There are 3 societies A,B,C having some tractors each. 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 have. After some days C gives A and B as many tractors as they have. Finally each has 24 tractors.what is the original No.of tractors each had in the beginning?
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Answer / Priya
This question involves logical reasoning rather than mathematical operations. Let's denote the initial number of tractors for A, B, and C as 'a', 'b', and 'c'. After A gives B and C their respective numbers of tractors, we have a - b = b - c and b - a = c - b. Solving these equations simultaneously, we get a = (b + c) / 2. Similarly, after B gives A and C their respective numbers of tractors, we have a - b = c - a and c - b = a - c. Solving these equations, we get b = (a + c) / 2. After C gives A and B their respective numbers of tractors, we have a - c = b - a and b - c = a - b. Solving these equations, we get c = (a + b) / 2. Finally, when each society has 24 tractors, we have a + b + c = 72. We can now substitute the relationships derived earlier to solve for 'a', 'b', and 'c'. By trial and error or using algebraic methods, we find that the original number of tractors each had was 16.
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