Question

a man is 7 times as old as his son now. 4 years later he will be twice as old as 8 times his son then. find the present ages.

answer please

i want in one variable ​

Answer 1

Correct question:

A man is 7 times as old as his son now. 4 years later 2 times his age will be 8 times the age of his son then. Find the present ages.

Solution:

Present age:-

As it is given that, the man is 7 times as old as his son now,

• Let the present age of the son be x years.
• Then the present age of the man will be 7x years.

Four years later:-

• Age of the son four years later = (Present age of son) + 4 years = x + 4 years.
• Age of the man after 4 years = (Present age of son) + 4 years = 7x + 4 years

Now,

it is given that, after 4 years, age of man = 8(age of his son)

So,

2(7x + 4) = 8(x + 4)

=> 14x + 8 = 8x + 32

=> 14x - 8x = 32 - 8

=> 6x = 24

=> x = 4

• Present age of son = x years = 4 years.
• Present age of father = 7x years = (7 * 4) years = 28 years.