Originally posted by ketch90
This might be the one -- not sure but give it a try.
Bob and John form a team together. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. John is as old as Bob was when John was half as old as he will become over ten years.
How old are Bob and John?
[when john was half as old as he will become over ten years] = [when john was 10]
so john is as old as Bob was when john was 10, from the second sentence.
let John(current) = x, and bob(current) = y. then x = |10+y-x| i think we can assume from the problem that bob is the older brother from the first sentence [Bob is as old as John will be
... ?] so since y > x, we know x = 10 + y - x. now note: y-x = x-10 (this will be useful later)
when bob was 1/2(x+y), john was [1/2(x+y) - (x-10)] years old. when bob is twice as big as that 2[1/2(x+y)-(x-10)], john will be 2[1/2(x+y)-(x-10)] - (x-10) years old. and this is bob's current age, according to the first sentence: so
y = 2[1/2(x+y)-(x-10)] - (x-10) = (x+y) - 2(x-10) - (x-10) = y -2x +30
or 0 = -2x +30 -> x=15. but then plugging that into the earlier equation x = 10 + y - x, we see that:
15 = 10+y - 15, or 30 = 10+y, or y = 20. thus john is 15 and bob is 20.
the key here is that the difference between their ages is constant, and relating everything to their current ages. then at its worst it's a system of two equations in two variables and can easily be solved.
good problem! cheers.