Question

The father’s age is six times his son’s age. Four years hence, the

age of the father will be four times his son’s age. The present ages,

in years, of the son and the father are, respectively
Please work out the problem.....otherwise reported​

Answer 1

Answer:

LET THE FATHER AGE IS X years AND THE PRESENT AGE OF SON IS Y YEARS

ACCORDING TO THE QUESTION

X=6Y

AFTER FOUR YEARS

X+4=4(Y+4)

HENCE WE GET TWO EQUATION

X=6Y...........(equation 1)

X+4=4Y+16..........(equation 2)

put X=6Y IN EQUATION 2

6Y-4Y=12

Y= 6 YEARS

AND X = 6Y =6×6= 36 YEARS

PRESENT AGE OF SON =6 YEARS

AND PRESENT AGE OF FATHER =36 YEARS ANSWER

Answer 2

Let the present age of father's be x years and present age of son's be y years.

According to the problem,

$x = 6y$

After 4 years

$x + 4 = 4(y + 4)$

Hence we get two equations

$x + 6y................(1)$

$x + 4 = 4(y + 4)...............(2)$

Simplifying eq (2)

$x + 4 = 4y + 16$

$x - 4y = 12$

Put X = 6y in eq (2) , we get

$6y - 4y = 12$

$2y = 12$

$y =12 \div 2 = 6 \: years$

$and \: x = 6y = 6 \times 6 = 36 \: years$

Present age of son = 6 years

Present age of father = 36 years

If you satisfy on my answer please mark it BRAINLIEST ✏️✏️

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