Mr Daniels Maths
Surds:Division

Set 1

Set 2

Set 3

Q1) \(\sqrt 9 \over{ \sqrt{ 3}} \) = [ \(\sqrt{3}\)]

Q1) \(2 \sqrt 90 \over{ \sqrt 10} \) = [ \(6\)]

Q1) \(20 \sqrt 16 \over{ 5 \sqrt 4} \) = [ \(8\)]

Q2) \(\sqrt 70 \over{ \sqrt{ 7}} \) = [ \(\sqrt{10}\)]

Q2) \(3 \sqrt 48 \over{ \sqrt 6} \) = [ \(6\sqrt{2}\)]

Q2) \(4 \sqrt 8 \over{ 2 \sqrt 4} \) = [ \(2\sqrt{2}\)]

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

Q3) \(4 \sqrt 8 \over{ \sqrt 8} \) = [ \(4\)]

Q3) \(10 \sqrt 8 \over{ 5 \sqrt 4} \) = [ \(2\sqrt{2}\)]

Q4) \(\sqrt 21 \over{ \sqrt{ 7}} \) = [ \(\sqrt{3}\)]

Q4) \(3 \sqrt 36 \over{ \sqrt 9} \) = [ \(6\)]

Q4) \(15 \sqrt 16 \over{ 3 \sqrt 4} \) = [ \(10\)]

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

Q5) \(3 \sqrt 70 \over{ \sqrt 10} \) = [ \(3\sqrt{7}\)]

Q5) \(6 \sqrt 20 \over{ 2 \sqrt 4} \) = [ \(3\sqrt{5}\)]

Q6) \(\sqrt 8 \over{ \sqrt{ 4}} \) = [ \(\sqrt{2}\)]

Q6) \(3 \sqrt 5 \over{ \sqrt 1} \) = [ \(3\sqrt{5}\)]

Q6) \(4 \sqrt 10 \over{ 2 \sqrt 5} \) = [ \(2\sqrt{2}\)]

Q7) \(\sqrt 20 \over{ \sqrt{ 10}} \) = [ \(\sqrt{2}\)]

Q7) \(5 \sqrt 40 \over{ \sqrt 5} \) = [ \(10\sqrt{2}\)]

Q7) \(10 \sqrt 6 \over{ 5 \sqrt 3} \) = [ \(2\sqrt{2}\)]

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

Q8) \(5 \sqrt 16 \over{ \sqrt 2} \) = [ \(10\sqrt{2}\)]

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

Q9) \(\sqrt 20 \over{ \sqrt{ 5}} \) = [ \(2\)]

Q9) \(3 \sqrt 40 \over{ \sqrt 5} \) = [ \(6\sqrt{2}\)]

Q9) \(20 \sqrt 6 \over{ 4 \sqrt 2} \) = [ \(5\sqrt{3}\)]

Q10) \(\sqrt 72 \over{ \sqrt{ 9}} \) = [ \(2\sqrt{2}\)]

Q10) \(3 \sqrt 81 \over{ \sqrt 9} \) = [ \(9\)]

Q10) \(6 \sqrt 20 \over{ 2 \sqrt 5} \) = [ \(6\)]