Mr Daniels Maths
Mixed and Improper Conversions

Set 1

Set 2

Set 3

Q1) 5\(\frac{1}{3}\)= [ \(16\over3\) ]

Q1) 10\(\frac{5}{6}\)= [ \(65\over6\) ]

Q1) 11\(\frac{9}{10}\)= [ \(119\over10\) ]

Q2) 3\(\frac{1}{3}\)= [ \(10\over3\) ]

Q2) 5\(\frac{3}{5}\)= [ \(28\over5\) ]

Q2) 7\(\frac{7}{12}\)= [ \(91\over12\) ]

Q3) 5\(\frac{2}{3}\)= [ \(17\over3\) ]

Q3) 10\(\frac{1}{6}\)= [ \(61\over6\) ]

Q3) 13\(\frac{3}{10}\)= [ \(133\over10\) ]

Q4) 6\(\frac{1}{3}\)= [ \(19\over3\) ]

Q4) 4\(\frac{2}{5}\)= [ \(22\over5\) ]

Q4) 5\(\frac{8}{11}\)= [ \(63\over11\) ]

Q5) 6\(\frac{2}{3}\)= [ \(20\over3\) ]

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

Q5) 4\(\frac{1}{12}\)= [ \(49\over12\) ]

Q6) 4\(\frac{2}{3}\)= [ \(14\over3\) ]

Q6) 5\(\frac{1}{5}\)= [ \(26\over5\) ]

Q6) 11\(\frac{5}{11}\)= [ \(126\over11\) ]

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

Q7) 4\(\frac{4}{5}\)= [ \(24\over5\) ]

Q7) 16\(\frac{5}{8}\)= [ \(133\over8\) ]

Q8) 5\(\frac{1}{2}\)= [ \(11\over2\) ]

Q8) 7\(\frac{2}{5}\)= [ \(37\over5\) ]

Q8) 7\(\frac{7}{8}\)= [ \(63\over8\) ]

Q9) 4\(\frac{1}{3}\)= [ \(13\over3\) ]

Q9) 12\(\frac{3}{5}\)= [ \(63\over5\) ]

Q9) 13\(\frac{1}{8}\)= [ \(105\over8\) ]

Q10) 6\(\frac{1}{2}\)= [ \(13\over2\) ]

Q10) 13\(\frac{4}{5}\)= [ \(69\over5\) ]

Q10) 4\(\frac{9}{10}\)= [ \(49\over10\) ]