Question

five years ago a man was seven times as old his son five years hence the father will be three times as old as his son find there present age​

Answer 1

Answer:

Father's present age is 40 years and son's present age is 10 years.

Step-by-step explanation:

Given :-

• 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.

To find :-

• Their present ages.

Solution :-

Let the present age of father be x years and the present age of son be y years.

5 years ago,

• Father's age = (x-5) years
• Son's age = (y-5) years

According to the question,

x-5 = 7(y-5)

→ x-5 = 7y -35

→ x = 7y-35+5

→ x = 7y-30..................(i)

After 5 years,

• Father's age = (x+5) years
• Son's age = (y+5) years

According to the question,

x+5 = 3(y+5)

→ x+5 = 3y+15

• Put x = 7y-30 from eq (i).

→ 7y-30+5 = 3y+15

→ 7y - 3y = 15+30-5

→ 4y = 40

→ y = 10

Now put y = 10 in eq(i) for getting the value of x.

x = 7y-30

→ x = 7×10-30

→ x = 40

Therefore,

★ Present age of father is 40 years.

★ Present age of son is 10 years.

Answer 2

Answer:

sons age: 10

father's age: 40

Step-by-step explanation:

let the present age of man be x.

let the present age of his son be y.

five years ago......

father's age= x-5

sons age=y-5

x-5=7(y-5)

x-5=7y-35

x=7y-35+5

x=7y-30(equation 1)

five years hence....

father's age= x+5

sons age= y+5

x+5=3(y+5)

x+5=3y+15(equation 2)

substituting value of x from equation 1 into equation 2

(7y-30)+5=3y+15

7y-25=3y+15

7y-3y=15+25

4y=40

y=40/4

y=10

this implies....

sons age=y= 10 years

father's age= x=7y-30

=70-30

father's age= 40 years....

hope it helps you:))