Question: Is average of the largest and smallest of 4 given numbers <strong>GREATER</strong> than the average of the 4 numbers?<br>Statement-I:<br>The difference between the largest and the second largest numbers is less than the difference between the second smallest and the smallest of the numbers.<br>Statement-II:<br>The difference between the largest and the smallest numbers is greater than the difference between the second largest and the second smallest of the numbers.

Consider the following for the next ten (10) items that follow :<br>Mark option (a) if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.<br>Mark option (b) if the question can be answered by using either statement alone.<br>Mark option (c) if the question can be answered by using both the statements together, but cannot be answered using either statement alone.<br>Mark option (d) if the question cannot be answered even by using both the statements together.

  1. A. The question can be answered by using one of the statements alone, but cannot be answered using the other statement alone
  2. B. The question can be answered by using either statement alone
  3. C. The question can be answered by using both the statements together, but cannot be answered using either statement alone
  4. D. The question cannot be answered even by using both the statements together

Correct Answer: A. The question can be answered by using one of the statements alone, but cannot be answered using the other statement alone

Explanation

Let the numbers be a \lt b \lt c \lt d. The question asks if \frac{a+d}{2} \gt \frac{a+b+c+d}{4}, which simplifies to a+d \gt b+c. Statement-I: (d-c) \lt (b-a) \implies a+d \lt b+c. This gives a definitive "No", meaning it is sufficient. Statement-II: (d-a) \gt (c-b) \implies d+b \gt a+c, which does not directly tell us if a+d \gt b+c. Hence, Statement-I alone is sufficient.

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