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Puzzle about a boy and a girl?
About a boy and a girl who both are at least seven years old we are told:
The boy is just as old as the girl will be when the boy turns twice as old as the girl was at the time when the boy had half the age of the sum of their current ages. Every age mentioned is an integer. How old are they.
I am looking for an elegant answer to the question as I already know the answer...
@Saurabh D: Your answer, although correct, is not very informative as to the method. Did you use an elegant method, please let us all know!!!
@Brian&Josh: Not bad! I like the reader-friendly approach of Brian's and the brevity and precision of Josh's answer. Josh is a little ahead mainly in one way. He realizes that the girls age must be a multiple of 6. Brian discovered that it would be a multiple of 3 thus doubling the trial and error part of the analysis.
The way I usually explain it is that we have a system of five equations in six unknown. The six unknown being b1, b2, b3 and g1, g2, g3 as the ages of the boy and the girl in the past, present and future respectively. Three equations explicitly stemming from the question:
b2 = g3 (the boy is just as old as the girl will be)
b3 = 2g1 (when the boy turns twice as old as the girl was)
b1 = (b2+g2)/2
then adding two implicitly known relations:
b2-b1 = g2-g1 (equal time went by between current and past ages)
b3-b2 = g3-g2 (equal time will be between furture and present ages)
And then to the messy part, I oftentimes leave out, namely eliminating five out of six un
(continued since I got cut off)
known from the above to get:
b1 = 7k/6 and b2 = 4k/3 and b3 = 5k/3
g1 = 5k/6 and g2 = k and g3 = 4k/3
And we see that all ages are integers if only g2 = k is divisible by 6.
I am not sure if Josh ensured that all ages, including the boys three ages, were integers, but they are. I always enjoy the part where the boys age of 24 is rejected to reach a unique conclusion!
3 Answers
- ?Lv 68 years agoFavorite Answer
t = when boy had half the age of B(0) + G(0), so
=> B(t) = (B(0)+G(0))/2
s = when boy was twice as old as girl at time t
=> B(s) = 2G(t)
0 = current time, boy is as old as girl at time s
B(0) = G(s)
Note also that B(r)-G(r) = K for some constant K. This lets us eliminate, say, B(r) from each of the above equations. Doing so gives
K/2 + G(t) = G(0)
K+G(s) = 2G(t)
K+G(0) = G(s)
This gives G(0) = K/2 + G(t) = K/2 + (K+G(s))/2 = K + (K+G(0))/2, so G(0) = 3K, G(t) = 5K/2, G(s) = 4K. In particular K must be divisible by 2, so G(0) must be divisible by 6. Since G(0) >= 7 and she is a girl, we have G(0) = 12 or G(0) = 18. But if G(0) = 18 then B(0) = K + G(0) = 6+18=24, so he's not a boy anymore. Thus G(0) = 12, forcing B(0) = 16.
Edit: Strictly speaking I just showed that if there is a solution, the girl is 12 and the boy is 16. But, it's very easy to check with what I derived; the G's are integers at a glance, and then the B's are from the first few formulas.
- BrianLv 78 years ago
This probably isn't the elegant solution you want, but.....
Let x and y be the current ages of the boy and girl, respectively.
Let X be the age of the boy when he was half the age of the sum of their current
ages, i.e., X = (x + y)/2. With Y being the age of the girl at this time, we have
that Y = X - (x - y), since the boy and girl will always be separated by the same
number of years. (I'm thinking that the boy is older than the girl for now, but that
will come out in the end.)
Now let Y' be the age of the girl when the boy turns twice as old as the girl was
when the boy turned X. Letting the boy's age be X' on this occasion, we have
that X' = 2Y = 2*(X - (x - y)) = 2*((1/2)*(x + y) - x + y) = x + y - 2x + 2y = -x + 3y.
But x = Y' = X' - (x - y) = -x + 3y - x + y, and so 3x = 4y.
So at least we know now that the boy is older than the girl.
Now we need to attend to the condition that both x >= 7 and y >= 7.
In order to have N = 3x = 4y we need N to be a common multiple of both 3 and 4
and for (N/4) >= 7. These conditions are satisfied by N = 36, 48, 60, 72, .....
For N = 36 we have x = 12 and y = 9, but this would make X = (x + y)/2 = 21/2,
which is not an integer, the last of the conditions that needs to be satisfied.
For N = 48 we have x = 16 and y = 12, which do satisfy all the conditions.
For N = 60 we have x = 20 and y = 15, making X a non-integer.
For N = 72 we have x = 24 and y = 18, which do satisfy the conditions, but
the boy is now a man so this doesn't count.
Thus the boy is 16 years old and the girl is 12 years old.
