Question 10096: Five years ago , a father awas twice as old as his son. In 4 years'time the sum of their ages will be 78. Find their present ages.
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! ok, asked to find present ages, so define:
father currently = x
son currently = y
So,
five years ago, father was x-5
five years ago, son was y-5
and,
in 4 years time, father will be x+4
in 4 years time, son will be y+4
Right then, we need 2 equations since we have 2 unknowns:
1. 5 years ago, father was twice as old as son:
--> (x-5) = 2(y-5)
--> x-5 = 2y-10
--> x-2y = -5
2. in 4 years time, sum of ages = 78
--> (x+4) + (y+4) = 78
--> x+y +8 = 78
--> x+y=70
So, now solve the 2 equations:
x-2y = -5
x+y = 70
Subtract them to leave -3y = -75 --> y = 25, so son is now 25.
put this info into one of the 2 equations: pick either. I shall use x+y=70 --> means x must be 70-25, so father is now 45.
NOW, check!
5 years ago, ages were 40 and 20... CORRECT doubled age!
4 years time, ages will be 49 and 29...add together to make 78!
jon.
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