Mr Daniels Maths
Surds:Division

Set 1

Set 2

Set 3

Q1) \(\sqrt 36 \over{ \sqrt{ 6}} \) = [ \(\sqrt{6}\)]

Q1) \(4 \sqrt 10 \over{ \sqrt 10} \) = [ \(4\)]

Q1) \(25 \sqrt 6 \over{ 5 \sqrt 2} \) = [ \(5\sqrt{3}\)]

Q2) \(\sqrt 12 \over{ \sqrt{ 2}} \) = [ \(\sqrt{6}\)]

Q2) \(3 \sqrt 63 \over{ \sqrt 7} \) = [ \(9\)]

Q2) \(12 \sqrt 9 \over{ 4 \sqrt 3} \) = [ \(3\sqrt{3}\)]

Q3) \(\sqrt 4 \over{ \sqrt{ 2}} \) = [ \(\sqrt{2}\)]

Q3) \(2 \sqrt 30 \over{ \sqrt 3} \) = [ \(2\sqrt{10}\)]

Q3) \(12 \sqrt 20 \over{ 3 \sqrt 4} \) = [ \(4\sqrt{5}\)]

Q4) \(\sqrt 90 \over{ \sqrt{ 9}} \) = [ \(\sqrt{10}\)]

Q4) \(2 \sqrt 36 \over{ \sqrt 4} \) = [ \(6\)]

Q4) \(4 \sqrt 15 \over{ 2 \sqrt 3} \) = [ \(2\sqrt{5}\)]

Q5) \(\sqrt 8 \over{ \sqrt{ 2}} \) = [ \(2\)]

Q5) \(3 \sqrt 24 \over{ \sqrt 3} \) = [ \(6\sqrt{2}\)]

Q5) \(8 \sqrt 15 \over{ 2 \sqrt 5} \) = [ \(4\sqrt{3}\)]

Q6) \(\sqrt 7 \over{ \sqrt{ 1}} \) = [ \(\sqrt{7}\)]

Q6) \(4 \sqrt 60 \over{ \sqrt 6} \) = [ \(4\sqrt{10}\)]

Q6) \(10 \sqrt 10 \over{ 5 \sqrt 2} \) = [ \(2\sqrt{5}\)]

Q7) \(\sqrt 20 \over{ \sqrt{ 5}} \) = [ \(2\)]

Q7) \(3 \sqrt 15 \over{ \sqrt 3} \) = [ \(3\sqrt{5}\)]

Q7) \(6 \sqrt 9 \over{ 2 \sqrt 3} \) = [ \(3\sqrt{3}\)]

Q8) \(\sqrt 4 \over{ \sqrt{ 1}} \) = [ \(2\)]

Q8) \(2 \sqrt 90 \over{ \sqrt 10} \) = [ \(6\)]

Q8) \(4 \sqrt 8 \over{ 2 \sqrt 4} \) = [ \(2\sqrt{2}\)]

Q9) \(\sqrt 72 \over{ \sqrt{ 9}} \) = [ \(2\sqrt{2}\)]

Q9) \(3 \sqrt 6 \over{ \sqrt 3} \) = [ \(3\sqrt{2}\)]

Q9) \(16 \sqrt 12 \over{ 4 \sqrt 4} \) = [ \(4\sqrt{3}\)]

Q10) \(\sqrt 8 \over{ \sqrt{ 4}} \) = [ \(\sqrt{2}\)]

Q10) \(4 \sqrt 2 \over{ \sqrt 1} \) = [ \(4\sqrt{2}\)]

Q10) \(25 \sqrt 6 \over{ 5 \sqrt 3} \) = [ \(5\sqrt{2}\)]