Two years ago father was 5 times as old as his son two years laters his age will be 8 more than 3 times of the age of his son.find the precent age.
Asked by mukul rao | 17th Jun, 2013, 12:58: PM
Expert Answer:
Let the present age of the father be x years and present age of son be y years
Two year ago the age of father = x-2 years
Two year ago the age of son = y-2 year
Case I: x-2 = 5(y-2)
x-2 = 5y-10
x-5y = -8
5y-x = 8
5y-8 = x
Case II: x+2 =3(y+2) +8
Putting the value of y
5y-8+2=3y+6+8
2y = 20y = 10 years
Putting the value of y in equation 1
X = 5(10)-8 = 42years
Present age of father =42 years
Son's present age = 10 years
Answered by | 17th Jun, 2013, 05:03: PM
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