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

Step 1: Set up variables and the final condition.
Let A, B and C start with \( a \), \( b \), and \( c \) tractors. After three rounds of doubling, each ends with 24 tractors.

Step 2: Work out A's tractors after all three rounds.
A only gives tractors away and never receives any, so after the full process A ends with \( 4(a-b-c) = 24 \), which gives \( a - b - c = 6 \).

Step 3: Work out B's count after all three rounds.
Tracking B through each round gives the final amount \( 6b - 2a - 2c = 24 \), which simplifies to \( 3b - a - c = 12 \).

Step 4: Combine the two equations.
Since \( a = 6 + b + c \), substitute into \( 3b - a - c = 12 \) to get \( 3b - (6+b+c) - c = 12 \), so \( 2b - 2c = 18 \), giving \( b - c = 9 \).

Step 5: Use C's final equation to solve the system.
Tracking C gives \( -a - b + 7c = 24 \). Substituting \( a = 6+b+c \) and \( b = c+9 \) into this equation reduces it to \( 4c = 48 \), so \( c = 12 \).

Step 6: Back-substitute to find b and a.
\( b = c + 9 = 21 \), and \( a = 6 + b + c = 6 + 21 + 12 = 39 \).

Final Answer:
The starting counts are A = 39, B = 21, C = 12. \[ \boxed{A=39,\ B=21,\ C=12} \]