How Many People Are In The Room?

Relevant wiki: Mean (Average)

Let the average age of the people in the room = $x$

Let Anne / Beth's age = $a$

Let the people in the room = $n$

Now, we know that:

$\frac{nx + a}{n+1} = x + 4$

This simplifies to:

$nx + a = (x + 4)(n+1)$

$nx + a = nx + 4n + x + 4$

$a = 4n + x + 4$

We also know that:

$\frac{nx + 2a}{n+2} = x + 7$

This simplifies to:

$nx + 2a = (x + 7)(n+2)$

$2a = 7n + 2x + 14$ .

If we substitute in the value of $a$ determined previously, then the resulting equation is:

$2(4n + x + 4) = 7n + 2x + 14$

$8n + 2x + 8 = 7n + 2x + 14$

$n + 8 = 14$

$n = \boxed{6}$

Let x = number of people in the room beforehand

  z=total age of people without Anne and Beth

  y=average age of people beforehand

  w=age of Anne=age of Beth

eq. 1: z = xy

eq. 2: z + w = (x + 1) (y + 4) OR z + w = xy + 4x + y + 4

eq. 3: z + 2w = (x + 2) (y + 7) OR z + 2w = xy + 7x + 2y + 14

simplifying eqs. 2 and 3 by elimination, we have

eq. 4: z = xy + x - 6

simplifying eqs. 1 and 4 also by elimination, we have

        0 = x - 6

        x = 6

This problem is way easier if, at first, you assume that the original average age in the room is 0. You can do so without loss of generality.

Let Anne's age be $a$ and Beth's age be $b$ and let there be $n$ people in the room before Anne and Beth enter. Then the average after Anne enters the room is 4, so

$\frac{0+a}{n+1} = 4$

Which implies

$a = 4\cdot (n+1)$ .

Then, Beth enters the room and the average becomes $4+3 = 7$ , so

$\frac{0+a+b}{n+2} = 7$

Since Beth is the same age as Anne, $b = a = 4\cdot (n+1)$ , substituting:

$\frac{2 \cdot 4 \cdot (n+1)}{n+2} = 7$

Simplified:

$\frac{n+1}{n+2} = \frac{7}{8}$ , to which n = 6 is the obvious (and only) solution.

Thus, there must be 6 people in the room before Anne and Beth enter.

Why can you assume that the average is 0?

If the initial average age in the room were not 0, but, say, $x$ , then we could just subtract $x$ from all the ages of the people in the room, which would then mean the new average is 0. Then we could do the same calculation and add $x$ to all ages after we've figured out the ages. Intuitively, the absolute value of the ages doesn't matter, just the age differences do, as implied by the question not mentioning any absolute values (which means the solution must be independent of what the exact ages are).