In some ways I'm the complete opposite of prudie -- I find the algebra in these problems less stimulating than the set-up, since by that point you're following rules to such an extent that, in effect, the algebra is just doing itself and you're along for the ride.
In that sense, this is the same problem as the previous series of sweets questions, because the set-up is the same.
How do I read these questions? When I'm looking at it, I confess, my eyes glazed over most of the details, and I was just looking at various key words. In rough order, these were:
1. "there are some red and green sweets", where the only part I was really caring about was that there are TWO types of thing (again, I capitalise for emphasis not to shout).
2. Then I see "the ratio of..." and more or less stop reading, because I understand that this will be useful information later to relate the number of each type of thing to the total. I'll think about how to do that later.
3. After that, I read "two ... taken at random and eaten", so I recognise this is the exact same set-up as before. The "eaten" here is important because that means that the number of total sweets changes, so this is a "without replacement probability" problem.
4. Then I see "they're both [one type]", and again I stop, I don't care about the number.
So my breakdown of the problem is "two types... in some ratio... taken at random and [not replaced] ... both one type ... solve."
It's obviously not the *only* way you can solve this problem, but what I am trying to emphasise is the key information needed to get into this.
My advice to your grandson, if he's floundering in trying to get into the algebra, is to try and seek these key words. So the aim is to relate to what you've seen before, but *also* to understand how to use what you've seen before in a new way. So:
1. You have R red sweets and G green sweets.
2. The probability of eating two of the same sweet in a row is
(G/N)*(G-1)/(N-1)
if green (or (R/N)*(R-1)/(N-1) if red), and we are told that
(G/N)*((G-1)/(N-1)) = 0.35 = 7/20
Again, because I obsess about generality, I barely care about the actual probability, but that is a matter of taste, and if it helps to go concrete then by all means ignore my choice here.
I think it helps, though, to rearrange at this point and forget what else we know about red/green/total information. You get:
20G(G-1) = 7N(N-1) (1)
and a key point I'm trying to emphasise is that you can get to this point for literally all "two sweets" problems with no real "thought". The only question you now have to ask is "what is G in terms of N" (or vice versa).
Using the information given:
R + G = N
R:G = 2:3 => R = 2G/3
=> (2/3) G + G = (5/3) G = N
and now substitute in to (1). The algebra from there is more of the same, another quadratic equation in G, with some whole-number solution, that will be somewhere in the teens probably, so that N is in the region of 20-something. In this particular case, it's probably helpful to note that if R:G = 2:3 and R and G are both whole numbers, then N is a multiple of 5, ie your candidate solutions are N=15, N= 20, N=25, N=30 etc., although I wouldn't try using this method in the actual exam, more as a sanity check.
I leave it to Prudie, or anyone else who's interested, to do the algebra and confirm the solution (N=25), but as I understand from your last question, the algebra itself isn't the issue, so I don't think it's helpful for me to provide it. That's why I've burbled on and on at length about my approach to setting up the problem. It boils down in the end to seeing enough key words to write down the formula (1) above, and then think about how to relate Y and N for the specific problem.