If A + D = B + C, A + E = C + D, 2C < A + E and 2A > B + D, then:

The correct answer is Option (2) → B > A > D > C > E

Each pair of letters is compared by combining the given relations so that the other letters cancel. The relations are numbered:

• \(A+D=B+C\)        (1)

• \(A+E=C+D\)        (2)

• \(2C\lt A+E\)        (3)

• \(2A\gt B+D\)        (4)

C and D (putting (2) into (3)):

• \(2C\lt C+D\)

• \(C\lt D\)

A and B (from (1)):

• \(A-B=C-D\), which is negative because \(C\lt D\)

• \(A\lt B\)

A and D (from (1), \(B=A+D-C\); putting this into (4)):

• \(2A\gt A+D-C+D\)

• \(A-D\gt D-C\), and \(D-C\) is positive

• \(A\gt D\)

C and E (from (2), \(E=C+D-A\)):

• \(E-C=D-A\), which is negative because \(A\gt D\)

• \(E\lt C\)

Joining the results: \(B\gt A\), \(A\gt D\), \(D\gt C\) and \(C\gt E\), which gives B > A > D > C > E.