maths question?

(a + b + c + d) / 4 = k
a + b + c + d = 4k

(a + b + 2b + c + b + d + b) / 4 =
(a + 5b + c + d) / 4 =
[4b + (a + b + c + d)] / 4
(4b + 4k) / 4
4(b + k) / 4
b + k
..........................
(ab + b^2 + cb + db) / 4
b(a + b + c + d) / 4
(b * 4k) / 4
4bk / 4
bk
............................
if the mode is 3, then x = 3 (or y = 3)
3, 3, 4, 5, y , 12

(3 + 3 + 4 + 5 + y + 12) / 6 = 6
27 + y = 36
y = 9

if x = 3, y = 9
if y = 3, x = 9

For 1a), let's reason about the new mean from the definition of mean:

k = 1/4 [a + b + c + d]

Then the new mean, k', is:
k' = 1/4 [(a + b) + (2b) + (c + b) + (d + b)]
= 1/4 [a + b + c + d + 4b]

Notice that this expression is the original mean plus 1/4 * 4b more.
So, k' = k + b.

1b) can be solved the same way.

2) Either x or y is 3, given that 3 occurs once already. It is trivial, then, to calculate the other value based on the definition of mean. That is,
1/6 [ 3 + 4 + 5 + 12 + x + y] = 6
x + y = 12

Note that there are two solutions, since we may NOT assume that x < y. To do so would imply that the mode is not 3. Thus, the possible solutions are {x, y} = {3, 9} or {9, 3}

5