Question

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

Answer 1

Answer:

Let the father's age be x and the son's age be y

Five years hence:

• Father's age = x+5
• Son's age = y+5

Given: x+5= 3(y+5)

That is x - 3y = 10 .... (1)

Five years ago:

• Father's age = x-5
• Son's age = y - 5

Given: x-57(y-5)

That is x - 7y=-30 .... (2)

Solving the two simultaneous equations we have

x = 40 and y = 10

Thus the present ages of father and son are 40 and

Thus the present ages of father and son are 40 and10.

Hope it helps!

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Answer 2

Step-by-step explanation:

Given :

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

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

To find :

Their present ages.

Solution :

Let the present ages of the Father and the Son be x and y years.

1st condition : [Hence/Future]

=> After 5 years man’s age = x + 5  

=> After 5 years ago son’s age = y + 5  

According to the question :

=> x + 5 = 3(y + 5)

=> x + 5 = 3y + 15

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

2nd condition : [Ago/Past]

=> 5 years ago man’s age = x - 5

=> 5 years ago son’s age = y - 5  

According to the question :

=> x - 5 = 7(y - 5)

=> x - 7y = -30 -- (2)

Subtracting (2) from (1) :

=> x - 7y = -30

=> x - 3y = 10

• 4y = 40
• y = 10

Substitute the value of y in (1) :

=> x - 3y = 10

=> x - 3(10) = 10

=> x - 30 = 10

=> x = 40

Therefore :

Father's present age = 40 years.

Son’s present age = 10 years.

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