Problem 1a: If 4 glasses of a mixture needs 1 cup of sugar how many cups of sugar are needed for 5 glasses?
This one is easy and makes sense. It's just simply $\frac{1}{4}*5$ Now taking it a notch higher:
Problem 1b: If 4 glasses of a mixture needs 1 cup of sugar and 6 glasses need 2 cups, how cups of sugar are needed for 7 glasses?
Geometrically you can solve the problem by plotting a line between (4,1) and (6,2); calculate the slope/intercept and get the answer. But how can I do this (algebraically) from a 5th grader's perspective?
Problem 1c: (A variation) If 4 glasses need 1 cup sugar, 6 glasses need 2 and 8 glasses need 4 then how many cups of sugar do I need for 7 glasses?
Geometrically, you get a triangular region and there are two solutions to the problem! How could a 5th grader solve this variation algebraically? Geometrically it's easy, but since there are two answers, how can a 5th grader interpret the solution since there isn't a unique solution?
(Background: my nephew came to me with #2 and wasn't aware of slope/intercepts but was struggling with solving it using simple algebra. Maybe I'm overlooking something but it seems to be a bit tricky for a 5th grader. What are some good ways of approaching these sets of problems?)
I'm not sure there is a good answer to 1b. Probably the answer you give is the best available. For a 5th grader, the last 2 glasses of water needed 1 cup, so one more glass will need 1/2 cup more for a total of 2 1/2. For an algebra student you could fit a quadratic through (0,0), (4,1), and (6,2), which gives approximately (7,2.6264).
For 1c I would do the same and say 7 is halfway from 6 to 8, so I need halfway between 2 and 4 cups of sugar, getting 3. But I'm not sure I believe that answer either.