The payload rating (PR) of a truck is the truck's recommended load weight, which is specified as a number of...

Phase 1: Owning the Dataset

Understanding the Problem

We have a truck with a Payload Rating \((\mathrm{PR})\) that represents its recommended load weight in tonnes. The key constraint is that exactly 3 out of 7 loads exceeded this \(\mathrm{PR}\).

Visual Representation

Let's create a number line showing the load weights:

Load weights:  50    51    52    52    54    54    56
              |-----|-----|-----|-----|-----|-----|-----|

Phase 2: Understanding the Question

Key Insight

If exactly 3 loads are greater than \(\mathrm{PR}\), then exactly 4 loads must be less than or equal to \(\mathrm{PR}\). This gives us a clear criterion for finding valid \(\mathrm{PR}\) values.

What We're Looking For

• Least \(\mathrm{PR}\): The smallest whole number of tonnes where exactly 3 loads exceed it
• Greatest \(\mathrm{PR}\): The largest whole number of tonnes where exactly 3 loads exceed it

Phase 3: Finding the Answer

Systematic Check

Let's position \(\mathrm{PR}\) at different values and count how many loads exceed it:

If \(\mathrm{PR} = 51\):
Loads \(\leq 51\): 50, 51 (2 loads)
Loads \(> 51\): 52, 52, 54, 54, 56 (5 loads) × Not 3

If \(\mathrm{PR} = 52\):
Loads \(\leq 52\): 50, 51, 52, 52 (4 loads)
Loads \(> 52\): 54, 54, 56 (3 loads) ✓ Exactly 3!

If \(\mathrm{PR} = 53\):
Loads \(\leq 53\): 50, 51, 52, 52 (4 loads)
Loads \(> 53\): 54, 54, 56 (3 loads) ✓ Exactly 3!

If \(\mathrm{PR} = 54\):
Loads \(\leq 54\): 50, 51, 52, 52, 54, 54 (6 loads)
Loads \(> 54\): 56 (1 load) × Not 3

Key Finding

The \(\mathrm{PR}\) can only be 52 or 53 tonnes to satisfy the constraint of exactly 3 loads exceeding it.

Phase 4: Solution

Final Answer:

• Least possible \(\mathrm{PR}\): 52 tonnes
• Greatest possible \(\mathrm{PR}\): 53 tonnes

These are the only two values that result in exactly 3 loads being greater than the truck's \(\mathrm{PR}\), satisfying all given constraints.