A fruit seller sells 45% of the oranges that he has along with one more orange to a customer. He then sells 20% of the remaining oranges and 2 more oranges to a second customer. He then sells 90% of the now remaining oranges to a third customer and is still left with 5 oranges. How many oranges did the fruit seller have initially?

Calculation:

Let the initial oranges with the fruit seller be x.

1st selling = 0.45x + 1

Remaining = x - (0.45x + 1) = 0.55x - 1

2nd selling = \(1\over 5\) × (0.55x - 1) = 0.11x - 0.2 + 2 = 0.11x + 1.8

Remaining after second selling = 0.55x - 1 - (0.11x + 1.8) = 0.55x - 0.11x - 1 - 1.8 = 0.44x - 2.8

3rd selling = 90% × (0.44x - 2.8)

Remaining after 3rd selling = 0.1 × (0.44x - 2.8) = 0.044x - 0.28

According to the question-

⇒ 0.044x - 0.28 = 5

⇒ 0.044x = 5.28

⇒ x = \(5.28\over 0.044\) = 120

∴ The number of oranges was 120. 

Alternate Method

At last, he sells 90% of the remaining oranges after selling the oranges to a second customer, then he has 10% of the remaining oranges.

10% of the remaining oranges after selling the oranges to the second customer = 5

So remaining oranges after selling the oranges to the second customer = 100% of the remaining oranges after selling the oranges to the second customer = 50 oranges

He gave 2 extra oranges to the second customer, so the remaining oranges = 50 + 2

He sells 20% of the remaining oranges to the second customer, so he has 80% of the remaining oranges = 52

100% of remaining oranges after selling the oranges to the first customer = (52/4) * 5 = 65 oranges

He gave 1 extra orange to the first customer, so the total oranges after selling 45% of the oranges = 65 + 1 = 66 oranges

(100% - 45% = 55%) of total oranges = 66

so

100% of oranges = (66/55) * 100 = 120 oranges