Mr Daniels Maths
Mixed and Improper Conversions

Set 1

Set 2

Set 3

Q1) 5\(\frac{1}{2}\)= [ \(11\over2\) ]

Q1) 7\(\frac{3}{5}\)= [ \(38\over5\) ]

Q1) 10\(\frac{9}{11}\)= [ \(119\over11\) ]

Q2) 4\(\frac{1}{3}\)= [ \(13\over3\) ]

Q2) 7\(\frac{5}{6}\)= [ \(47\over6\) ]

Q2) 4\(\frac{5}{11}\)= [ \(49\over11\) ]

Q3) 6\(\frac{1}{3}\)= [ \(19\over3\) ]

Q3) 13\(\frac{2}{5}\)= [ \(67\over5\) ]

Q3) 16\(\frac{5}{8}\)= [ \(133\over8\) ]

Q4) 6\(\frac{2}{3}\)= [ \(20\over3\) ]

Q4) 8\(\frac{4}{5}\)= [ \(44\over5\) ]

Q4) 12\(\frac{4}{9}\)= [ \(112\over9\) ]

Q5) 6\(\frac{1}{2}\)= [ \(13\over2\) ]

Q5) 4\(\frac{1}{6}\)= [ \(25\over6\) ]

Q5) 13\(\frac{2}{9}\)= [ \(119\over9\) ]

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

Q6) 9\(\frac{2}{5}\)= [ \(47\over5\) ]

Q6) 7\(\frac{7}{10}\)= [ \(77\over10\) ]

Q7) 3\(\frac{2}{3}\)= [ \(11\over3\) ]

Q7) 5\(\frac{3}{5}\)= [ \(28\over5\) ]

Q7) 11\(\frac{9}{10}\)= [ \(119\over10\) ]

Q8) 3\(\frac{1}{3}\)= [ \(10\over3\) ]

Q8) 8\(\frac{3}{5}\)= [ \(43\over5\) ]

Q8) 6\(\frac{5}{12}\)= [ \(77\over12\) ]

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

Q9) 7\(\frac{4}{5}\)= [ \(39\over5\) ]

Q9) 14\(\frac{7}{9}\)= [ \(133\over9\) ]

Q10) 5\(\frac{1}{3}\)= [ \(16\over3\) ]

Q10) 12\(\frac{4}{5}\)= [ \(64\over5\) ]

Q10) 12\(\frac{8}{11}\)= [ \(140\over11\) ]