Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

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

Q1) \(61\over5\)= [ 12\(\frac{1}{5}\)]

Q1) \(77\over9\) = [ 8\(\frac{5}{9}\)]

Q2) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q2) \(44\over5\)= [ 8\(\frac{4}{5}\)]

Q2) \(133\over8\) = [ 16\(\frac{5}{8}\)]

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

Q3) \(41\over5\)= [ 8\(\frac{1}{5}\)]

Q3) \(91\over8\) = [ 11\(\frac{3}{8}\)]

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

Q4) \(43\over5\)= [ 8\(\frac{3}{5}\)]

Q4) \(49\over10\) = [ 4\(\frac{9}{10}\)]

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

Q5) \(51\over5\)= [ 10\(\frac{1}{5}\)]

Q5) \(140\over11\) = [ 12\(\frac{8}{11}\)]

Q6) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q6) \(21\over5\)= [ 4\(\frac{1}{5}\)]

Q6) \(91\over10\) = [ 9\(\frac{1}{10}\)]

Q7) \(10\over3\)= [ 3\(\frac{1}{3}\) ]

Q7) \(63\over5\)= [ 12\(\frac{3}{5}\)]

Q7) \(112\over11\) = [ 10\(\frac{2}{11}\)]

Q8) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q8) \(67\over5\)= [ 13\(\frac{2}{5}\)]

Q8) \(112\over9\) = [ 12\(\frac{4}{9}\)]

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

Q9) \(34\over5\)= [ 6\(\frac{4}{5}\)]

Q9) \(133\over10\) = [ 13\(\frac{3}{10}\)]

Q10) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q10) \(48\over5\)= [ 9\(\frac{3}{5}\)]

Q10) \(35\over9\) = [ 3\(\frac{8}{9}\)]