Apples can be packed in sets of 10 apples in a box of type A or 25 apples in a…

A linear equation of the form px + qy = N has a family of non-negative integer solutions that can be counted by reducing it using the GCD of p and q, isolating one variable as a function of a parameter, and finding how many parameter values keep both variables non-negative — the valid values then form an arithmetic progression (AP), whose term count is (last minus first) divided by common difference, plus 1.

1. Let a = number of type A boxes (10 apples each) and b = number of type B boxes (25 apples each); then 10a + 25b = 1200.

2. Divide throughout by 5 (the GCD of 10, 25 and 1200): 2a + 5b = 240.

3. For a to be a whole number, 5b must be even, so b itself must be even; write b = 2m for m = 0, 1, 2, ...

4. Substituting gives 2a = 240 − 10m, so a = 120 − 5m.

5. Requiring a ≥ 0 gives m ≤ 24, and b ≥ 0 gives m ≥ 0, so m can take any integer value from 0 to 24.

6. The number of valid combinations equals the number of integers in this range: (24 − 0) ÷ 1 + 1 = 25.

Checking the two extremes confirms the range: m = 0 gives (a, b) = (120, 0), and 10(120) + 25(0) = 1200, which checks out; m = 24 gives (a, b) = (0, 48), and 25(48) = 1200, which also checks out. Both boundary combinations satisfy the equation, so all 25 intermediate values of m are valid too.

So exactly 25 different combinations of type A and type B boxes can pack the 1200 apples.