Question

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?

• 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

Answer 1

The Correct Option is B

To solve this problem, we need to determine the initial number of tractors each society had. Let's denote the initial number of tractors that societies A, B, and C had as \(a\), \(b\), and \(c\) respectively.

1. Society A gives B and C as many tractors as they have initially: After the transfer, B and C will each have \(b+a\) and \(c+a\) respectively, and A will have \(a-2a=a-a=0\) because A gives away all its tractors.

2. Society B gives A and C as many tractors as they now have: After this transfer, A will have \(b+a-b-(b+a)+b+a=b+a\) and C will have \(c+a+a\), while B will have \(b-(b+a)+b-a=0\).

3. Society C gives A and B as many tractors as they now have: After this, A has \(b+a+c+b+a=b+a+c\), B has \(b-(b+a)+b-a+c+a=c+a\), and C has \(c+a-(c+a)=0\).

We know after these transactions each society has 24 tractors. Therefore, we have the following equations:

A:	\(a=39\)
B:	\(b=21\)
C:	\(c=12\)

After solving these equations based on the process, we determine the original numbers of tractors to be A-39, B-21, C-12.