Problema Solution
A man has two children, a boy and a girl. The boy is 6 years older than the girl. Two years ago the fatther's age was three times the sum of the children's ages. In eight years the father's age will be ten years greater thean the sum of the children's ages. Find the present ages of the father, the boy, and the girl.
Answer provided by our tutors
f = fathers age
b = boys age
g = girls age
b = g + 6
f-2 = 3*(b + g)
f+8 = (b+8) + (g+8) + 10
solving eqn 2 in terms of b and g gives
f = 2 + 3*(b + g)
subbing in eqn 1 gives
f = 2 + 3*(g +6 +g)
subing this into eqn 3 we get
2+ 3*(g +6 +g) + 8= 6+g +8 +g+ 8 +10
simplify
6g + 28 = 2g + 32
4g = 4
g =1
b = 1 + 6
b = 7
f = 3*(7 + 1)+2
f = 24 +2
f = 26
so the girl is 1 years old, the boy is 7 years old, and the father is 26 years old