Mr Daniels Maths
Surds:Division

Set 1

Set 2

Set 3

Q1) \(\sqrt 24 \over{ \sqrt{ 6}} \) = [ \(2\)]

Q1) \(5 \sqrt 12 \over{ \sqrt 3} \) = [ \(10\)]

Q1) \(4 \sqrt 3 \over{ 2 \sqrt 1} \) = [ \(2\sqrt{3}\)]

Q2) \(\sqrt 72 \over{ \sqrt{ 9}} \) = [ \(2\sqrt{2}\)]

Q2) \(5 \sqrt 3 \over{ \sqrt 1} \) = [ \(5\sqrt{3}\)]

Q2) \(9 \sqrt 12 \over{ 3 \sqrt 3} \) = [ \(6\)]

Q3) \(\sqrt 6 \over{ \sqrt{ 3}} \) = [ \(\sqrt{2}\)]

Q3) \(5 \sqrt 18 \over{ \sqrt 3} \) = [ \(5\sqrt{6}\)]

Q3) \(9 \sqrt 4 \over{ 3 \sqrt 1} \) = [ \(6\)]

Q4) \(\sqrt 28 \over{ \sqrt{ 4}} \) = [ \(\sqrt{7}\)]

Q4) \(3 \sqrt 16 \over{ \sqrt 8} \) = [ \(3\sqrt{2}\)]

Q4) \(25 \sqrt 8 \over{ 5 \sqrt 4} \) = [ \(5\sqrt{2}\)]

Q5) \(\sqrt 18 \over{ \sqrt{ 9}} \) = [ \(\sqrt{2}\)]

Q5) \(4 \sqrt 36 \over{ \sqrt 9} \) = [ \(8\)]

Q5) \(4 \sqrt 12 \over{ 2 \sqrt 4} \) = [ \(2\sqrt{3}\)]

Q6) \(\sqrt 90 \over{ \sqrt{ 10}} \) = [ \(3\)]

Q6) \(3 \sqrt 10 \over{ \sqrt 10} \) = [ \(3\)]

Q6) \(25 \sqrt 9 \over{ 5 \sqrt 3} \) = [ \(5\sqrt{3}\)]

Q7) \(\sqrt 50 \over{ \sqrt{ 5}} \) = [ \(\sqrt{10}\)]

Q7) \(4 \sqrt 4 \over{ \sqrt 1} \) = [ \(8\)]

Q7) \(10 \sqrt 12 \over{ 5 \sqrt 4} \) = [ \(2\sqrt{3}\)]

Q8) \(\sqrt 50 \over{ \sqrt{ 10}} \) = [ \(\sqrt{5}\)]

Q8) \(3 \sqrt 6 \over{ \sqrt 1} \) = [ \(3\sqrt{6}\)]

Q8) \(20 \sqrt 6 \over{ 5 \sqrt 2} \) = [ \(4\sqrt{3}\)]

Q9) \(\sqrt 15 \over{ \sqrt{ 3}} \) = [ \(\sqrt{5}\)]

Q9) \(4 \sqrt 60 \over{ \sqrt 10} \) = [ \(4\sqrt{6}\)]

Q9) \(8 \sqrt 6 \over{ 2 \sqrt 3} \) = [ \(4\sqrt{2}\)]

Q10) \(\sqrt 18 \over{ \sqrt{ 6}} \) = [ \(\sqrt{3}\)]

Q10) \(2 \sqrt 27 \over{ \sqrt 3} \) = [ \(6\)]

Q10) \(25 \sqrt 12 \over{ 5 \sqrt 4} \) = [ \(5\sqrt{3}\)]