Mr Daniels Maths
Fraction Cross Cancellation

Set 1

Set 2

Set 3

Q1) \(2\over3\) x \(6\over10\) = [ \(\frac{2}{5}\)]

Q1) \(2\over3\) x \(6\over8\) \(\div\) \(13\over24\)= [ \(\frac{12}{13}\)]

Q1) \(2\over3\) x \(6\over7\) - \(2\over7\)= [ \(\frac{2}{7}\)]

Q2) \(2\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{4}{9}\)]

Q2) \(2\over3\) x \(6\over8\) \(\div\) \(14\over16\)= [ \(\frac{4}{7}\)]

Q2) \(2\over3\) x \(6\over9\) x \(27\over13\) - \(6\over9\)= [ \(\frac{10}{39}\)]

Q3) \(3\over4\) x \(8\over9\) = [ \(\frac{2}{3}\)]

Q3) \(2\over3\) x \(9\over10\) \(\div\) \(15\over20\)= [ \(\frac{4}{5}\)]

Q3) \(2\over3\) x \(9\over10\) + \(3\over10\)= [ \(\frac{9}{10}\)]

Q4) \(2\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{4}{9}\)]

Q4) \(2\over4\) x \(8\over9\) \(\div\) \(11\over18\)= [ \(\frac{8}{11}\)]

Q4) \(2\over3\) x \(6\over9\) x \(18\over11\) - \(7\over9\)= [ -\(\frac{5}{99}\)]

Q5) \(2\over3\) x \(6\over9\) = [ \(\frac{4}{9}\)]

Q5) \(2\over3\) x \(6\over10\) x \(30\over13\)= [ \(\frac{12}{13}\)]

Q5) \(2\over3\) x \(6\over7\) + \(3\over7\)= [ 1]

Q6) \(2\over3\) x \(9\over10\) = [ \(\frac{3}{5}\)]

Q6) \(2\over3\) x \(6\over10\) \(\div\) \(13\over20\)= [ \(\frac{8}{13}\)]

Q6) \(2\over3\) x \(6\over10\) x \(20\over15\) - \(2\over10\)= [ \(\frac{1}{3}\)]

Q7) \(2\over3\) x \(6\over8\) = [ \(\frac{1}{2}\)]

Q7) \(2\over3\) x \(6\over7\) x \(14\over9\)= [ \(\frac{8}{9}\)]

Q7) \(3\over4\) x \(8\over10\) - \(6\over10\)= [ 0]

Q8) \(2\over3\) x \(6\over7\) = [ \(\frac{4}{7}\)]

Q8) \(2\over3\) x \(6\over8\) \(\div\) \(13\over16\)= [ \(\frac{8}{13}\)]

Q8) \(2\over4\) x \(8\over10\) + \(6\over10\)= [ 1]

Q9) \(2\over3\) x \(6\over10\) = [ \(\frac{2}{5}\)]

Q9) \(3\over4\) x \(8\over9\) \(\div\) \(14\over18\)= [ \(\frac{6}{7}\)]

Q9) \(2\over4\) x \(8\over9\) x \(27\over14\) - \(6\over9\)= [ \(\frac{4}{21}\)]

Q10) \(2\over3\)  \(\div\) \(9\over6\) =   [ \(\frac{4}{9}\)]

Q10) \(2\over4\) x \(8\over9\) \(\div\) \(9\over18\)= [ \(\frac{8}{9}\)]

Q10) \(2\over4\) x \(8\over10\) x \(30\over14\) - \(4\over10\)= [ \(\frac{16}{35}\)]