\(0 . \overline{42} \times(0.1 \overline{6} \div 0 . \overline{7})\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(0 . \overline{09}\)
Each recurring decimal is first written as a fraction: a repeating block goes over as many 9s as it has digits, and a digit that does not repeat adds a 0 after the 9s.
First number, a block of two digits over 99: \(\displaystyle 0.\overline{42}=\frac{42}{99}=\frac{14}{33}\)
Second number, 16 less the 1 that does not repeat, over 90: \(\displaystyle 0.1\overline{6}=\frac{16-1}{90}=\frac{15}{90}=\frac{1}{6}\)
Third number: \(\displaystyle 0.\overline{7}=\frac{7}{9}\)
Working out the bracket first:
Division in the bracket: \(\displaystyle \frac{1}{6}\div\frac{7}{9}=\frac{1}{6}\times\frac{9}{7}=\frac{3}{14}\)
Multiplied by the first number: \(\displaystyle \frac{14}{33}\times\frac{3}{14}=\frac{3}{33}=\frac{1}{11}\)
Back to a recurring decimal: \(\displaystyle \frac{1}{11}=\frac{9}{99}=0.\overline{09}\)