Question

two years a go man was 6 times as old as his daughter. In 18 years, he will be twice as old as his daughter. What are their present ages

Answer 1

If the man's age was 6 times as old as his daughter and will be twice after 18 years then the present age of daughter is 7 and her father is 32.

Given that age of a man was 6 times as old as his daughter and in 18 years, he will be twice as old as his daughter.

We are required to find the present ages of both daughter and her father.

Let the age of daughter before 2 years be x.

The equations are as under:

In this way the father's age was 6x.

Present age of daughter be x+2,

Present age of her father be 6x+2.

After 18 years the age of daughter be x+2+18=x+20

After 18 years the age of her father be 6x+2+18=6x+20

It is also given that the age of her father will be 2 times the age of the daughter. So, 2(x+20)=6x+20.

2x+40=6x+20

40-20=6x-2x

20=4x

x=5

Present age of daughter=5+2=7 years.

Present age of her father=6*5+2=30+2=32 years.

Hence if the man's age was 6 times as old as his daughter and will be twice after 18 years then the present age of daughter is 7 and her father is 32.

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