Question 1039147: A MAN IS FOUR TIME AS OLD AS HIS SON, FOUR YEARS AGO THE SUM OF THEIR AGES WAS 48 YEARS. FIND THEIR PRESENT AGES. Found 2 solutions by jorel555, Alan3354:Answer by jorel555(1290) (Show Source):
You can put this solution on YOUR website! Let n be the son. Then the father is 4n. Four years ago, their ages were n-4 and
4n-4. So
n-4+4n-4=48
5n=56
n=56/5=11.2
4n=44.8
The son is 11.2 years, and the father is 44.8 years!!!!!!!!!!!
You can put this solution on YOUR website! A MAN IS FOUR TIME AS OLD AS HIS SON, FOUR YEARS AGO THE SUM OF THEIR AGES WAS 48 YEARS. FIND THEIR PRESENT AGES.
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The sum of their ages now is 48+4+4 = 56.
M = 4S
M + S = 56
4S + S = 56
S = 11.2 - Son's age
M = 44.8 - Man's age
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