Question

41. There are three societies A, B and C, each having some tractors. A gives B and C as many tractors as they already have. After some days, B gives A and C as many tractors as they then have. After some more days, C gives A and B as many tractors as they then have. Finally, each society has 24 tractors. What was the original number of tractors each had at the start?

• A. A-29, B-21, C-12
• B. A-39, B-21, C-12
• C. A-21, B-12, C-29
• D. A-21, B-12, C-39

Hint:

Work backward from the final 24-24-24 using the fact that each round exactly doubles two of the three amounts.

Answer 1

The Correct Option is B

Instead of deriving the equations, we can just test option B forward through the three rounds and see if it lands on 24, 24, 24.

1. A-29, B-21, C-12: starting with 29, after A gives out (21+12)=33 tractors, A would go negative, so this start is impossible.
2. A-39, B-21, C-12: start $a=39, b=21, c=12$. Round 1 (A gives): $b$ becomes $42$, $c$ becomes $24$, $a = 39-33=6$. Round 2 (B gives): $a$ becomes $12$, $c$ becomes $48$, $b = 42-30=12$. Round 3 (C gives): $a$ becomes $24$, $b$ becomes $24$, $c=48-24=24$. All three land on 24, so this option works.
3. A-21, B-12, C-29: simulating these numbers forward does not reproduce the same 24-24-24 finish, so it is wrong.
4. A-21, B-12, C-39: this also fails to end at 24-24-24 when simulated forward, so it is wrong.

Simulating option B's numbers round by round is a direct check that avoids solving the algebra at all.

Let's summarize:

• Forward simulation of $a=39,b=21,c=12$ gives $24,24,24$ after three rounds.
• No other option reproduces this finish.

So the starting numbers are A-39, B-21, C-12, option B.