Question

A. 48

B. 44
C. 52
D. 40

6. The sum of the ages of a father and his son is 74. Four years ago the
father's age was twice as old as his son. How old is the father now?​

Answer 1

A (48)

Explanation:

let the father's age = x and his son's age = y

according to the question,

the sum of their ages is 74

×+y=74......equation 1

four years ago (i.e x-4 & y-4) father's age was twice as old as his son

x-4=2(y-4)

or, x-4=2y-8

x-2y=-4.......equation 2

now, in equation 1

x+y=74

hence,

x=74-y

substituting the values in equation 2

x-2y=-4

(74-y)-2y=-4

74-y-2y=-4

74-3y=-4

74+4=3y

3y=78

y=78/3

y=26

Hence the kid is 26 years old

x+y=74

x+26=74

x=74-26

x=48

Hence the dad is 48 years old

Answer 2

The correct answer is: Option A: 48

Explanation:

Let x be the present age of father and y be the present age of son

Given that

"The sum of the ages of a father and his son is 74"

`x+y = 74\ \ \ Eqn\ 1`

Now

x-4 will be the age of father 4 years ago

y-4 will be the age of son 4 years ago

The equation will be:

`x-4 = 2(y-4)\\x-4 = 2y-8\\x-2y-4 =-8\\x-2y = -8+4\\x-2y = -4\ \ \ Eqn\ 2`

From equation 1:

`x = 74-y`

Putting in equation 2

`74-y-2y = -4\\74-3y = -4\\-3y = -4-74\\-3y = -78\\(-3y)/(-3) = (-78)/(-3)\\y = 26`

Put y = 26 in equation 1

`x+26 = 74\\x = 74-26\\x = 48`

The father is 48 years old now.

Hence,

The correct answer is: Option A: 48