Simplify \((0.\overline{1})^2 \times \left\{1-9\times (0.1\overline{6})^2\right\}\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{1}{108}\)
Change each recurring decimal into a fraction first; the rest is plain arithmetic.
One recurring digit over one 9: \(\displaystyle0.\overline{1}=\frac{1}{9}\), so \(\displaystyle(0.\overline{1})^2=\frac{1}{81}\)
One fixed and one recurring digit: \(\displaystyle0.1\overline{6}=\frac{16-1}{90}=\frac{15}{90}=\frac{1}{6}\), so \(\displaystyle(0.1\overline{6})^2=\frac{1}{36}\)
Inside the bracket: \(\displaystyle1-9\times\frac{1}{36}=1-\frac{1}{4}=\frac{3}{4}\)
The whole expression: \(\displaystyle\frac{1}{81}\times\frac{3}{4}=\frac{3}{324}=\frac{1}{108}\)
Tip: A decimal with a non-recurring part, such as \(0.1\overline{6}\), becomes (all the digits − the non-recurring digits) over as many 9s as recurring digits followed by as many 0s as non-recurring digits: \(\displaystyle\frac{16-1}{90}\).