Question

5 years ago A man was 7 times as old as his son 5 years hence the father will be 3 times as old as his son find their present ages

Answer 1

Heya !!!!


Let the present age of son and father be X and Y year's.



Before 5 years age of father = (X-5) years


Before 5 years age of son = (Y-5) years



According to question,




(X-5) = 7(Y-5)


X - 5 = 7Y - 35


X - 7Y = -35+5


X - 7Y = -30 --------(1)


After 5 years age of father = (X+5) years


After 5 years age of son = (Y+5) years.


According to question,


X+5 = 3(Y+5)


X + 5 = 3Y + 15


X - 3Y = 15-5


X - 3Y = 10 --------(2)


From equation (1) we get,



X - 7Y = -30



X = -30 + 7Y -----------(3)


Putting the value of X in equation (2)


X - 3Y = 10


-30 + 7Y - 3Y = 10


4Y - 30 = 10


4Y = 10 +30


4Y = 40


Y = 40/4 = 10 years


Putting the value of Y in equation (3)


X = -30 + 7Y => -30 + 7 × 10 = -30 + 70


X = 40 years.



Age of son = Y = 10 years

and,



Age of father = X = 40 years.




HOPE IT WILL HELP YOU....... :-)

Answer 2

Let present age of man be X years
Let present age of his son be y years

Age five years ago

(X-5). = 7 (Y-5)
It becomes ,
X-5=7y-35
X-5-7y+35
X-7y+30...........(1)


Age five years hence

(X+5). =3 ( Y+5)
It becomes
X+5=3y+15
X-3y-10..........(2)


Solving these equation by elimination method

X-7y+30=0
X-3y-10=0
- +. +
X gets cancelled out and we are left with

-4y+40=0
Or y=10

Put value of y in equation (1)

X-70+30=0
Or x=40


Hope it helps you

Pls mark brainliest