Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q1) \(59\over6\)= [ 9\(\frac{5}{6}\)]

Q1) \(91\over11\) = [ 8\(\frac{3}{11}\)]

Q2) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

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

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

Q3) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q3) \(28\over5\)= [ 5\(\frac{3}{5}\)]

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

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

Q4) \(67\over6\)= [ 11\(\frac{1}{6}\)]

Q4) \(133\over12\) = [ 11\(\frac{1}{12}\)]

Q5) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

Q5) \(41\over6\)= [ 6\(\frac{5}{6}\)]

Q5) \(91\over11\) = [ 8\(\frac{3}{11}\)]

Q6) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q6) \(38\over5\)= [ 7\(\frac{3}{5}\)]

Q6) \(49\over10\) = [ 4\(\frac{9}{10}\)]

Q7) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q7) \(36\over5\)= [ 7\(\frac{1}{5}\)]

Q7) \(133\over10\) = [ 13\(\frac{3}{10}\)]

Q8) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

Q8) \(63\over5\)= [ 12\(\frac{3}{5}\)]

Q8) \(56\over11\) = [ 5\(\frac{1}{11}\)]

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

Q9) \(44\over5\)= [ 8\(\frac{4}{5}\)]

Q9) \(49\over10\) = [ 4\(\frac{9}{10}\)]

Q10) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

Q10) \(67\over5\)= [ 13\(\frac{2}{5}\)]

Q10) \(91\over11\) = [ 8\(\frac{3}{11}\)]