Full explanation
We are given:
- Apples = 30c
- Bananas = 40c
- Oranges = 50c
- Total spent = €2 = 200c
Let: a = number of apples, b = number of bananas, o = number of oranges
Then: 30a+40b+50o=200 And all a,b,o are non-negative integers.
Let's evaluate each statement:
1. He cannot have exactly 3 apples. Assume a=3, so: 30×3=90c Leftover: 200−90 = 110c So, 40b+50o = 110 If we put an integer to the number of banana, it should give an integer solution which specify the number of oranges. If not, it is true that he cannot have exactly 3 apples. If b = 1, o = (70)/(50) = 1.4 (Not an integer) If b = 2, o = 0.6 (Not an integer) If b = 3, o = minus value No integer solution → Statement 1 is correct.
2. He must have at least one banana. Try if it's possible with no bananas, b=0: Then 30a+50o=200 If we try smallest value first : a = 0, then, 50o = 200 → o = 4 (an integer solution) So he can have no bananas → Statement 2 is incorrect.
3. He has 4, 5 or 6 fruits. Check all valid combinations that solve: 30a+40b+50o = 200 If a=0,b=0,o=4 → 5o × 4 = 200,Total = 4 fruits If a=2,b=1,o=2 : 60+40+100=200 → Total = 5 fruits If a=4,b=2,o=0: 120+80=200 → Total = 6 fruits Valid totals: 4, 5, 6 → Statement 3 is correct
4. If he has all 3 types of fruit, he must have fewer apples than bananas and oranges combined.
We can check the calculation of the statement 3. In the 5 fruit combination which contains all 3 types of fruit, the number of apples is indeed fewer than that of bananas and oranges combined.
Therefore, this statement 4 is correct.
Correct statements: 1, 3, and 4