Age Problem

Question

Mark is a father of five kids, four girls and one boy. The kids ages are three years apart and the youngest girl are seven times younger than her older sister. On the other hand, the youngest girl is four times younger than her brother. What is Mark's age now if I'm three years he will be nine times older than his son was three years ago.

Comments

....o(0_0)o THERE IS NO USE TO ANSWERING THAT QUESTION!

Answer 1

Let Mark's age be M. Let the age of his son be B (boy) now. 3 years ago his son was B-3 years old, and in 3 years' time Mark will be M+3 years old, so M+3=9(B-3); M+3=9B-27, M=9B-30.

Let the youngest girl's age be Y, so B=4Y. 

If we put the children's ages in order we have a number of possibilities:

1. Y G G G 4Y
2. Y G G 4Y G
3. Y G 4Y G G
4. Y 4Y G G G

One of the G's (girls) is 7Y and therefore older than B=4Y. So the boy can't be the eldest. So we eliminate (1).

What about (2)? The ages are 3 years apart so if the eldest G=7Y then 7Y-4Y=3 and 3Y=3 making Y=1, the age of the youngest girl. That makes the ages 1 4 7 10 14 which actually fits (4). So the boy is 4, the youngest is 1 and the middle girls is 7. M=9B-30=6 which cannot be! That seems to eliminate (2).

Now (3): 7Y-4Y=6, 3Y=6, Y=2: the ages are 2 5 8 11 14, the boy is 8 and the youngest is 2. M=9B-30=42.

That seems reasonable: Mark is 42, his son is 8 and his 4 daughters are 2, 5, 11 and 14.