Mr Daniels Maths
Conversions: Mixed to Improper fractions

Set 1

Set 2

Set 3

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

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

Q1) 6\(\frac{5}{12}\)= [ \(77\over12\) ]

Q2) 3\(\frac{2}{3}\)= [ \(11\over3\) ]

Q2) 10\(\frac{1}{6}\)= [ \(61\over6\) ]

Q2) 5\(\frac{4}{9}\)= [ \(49\over9\) ]

Q3) 6\(\frac{1}{3}\)= [ \(19\over3\) ]

Q3) 11\(\frac{4}{5}\)= [ \(59\over5\) ]

Q3) 3\(\frac{9}{11}\)= [ \(42\over11\) ]

Q4) 4\(\frac{1}{3}\)= [ \(13\over3\) ]

Q4) 13\(\frac{4}{5}\)= [ \(69\over5\) ]

Q4) 10\(\frac{2}{11}\)= [ \(112\over11\) ]

Q5) 5\(\frac{2}{3}\)= [ \(17\over3\) ]

Q5) 5\(\frac{3}{5}\)= [ \(28\over5\) ]

Q5) 3\(\frac{2}{11}\)= [ \(35\over11\) ]

Q6) 3\(\frac{1}{3}\)= [ \(10\over3\) ]

Q6) 10\(\frac{5}{6}\)= [ \(65\over6\) ]

Q6) 11\(\frac{5}{11}\)= [ \(126\over11\) ]

Q7) 4\(\frac{2}{3}\)= [ \(14\over3\) ]

Q7) 8\(\frac{1}{5}\)= [ \(41\over5\) ]

Q7) 7\(\frac{7}{11}\)= [ \(84\over11\) ]

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

Q8) 4\(\frac{2}{5}\)= [ \(22\over5\) ]

Q8) 6\(\frac{3}{10}\)= [ \(63\over10\) ]

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

Q9) 4\(\frac{3}{5}\)= [ \(23\over5\) ]

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

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

Q10) 5\(\frac{4}{5}\)= [ \(29\over5\) ]

Q10) 8\(\frac{3}{11}\)= [ \(91\over11\) ]