3 people, A, B and C, have integer ages. A total of 5 digits is required to construct the 3 ages and no digit is repeated.
28 years ago, B was twice the age that A was 4 years ago.
In 11 years time, A will be 3 times the age that C will be in 9 years time.
What are the ages of A, B and C ?
This one is confusing me, as it seems that if the present ages of A, B and C are X, Y and Z, then (Y-28) = 2(X-4), ie (Y-28) = (2X-8), so Y = (2X + 20) and therefore Y = X + 10. This would clearly mean that at least one digit of the ages is repeated, which is not allowed.
Please can someone advise, what am I overlooking ?
To my mind you will need to use both sets of criteria to then have simultaneous equations to complete the task.
You have only used '28 years ago' in your working out. Why then have they also given you 'In 11 years time'?
I agree with your figures factor, no unique solution unless we are missing part of the question.
By the way on (25,70 and 3) B would be 42 28 years ago, you only need to subtract 4 from A's age
BO was saying that in his/her opinion even the first rule would not lead to a valid solution, but he/she had made an error in jumping from Y=2x +20 does to y= x +10
Whoops ! In my defence O-Level Maths was 1977, still I shouldn't have made that daft error !
I also forgot to mention that A's age is a prime number, so there is a unique solution.
Thanks very much for all your help folks !
PS I think this is still a 'live' puzzle - the closing date is usually in the first half of the month, but does anyone know for sure ? That would explain why this month's puzzle isn't available yet online.
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