Question

Is the average of the largest and the smallest of four given numbers greater than the average of the four numbers? I. The difference between the largest and the second largest numbers is greater than the difference between the second smallest and the smallest numbers.
I II. The difference between the largest and the second largest numbers is less than the difference between the second largest and the second smallest numbers.

• A. If the question can be answered with the help of statement I alone
• B. If the question can be answered with the help of statement II alone
• C. If both, statement I and statement II are needed to answer the question
• D. If the question cannot be answered even with the help of both the statements

Hint:

Convert averages to algebraic inequalities to simplify reasoning.

Answer 1

The Correct Option is A

Let the numbers be arranged in increasing order: \( a<b<c<d \).
Then the average of the four numbers is \( \frac{a + b + c + d}{4} \), and the average of the smallest and largest is \( \frac{a + d}{2} \).
We need to determine whether \( \frac{a + d}{2}>\frac{a + b + c + d}{4} \).
Multiply both sides by 4:
\( 2(a + d)>a + b + c + d a + d>b + c \)
So, we must check whether \( a + d>b + c \).
Statement I talks about the spread of values between ends and second closest numbers, suggesting larger disparity at extremes, which may imply \( a + d>b + c \).
Hence, this alone may be sufficient to answer the question.
Statement II is comparative and vague in terms of how it impacts the overall sum of values. It doesn't clearly help in evaluating the inequality above.
Therefore, only Statement I is sufficient.