Question

Two years ago a man was 6 times as old as his son.In 18 years he is twice as old as his son. Determine their age? ​

Answer 1

Answer:

let m = man's present age

let d - daughter's

:

Write an equation for each statement, simplify as much as possible

:

Two years ago a man was six times as old as his daughter.

m- 2 = 6(d-2)

m - 2 = 6d - 12

m = 6d - 12 + 2

m = 6d - 10

:

In 18 years he will be twice as old as his daughter.

m + 18 = 2(d+18)

m + 18 = 2d + 36

m = 2d + 36 - 18

m = d + 18

From the 1st equation, replace m with (6d-10)

6d - 10 = 2d + 18

6d - 2d = 18 + 10

4d = 28

d = 28/4

d = 7 yrs is the daughter's age

Find Dad's age

m = 6(7) - 10

m = 42 - 10

m = 32 yrs is Dad's age

:

:

Check solutions in the statement:

"In 18 years he will be twice as old as his daughter."

32 + 18 = 2(7 + 18)

50 = 2(25)

Answer 2

Given :-

Age of the man 2 years ago = 6 × Age of his son

Age of the man in 18 years = 2 × Age of his son

To Find :-

Age of the man.

Age of his son.

Solution :-

Let the age of the man and his son be x and y respectively.

We know that,

Two years ago a man was six times as old as his son.

$\sf x- 2 = 6(y-2)$

$\sf x - 2 = 6y - 12$

$\sf x = 6y - 12 + 2$

$\sf x= 6y - 10 \qquad ...(1)$

Given that,

Age of the man in 18 years = 2 × Age of his son

$\sf x + 18 = 2(y+18)$

$\sf x + 18 = 2y + 36$

$\sf x = 2y + 36 - 18$

$\sf x = y + 18 \qquad ...(2)$

Substituting the equation (1) and (2),

$\sf 6y - 10 = 2y + 18$

$\sf 6y - 2y = 18 + 10$

$\sf 4y = 28$

$\sf y=\dfrac{28}{4}$

$\sf y=7$

Age of the son = y = 7 years

Finding the age of man,

$\sf x = 6(7) - 10$

$\sf x = 42 - 10$

$\sf x=32$

Therefore, the age of man and son is 32 and 7 years respectively.