From certain number of apples, a man sells 1/3 of the apples to the first customer. He sells 1/2 of the remaining apples to the second customer, and 1/3 of the remaining apples plus 5 to the third customer. he then finds himself left with 3 apples. How many apples did the man have initially?

Correct option is C

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Solution:

Let the initial number of apples be x 

For  First Customer:

- The man sells 1/3 of the apples to the first customer.

- Apples remaining =  x - $$\frac{x}{3}$$​ = $$\frac{2x}{3}$$​

For the Second Customer:

- The man sells 1/2 of the remaining apples to the second customer.

- Apples remaining =  $$\frac{2x}{3} - \frac{1}{2} \times \frac{2x}{3} = \frac{x}{3}$$​​

For the Third Customer:

- The man sells 1/3 of the remaining apples plus 5 to the third customer.

- Since he is left with 3 apples after this sale, we set up the following equation:

$$\frac{1}{3} \times \frac{x}{3} + 5 + 3 = \frac{x}{3}\\ \ \\ \frac{x}{9} +8 = \frac x3 \\ \ \\ \frac{3x-x}{9} = 8 \\ \ \\ 2x = 72 \\ \ \\ x = 36$$​