Mr Daniels Maths
Conversions: Mixed to Improper fractions

Set 1

Set 2

Set 3

Q1) 5\(\frac{2}{3}\)= [ \(17\over3\) ]

Q1) 12\(\frac{4}{5}\)= [ \(64\over5\) ]

Q1) 6\(\frac{5}{12}\)= [ \(77\over12\) ]

Q2) 5\(\frac{1}{3}\)= [ \(16\over3\) ]

Q2) 5\(\frac{2}{5}\)= [ \(27\over5\) ]

Q2) 9\(\frac{6}{11}\)= [ \(105\over11\) ]

Q3) 3\(\frac{2}{3}\)= [ \(11\over3\) ]

Q3) 4\(\frac{2}{5}\)= [ \(22\over5\) ]

Q3) 10\(\frac{1}{9}\)= [ \(91\over9\) ]

Q4) 3\(\frac{2}{3}\)= [ \(11\over3\) ]

Q4) 6\(\frac{3}{5}\)= [ \(33\over5\) ]

Q4) 9\(\frac{6}{11}\)= [ \(105\over11\) ]

Q5) 4\(\frac{1}{3}\)= [ \(13\over3\) ]

Q5) 9\(\frac{2}{5}\)= [ \(47\over5\) ]

Q5) 9\(\frac{5}{8}\)= [ \(77\over8\) ]

Q6) 5\(\frac{2}{3}\)= [ \(17\over3\) ]

Q6) 4\(\frac{2}{5}\)= [ \(22\over5\) ]

Q6) 13\(\frac{1}{8}\)= [ \(105\over8\) ]

Q7) 4\(\frac{2}{3}\)= [ \(14\over3\) ]

Q7) 10\(\frac{1}{5}\)= [ \(51\over5\) ]

Q7) 7\(\frac{7}{11}\)= [ \(84\over11\) ]

Q8) 4\(\frac{1}{3}\)= [ \(13\over3\) ]

Q8) 5\(\frac{5}{6}\)= [ \(35\over6\) ]

Q8) 11\(\frac{9}{10}\)= [ \(119\over10\) ]

Q9) 4\(\frac{2}{3}\)= [ \(14\over3\) ]

Q9) 9\(\frac{2}{5}\)= [ \(47\over5\) ]

Q9) 7\(\frac{7}{11}\)= [ \(84\over11\) ]

Q10) 5\(\frac{2}{3}\)= [ \(17\over3\) ]

Q10) 7\(\frac{4}{5}\)= [ \(39\over5\) ]

Q10) 14\(\frac{7}{8}\)= [ \(119\over8\) ]