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There are 4 numbers A, B, C and D. The sum of the number A, B, C is 47, the sum of B, C, D is 53, the sum of C, D and A is 55 and the sum of D, A, and B,

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Notice that in the given equations, each of the variables
is mentioned three times. So if get the total of all given equations we may get
thrice the value of a set of variables (A+B+C+D).
(A+B+C+D)+ (A+B+C+D) + (A+B+C+D) = 204
3(A+B+C+D) = 204
A+B+C+D = 68

The sum of all variables is 68.
If we subtract the given equation, which has one missing
variable, from the value of the set computed, we may get the value of the
missing variable. (A+B+C+D) - (A+B+C) =
D 68 - 47 =
D 21 = D (A+B+C+D) - (B+C+D) =
A 68
- 53 =
A 15 = A (A+B+C+D) - (A+D+C) =
B 68 - 55 = B 13 =
B (A+B+C+D) - (D+A+B) =
C 68 - 49 = C 19 =
C