Mr Daniels Maths
Fraction Cross Cancellation

Set 1

Set 2

Set 3

Q1) \(2\over4\) x \(8\over10\) = [ \(\frac{2}{5}\)]

Q1) \(2\over3\) x \(6\over8\) x \(16\over13\)= [ \(\frac{8}{13}\)]

Q1) \(2\over3\) x \(6\over8\) x \(24\over13\) - \(2\over8\)= [ \(\frac{35}{52}\)]

Q2) \(2\over3\)  \(\div\) \(10\over6\) =   [ \(\frac{2}{5}\)]

Q2) \(2\over3\) x \(6\over9\) \(\div\) \(11\over18\)= [ \(\frac{8}{11}\)]

Q2) \(2\over3\) x \(6\over10\) x \(30\over14\) + \(6\over10\)= [ 1\(\frac{16}{35}\)]

Q3) \(2\over3\) x \(6\over9\) = [ \(\frac{4}{9}\)]

Q3) \(2\over3\) x \(6\over7\) \(\div\) \(11\over14\)= [ \(\frac{8}{11}\)]

Q3) \(2\over3\) x \(9\over10\) - \(4\over10\)= [ \(\frac{1}{5}\)]

Q4) \(2\over4\)  \(\div\) \(10\over8\) =   [ \(\frac{2}{5}\)]

Q4) \(2\over3\) x \(6\over10\) \(\div\) \(12\over20\)= [ \(\frac{2}{3}\)]

Q4) \(3\over4\) x \(8\over9\) - \(5\over9\)= [ \(\frac{1}{9}\)]

Q5) \(2\over3\) x \(6\over8\) = [ \(\frac{1}{2}\)]

Q5) \(3\over4\) x \(8\over10\) \(\div\) \(13\over20\)= [ \(\frac{12}{13}\)]

Q5) \(3\over4\) x \(8\over9\) x \(18\over15\) - \(6\over9\)= [ \(\frac{2}{15}\)]

Q6) \(3\over4\)  \(\div\) \(10\over8\) =   [ \(\frac{3}{5}\)]

Q6) \(2\over3\) x \(6\over8\) x \(24\over13\)= [ \(\frac{12}{13}\)]

Q6) \(2\over3\) x \(6\over7\) x \(21\over13\) + \(2\over7\)= [ 1\(\frac{19}{91}\)]

Q7) \(2\over3\) x \(6\over9\) = [ \(\frac{4}{9}\)]

Q7) \(2\over3\) x \(6\over8\) x \(24\over14\)= [ \(\frac{6}{7}\)]

Q7) \(2\over3\) x \(6\over7\) - \(3\over7\)= [ \(\frac{1}{7}\)]

Q8) \(2\over3\)  \(\div\) \(10\over6\) =   [ \(\frac{2}{5}\)]

Q8) \(2\over3\) x \(6\over8\) x \(16\over10\)= [ \(\frac{4}{5}\)]

Q8) \(3\over4\) x \(8\over10\) - \(5\over10\)= [ \(\frac{1}{10}\)]

Q9) \(2\over4\)  \(\div\) \(10\over8\) =   [ \(\frac{2}{5}\)]

Q9) \(2\over4\) x \(8\over10\) x \(20\over10\)= [ \(\frac{4}{5}\)]

Q9) \(2\over4\) x \(8\over10\) x \(20\over13\) + \(5\over10\)= [ 1\(\frac{3}{26}\)]

Q10) \(2\over3\) x \(6\over10\) = [ \(\frac{2}{5}\)]

Q10) \(2\over3\) x \(6\over10\) x \(20\over13\)= [ \(\frac{8}{13}\)]

Q10) \(2\over4\) x \(8\over9\) x \(18\over12\) - \(4\over9\)= [ \(\frac{2}{9}\)]