Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

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

Q1) \(63\over5\)= [ 12\(\frac{3}{5}\)]

Q1) \(112\over9\) = [ 12\(\frac{4}{9}\)]

Q2) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q2) \(41\over6\)= [ 6\(\frac{5}{6}\)]

Q2) \(91\over8\) = [ 11\(\frac{3}{8}\)]

Q3) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

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

Q3) \(91\over12\) = [ 7\(\frac{7}{12}\)]

Q4) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q4) \(52\over5\)= [ 10\(\frac{2}{5}\)]

Q4) \(112\over9\) = [ 12\(\frac{4}{9}\)]

Q5) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q5) \(43\over6\)= [ 7\(\frac{1}{6}\)]

Q5) \(77\over10\) = [ 7\(\frac{7}{10}\)]

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

Q6) \(32\over5\)= [ 6\(\frac{2}{5}\)]

Q6) \(49\over11\) = [ 4\(\frac{5}{11}\)]

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

Q7) \(37\over6\)= [ 6\(\frac{1}{6}\)]

Q7) \(105\over8\) = [ 13\(\frac{1}{8}\)]

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

Q8) \(49\over5\)= [ 9\(\frac{4}{5}\)]

Q8) \(49\over12\) = [ 4\(\frac{1}{12}\)]

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

Q9) \(32\over5\)= [ 6\(\frac{2}{5}\)]

Q9) \(119\over11\) = [ 10\(\frac{9}{11}\)]

Q10) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

Q10) \(28\over5\)= [ 5\(\frac{3}{5}\)]

Q10) \(49\over8\) = [ 6\(\frac{1}{8}\)]