Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

Q1) \(51\over5\)= [ 10\(\frac{1}{5}\)]

Q1) \(133\over9\) = [ 14\(\frac{7}{9}\)]

Q2) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q2) \(33\over5\)= [ 6\(\frac{3}{5}\)]

Q2) \(49\over8\) = [ 6\(\frac{1}{8}\)]

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

Q3) \(47\over5\)= [ 9\(\frac{2}{5}\)]

Q3) \(126\over11\) = [ 11\(\frac{5}{11}\)]

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

Q4) \(42\over5\)= [ 8\(\frac{2}{5}\)]

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

Q5) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q5) \(43\over5\)= [ 8\(\frac{3}{5}\)]

Q5) \(35\over11\) = [ 3\(\frac{2}{11}\)]

Q6) \(10\over3\)= [ 3\(\frac{1}{3}\) ]

Q6) \(44\over5\)= [ 8\(\frac{4}{5}\)]

Q6) \(49\over12\) = [ 4\(\frac{1}{12}\)]

Q7) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q7) \(59\over6\)= [ 9\(\frac{5}{6}\)]

Q7) \(140\over11\) = [ 12\(\frac{8}{11}\)]

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

Q8) \(41\over5\)= [ 8\(\frac{1}{5}\)]

Q8) \(63\over8\) = [ 7\(\frac{7}{8}\)]

Q9) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q9) \(24\over5\)= [ 4\(\frac{4}{5}\)]

Q9) \(77\over10\) = [ 7\(\frac{7}{10}\)]

Q10) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q10) \(39\over5\)= [ 7\(\frac{4}{5}\)]

Q10) \(84\over11\) = [ 7\(\frac{7}{11}\)]