Mr Daniels Maths
Fraction Cross Cancellation

Set 1

Set 2

Set 3

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

Q1) \(3\over4\) x \(8\over9\) \(\div\) \(14\over18\)= [ \(\frac{6}{7}\)]

Q1) \(2\over3\) x \(6\over10\) x \(30\over14\) + \(6\over10\)= [ 1\(\frac{16}{35}\)]

Q2) \(2\over3\)  \(\div\) \(7\over6\) =   [ \(\frac{4}{7}\)]

Q2) \(2\over4\) x \(8\over9\) x \(18\over10\)= [ \(\frac{4}{5}\)]

Q2) \(2\over3\) x \(6\over8\) + \(4\over8\)= [ 1]

Q3) \(2\over4\) x \(8\over10\) = [ \(\frac{2}{5}\)]

Q3) \(2\over3\) x \(9\over10\) x \(20\over14\)= [ \(\frac{6}{7}\)]

Q3) \(2\over4\) x \(8\over10\) - \(3\over10\)= [ \(\frac{1}{10}\)]

Q4) \(2\over3\)  \(\div\) \(10\over6\) =   [ \(\frac{2}{5}\)]

Q4) \(2\over3\) x \(9\over10\) x \(20\over15\)= [ \(\frac{4}{5}\)]

Q4) \(2\over3\) x \(9\over10\) - \(3\over10\)= [ \(\frac{3}{10}\)]

Q5) \(3\over4\)  \(\div\) \(10\over8\) =   [ \(\frac{3}{5}\)]

Q5) \(2\over3\) x \(6\over10\) x \(20\over10\)= [ \(\frac{4}{5}\)]

Q5) \(3\over4\) x \(8\over10\) - \(3\over10\)= [ \(\frac{3}{10}\)]

Q6) \(3\over4\) x \(8\over10\) = [ \(\frac{3}{5}\)]

Q6) \(2\over3\) x \(6\over9\) x \(18\over12\)= [ \(\frac{2}{3}\)]

Q6) \(2\over4\) x \(8\over9\) + \(3\over9\)= [ \(\frac{7}{9}\)]

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

Q7) \(2\over4\) x \(8\over9\) \(\div\) \(14\over18\)= [ \(\frac{4}{7}\)]

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

Q8) \(2\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{4}{9}\)]

Q8) \(2\over3\) x \(6\over8\) x \(24\over15\)= [ \(\frac{4}{5}\)]

Q8) \(2\over3\) x \(9\over10\) + \(3\over10\)= [ \(\frac{9}{10}\)]

Q9) \(3\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{2}{3}\)]

Q9) \(2\over4\) x \(8\over9\) x \(27\over14\)= [ \(\frac{6}{7}\)]

Q9) \(2\over3\) x \(6\over10\) x \(20\over12\) + \(3\over10\)= [ \(\frac{29}{30}\)]

Q10) \(2\over3\)  \(\div\) \(10\over9\) =   [ \(\frac{3}{5}\)]

Q10) \(2\over3\) x \(6\over9\) x \(18\over11\)= [ \(\frac{8}{11}\)]

Q10) \(2\over3\) x \(9\over10\) - \(6\over10\)= [ 0]