This is gcf or lcm problem...Dont know which one, but
There are less than 45 apples, but the number is divisible by both 3 and 5. This will mean it is divible by 15 too. if 45 is not an option, we will look at the next greatest multiple of 15 under that which is 30. 30 is the greatest possible number of apples.
So this is the problem:
A fruit stall owner had some apples and oranges. He sold an equal number of apples and oranges. He had 1/3 of the apples and 5/9 of the oranges left. What fraction of the fruits were sold?
2/3 of the apples is equal to 4/9 of the oranges. So...
2A/3 = 4O/9, and A is total apples and O is total oranges.
You want (2A/3 + 4O/9) / (A + O).
Look, that all of us can write O as 6A /4 or 3A/2.
So.... that means, the expression turns into (2A/3 + 4(3A/2)/9) / (A + 3A/2).
This simplifies to 8/15!!! Lol. The answer is 8/15.