Math, asked by Anonymous, 9 months ago

5 years ago a father was 7 times as old as his son and after 5 years he will be thrice as old as his son find the present age of father.

Answers

Answered by Anonymous

Step-by-step explanation:

$age \: of \: the \: son = x$

$age \: of \: the \: father = 7x$

$present \: age \: = x + 5 \: and \: 7x + 5$

$future \: age \: of \: the \: son = x + 10$

$father = 7x + 10$

$7 x + 10 = 3(x + 10)$

$x = 5$

$7x = 35$

$x + 5 = 5 + 5 = 10$

$son \: age \: is \: 10$

$father \: age \: is \: 7x + 5 = 7 \times 5 + 5 = 40$

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Answered by pinjaraarifisha

Answer:

letfather's age = x

son's age = y

5 years hence, age of father = x+5

age of son = y+5

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

⇒ x =3y+10

5 years ago, age of father = x-5

age of son = y-5

So x-5 = 7(y-5)

⇒3y + 10 - 5 = 7y - 35

⇒ 4y = 40

⇒y = 10 age of son

x = 3y +10 = 40

son age = 10 year

son age = 10 yearfather's age= 40 year

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