Converting the following recurring decimals to common fractions in the simplest form: \( \begin{array}{lll}\text { (a) } 0, \dot{8} & \text { (b) } 0, \dot{6} \dot{3} & \text { (c) } 0, \dot{3} 1 \dot{2}\end{array} \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Extra Insights
To convert the recurring decimals to common fractions, here’s how you can do it! For \(0.\dot{8}\): Let \(x = 0.\dot{8}\). Then, multiplying by 10 gives \(10x = 8.\dot{8}\). Subtracting the original from this results in \(10x - x = 8\), which simplifies to \(9x = 8\). Therefore, \(x = \frac{8}{9}\). For \(0.\dot{6} \dot{3}\): Let \(y = 0.\dot{6} \dot{3}\). Then, multiplying by 100 gives \(100y = 63.\dot{6} \dot{3}\) and multiplying by 10 gives \(10y = 6.\dot{6} \dot{3}\). Now, subtracting gives \(100y - 10y = 63 - 6\), leading to \(90y = 57\) or \(y = \frac{57}{90}\). Simplifying gives \(y = \frac{19}{30}\). For \(0.\dot{3} 1 \dot{2}\): Let \(z = 0.\dot{3} 1 \dot{2}\). By multiplying by 1000 we get \(1000z = 312.\dot{3} 1 \dot{2}\), and also multiply by 10 gives \(10z = 3.\dot{3} 1 \dot{2}\). Subtracting these two gives \(1000z - 10z = 312 - 3\), leading us to \(990z = 309\) or \(z = \frac{309}{990}\). Simplifying this gives \(z = \frac{103}{330}\). So, the simplest forms are: (a) \(\frac{8}{9}\), (b) \(\frac{19}{30}\), (c) \(\frac{103}{330}\). Happy fraction converting!
