Math, asked by sumonamandal, 8 months ago

A father is 7 times old as his son. two years ago, the father was 13 times as old as his son. What is the present age of the father?

Answers

Answered by upendracachet
18

Answer:

Given that father was 13 times old as his son. x = 4. if the age of son is 4 then the age of father will be 7 * 4 = 28. Their present ages are son = 4 years and his father age = 28 years.

Attachments:
Answered by ElijahAF
20

At present

Age of the son = x years

Father's age = 7x years

2 years ago

Son's age = ( x - 2 ) years

Father's age = ( 7x - 2 ) years

According to the problem given,

7x - 2 = 13 ( x - 2 )

7x - 2 = 13x - 26

7x - 13x = -26 + 2

-6x = -24

x = ( -24 ) / ( -6 )

x = 4

Therefore ,

At present ,

Son age = x = 4years

Father's age = 7x = 7 × 4 = 28years

Hence, the father's age is 28 years

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