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.
Answers
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.
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.
Step-by-step explanation: