Question

five years ago a man was seven times as old as his son, the father will be three times as old as his son. Find their present age

Answer 1

let +nt age of son be x yrs
and +nt age of father=y yrs
ATQ, (x-5)7=y
=> ,7x-y=35----------1
=>3x-y=-15-----------2
=>changing the symbols,so
=>4x=50
=>x=50/4
so putting this in equation,1,
=>7×50/4-y=35
=>y=47.5 n x=12.5

Answer 2

☆Let the age of man be x
Let the age of his father be y

5 years ago,
his age=x-5
his son's age =y-5
A.T.Q
x-5=7(y-5)
x-5=7y-35
x=7y-30 (1)

now 5 yrs hence,
his age =x+5
his son's age=y+5
A.T.Q
x+5=3(y+5)
x+5=3y+15
(7y-30)+5=3y+15 [(frm (1)]
7y-30=3y+15-5
7y-30=3y+10
7y-3y=10+30
4y=40
y=40/4
y=10

now put x in (1)

x=(7×10)-30
x=70-30
x=40
so ,
man'sage =40
his son'sage=10

HOPE THIS WILL HELPS YU DEAR!
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