Maths problem solving today

Today, we had a bit of fun solving this problem. 

Mr Nath gave Mrs Tofa the alphabet saying each letter had a value. He told her she had to work out the value of each letter using two equations. He said:

A = 3 

B = 12 

C = 27 

D = 48

E = 75

F =108

Here are some strategies some of our class members came up with:

A = 3  1.5  + 1.5     (=3)

B = 12 2.5 +2.5 (5)(=12)

C = 27

D = 48

E = 75

A = 3  (1X3) = 3 (x 2) = 3

B = 12 (2x2) = 4 (x3) = 12

C = 18 

D = 48

E = 75

A = 3     3 - 2 = 1 + 2 = 3

B = 12   12 - 2 = 10 + 2 = 12 

C = 27   27 - 2 = 25 + 2 = 27

D = 48   48 - 2 = 46 + 2 = 48

E = 75   75 - 2 = 73 + 2 = 75

F = 108 108 - 2 = 106 + 2 = 108

Here is a solution we agreed on at the end

A = 3 ( 1 x 1)= 1 x 3=3

B = 12 (2 x 2)= 4 x 3=12

C = 27 (3 x 3)= 9 x 3=27

D = 48 (4 x 4)=16 x 3= 48

E = 75 (5 x 5)=25 x 3= 75

F =108 (6 x 6)= 36 x 3= 108

or

The number of the letter in the alphabet sequence squared X 3 = value of the letter. 

The class then had to work out the value of each letter of the alphabet and add the value of their names. It was quite fun to see the value of our names.