Math, asked by Anonymous, 1 month ago

Show Answer Type Questions (SLAB-I)
One
year ago, a man was 8 times as old as his son.
Now his age is equal to the square of his son's age
find their present age.​

Answers

Answered by Saby123
14

Question : One  year ago, a man was 8 times as old as his son.

Now his age is equal to the square of his son's age  . Find their present ages !

Solution : Let the present age of the son be x years .

So, the age of the son one year ago is (X - 1) years .

The man's age is 8 times his son's age now.

Age of the man 1 year ago > 8(x - 1) years

Current age of man > 8(x - 1) + 1 years

> 8x - 8 + 1

> 8x - 7 years .

This is equal to the square of the son's age .

Therefore

> 8x - 7 = x²

> x² - 8x + 7 = 0

> x² - 7x - x + 7 = 0

> x( x - 7 ) + 1( x - 7) = 0

> ( x + 1)( x - 7) = 0

x = 7 or -1 .

However the age can never be in negative .

So, the only possible value of x is 7 .

Thus,

The current age of the man is 49 years .

The current age of his son is 7 years.

This is the required answer .

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