Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q1) \(49\over6\)= [ 8\(\frac{1}{6}\)]

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

Q2) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q2) \(67\over6\)= [ 11\(\frac{1}{6}\)]

Q2) \(133\over10\) = [ 13\(\frac{3}{10}\)]

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

Q3) \(22\over5\)= [ 4\(\frac{2}{5}\)]

Q3) \(98\over9\) = [ 10\(\frac{8}{9}\)]

Q4) \(10\over3\)= [ 3\(\frac{1}{3}\) ]

Q4) \(43\over6\)= [ 7\(\frac{1}{6}\)]

Q4) \(28\over9\) = [ 3\(\frac{1}{9}\)]

Q5) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q5) \(68\over5\)= [ 13\(\frac{3}{5}\)]

Q5) \(133\over11\) = [ 12\(\frac{1}{11}\)]

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

Q6) \(52\over5\)= [ 10\(\frac{2}{5}\)]

Q6) \(112\over11\) = [ 10\(\frac{2}{11}\)]

Q7) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

Q7) \(59\over5\)= [ 11\(\frac{4}{5}\)]

Q7) \(42\over11\) = [ 3\(\frac{9}{11}\)]

Q8) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q8) \(27\over5\)= [ 5\(\frac{2}{5}\)]

Q8) \(49\over9\) = [ 5\(\frac{4}{9}\)]

Q9) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q9) \(57\over5\)= [ 11\(\frac{2}{5}\)]

Q9) \(91\over11\) = [ 8\(\frac{3}{11}\)]

Q10) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q10) \(21\over5\)= [ 4\(\frac{1}{5}\)]

Q10) \(98\over11\) = [ 8\(\frac{10}{11}\)]