Q. 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?
(a) Value-I < Value-II
(b) Value-II < Value-I
(c) Value-I = Value-II
(d) Cannot be determined due to insufficient data
Correct Answer : (c) Value-I = Value-II
UPSC Prelims 2025 CSAT
Explanation :
We have a set of 11 consecutive integers in each case.
Value-I: “Minimum average when the 11 numbers are consecutive integers ≥ –5.”
1. If the 11 numbers are consecutive and all ≥ –5, the smallest possible sequence is
−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5.
2. The average of 11 equally spaced numbers is the middle (6th) term. Here the 6th term is 0.
3. Any shift upward (starting at −4 or higher) would increase every term, hence increase the average.
⇒ Minimum average = 0.
So Value-I = 0.
Value-II: “Minimum product when the 11 numbers are consecutive non-negative integers.”
1. Let the sequence be k, k+1, …, k+10 with k ≥ 0.
2. If k = 0, the sequence is 0, 1, 2, …, 10, and the product is 0.
3. For any k ≥ 1, all terms are positive and the product > 0.
⇒ Minimum product = 0.
So Value-II = 0.
Comparison: Value-I = 0 and Value-II = 0, hence
Answer: (c) Value-I = Value-II.
