There are 3 societies a,b,c. a lent tractors to b and c as many as they had. After some time b gave as many tractors to a and c as many as they have. After sometime c did the same thing. At the end of this transaction each one of them had 24. Find the tractors each originally had.
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Answer / Shivam Kumar
Let A, B, and C be the number of tractors initially held by societies a, b, and c, respectively. Since society a gave as many tractors to b and c as they had, we have A = B + C. After giving away their tractors, society b has some number x tractors left, which it then gives equally to a and c. This means that (x/2) + (x/2) = A-x. Solving for x gives x = 12A - 24. Society c does the same thing as society b, giving away some number y tractors and receiving (y/2) back. This means that y + (y/2) = B - y. Solving for y gives y = 16B - 32. Since each one of them had 24 at the end, we have A + B + C = 3*24. Substituting our expressions for A and B from above, we get (B+C) + B + (B+C) = 72. Simplifying this equation gives 3B + 2C = 48. Since each one of them had 24 at the end, we have A + B + C = 3*24. Substituting our expression for C from above, we get (12A - 24) + B + (16B - 32) = 72. Solving this system of equations gives A = 8 and B = 10.
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