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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.
Least
Greatest
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.