Maths Question(Challenging)

You mean if A, B & C are the people

I'll assume all ages are integers

Then A*B*C*Mine=2450

and A+B+C=2*Your

Then the trick is to look for the prime factors for 2450

And 2450=2*5^2*7^2

Now A+B+C is even and all primes save one is even, so one of the three people has an even number age, the other two are odd

And 2450 has five total factors for four people, & Mine is odd

There are five possibilities of four integers to have a product of 2450

2*5*7*35
2*5*5*49
2*7*7*25
10*5*7*7
14*5*5*7

look at 2*5*7*35 sums
2+5+7=14 Your=7 Mine=35
2+5+35=42 Your=21 Mine=7
2+7+35=44 Your=22 Mine=5

2*5*5*49
2+5+5=12 Your=6 Mine=49
2+5+49=56 Your=28 Mine=5

10*5*7*7
10+5+7=22 Your=11 Mine=7
10+7+7=24 Your=12 Mine=5

14*5*5*7
14+5+5=12 Your=12 Mine=7
14+5+7=26 Your=13 Mine=5

I see no unique (or outstanding) solution, so I would make the conclusion that all ages are not integers greater then one. Moreover, I see further analysis is required and I should look at least one of the people couldn't speak (Since they had to be one(1)).

Rap

Further thinking on this problem--
as I said this cannot have an integral solutions so I started thinking about properties of numbers. Assumming all questioned were different ages.

Now the fundamental law of arithmetic says that all numbers are prime or the product of primes. But in this case the primes must have a relationship with the sums (called a forcing function). And the numbers where the prime products (including 1) is the sum are known as perfect numbers. And the simplest perfect number with three elements is 6 (1*2*3=1+2+3).
Now lets do some algebra. Lets use 1,2,3 as a distribution of n, so their ages are 1n,2n,3n, I'll call your age X and My age K (a misspelled Konstant)
So your ABCMine becomes
(1n)(2n)(3n)X=2450
6Xn^3=2450
&
(1n+2n+3n)=2K
6n=2K
n=K/3
so
6X(K/3)^3=2450
6/27X*K^3=2450
X*K^3=27*2450/6=3^3*2*5^2*7^2/(2*3)=3^2*5^2+7^2=105^2
X=105^2/K^3
So lets give it a try---if I say I'm 35 then X=105^2/35^3=9/35.
So you must be pretty young, because if we're the same age K=X and X^4=105^2
and X=sqrt(105) about 10+

Recommendation--If you want this to give reasonable numbers try 2450*105 (257250)

Rap

more hints to the question...
One of the persons celebrated a very important birthday this year which I celebrated 5 years ago.

Rap's first method is how I approached it (and got nowhere).

Perhaps the trick to the problem is that you (YOUR) are one of the three people I spoke to. In that case, YOUR has to appear as one of the three ages that sum to twice YOUR. There is only one instance of that:

2+5+7 = 2*7
YOUR = 7
MINE = 35

Well bearing in mind what vantackie just said the two ages must be 5 apart...

but then according to rapraps solutions the only ones 5 apart are 7 and 12... not very important