Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

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

Q1) \(27\over5\)= [ 5\(\frac{2}{5}\)]

Q1) \(63\over8\) = [ 7\(\frac{7}{8}\)]

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

Q2) \(29\over5\)= [ 5\(\frac{4}{5}\)]

Q2) \(35\over11\) = [ 3\(\frac{2}{11}\)]

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

Q3) \(31\over6\)= [ 5\(\frac{1}{6}\)]

Q3) \(63\over10\) = [ 6\(\frac{3}{10}\)]

Q4) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q4) \(41\over6\)= [ 6\(\frac{5}{6}\)]

Q4) \(35\over8\) = [ 4\(\frac{3}{8}\)]

Q5) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q5) \(38\over5\)= [ 7\(\frac{3}{5}\)]

Q5) \(119\over12\) = [ 9\(\frac{11}{12}\)]

Q6) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

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

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

Q7) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q7) \(63\over5\)= [ 12\(\frac{3}{5}\)]

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

Q8) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q8) \(49\over6\)= [ 8\(\frac{1}{6}\)]

Q8) \(133\over11\) = [ 12\(\frac{1}{11}\)]

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

Q9) \(52\over5\)= [ 10\(\frac{2}{5}\)]

Q9) \(133\over10\) = [ 13\(\frac{3}{10}\)]

Q10) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q10) \(26\over5\)= [ 5\(\frac{1}{5}\)]

Q10) \(70\over9\) = [ 7\(\frac{7}{9}\)]