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
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
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
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
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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