Problem: From a number of apples, a man sells half the number of existing apples plus 1 to the first customer, sells $\frac{1}{3}$ rd of  the remaining apples plus 1 to the second customer and $\frac{1}{5}$ th of the remaining apples pluse 1 to third custormer . He then finds that he has 3 apples \left. How many apples did he have originally? 

Correct Answer: Originally he had 20 apples.

Explanation:

Let, the number of apples be x

The man sells apples to the first customer =$\frac{x}{2}+1=\frac{x+2}{2}$

Remaining Apples = x-$\frac{x+2}{2}=\frac{2x-x-2}{2}=\frac{x-2}{2}$

He sells apples to the second customer = $\frac{1}{3}(\frac{x-2}{2})+1=\frac{x-2}{6}+1=\frac{x-2+6}{6}=\frac{x+4}{6}$

Remaining apples = $\frac{x-2}{2}-\frac{x-10}{15}=\frac{5(x-5)-(x+10)}{15}=\frac{5x-25-10}{15}$ =$\frac{4x-35}{15}$

According to the question,

$\frac{4x-35}{15}$=3

=> 4x-35=45

=> 4x = 45+35 =80

so, x= $\frac{80}{4}$=20