NIMCET 2009 — Reasoning PYQ

Given the equations and inequalities:

1. $A+B=C+D$

2. $B+D=2A$

3. D + E > A + B

4. C + D > A + E

Step 1: Compare A, B, and D

From equation (2): $B+D=2A$, which can be rewritten as $D-A=A-B$.

This implies that $A$ is the arithmetic mean of $B$ and $D$.

• If D > A, then A > B (Result: D > A > B).

• If B > A, then A > D (Result: B > A > D).

Step 2: Compare D and A using Inequality (3)

From inequality (3): D + E > A + B.

Substitute $B=2A-D$ from equation (2) into inequality (3):

D + E > A + (2A - D)

D + E > 3A - D

2D + E > 3A

This suggests $D$ is likely larger than $A$ to satisfy the condition alongside $E$.

Step 3: Compare E and B

From inequality (3): D + E > A + B.

From equation (1): $A+B=C+D$, so we can substitute:

D + E > C + D

E > C

Also, from inequality (4): C + D > A + E.

Since $C+D=A+B$, we substitute:

A + B > A + E

B > E

Combining these: B > E > C.

Step 4: Combine all findings

From Step 1 and Step 2, we established that if D > A, then A > B.

We also found B > E > C in Step 3.

Combining these sequences:

Final Conclusion:

The order D > A > B > E > C satisfies all given conditions.

Correct Option: (d)