Consider a set of 11 numbers:
Value-I = Minimum value of the average of the numbers of the set when they are consecutive integers ≥ –5.
Value-II = Minimum value of the product of the numbers of the set when they are consecutive non-negative integers.
Which one of the following is correct?

The correct answer is Value-I = Value-II.

Let us evaluate each of the two given values step-by-step.

Step 1: Calculate Value-I
We are given a set of 11 consecutive integers. Let these 11 consecutive integers be represented as:
$x,x+1,x+2,\dots ,x+10$
where $x\ge -5$.

The average of 11 consecutive integers is simply the middle (6th) term, which is:
$\text{Average}=x+5$
To find the minimum value of this average, we must take the smallest possible value for $x$, which is $x=-5$.

Substituting $x=-5$:
$\text{Value-I}=-5+5=0$

Step 2: Calculate Value-II
We are given a set of 11 consecutive non-negative integers (i.e., integers $\ge 0$).
To minimize the product of these 11 consecutive non-negative integers, the smallest possible set starts at 0:
$\{0,1,2,3,4,5,6,7,8,9,10\}$

Since 0 is included in this set of numbers, the product of all 11 numbers in the set is:
$\text{Value-II}=0\times 1\times 2\times \dots \times 10=0$

Step 3: Compare Value-I and Value-II
Comparing the two calculated values:
$\text{Value-I}=0$
$\text{Value-II}=0$
Therefore, $\text{Value-I}=\text{Value-II}$.