47. 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?

Question

Accepted Answer

The correct answer is option C. Value-I: Minimum value of the average of 11 consecutive integers ≥ –5.
To minimize the average of consecutive integers, we need to choose the smallest possible integers. The integers must be ≥ -5. So, the smallest 11 consecutive integers starting from -5 are: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5.
The sum of these numbers is (-5 + 5) + (-4 + 4) + ... + (-1 + 1) + 0 = 0.
The average = Sum / Number of terms = 0 / 11 = 0.
So, Value-I = 0.

Value-II: Minimum value of the product of 11 consecutive non-negative integers.
Non-negative integers are 0, 1, 2, 3, ... To minimize the product of 11 consecutive non-negative integers, we must include 0 in the set. The smallest 11 consecutive non-negative integers are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
The product of these numbers = 0 * 1 * 2 * ... * 10 = 0.
So, Value-II = 0.

Comparing Value-I and Value-II:
Value-I = 0
Value-II = 0
Therefore, Value-I = Value-II. This question tests the understanding of averages, products, and properties of integers, including negative and non-negative numbers.

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