Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

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

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

Q1) \(91\over9\) = [ 10\(\frac{1}{9}\)]

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

Q2) \(68\over5\)= [ 13\(\frac{3}{5}\)]

Q2) \(84\over11\) = [ 7\(\frac{7}{11}\)]

Q3) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q3) \(54\over5\)= [ 10\(\frac{4}{5}\)]

Q3) \(119\over11\) = [ 10\(\frac{9}{11}\)]

Q4) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q4) \(53\over6\)= [ 8\(\frac{5}{6}\)]

Q4) \(49\over9\) = [ 5\(\frac{4}{9}\)]

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

Q5) \(23\over5\)= [ 4\(\frac{3}{5}\)]

Q5) \(119\over10\) = [ 11\(\frac{9}{10}\)]

Q6) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q6) \(23\over6\)= [ 3\(\frac{5}{6}\)]

Q6) \(63\over8\) = [ 7\(\frac{7}{8}\)]

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

Q7) \(31\over5\)= [ 6\(\frac{1}{5}\)]

Q7) \(91\over11\) = [ 8\(\frac{3}{11}\)]

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

Q8) \(59\over5\)= [ 11\(\frac{4}{5}\)]

Q8) \(140\over11\) = [ 12\(\frac{8}{11}\)]

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

Q9) \(22\over5\)= [ 4\(\frac{2}{5}\)]

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

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

Q10) \(31\over6\)= [ 5\(\frac{1}{6}\)]

Q10) \(133\over11\) = [ 12\(\frac{1}{11}\)]