Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

Q1) \(13\over2\)= [ 6\(\frac{1}{2}\) ]

Q1) \(21\over5\)= [ 4\(\frac{1}{5}\)]

Q1) \(105\over8\) = [ 13\(\frac{1}{8}\)]

Q2) \(19\over3\)= [ 6\(\frac{1}{3}\) ]

Q2) \(43\over6\)= [ 7\(\frac{1}{6}\)]

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

Q3) \(17\over3\)= [ 5\(\frac{2}{3}\) ]

Q3) \(37\over6\)= [ 6\(\frac{1}{6}\)]

Q3) \(140\over11\) = [ 12\(\frac{8}{11}\)]

Q4) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

Q4) \(32\over5\)= [ 6\(\frac{2}{5}\)]

Q4) \(98\over9\) = [ 10\(\frac{8}{9}\)]

Q5) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q5) \(22\over5\)= [ 4\(\frac{2}{5}\)]

Q5) \(91\over9\) = [ 10\(\frac{1}{9}\)]

Q6) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q6) \(26\over5\)= [ 5\(\frac{1}{5}\)]

Q6) \(98\over11\) = [ 8\(\frac{10}{11}\)]

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

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

Q7) \(91\over8\) = [ 11\(\frac{3}{8}\)]

Q8) \(11\over3\)= [ 3\(\frac{2}{3}\) ]

Q8) \(33\over5\)= [ 6\(\frac{3}{5}\)]

Q8) \(49\over8\) = [ 6\(\frac{1}{8}\)]

Q9) \(16\over3\)= [ 5\(\frac{1}{3}\) ]

Q9) \(65\over6\)= [ 10\(\frac{5}{6}\)]

Q9) \(42\over11\) = [ 3\(\frac{9}{11}\)]

Q10) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

Q10) \(61\over5\)= [ 12\(\frac{1}{5}\)]

Q10) \(91\over10\) = [ 9\(\frac{1}{10}\)]