Colin paid $12 for 4 buns and 4 cakes. 3 buns cost as much as 2 cakes. What was the total cost of 6 buns and 9 cakes?

Answer 1

4b + 4c = 12 or b + c = 3
and 3b = 2c or 6b = 4c
sub that back into the first equation
4b + 6b = 12
10b = 12
b = 12/10 = 1.2 ---> so 1 bun costs $1.20
then 1.2 + c = 3 -----> c = $1.80

6 buns + 9cakes = 6*1.2 + 9*1.8 = $23.40

Not sure if this algebraic solution is suitable for grade 5
Don't know what method you have learned for this type of problem.