Math, asked by blah94, 3 months ago

the age of John's father is three times the age of John five years ago, the age of the father was four-time John's age that time. find their present ages

Answers

Answered by itsesarthak
2

Answer:

John’s Father = F

John = J

sister Alice = A

F = 5*J

J = 2*A

In two years the sum of their ages (everyone’s age + 2):

(F+2) + (J + 2) + (A + 2) = 58

we get three equations:

F = 5*J

J = 2*A

F + J + A + 6 = 58

We rewrite J = 2*A as A = J/2

F = 5*J

A = J/2

F + J + A = 58 - 6

We apply F and A values in latest equation:

5*J + J + J/2 = 52

sum

6.5*J = 52

J = 8

Therefore John’s Ages is: 8 years

(Father’s = 8*5 = 40)

(Sister = 8/2 = 4)

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