Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q1) \(44\over5\)= [ 8\(\frac{4}{5}\)]

Q1) \(133\over11\) = [ 12\(\frac{1}{11}\)]

Q2) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q2) \(43\over5\)= [ 8\(\frac{3}{5}\)]

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

Q3) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

Q3) \(35\over6\)= [ 5\(\frac{5}{6}\)]

Q3) \(91\over8\) = [ 11\(\frac{3}{8}\)]

Q4) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q4) \(28\over5\)= [ 5\(\frac{3}{5}\)]

Q4) \(77\over8\) = [ 9\(\frac{5}{8}\)]

Q5) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q5) \(39\over5\)= [ 7\(\frac{4}{5}\)]

Q5) \(105\over8\) = [ 13\(\frac{1}{8}\)]

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

Q6) \(31\over5\)= [ 6\(\frac{1}{5}\)]

Q6) \(140\over9\) = [ 15\(\frac{5}{9}\)]

Q7) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q7) \(69\over5\)= [ 13\(\frac{4}{5}\)]

Q7) \(63\over10\) = [ 6\(\frac{3}{10}\)]

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

Q8) \(53\over5\)= [ 10\(\frac{3}{5}\)]

Q8) \(133\over8\) = [ 16\(\frac{5}{8}\)]

Q9) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

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

Q9) \(56\over9\) = [ 6\(\frac{2}{9}\)]

Q10) \(10\over3\)= [ 3\(\frac{1}{3}\) ]

Q10) \(58\over5\)= [ 11\(\frac{3}{5}\)]

Q10) \(35\over9\) = [ 3\(\frac{8}{9}\)]