₦airaland Forum

What is the combined weight of the dog, cat and rabbit?

17+7+3=27

combined weight of the rabbit, cat and dog is 27kg

rabbit = r
cat = c
dog = d

r + c = 10
d + r = 20
d + c = 24
r+c+d+r+d+c = 10+20+24
2r + 2c + 2d = 10 + 20 + 24
r + c + d = (10 + 20 + 24 )/2 = 5 + 10 + 12 = 27

27kg

goofyone:
rabbit = r
cat = c
dog = d

r + c = 10
d + r = 20
d + c = 24
r+c+d+r+d+c = 10+20+24
2r + 2c + 2d = 10 + 20 + 24
r + c + d = (10 + 20 + 24 )/2 = 5 + 10 + 12 = 27

27kg

Hmmm different figures and workings but the same answers...crosscheck your solvings
r=rat
c=cat
d=dog
r+c=10...eqn1
r+d=20...eqn2
c+d=24...eqn3
r+c+d=?
From eqn1, r=10-c
substitute r for (10-c) into eqn2
(10-c)+d=20
-c+d=20-10
d-c=10...eqn4
Subtract eqn4 from eqn3
(d+c=24)-(d-c=10)
(d-d)+c--c)=24-10
c+c=14
c=7
From eqn1...r+c=10
r+7=10
r=3
From eqn2...r+d=20
3+d=20
d=17
Total= 7+3+17=27

Valued:
Hmmm different figures and workings but the same answers...crosscheck your solvings
r=rat
c=cat
d=dog
r+c=10...eqn1
r+d=20...eqn2
c+d=24...eqn3
r+c+d=?
From eqn1, r=10-c
substitute r for (10-c) into eqn2
(10-c)+d=20
-c+d=20-10
d-c=10...eqn4
Subtract eqn4 from eqn3
(d+c=24)-(d-c=10)
(d-d)+c--c)=24-10
c+c=14
c=7
From eqn1...r+c=10
r+7=10
r=3
From eqn2...r+d=20
3+d=20
d=17
Total= 7+3+17=27

Even though absolutely correct, that's just too long for a solution. Simple is always better.