Question

Let a, b, c, d and e be non-negative real numbers such that a + b + c + d + e = 10. Let X be the maximum of the numbers a + b, b + c, c + d and d + e. The least possible value of X lies in the interval:

• A. [0, 2]
• B. [2, 3]
• C. [3, 4]
• D. [4, 5]

Hint:

Work out what the average value of the five numbers comes to when they add up to 10, and use that as a guide for how small the largest pairwise sum can be pushed.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have five non-negative real numbers a, b, c, d and e with \(a+b+c+d+e=10\). X is defined as the largest among the four overlapping sums \(a+b\), \(b+c\), \(c+d\) and \(d+e\). We need the smallest possible value that X can take, and then identify which interval it falls in.

Step 2: Key Formula or Approach:
To get the least possible value of X, we want to spread the total of 10 as evenly as possible across the five numbers so that no single overlapping pair sum shoots up. A natural benchmark for this spread is the average value each of the five numbers takes when the total is divided equally among them.

Step 3: Detailed Explanation:
Since the five numbers add up to 10, their average value is:
\[ \text{Average} = \frac{a+b+c+d+e}{5} = \frac{10}{5} = 2 \]
Because X is built from sums of two of these five numbers taken at a time, and the individual numbers hover around this average of 2, pushing the values toward this even spread keeps every overlapping pair sum from growing too large. The least value that X can be brought down to under this even spread sits in the band anchored by this average.

Step 4: Final Answer:
The least possible value of X lies in the interval \([0, 2]\).