Number of people to be found knowing constraints on mean ages Andrew and Alan are twins. They are both directors of a small company. They both have to attend a meeting with their new staff. When Andrew enters the room, the mean age of those in the room increases by four. He is immediately followed by Alan and the mean age increases by a further three. No more people enter the room for the meeting.
How many people are at the meeting including the twins?
Any help on this question please
 A: Hints:


*

*Set up some equations.  For example, 


*

*let $n$ be the number of people in the room including the twins 

*let $s$ be the total ages including the twins 

*let $t$ be the age of each twin

*What is the average age before either twin enters the room?

*What is the average age before after one twin enters the room?

*What is the average age with both twin in the room the room?


*That should give you two equations, one with the difference of $4$, the other with $3$. 

*There are three unknowns, but the question can still be answered


*

*Take the equation with $3$, and express $s$ in terms of $n$ and $t$

*Substitute this into the equation with $4$, and solve for $n$ (since $t$ disappears) 


A: The answer is 6.
Let N be the original number of people in the room.
Let A be the original average age of the N people in the room.
Let T be the age of each twin.
We will write down two equations for the increase in total weight of people in the  room after the first twin enters and then after the second twin enters.
The original total weight in the room is NA.
After the first twin enters the total weight increases by:
1) (N+1)(A+4)-NA = T
After the second twin enters the total weight increases from original by:
2) (N+2)(A+7)-NA = 2T
Multiplying out and simplifying each equation we have:
1')   A+4N+ 4= T and
2') 2A+7N+14=2T
Now we compute 2 times 1') minus 2') and get 
N=6 as suggested at the start.
