Mr Daniels Maths
Mixed and Improper Conversions

Set 1

Set 2

Set 3

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

Q1) 7\(\frac{2}{5}\)= [ \(37\over5\) ]

Q1) 9\(\frac{11}{12}\)= [ \(119\over12\) ]

Q2) 4\(\frac{1}{3}\)= [ \(13\over3\) ]

Q2) 6\(\frac{5}{6}\)= [ \(41\over6\) ]

Q2) 7\(\frac{7}{11}\)= [ \(84\over11\) ]

Q3) 6\(\frac{2}{3}\)= [ \(20\over3\) ]

Q3) 6\(\frac{1}{5}\)= [ \(31\over5\) ]

Q3) 5\(\frac{8}{11}\)= [ \(63\over11\) ]

Q4) 3\(\frac{2}{3}\)= [ \(11\over3\) ]

Q4) 9\(\frac{3}{5}\)= [ \(48\over5\) ]

Q4) 8\(\frac{10}{11}\)= [ \(98\over11\) ]

Q5) 4\(\frac{2}{3}\)= [ \(14\over3\) ]

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

Q5) 6\(\frac{4}{11}\)= [ \(70\over11\) ]

Q6) 5\(\frac{2}{3}\)= [ \(17\over3\) ]

Q6) 7\(\frac{4}{5}\)= [ \(39\over5\) ]

Q6) 3\(\frac{2}{11}\)= [ \(35\over11\) ]

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

Q7) 9\(\frac{1}{5}\)= [ \(46\over5\) ]

Q7) 14\(\frac{7}{9}\)= [ \(133\over9\) ]

Q8) 5\(\frac{1}{2}\)= [ \(11\over2\) ]

Q8) 4\(\frac{1}{6}\)= [ \(25\over6\) ]

Q8) 12\(\frac{8}{11}\)= [ \(140\over11\) ]

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

Q9) 9\(\frac{1}{6}\)= [ \(55\over6\) ]

Q9) 9\(\frac{6}{11}\)= [ \(105\over11\) ]

Q10) 6\(\frac{1}{3}\)= [ \(19\over3\) ]

Q10) 4\(\frac{5}{6}\)= [ \(29\over6\) ]

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