SOLUTION: a father is twice older than his son and the sum of their ages is 48.How old is each?

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Question 567885: a father is twice older than his son and the sum of their ages is 48.How old is each?
Found 3 solutions by Alan3354, Gurajan, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
a father is twice older than his son and the sum of their ages is 48.How old is each?
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Twice = 2 times
2 times older = 3 times as old
--> 12 & 36

Answer by Gurajan(1)   (Show Source): You can put this solution on YOUR website!
Let age of son be x years.
Age of father's be y years
Now according to ques.
x+y=48
Y=2x
So eq.becomes,
x+y=48
2x-y=0
Adding these eq.
3x=48
x=16
y=2x =2×16=32
Age of son is 16
Age of father is 32.
Ik

Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
Let the son's age is x years.

Then the father's age is 3x years,  (twice older means "three times old as") and the condition says that

x + 3x = 48.


Hence,  4x = 48  and x =  = 12.


Answer.  The son is 12 years old.  The father is 36 years old.


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