Math, asked by harishshivaraj56780, 1 month ago

A FATHER IS 7 TIMES AS OLD AS HIS SON . THREE YEARS AGO THE FATHER WAS THIRTEEN
TIMES AS OLD AS HIS SON. WHAT ARE THERE PRESENT AGES.?

Answers

Answered by amreenksjbms
2

Answer:

Solution

verified

Verified by AMREEN

Let the present age of the son be x and that to of his father which is 7 times =7x

Then, 2 years ago their ages :

(x-2) of son

(7x-2) of father

Given, 13(x−2)=7x−2

Then,

13x−26=7x−2

=13x−7x=26−2

6x=24

or x=4 and that of his father = 28 years

age of son=x=4years

age of father=7x=28years//

28 IS THE ANSWER

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