A father age is four times of his elder son and five times that of his younger son; when hte elder son has lived to three times of his present age, hte father's age will exceed twice that of his younger son by four years.Find the present ages.
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Answer:
Father's age is f, son's ages are x and y, with x > y.
Equations are:
f = 4x
f = 5y
f+n = 2(y+n) + 3
x+n = 3x
(n = number of years to go until elder son is 3 times his present age. )
Solve these for n, f, x, y.
f+2x = 2(y+2x) + 3
f = 2y+2x + 3
but y = f/5 and x = f/4
f = 2f/5 + 2f/4 + 3
So f = 30
x = 7.5, y = 6, n = 15
Answers: father is 30, sons are 7.5 and 6.
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