Question

five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.​

Answer 1

ANSWER:

age of the man = 40 years

age of his son = 10 years

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Step-by-step explanation:

let the age of man be x

and tge age of his son be y

now,

given that,

five years ago a man was seven times as old as his son.

five years ago = x - 5 and y - 5

ACCORDING TO THE QUESTION,

x - 5 = 7(y - 5)

x - 5 = 7y - 35

x - 7y = -35 + 5

x - 7y = -30. .....(1)

now,

also,

given that,

Five years hence, the father will be three times as old as his son.

five years hence = x + 5 and y + 5

ACCORDING TO THE QUESTION,

x + 5 = 3( y + 5)

x + 5 = 3y + 15

x - 3y ! 15 - 5

x - 3y = 10 .....(2)

now,

we have ,

eqn (1) = x - 7y = -30

(2) = x - 3y = 10

substracting (2) from (1)

(1) - (2)

x - 7y - (x - 3y) = -30 - 10

x - 7y - x + 3y = -40

-4y = -40

y = -40/-4

y = 10

now,

putting the value of x on (1)

x - 7y = -30

x - 7(10) = -30

x - 70 = -30

x = -30 + 70

x = 40

so,

age of the man = 40 years

age of his son = 10 years

Answer 2

SOLUTION :

Let the age of man be x

The age of his son be y

Five years hence,

Age of the man = x + 5

Age of the son = y + 5

It is said that the father will be three times as old as his son.

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x = 3y + 15 - 5

x = 3y + 10 ------(1)

Five years ago a man was seven times as old as son.

Then,

Age of man = x - 5

Age of his son = y - 5

According to the problem,

x - 5 = 7(y - 5)

(3y + 10) - 5 = 7y - 35 [ from eq - 1 ]

3y + 5 = 7y - 35

3y - 7y = -35 - 5

-4y = -40

y = 40/4

y = 10

The age of his son is 10 years.

Substitute y in eq - 1

x = 3y + 10

x = 3(10) + 10

x = 30 + 10

x = 40

Therefore,

The age of man = x = 40 years.

The age of man = x = 40 years. The age of son = y = 10 years.