Math, asked by goku93, 1 year ago

Two years ago father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages.

Answers

Answered by shreeram9
5
let present age of the father be x
and present age of his son be y

according to the problem,

x-2 = 3(y-2)
2(x+2)=5(y+2)

x-2 = 3y-6
x-3y = -4
x = 3y -4.....(1)

2x+4 = 5y+10
2x -5y = 6......(2)

from (1)
6y -8 -5y = 6
y = 14

substitute y =14 in (1)
x = 42 - 4
x =38

present age of father = 38
present ave of his son =14
Answered by shika7pearl85
11
let the present age of father be x
and,
let the present age of son be y
a/q
two yrs. ago, x-2 =y-2
2(x+2)= 5(y+2)
x-2=3y-6
x-3y=-4
2x+4=5y+10
2x-5y=6
now,
6y-8-5y=6
y=14
x=42-4=38
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