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?

Let the no. of tractors with A, B, and C be a, b, and c respectively.

After A gives out tractors to both B and C, the number of tractors with B becomes 2b and that with C becomes 2c. The remaining tractors with A becomes a-(b+c).

After B's turn to give out tractors, A has 2{a-(b+c}, C has 2(2c)= 4c, and B will be left with 2b-(a-b-c)-2c tractors.

Finally after C's turn, A has 4{a-(b+c)} = 4(a-b-c) tractors, B has 4b-2(a-b-c)-4c = 4b-2a+2b+2c-4c=6b-2a-2c tractors. And C is left with 4c-2(a-b-c)-{2b-(a-b-c)-2c}= 4c-2a+2b+2c-(2b-a+b+c-2c)= 4c-2a+2b+2c-2b+a-b+c= -a-b+7c tractors

It is given that they are all left with 24 tractors in the end.

For A, 4(a-b-c)=24

=> a-b-c=6

=> a= 6+b+c.

For B, 6b-2a-2c=24

=> 3b-a-c=12

=>3b-6-b-c-c=12

=>2b-2c=18

=>b-c=9

=>b=c+9.

For C, -a-b+7c=24

=>-6-b-c-c-9+7c=24

=>5c-b=39

=>5c-c-9=39

=>4c=48

.'. c=12.

b=c+9 

.'. b=21.

a=6+b+c

=> a= 6+21+12

.'. a=39.