Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q1) \(35\over6\)= [ 5\(\frac{5}{6}\)]

Q1) \(98\over9\) = [ 10\(\frac{8}{9}\)]

Q2) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q2) \(26\over5\)= [ 5\(\frac{1}{5}\)]

Q2) \(133\over11\) = [ 12\(\frac{1}{11}\)]

Q3) \(10\over3\)= [ 3\(\frac{1}{3}\) ]

Q3) \(23\over5\)= [ 4\(\frac{3}{5}\)]

Q3) \(105\over11\) = [ 9\(\frac{6}{11}\)]

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

Q4) \(21\over5\)= [ 4\(\frac{1}{5}\)]

Q4) \(70\over11\) = [ 6\(\frac{4}{11}\)]

Q5) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q5) \(69\over5\)= [ 13\(\frac{4}{5}\)]

Q5) \(63\over8\) = [ 7\(\frac{7}{8}\)]

Q6) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q6) \(65\over6\)= [ 10\(\frac{5}{6}\)]

Q6) \(112\over9\) = [ 12\(\frac{4}{9}\)]

Q7) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q7) \(23\over6\)= [ 3\(\frac{5}{6}\)]

Q7) \(77\over10\) = [ 7\(\frac{7}{10}\)]

Q8) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

Q8) \(56\over5\)= [ 11\(\frac{1}{5}\)]

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

Q9) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

Q9) \(38\over5\)= [ 7\(\frac{3}{5}\)]

Q9) \(119\over8\) = [ 14\(\frac{7}{8}\)]

Q10) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q10) \(43\over6\)= [ 7\(\frac{1}{6}\)]

Q10) \(119\over12\) = [ 9\(\frac{11}{12}\)]