Question

five years ago a man has 7 times as old as his son . five years hence , the father will be three times as old as his son . find their present age .​

Answer 1

Answer:

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

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Let x be the age of the son, 5 years ago father's age = 7x

Let the present age of

Son = x + 5

Father = 7x + 5

Ages 5yrs later

Son = x + 5 + 5 = x + 10

Father = 7x + 5 + 5 = 7x + 10

According to question,

Father is 3 times as old as his son

➨ 7x + 10 = 3 ( x + 10)

➨ 7x + 10 = 3x + 30

➨ 7x - 3x = 30 - 10

➨ 4x = 20

➨ x = 5

Present age of Son = x + 5 = 5 + 5 = 10yrs

Present age of Father = 7x+ 5 = 7 × 5 + 5 = 40yrs

Therefore, the ages of father and son are 40 yrs and 10 yrs.