Question

A set of 51 numbers in such that if we remove the fist number, the average of the remaining numbers will be reduced by 1 from the average of all the numbers, if we remove the first and second, average be reduced by 2 from the original average of 51 numbers and so on. If average of all the 51 numbers is 1000, then which of the following is true

• A. The numbers can be uniquely determined using the given data
• B. There exist many sets of numbers satisfying the given conditions
• C. Such a set of numbers cannot exist
• D. Data insufficient to say anything about these numbers

Answer 1

Answer:

A. The numbers can be uniquely determined using the given data

Understanding the Problem

• The question describes a scenario involving a set of 51 numbers with specific properties related to their average when elements are removed from the beginning.

• We are given that the average of all 51 numbers is 1000.

• The core of the problem lies in how the average changes as the first number, then the first two, and so on, are removed.

Key Mathematical Concepts

• Average: The sum of a set of numbers divided by the count of the numbers.

• Sum from Average: The sum of a set of numbers can be calculated by multiplying the average by the count of numbers (Sum = Average * Count).

Analyzing the Given Conditions

• Let $S_{51}$ be the sum of all 51 numbers, and $A_{51}$ be their average. We know $A_{51} = 1000$. Therefore, $S_{51} = 51 \times 1000 = 51000$.

• When the first number ($n_1$) is removed, there are 50 numbers remaining. The average of these 50 numbers ($A_{50}$) is $A_{51} - 1$.

• The sum of these 50 numbers is $S_{50} = S_{51} - n_1$.

• So, $A_{50} = \frac{S_{50}}{50} = \frac{S_{51} - n_1}{50}$.

• We are given $A_{50} = A_{51} - 1 = 1000 - 1 = 999$.

• Substituting the values: $999 = \frac{51000 - n_1}{50}$.

• $999\times 50 = 51000 - n_1$

• $\rightarrow 49950 = 51000 - n_1 $

• $\rightarrow n_1 = 51000 - 49950 = 50$.

• Continuing this logic for subsequent removals:

• When the first two numbers ($n_1, n_2$) are removed, the average of the remaining 49 numbers ($A_{49}$) is $A_{51} - 2$.

• $A_{49} = 1000 - 2 = 998$.

• The sum of these 49 numbers is $S_{49} = S_{51} - n_1 - n_2$.

• $A_{49} = \frac{S_{49}}{49} = \frac{S_{51} - n_1 - n_2}{49}$.

• $998 = \frac{51000 - 50 - n_2}{49} $

• $\rightarrow 998 \times 49 = 50950 - n_2 $

• $\rightarrow 48902 = 50950 - n_2 $

• $\rightarrow n_2 = 50950 - 48902 = 2048$.

• This pattern continues. For each number removed from the front, the average decreases by 1, and the count of numbers decreases by 1.