Match List-I with List-II.
| List-I (Recurring decimal) | List-II (Fraction) | ||
|---|---|---|---|
| (A) | \(0.\overline{3}\) | (I) | \(\frac{5}{11}\) |
| (B) | \(0.1\overline{6}\) | (II) | \(\frac{1}{6}\) |
| (C) | \(0.\overline{27}\) | (III) | \(\frac{1}{3}\) |
| (D) | \(0.\overline{45}\) | (IV) | \(\frac{3}{11}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
A pure recurring decimal takes the repeating block over as many 9s as it has digits (\(0.\overline{ab}=\displaystyle\frac{ab}{99}\)); for \(0.a\overline{b}\), use \(\displaystyle\frac{ab-a}{90}\).
(A) \(0.\overline{3}\) — this matches (III) \(\frac{1}{3}\). \(\displaystyle\frac{3}{9}=\displaystyle\frac{1}{3}\)
(B) \(0.1\overline{6}\) — this matches (II) \(\frac{1}{6}\). \(\displaystyle\frac{16-1}{90}=\displaystyle\frac{15}{90}=\displaystyle\frac{1}{6}\)
(C) \(0.\overline{27}\) — this matches (IV) \(\frac{3}{11}\). \(\displaystyle\frac{27}{99}=\displaystyle\frac{3}{11}\)
(D) \(0.\overline{45}\) — this matches (I) \(\frac{5}{11}\). \(\displaystyle\frac{45}{99}=\displaystyle\frac{5}{11}\)