Question

7.
3. Two years ago, a man was five times as old as his
son. Two years later, his age will be 8 more than
three times the age of his son. Find their
present ages.
​

Answer 1

Answer:

Let the age of a man = x years

And the age of his son = y years

Two years ago, Man’s age = (x – 2) years

Son’s age = (y – 2) years

According to the question,

(x – 2) = 5(y – 2)

⇒ x – 2 = 5y – 10

⇒ x = 5y – 10 + 2

⇒x = 5y – 8 ...(i)

Two years later,

Father’s age = (x + 2) years

Son’s age = (y + 2) years

According to the question,

(x + 2) = 8 + 3(y + 2)

⇒ x + 2 = 8 + 3y + 6

⇒ x = 3y + 12 …(ii)

From Eq. (i) and (ii),

we get

5y – 8 = 3y + 12

⇒ 5y – 3y = 12 + 8

⇒ 2y = 20

⇒ y = 10

On putting the value of y = 11 in Eq. (i),

we get

x = 5(10) – 8

⇒ x = 50 – 8

⇒ x = 42

Answer 2

Explanation:

let fathers age be x

sons age y

2 yrs ago :

x-2=5(y-2)

2 yrs later:

x+2=8+3(y+2)

So,

5y-8+2=8+3y+6

2y=20

y=10

So, sons age is 10

father= x=50-8=42

hope it helps