Mr Daniels Maths
Conversions: Improper to Mixed fractions

Set 1

Set 2

Set 3

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

Q1) \(67\over6\)= [ 11\(\frac{1}{6}\)]

Q1) \(70\over9\) = [ 7\(\frac{7}{9}\)]

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

Q2) \(67\over5\)= [ 13\(\frac{2}{5}\)]

Q2) \(140\over11\) = [ 12\(\frac{8}{11}\)]

Q3) \(14\over3\)= [ 4\(\frac{2}{3}\) ]

Q3) \(62\over5\)= [ 12\(\frac{2}{5}\)]

Q3) \(126\over11\) = [ 11\(\frac{5}{11}\)]

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

Q4) \(52\over5\)= [ 10\(\frac{2}{5}\)]

Q4) \(56\over9\) = [ 6\(\frac{2}{9}\)]

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

Q5) \(46\over5\)= [ 9\(\frac{1}{5}\)]

Q5) \(105\over8\) = [ 13\(\frac{1}{8}\)]

Q6) \(20\over3\)= [ 6\(\frac{2}{3}\) ]

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

Q6) \(119\over11\) = [ 10\(\frac{9}{11}\)]

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

Q7) \(37\over6\)= [ 6\(\frac{1}{6}\)]

Q7) \(119\over8\) = [ 14\(\frac{7}{8}\)]

Q8) \(13\over3\)= [ 4\(\frac{1}{3}\) ]

Q8) \(42\over5\)= [ 8\(\frac{2}{5}\)]

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

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

Q9) \(59\over5\)= [ 11\(\frac{4}{5}\)]

Q9) \(77\over8\) = [ 9\(\frac{5}{8}\)]

Q10) \(11\over2\)= [ 5\(\frac{1}{2}\) ]

Q10) \(41\over5\)= [ 8\(\frac{1}{5}\)]

Q10) \(77\over9\) = [ 8\(\frac{5}{9}\)]