Question

5 years ago a man was 7 times as old as his son after five years he will thrice as old as his son. find their present ages.​

Answer 1

Step-by-step explanation:

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years

Answer 2

Answer:

Sons age 5 years ago=x father's age = 7x

their present age = x+5 and 7x+5 resp

Sons age 5 years hence = x+10(since x is sons age 5 years ago from present age and we are talking of his age 5 years hence from his present age so 5+5=10)

 

fathers age 5 years hence =3(x+10)

hence the fathers present age will be 3(x+10)-5

also fathers present age =7x+5

therefore 7x+5=3(x+10)-5

7x+5=3x+30-5

7x-3x=25-5

4x=20

x=5

 

So, son's present age = 10years and fathers, present age = 40 years

Step-by-step explanation:

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