Mr Daniels Maths
Fraction Cross Cancellation

Set 1

Set 2

Set 3

Q1) \(2\over3\)  \(\div\) \(10\over9\) =   [ \(\frac{3}{5}\)]

Q1) \(3\over4\) x \(8\over9\) x \(18\over13\)= [ \(\frac{12}{13}\)]

Q1) \(3\over4\) x \(8\over10\) + \(6\over10\)= [ 1\(\frac{1}{5}\)]

Q2) \(3\over4\) x \(8\over10\) = [ \(\frac{3}{5}\)]

Q2) \(2\over3\) x \(6\over7\) \(\div\) \(12\over14\)= [ \(\frac{2}{3}\)]

Q2) \(2\over4\) x \(8\over9\) x \(18\over15\) - \(4\over9\)= [ \(\frac{4}{45}\)]

Q3) \(3\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{2}{3}\)]

Q3) \(2\over3\) x \(6\over8\) \(\div\) \(14\over24\)= [ \(\frac{6}{7}\)]

Q3) \(2\over3\) x \(9\over10\) + \(5\over10\)= [ 1\(\frac{1}{10}\)]

Q4) \(2\over3\) x \(6\over7\) = [ \(\frac{4}{7}\)]

Q4) \(2\over4\) x \(8\over10\) \(\div\) \(10\over20\)= [ \(\frac{4}{5}\)]

Q4) \(2\over3\) x \(6\over10\) x \(20\over13\) - \(4\over10\)= [ \(\frac{14}{65}\)]

Q5) \(2\over4\) x \(8\over10\) = [ \(\frac{2}{5}\)]

Q5) \(2\over3\) x \(6\over9\) \(\div\) \(14\over27\)= [ \(\frac{6}{7}\)]

Q5) \(2\over3\) x \(9\over10\) + \(4\over10\)= [ 1]

Q6) \(2\over3\)  \(\div\) \(9\over6\) =   [ \(\frac{4}{9}\)]

Q6) \(2\over4\) x \(8\over10\) x \(30\over14\)= [ \(\frac{6}{7}\)]

Q6) \(2\over3\) x \(6\over8\) - \(4\over8\)= [ 0]

Q7) \(2\over3\)  \(\div\) \(8\over6\) =   [ \(\frac{1}{2}\)]

Q7) \(2\over3\) x \(6\over9\) \(\div\) \(10\over18\)= [ \(\frac{4}{5}\)]

Q7) \(2\over3\) x \(6\over10\) - \(3\over10\)= [ \(\frac{1}{10}\)]

Q8) \(2\over3\)  \(\div\) \(7\over6\) =   [ \(\frac{4}{7}\)]

Q8) \(2\over3\) x \(6\over8\) x \(16\over15\)= [ \(\frac{8}{15}\)]

Q8) \(3\over4\) x \(8\over9\) - \(2\over9\)= [ \(\frac{4}{9}\)]

Q9) \(2\over3\) x \(6\over9\) = [ \(\frac{4}{9}\)]

Q9) \(2\over3\) x \(6\over9\) x \(27\over14\)= [ \(\frac{6}{7}\)]

Q9) \(3\over4\) x \(8\over10\) x \(20\over15\) - \(6\over10\)= [ \(\frac{1}{5}\)]

Q10) \(2\over4\)  \(\div\) \(10\over8\) =   [ \(\frac{2}{5}\)]

Q10) \(2\over4\) x \(8\over9\) x \(18\over12\)= [ \(\frac{2}{3}\)]

Q10) \(2\over3\) x \(6\over7\) x \(21\over14\) - \(7\over7\)= [ -\(\frac{1}{7}\)]