Question

17.Five years ago, a man was 7 times as old as his son. Five years hence, he will
be 3 times as old as his son. Find their present ages.​

Answer 1

Answer:

son=10yrs. father=40yrs

Step-by-step explanation:

let son's age be 'a' and father's age be 'b'

5 yrs earlier

son's age = a-5

father's age = b-5

given that father was 7 times older than his son therefore

7(a-5)=b-5

7a-35=b-5

7a=b-5+35

7a=b+30

b=7a-30

given that 5 yrs later father is 3 times older than his son . therefore

son's age=a+5

father's age=b+5

3(a+5)=b+5

3a+15=b+5

substituting the value of b

3a +15=7a-30+5

3a+15=7a-25

25+15=7a-3a

40=4a

a=40/4

a=10

son's age=a=10yrs

father'sage=b=7a-30=7(10)-30=40yrs

hope it helps..