Question

Five years hence, a man's age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present age

Answer 1

Answer:

Step-by-step explanation:

let the age of man be x

let age of his father be y

5 years ago,

his age=x-5

his son's age=y-5

x-5=7(y-5)

x-5=7y-35

x=7y-30-(1)

5 years hence,

his age=x+5

his son's age=y+5

x+5=3(y+5)

x+5=3y+15

(7y-30)+5=3y+15(from (1))

7y-25=3y+15

7y-3y=25+15

4y=40

y=10

now put x in (1)

x=(7×10)-30

x=70-30

x=40

therefore,

man's age=40 and his son's age=10

Answer 2

Answer:

Step-by-step explanation:

Given :-

Five years hence, a man's age will be three times the age of his son.

Five years ago, the man was seven times as old as his son.

To Find  :-

Their Present ages

Solution :-

Let the present age of the man be x years

And His son's  be y years.  

1st Equation :-

After 5 years man’s age = x + 5  

After 5 years ago son’s age = y + 5  

As per the question  

⇒ x + 5 = 3(y + 5)

⇒ x – 3y = 10 ……………(i)  

2nd Equation

5 years ago man’s age = x – 5  

5 years ago son’s age = y – 5  

As per the question

⇒ x – 5 = 7(y – 5)

⇒ x – 7y = -30 …….(ii)

Subtracting (ii) from (i), we have  

⇒ 4y = 40  

⇒ y = 10  

Putting y value in Eq (i)

⇒ x – 3y = 10

⇒ x – 3 × 10 = 10  

⇒ x = 10 + 30

⇒ x = 40

Hence, man’s present age = 40 years and son’s present age = 10 years.