Average property value with symmetric deviation: The average property value of A, B, and C is ₹11111 crore. The property of C is as much less than the mean of A and C as the property of A is greater than that same mean (i.e., A and C are equally spaced around their own mean). What is the property of B?

Introduction / Context:
The problem features symmetry about the mean for A and C. It asks for B given that the overall average of A, B, C is fixed.

Given Data / Assumptions:

• Average of A, B, C = ₹11111 crore
• A and C are equidistant around their mean (their deviations cancel)

Concept / Approach:
Let m = (A + C) / 2. If A is m + d and C is m - d, then A + C = 2m regardless of d. The total sum A + B + C = 3 * 11111, so B is determined solely by the overall average, independent of d.

Step-by-Step Solution:
Sum of all three = 3 * 11111 = 33333A + C = 2m (unknown, but fixed by their mean)B = 33333 - (A + C)Since A and C are symmetric, no net shift from their mean; the only value ensuring the overall average is B = 11111

Verification / Alternative check:
Pick any symmetric pair (m ± d); their sum is 2m. Choosing B = 11111 preserves the overall average of 11111.

Why Other Options Are Wrong:
Values like 110 crore or 21410 crore would break the overall mean of 11111 unless A + C were contrived accordingly, contradicting the symmetry context described.

Common Pitfalls:
Overcomplicating the symmetric condition; the key is that the overall average fixes B directly to the same average amount.

Final Answer:
11111 crore