Tale of 3 Adult Pigs

From the question, when writing the work done in 1 day for each pig duo, we can set up system of equations as shown:

$\dfrac{1}{a} +\dfrac{1}{b} =\dfrac{7}{12}$

$\dfrac{1}{b}+\dfrac{1}{c} = \dfrac{9}{12}$

$\dfrac{1}{c} + \dfrac{1}{a} = \dfrac{10}{12}$

Solving for the variables, we will get $a=3$ , $b=4$ , $c=2$ .

Thus, $a+b+c = \boxed{9}$

The first thing we need to do is to ascertain each pig’s rate of work, meaning how much work can each pig do per day. We are  told that A & B can work together to build 7 houses in 12 days. This means that in one day, A & B can build 7/12th of a house. Likewise, in one day, B & C can build 9/12th of a house and C & A can build 10/12th house We are also told that each individual pigs A, B and C can independently build each house on their own in a, b and c days. This means that the rate of work done per day by each pig is 1/a, 1/b and 1/c . Putting this information together, 1/a+1/b=7/12 ( equation 1) 1/b+1/c=9/12 (equation 2) 1/a+1/c=10/12 (equation 3) Now, we can solve the three unknowns, a, b and c, with substitution. Take Equation 2 minus Equation 1. 1/c – 1/a= 9/12 – 7/12 ( equation 4) = 2/12 Take Equation 4 plus Equation 3. 1/c + 1/c = 12/12 2/c=1, c= 2 From Equation 3, substitute c=2 into the equation 1/a+1/2=10/12 1/a= 10/12- 1/2= 10/12-6/12= 4/12=1/3 Therefore, 1/a=1/3 or a=3 From Equation 1, substitute a=3 into the equation 1/3+1/b=7/12 1/b= 7/12-4/12 Therefore, 1/b=3/12=1/4 or b=4 Hence, a+b+c= 3+4+2=9