Question

One year ago a man age is eight times the age of his son. At present man age is the squar of sons age. Find the present age.

Answer 1

Answer:

Fathers age is 49 and sons age is 7

Step-by-step explanation:

Let mans age be x and sons age be y.

x = y^2

x-1 = 8(y-1)

y^2 - 1 = 8y-8

y^2 - 8y +7 = 0

y^2 -y - 7y +7 = 0

y(y-1) - 1(y-1) = 0

(y-1)(y-7) = 0

Y = 1 ; y = 7

As one year ago, the age is mentioned,

Y = 7

X = Y^2 = 49

Answer 2

Hi there !
Here's the answer:

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Let the present age of Man = x
and Son's age be y

Given,
At this time,
Man's Age = (His son's age)²

=> x = y² --------(1)

One year Ago,
Age of Man = x-1
Age of His son = y-1

Given,
Age of Man = 8×(His Son's Age)

=> x-1 = 8(y-1)

=> x-1 = 8y-8

=> x -8y + 7= 0 ------(2)

Substitute Eq(1) in Eq(2)

=> y²-8y+ 7= 0

Factorise the equation

=> y² - y - 7y + 7 = 0

=> y(y-1)-7(y-1) = 0

=> (y-1)(y-7) = 0

=> y = 1 or y = 7

y can't be 1
If son's present age = 1
One year ago, son's age = 0

The condition ,
Father's age one year ago = 8 times of His son, will not be satisfied

•°• y = 7

The son's present Age = 7 years
Father's present Age = 7² = 49 years

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