 16.

Teacher's Note:
1. Reduce the fraction by the H.C.F. 9.
2. Add only after reducing.

Answer: (c) 7

Explanation:
1. The root is real when 9 − (n − 2)² ≥ 0, i.e. |n − 2| ≤ 3.
2. So −1 ≤ n ≤ 5, giving n = −1, 0, 1, 2, 3, 4, 5, which is 7 values.

Teacher's Note:
1. A square root of a negative number is not real.
2. List the integers in the interval.

Answer: (b) 7

Explanation:
1. 1 + 2 + 3 + 4 = 10, so we need √(10 + x) to be an integer.
2. 10 + x must be a perfect square between 11 and 109: 16, 25, 36, 49, 64, 81, 100.
3. These give x = 6, 15, 26, 39, 54, 71, 90, a total of 7.

Teacher's Note:
1. List perfect squares in the allowed range.
2. Subtract 10 from each.

Answer: (b) 0.01234

Explanation:
1. √625 = 25, 704 + 25 = 729, √729 = 27; √484 = 22; 22 + 27 = 49 and √49 = 7.
2. The numerator is √(954 + 7) = √961 = 31.
3. So 31/0.00155 = 0.02/(0.000081 x), giving x = 0.02 × 0.00155/(0.000081 × 31) ≈ 0.01234.

Teacher's Note:
1. Work out the nested root from the inside outwards.
2. Keep track of the powers of ten carefully.

Answer: (c) \( n+2\sqrt n+1 \)

Explanation:
1. Let n = a². The next perfect square is (a + 1)² = a² + 2a + 1.
2. Since a = √n, this equals \( n+2\sqrt n+1 \).

Teacher's Note:
1. Write n as a² and add 1 to a.
2. Test with n = 4: next square 9 = 4 + 4 + 1.

Answer: (a) \( 2-\sqrt2 \)

Explanation:
1. Let y = √(x + √(x + ...)). Then y² = x + y and the given equation says x + y = 2.
2. So y² = 2 and y = √2 (positive).
3. x = 2 − y = \( 2-\sqrt2 \).

Teacher's Note:
1. The part inside the root repeats itself.
2. The square root is taken as positive.

Answer: (d) 9

Explanation:
1. A two-digit multiple of 13 beginning with 9 is only 91, so the next digit is 1.
2. Multiples beginning with 1: 13, so the next digit is 3; beginning with 3: 39, so the next is 9; beginning with 9: 91 again.
3. N = 913913913... repeats in blocks of 3 digits. 100 = 3 × 33 + 1, so the last digit is the first digit of the block, 9.

Teacher's Note:
1. Each pair of consecutive digits is a multiple of 13 (like 26, 65, 52).
2. Look for the repeating block and use the remainder.

Answer: (c) \( \frac{1}{2008} \)

Explanation:
1. Let x = 1/2 + ... + 1/2007 and y = 1 + 1/2 + ... + 1/2007, so y − x = 1.
2. The expression is (x + 1/2008)y − (y + 1/2008)x = xy + y/2008 − xy − x/2008 = (y − x)/2008.
3. Hence it equals \( \frac{1}{2008} \).

Teacher's Note:
1. Name the two repeated sums x and y.
2. The product terms cancel.

Answer: (a) 70

Explanation:
1. Let A = 1/(1/x + 1/2) = 2x/(2 + x).
2. Then 1/(A + A) = (2 + x)/(4x), and the two such terms in the big denominator add to (2 + x)/(2x).
3. So the whole expression is 2x/(2 + x) = x/36.
4. Cross-multiplying: 72 = 2 + x, so x = 70.

Teacher's Note:
1. Simplify from the innermost fraction outwards.
2. Cross-multiply at the end and cancel x.

Answer: (b) \( \left(\frac2n\right)^2 \)

Explanation:
1. The factors are (2/3)², (3/4)², (4/5)², ..., ((n − 1)/n)².
2. Everything cancels in the product except the first numerator and the last denominator: (2/n)².
3. The value is \( \left(\frac2n\right)^2 \).

Teacher's Note:
1. This is a telescoping product.
2. Test with n = 4: (2/3 × 3/4)² = (1/2)² = (2/4)².

Answer: (d) 4

Explanation:
1. 7 + 4√3 = (2 + √3)², so √x = 2 + √3.
2. 1/√x = 1/(2 + √3) = 2 − √3.
3. √x + 1/√x = (2 + √3) + (2 − √3) = 4.

Teacher's Note:
1. Spot x as a perfect square of a surd: 4 + 3 + 2·2·√3.
2. Then rationalise the reciprocal.

Answer: (a) 99

Explanation:
1. x = (√3 − √2)²/(3 − 2) = 5 − 2√6 and y = 5 + 2√6.
2. x² = 49 − 20√6, y² = 49 + 20√6 and xy = 1.
3. x² + xy + y² = 49 + 1 + 49 = 99.

Teacher's Note:
1. Rationalise x and y; the root terms cancel in x² + y².
2. xy = 1 because they are reciprocals.

Answer: (b) 3

Explanation:
1. Rationalise each term: 1/(√(k+1) + √k) = √(k+1) − √k.
2. The sum telescopes: (√2 − 1) + (√3 − √2) + ... + (√100 − √99) = √100 − 1 = 9.
3. The square root of 9 is 3.

Teacher's Note:
1. Every term is a difference of consecutive square roots.
2. Only the first and last terms survive.

Answer: (b) \( \sqrt[4]{27} \)

Explanation:
1. ⁴√48 = ⁴√(16 × 3) = 2 ⁴√3.
2. Multiplying ⁴√3 by ⁴√27 gives ⁴√81 = 3, which is rational.
3. The simplest factor is \( \sqrt[4]{27} \).

Teacher's Note:
1. Remove the perfect fourth power 16 first.
2. You need the missing power of 3 to make 3⁴.

Answer: (b) \( s\le0 \)

Explanation:
1. Write in primes: numerator 2^(r+s)3^(r+s) × 2^2(r−s)3^(r−s); denominator 2^3r × 3^2r+4s.
2. The exponent of 2 is (r + s) + 2r − 2s − 3r = −s; the exponent of 3 is (r + s) + (r − s) − 2r − 4s = −4s.
3. The value is 2^−s × 3^−4s, an integer exactly when s ≤ 0.

Teacher's Note:
1. Reduce to prime powers.
2. r cancels completely, so only s matters.

Answer: (a) \( c < a < b \)

Explanation:
1. 2^a > 2^2c gives a > 2c, so a > c (c is positive).
2. 3^b > 3^2a gives b > 2a, so b > a.
3. Hence c < a < b.

Teacher's Note:
1. Make the bases equal: 4 = 2², 9 = 3².
2. For bases larger than 1, the bigger exponent gives the bigger power.

Answer: (b) 3

Explanation:
1. Take 2²⁰⁰⁵ common: 2²⁰⁰⁵(2³ − 2² − 2 + 1).
2. The bracket is 8 − 4 − 2 + 1 = 3.
3. So k = 3.

Teacher's Note:
1. Factor out the smallest power.
2. 2²⁰⁰⁸ = 2³ × 2²⁰⁰⁵.

Answer: (a) \( 0 \)

Explanation:
1. Let 2^x = 3^y = 6^−z = k. Then 2 = k^1/x, 3 = k^1/y and 6 = k^−1/z.
2. Since 2 × 3 = 6: k^{1/x + 1/y} = k^−1/z, so 1/x + 1/y = −1/z.
3. Therefore 1/x + 1/y + 1/z = 0.

Teacher's Note:
1. Express 2, 3 and 6 as powers of k.
2. 6 = 2 × 3 links the three exponents.

Answer: (c) \( 25\sqrt5 \)

Explanation:
1. 4^x(1 − 1/4) = 24, so 4^x × 3/4 = 24 and 4^x = 32.
2. 2^2x = 2⁵ gives 2x = 5, x = 5/2.
3. (2x)^x = 5^5/2 = 25√5, i.e. \( 25\sqrt5 \).

Teacher's Note:
1. Take 4^{x−1} common.
2. Note that 2x = 5, which is neat.

Answer: (c) \( 2^{71} \)

Explanation:
1. Take 2⁷¹ common: 2⁷¹(2² − 2 − 1).
2. The bracket is 4 − 2 − 1 = 1.
3. The value is 2⁷¹.

Teacher's Note:
1. Factor out the smallest power of 2.
2. Check with small exponents: 2³ − 2² − 2¹ = 2¹.

Answer: (b) \( \frac1x \)

Explanation:
1. x = −1/2.
2. (a) 2^1/x = 2⁻² = 0.25; (b) 1/x = −2; (c) 2^x = 2^−1/2 = 1/√2 ≈ 0.707; (d) 1/√(1/2) = √2 ≈ 1.414.
3. The smallest value is −2, option (b).

Teacher's Note:
1. Evaluate every option at x = −1/2.
2. A negative number is smaller than all the positive ones.

Answer: (c) \( \frac{7}{48} \)

Explanation:
1. 2^x = 2^2y = 2^3z, so x = 2y = 3z. Then x = 3z and y = 3z/2.
2. 1/(2x) = 1/(6z), 1/(4y) = 1/(6z), so the left side is 3 × 1/(6z) = 1/(2z).
3. 1/(2z) = 24/7 gives z = 7/48, i.e. \( \frac{7}{48} \).

Teacher's Note:
1. Express x and y in terms of z.
2. All three terms become equal to 1/(6z).

Answer: (b) 2

Explanation:
1. (b/a)^x−3 = (a/b)^−(x−3) = (a/b)^3−x.
2. Equal bases give x − 1 = 3 − x, so 2x = 4 and x = 2.

Teacher's Note:
1. Turn the right side into the same base a/b.
2. Then equate the exponents.

Answer: (a) 1

Explanation:
1. (a^x)^y = b^y = c, so a^xy = c.
2. Then (a^xy)^z = c^z = a, so a^xyz = a¹.
3. Hence xyz = 1.

Teacher's Note:
1. Substitute one relation into the next.
2. a^{xyz} = a^1 gives xyz = 1.

Answer: (b) \( \left(-\frac52\right)(\sqrt3+\sqrt5) \)

Explanation:
1. Multiply by (√3 + √5)/(√3 + √5): the denominator is 3 − 5 = −2.
2. The result is 5(√3 + √5)/(−2) = \( -\frac52(\sqrt3+\sqrt5) \).

Teacher's Note:
1. The product (a − b)(a + b) = a² − b² gives −2 here.
2. Watch the negative sign.

Answer: (d) \( 2^{99} \)

Explanation:
1. Subtract exponents: 2¹⁰⁰/2¹ = 2⁹⁹.
2. The value is 2⁹⁹.

Teacher's Note:
1. Dividing by 2 lowers the exponent by 1.
2. Do not halve the exponent.