Mr Daniels Maths
Mixed and Improper Conversions

Set 1

Set 2

Set 3

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

Q1) 13\(\frac{2}{5}\)= [ \(67\over5\) ]

Q1) 11\(\frac{9}{10}\)= [ \(119\over10\) ]

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

Q2) 10\(\frac{3}{5}\)= [ \(53\over5\) ]

Q2) 7\(\frac{7}{12}\)= [ \(91\over12\) ]

Q3) 3\(\frac{1}{3}\)= [ \(10\over3\) ]

Q3) 12\(\frac{4}{5}\)= [ \(64\over5\) ]

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

Q4) 6\(\frac{1}{2}\)= [ \(13\over2\) ]

Q4) 4\(\frac{5}{6}\)= [ \(29\over6\) ]

Q4) 15\(\frac{5}{9}\)= [ \(140\over9\) ]

Q5) 6\(\frac{1}{3}\)= [ \(19\over3\) ]

Q5) 9\(\frac{3}{5}\)= [ \(48\over5\) ]

Q5) 5\(\frac{4}{9}\)= [ \(49\over9\) ]

Q6) 6\(\frac{1}{3}\)= [ \(19\over3\) ]

Q6) 12\(\frac{4}{5}\)= [ \(64\over5\) ]

Q6) 13\(\frac{1}{8}\)= [ \(105\over8\) ]

Q7) 5\(\frac{1}{2}\)= [ \(11\over2\) ]

Q7) 12\(\frac{1}{5}\)= [ \(61\over5\) ]

Q7) 9\(\frac{11}{12}\)= [ \(119\over12\) ]

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

Q8) 9\(\frac{2}{5}\)= [ \(47\over5\) ]

Q8) 10\(\frac{2}{11}\)= [ \(112\over11\) ]

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

Q9) 9\(\frac{2}{5}\)= [ \(47\over5\) ]

Q9) 7\(\frac{7}{11}\)= [ \(84\over11\) ]

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

Q10) 5\(\frac{5}{6}\)= [ \(35\over6\) ]

Q10) 10\(\frac{2}{11}\)= [ \(112\over11\) ]