Question

2 years ago father's age is 7 time more than son's age. after three years the sum of their age is 50 . what is their present age ​

Answer 1

Answer:

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Step-by-step explanation:

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

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of father

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,13x−26=7x−2

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,13x−26=7x−2=13x−7x=26−2

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,13x−26=7x−2=13x−7x=26−26x=24

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,13x−26=7x−2=13x−7x=26−26x=24or x=4 and that of his father = 28 years

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,13x−26=7x−2=13x−7x=26−26x=24or x=4 and that of his father = 28 yearsageofson=x=4years

Let the present age of the son be x and that to of his father which is 7 times =7xThen, 2 years ago their ages :(x-2) of son (7x-2) of fatherGiven, 13(x−2)=7x−2Then,13x−26=7x−2=13x−7x=26−26x=24or x=4 and that of his father = 28 yearsageofson=x=4yearsageoffather=7x=28years

Answer 2

Let, the present age of father be '$x$' and the present age of son be '$y$'.

Given,

$2$ years ago father's age is $7$ times more than son's age.

Two years ago, father's age is $(x-2)$ years and son's age is $(y-2)$ years.

According to above data:

$(x-2)=(y-2)\times7$

$x-2=7y-14$

$x=7y-14+2$

$x=7y-12$.    ______$(1)$

Given, after three years sum of their ages is $50$.

After three years, father's age is $(x+3)$ years and son's age is $(y+3)$ years.

According to above data:

$(x+3)+(y+3)=50$

$x+y+6=50$

From equation $(1)$, $x=7y-12$.

$(7y-12)+y+6=50$

$8y-12+6=50$

$8y-6=50$

$8y=50+6$

$8y=56$

$y=\frac{56}{8}$

$y=7$ years.

Hence, the present age of the son is $7$ years.

From equation $(1)$, $x=7y-12$

$x=(7\times7)-12$

$x=49-12$

$x=37$ years.

Therefore, the present age of the father is $37$ years.