Question

10 years Hence a man's age will be twice the age of his son ten years ago the man was four times as old as his son find their present ages ​

Answer 1

Step-by-step explanation:

current age of man and son be X and y respectively

10 years hence their age will be X+10

and y+10

10 years ago their age will be X-10

and y-10

equations are

X+10 =2(y+10)

X+10=2y+20

x-2y=20-10

x-2y=10. (eqn.1)

x-10=4(y-10)

x-10=4y-40

x-4y=-40+10

x-4y=-30. (eqn.2)

epn.2 - eqn1

X-4Y = -30

X-2Y = 10

(-) (+). (-)

______________

-2Y =-40

Y=20

X-4Y= -30

X= -30+4(20)

= -30+80

X=50

so,age of father is 50 and his son is 20

Answer 2

Answer:

Let the son's age be x

Son = x

Man = 2x

10 years ago:

Son = x - 10

Man = 2x - 10

The man was 4 times as old as his son:

2x - 10 = 4 (x - 10)

Solve x:

2x - 10 = 4 (x - 10)

2x - 10 = 4x - 40

4x - 2x = 40 - 10

2x = 30

x = 15

Find their age:

Son = x = 15

Father = 2x = 2(15) = 30

Answer: The son is 15 years old and the father is 30 years old.