Answer :
To solve this problem, we need to determine how many palettes of roofing material the elevator can carry in addition to the operator's weight.
1. Start by considering the total weight that the elevator can carry, which is 2,320 pounds.
2. Take into account the weight of the elevator operator, which is 195 pounds. Subtract this weight from the elevator's total capacity to find out how much weight is left for the roofing material:
[tex]\[
2320 - 195 = 2125 \text{ pounds}
\][/tex]
So, 2125 pounds are available for the roofing material.
3. Each palette of roofing material weighs 425 pounds. To find out how many palettes can fit within the remaining capacity, divide the available weight by the weight of one palette:
[tex]\[
\text{Number of palettes} = \frac{2125}{425} \approx 5
\][/tex]
4. Since you cannot carry a fraction of a palette and have to work with whole palettes, the maximum number of complete palettes that can be loaded is 5.
Therefore, the maximum number of palettes of roofing material that the elevator can carry in one trip is 5.
1. Start by considering the total weight that the elevator can carry, which is 2,320 pounds.
2. Take into account the weight of the elevator operator, which is 195 pounds. Subtract this weight from the elevator's total capacity to find out how much weight is left for the roofing material:
[tex]\[
2320 - 195 = 2125 \text{ pounds}
\][/tex]
So, 2125 pounds are available for the roofing material.
3. Each palette of roofing material weighs 425 pounds. To find out how many palettes can fit within the remaining capacity, divide the available weight by the weight of one palette:
[tex]\[
\text{Number of palettes} = \frac{2125}{425} \approx 5
\][/tex]
4. Since you cannot carry a fraction of a palette and have to work with whole palettes, the maximum number of complete palettes that can be loaded is 5.
Therefore, the maximum number of palettes of roofing material that the elevator can carry in one trip is 5.
