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 plus $1$ to the third customer. He then finds that he has $3$ apples left. How many apples did he have originally?

Concept & Strategy The most efficient way to solve successive division problems (like selling fractions of remaining amounts) is the **Working Backwards** approach. By starting with the final remaining quantity and reversing the operations, we can easily find the original amount without setting up complex algebraic equations. Step-by-Step Solution * **Final state:** $3$ apples are left. * **Before the third customer:** The man sold $\frac{1}{5}$th of his apples plus $1$, leaving him with $3$. Let the amount before the 3rd customer be $y$. Equation: $y - (\frac{1}{5}y + 1) = 3$ $\frac{4}{5}y - 1 = 3 \Rightarrow \frac{4}{5}y = 4 \Rightarrow y = 5$ * **Before the second customer:** He sold $\frac{1}{3}$rd plus $1$, leaving him with $5$. Let the amount before the 2nd customer be $z$. Equation: $z - (\frac{1}{3}z + 1) = 5$ $\frac{2}{3}z - 1 = 5 \Rightarrow \frac{2}{3}z = 6 \Rightarrow z = 9$ * **Before the first customer (Original Amount):** He sold half plus $1$, leaving him with $9$. Let the original amount be $x$. Equation: $x - (\frac{1}{2}x + 1) = 9$ $\frac{1}{2}x - 1 = 9 \Rightarrow \frac{1}{2}x = 10 \Rightarrow x = 20$ Exam Strategy & Shortcut **Option Elimination (Reverse Check)**: Instead of writing algebraic equations, quickly test the options! Let's test Option (c) $20$. * Start: $20$ * Sell half plus $1$: $10 + 1 = 11$. Remaining: $9$. * Sell one-third plus $1$: $\frac{1}{3}(9) + 1 = 3 + 1 = 4$. Remaining: $5$. * Sell one-fifth plus $1$: $\frac{1}{5}(5) + 1 = 1 + 1 = 2$. Remaining: $3$. This perfectly matches the given $3$ remaining apples. This method is visually intuitive and much faster during exams. Common Pitfall Students often try to set up a single, massive equation in terms of $x$ right from the beginning. This leads to heavy nested fractions, high chances of arithmetic errors, and wastes valuable exam time. Always work step-by-step backwards or test options. Final Answer Therefore, the correct answer is **20**.