Question

The son is 22 years younger than his father, in 4 years the father is 3 times as old as the son. How old are father and son now? How many years ago was the father 12 times his son?​

Answer 1

Step-by-step explanation:

Let the son's present age be x years

$\therefore$ Father's present age = (x + 22) years.

Four years after:

Son's age = (x + 4)

Father's age = (x + 22 + 4) =(x + 26) years.

According to the given condition:

Father's age = 3 times the son's age

$\therefore \: x + 26 = 3(x + 4) \\ \therefore \: x + 26 = 3x + 12 \\ \therefore \: 26 - 12= 3x - x \\ \therefore \: 14= 2 x \\ \therefore \: x = \frac{14}{2} \\ \huge \red{ \boxed{ \therefore \: x = 7}} \\ \\ \implies \: x + 22 = 7 + 22 \\ \huge\red{ \boxed{ \therefore \:( x + 22)= 29}}$

Hence, now :

Father's age = 29 years

Son's age = 7 years.

Let us move to the next part of the question:

Let y years ago father's age was 12 times his son.

y years ago

son's age = (7 - y) years

Father's age = (29 - y) years

$\therefore\: 29 - y = 12(7 - y)\\\\\therefore\: 29 - y = 84 - 12y\\\\\therefore\: 12 y - y = 84 - 29\\\\\therefore\: 11y = 55\\\\\therefore\: y = \frac{55}{11}\\\\\huge\orange{\boxed {\therefore\: y = 5}}$

Thus, 5 years ago father's age was 12 times than his son.

Answer 2

Answer:

Step-by-step explanation:

Let son's age be x

Let father's age be y

Presently,

y-x =22...... First equation

In 4 years time,

y+4=3(x+4)

y+4=3x+12

Y-3x=8.... Second equation

Writing put both equations,

y-x=22

y-3x=8

Using elimination method to eliminate y by adding up the two equations,

-2x=-14

Divide -14 by the coefficient of x

x=-14÷-2 =7

If x =7,

y=7+22=29yrs

Presently, father is 29years while son is 7years

How many years ago was the father 12 times the son's age?

Let the number of years be p

29-p=12(7-p)

29-p=84-12p

Group into like terms

-p+12p=84-29

11p=55

P=55/11=5

5years ago the father was twelve times the son's age