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The payload rating (PR) of a truck is the truck's recommended load weight, which is specified as a number of tonnes (t), where \(1 \text{ tonne} = 1{,}000 \text{ kilograms}\). A certain truck's PR is a whole number of tonnes. The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR. The weights of these 7 loads, in tonnes, are as follows: \(50, 51, 52, 52, 54, 54, 56\).
Select the Least and the Greatest possible values for the truck's PR, in tonnes. Make only two selections, one in each column.
49
50
51
52
53
54
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 \(\mathrm{PR}\).
Let's create a number line showing the load weights:
Load weights: 50 51 52 52 54 54 56
|-----|-----|-----|-----|-----|-----|-----|
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.
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
The \(\mathrm{PR}\) can only be 52 or 53 tonnes to satisfy the constraint of exactly 3 loads exceeding it.
Final Answer:
These are the only two values that result in exactly 3 loads being greater than the truck's \(\mathrm{PR}\), satisfying all given constraints.