Mr Daniels Maths
Fraction Cross Cancellation

Set 1

Set 2

Set 3

Q1) \(2\over3\)  \(\div\) \(9\over6\) =   [ \(\frac{4}{9}\)]

Q1) \(2\over3\) x \(6\over7\) x \(14\over12\)= [ \(\frac{2}{3}\)]

Q1) \(2\over3\) x \(6\over7\) x \(21\over15\) - \(7\over7\)= [ -\(\frac{1}{5}\)]

Q2) \(3\over4\)  \(\div\) \(10\over8\) =   [ \(\frac{3}{5}\)]

Q2) \(2\over3\) x \(6\over7\) x \(14\over13\)= [ \(\frac{8}{13}\)]

Q2) \(2\over3\) x \(9\over10\) + \(5\over10\)= [ 1\(\frac{1}{10}\)]

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

Q3) \(2\over3\) x \(6\over10\) \(\div\) \(9\over20\)= [ \(\frac{8}{9}\)]

Q3) \(2\over4\) x \(8\over10\) x \(20\over9\) - \(5\over10\)= [ \(\frac{7}{18}\)]

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

Q4) \(2\over4\) x \(8\over10\) \(\div\) \(9\over20\)= [ \(\frac{8}{9}\)]

Q4) \(2\over4\) x \(8\over9\) x \(18\over14\) - \(7\over9\)= [ -\(\frac{13}{63}\)]

Q5) \(2\over3\)  \(\div\) \(10\over9\) =   [ \(\frac{3}{5}\)]

Q5) \(2\over3\) x \(6\over8\) \(\div\) \(10\over16\)= [ \(\frac{4}{5}\)]

Q5) \(2\over3\) x \(6\over8\) x \(24\over15\) + \(6\over8\)= [ 1\(\frac{11}{20}\)]

Q6) \(2\over3\) x \(6\over8\) = [ \(\frac{1}{2}\)]

Q6) \(2\over3\) x \(6\over8\) x \(24\over15\)= [ \(\frac{4}{5}\)]

Q6) \(2\over3\) x \(6\over8\) x \(16\over10\) - \(5\over8\)= [ \(\frac{7}{40}\)]

Q7) \(2\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{4}{9}\)]

Q7) \(2\over4\) x \(8\over10\) \(\div\) \(12\over20\)= [ \(\frac{2}{3}\)]

Q7) \(2\over3\) x \(9\over10\) - \(4\over10\)= [ \(\frac{1}{5}\)]

Q8) \(3\over4\)  \(\div\) \(9\over8\) =   [ \(\frac{2}{3}\)]

Q8) \(2\over3\) x \(6\over7\) \(\div\) \(12\over14\)= [ \(\frac{2}{3}\)]

Q8) \(2\over3\) x \(6\over7\) x \(21\over15\) - \(3\over7\)= [ \(\frac{13}{35}\)]

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

Q9) \(2\over3\) x \(9\over10\) x \(20\over13\)= [ \(\frac{12}{13}\)]

Q9) \(2\over3\) x \(6\over9\) - \(4\over9\)= [ 0]

Q10) \(3\over4\) x \(8\over10\) = [ \(\frac{3}{5}\)]

Q10) \(3\over4\) x \(8\over10\) x \(20\over15\)= [ \(\frac{4}{5}\)]

Q10) \(2\over4\) x \(8\over9\) - \(4\over9\)= [ 0]