Three animal friends!!

Let the weight of the dog be D , the weight of rabbit be R and the weight of the cat be C .

According to the question, C+R=10 kg

And, R+D= 20 kg

And, D+C=24 kg

Now, add all the equations above, i.e, C + R + R + D + D + C = 10 + 20 + 24 kg .

I.e, 2( C+R+D )= 54 kg

Or, C+R+D = $\displaystyle\frac{54}{2}$ = 27 kg

Let the weights of dog be x cat be y and rabbit be z. According to the problem: $y+z=10$ & $x+z=20$ & $x+y=24$ . $x+z=20\quad \therefore \quad z=20-x\rightarrow Eq\quad 1$ .

Let $y+z=10\rightarrow Eq\quad 2$ . Substitute Eq1 in Eq2:- We will get: $y-x=-10$ . Add $y-x=-10\quad \& \quad y+x=24$ and we will get $y=7$ .

On Substituting $y=7$ in the other equations we will get finally $x=17\quad \& \quad y=7\quad \& \quad z=3$ .

Add $x,y,z$ to the the sum of weights of the rabbit, dog & cat is $\boxed{27 kg}$ .

let the notations for dog , cat and rabbit be d , c & r

hence c+r =10

d+r= 20

d+c= 24

also d+c = 20 +4

i.e d+c = d+r +4

i.e c= r+ 4

i.e from 1 we get r= 3

from 2 d= 17 hence c = 7

total weight= 27

half of the total of the 3 given weights = half of (10+20+24) = half of 54 = 27.