# Who's the oldest?

Albert, Brian and Charles are brothers with different ages.

Albert has twice the age Brian will have when Charles has twice the age Albert has now.

Who's the oldest?

• The brothers are all alive? I.e. some aren't waiting to be born? – corsiKa Jun 15 '15 at 15:30
• I leave it up to you. – Masclins Jun 15 '15 at 15:39

They have A,B,C initially.

P1: Charles will have twice the age Albert has now:

It will be the year Charles will be 2A. Year difference is 2A-C

P2: Brian will have P1:

Add 2A-C years to Brian's initial age and his age will be 2A+B-C

P3: Albert has twice the age P2

Multiply Brian's age with 2. and Albert's initial age A = 4A+2B-2C

which means C = (3A+2B)/2 thus Charles is the eldest.

• Beat me to it by 20 seconds! Have an upvote. – frodoskywalker Jun 15 '15 at 8:49

There are 2 points in time:

1. First, they are $a$, $b$, and $c$ years old

2. After $n>0$ years, they are $a+n$, $b+n$, and $c+n$ years old

The equations are:

1. $$a=2(b+n)$$
2. $$c+n=2a$$

From equation 1 we get $a>b$ because $n>0$.

Equation 2: $$c+n=a+a$$ Replace first $a$:

$$c+n=2(b+n) +a$$

$$c=2b +a+n$$

We get $c>a$.

Solution: $$c>a>b$$

I'll try a different approach which I think is also correct.

Assuming the answer is consistent with every possible age of A, B and C one may take the option in which the "time when Charles will have twice the age Albert has now" is actually now. In that case C is twice the age of A and A is twice the age of B making C the oldest. So for example C = 4, A = 2, B = 1. This does satisfy the condition.

At every point in time (since the birth of the youngest), the order is the same. What was true will be true.

Thus, instantly: $A>B$, $C>A$.

$C$ is the oldest.

The other answers have derived the equation

$$c=\frac{3a+2b}{2}$$

Clearly, if $a$ and $b$ are positive, then it is obvious that $c$ is the greatest.

But what if Albert and/or Brian are waiting to be born still?

Let us say that Albert will be born in 2 years, or $a=-2$ and that Brian will be born in 4 years, or $b=-4$. We can then conclude that $$c = \frac{3*(-2) + 2*(-4)}{2} = \frac{-6 + -8}{2} = -7$$. Thus we can say that Albert is the oldest in this scenario. We could just as easily concoct a scenario where Brian is older than Albert and the math shows that, as long as they aren't born yet, we'll be fine.

So if we accept that some of these souls are still waiting to be born, then we can not conclusively say whether or not any of them are the oldest with the information given.

A=2*(B+(2A-C)
A=2B+4A-2C
A=(2*(C-B))/3
A=C*2/3-B*2/3
Then: C>B (because A>0)
& A<C

Finally: C>B and C>A