NIMCET 2026 — Reasoning PYQ

To solve this, we define the sum of the four quantities as $A+B+C+D=10600$. We then convert the average information into equations:

1. If we remove A, the average of the remaining numbers is 1000:

   $$\frac{B+C+D}{3}=1000⟹B+C+D=3000$$

2. If we remove B, the average of the remaining numbers is 3220:

   $$\frac{A+C+D}{3}=3220⟹A+C+D=9660$$

3. If we remove C, the average of the remaining numbers is 3180:

   $$\frac{A+B+D}{3}=3180⟹A+B+D=9540$$

Now, we add these three derived equations together:

$$(B+C+D)+(A+C+D)+(A+B+D)=3000+9660+9540$$

$$2A+2B+2C+3D=22200$$

Since we know that $A+B+C+D=10600$, we can express the sum $(A+B+C)$ as $(10600-D)$. Substituting this into the combined equation:

$$2(A+B+C)+3D=22200$$

$$2(10600-D)+3D=22200$$

Expanding and solving for D:

$$21200-2D+3D=22200$$

$$21200+D=22200$$

$$D=22200-21200=1000$$

Conclusion: The value of D is 1000, which corresponds to option 4.