$A$ gives $B$ as many rupees as $B$ has and $C$ as many rupees as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many rupees as each then has. $C$, similarly, then gives $A$ and $B$ as many rupees as each then has. If each finally has ₹ 16, with how many rupees does $A$ start?

## Concept & Logic This is a classic "working backwards" problem. Since the final state and the operational rules (doubling the other people's money) are known, reversing the operations step-by-step from the end to the beginning is the most efficient strategy. "Giving someone as many rupees as they have" is exactly the same as multiplying their current amount by 2. Therefore, when working backwards, the people who *received* money will have their amounts halved. ## Step-by-Step Solution * **Final State (After C's turn):** * $A = 16$ * $B = 16$ * $C = 16$ * (Notice the total money in the system is always $16 + 16 + 16 = 48$) * **Reverse Step 1 (Before C gave to A and B):** * $C$ doubled $A$ and $B$'s money to reach 16. So before this, $A$ and $B$ must have had half. * $A = 16 \div 2 = 8$ * $B = 16 \div 2 = 8$ * $C$ gave away $8$ to $A$ and $8$ to $B$ (Total 16). So $C$ must have had $16 + 16 = 32$. * State is now: **$A=8, B=8, C=32$** * **Reverse Step 2 (Before B gave to A and C):** * $B$ doubled $A$ and $C$'s money to reach 8 and 32. * $A = 8 \div 2 = 4$ * $C = 32 \div 2 = 16$ * $B$ gave away $4$ to $A$ and $16$ to $C$ (Total 20). So $B$ must have had $8 + 20 = 28$. * State is now: **$A=4, B=28, C=16$** * **Reverse Step 3 (Before A gave to B and C - Initial State):** * $A$ doubled $B$ and $C$'s money to reach 28 and 16. * $B = 28 \div 2 = 14$ * $C = 16 \div 2 = 8$ * $A$ gave away $14$ to $B$ and $8$ to $C$ (Total 22). So $A$ must have had $4 + 22 = 26$. * State is now: **$A=26, B=14, C=8$** ## Exam Strategy & Shortcut Use a quick table and the rule: "Total is always 48. To reverse a turn, halve the receivers, and subtract their new sum from 48 to find the giver." | Turn | A | B | C | | :--- | :--- | :--- | :--- | | **End** | 16 | 16 | 16 | | **Pre-C** | 8 | 8 | 48 - (8+8) = 32 | | **Pre-B** | 4 | 48 - (4+16) = 28 | 16 | | **Start**| 48 - (14+8) = 26 | 14 | 8 | This takes less than 30 seconds to scribble on scratch paper. ## Common Pitfall Trying to build forward-moving algebraic equations. You end up with fractions and nested parentheses like $2(2(a - b - c) - 2b) = 16$, which guarantees calculation errors under time pressure. Always reverse-engineer terminal state problems. ## Final Answer Therefore, the correct answer is **26**.