Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q1) \(26\over5\)= [ 5\(\frac{1}{5}\)]

Q1) \(77\over8\) = [ 9\(\frac{5}{8}\)]

Q2) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q2) \(31\over5\)= [ 6\(\frac{1}{5}\)]

Q2) \(133\over9\) = [ 14\(\frac{7}{9}\)]

Q3) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q3) \(67\over6\)= [ 11\(\frac{1}{6}\)]

Q3) \(119\over10\) = [ 11\(\frac{9}{10}\)]

Q4) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q4) \(36\over5\)= [ 7\(\frac{1}{5}\)]

Q4) \(119\over8\) = [ 14\(\frac{7}{8}\)]

Q5) \(10\over3\)= [ 3\(\frac{1}{3}\) ]

Q5) \(56\over5\)= [ 11\(\frac{1}{5}\)]

Q5) \(70\over9\) = [ 7\(\frac{7}{9}\)]

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

Q6) \(67\over5\)= [ 13\(\frac{2}{5}\)]

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

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

Q7) \(61\over6\)= [ 10\(\frac{1}{6}\)]

Q7) \(35\over9\) = [ 3\(\frac{8}{9}\)]

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

Q8) \(62\over5\)= [ 12\(\frac{2}{5}\)]

Q8) \(63\over11\) = [ 5\(\frac{8}{11}\)]

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

Q9) \(49\over5\)= [ 9\(\frac{4}{5}\)]

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

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

Q10) \(57\over5\)= [ 11\(\frac{2}{5}\)]

Q10) \(98\over9\) = [ 10\(\frac{8}{9}\)]