A shipping container will be used to transport several 50-kilogram crates across the country by rail. The greatest weight that can be loaded into the container is 25000 kilograms. Other shipments weighing 6900 kilograms have already been loaded into the container. Which inequality can be used to determine z, the greatest number of 50-kilogram crates that can be loaded onto the shipping container?

Answer

50(6900 + x) ≥ 25000

25000 ≤ 50x + 6900

25000 ≥ 50x +6900

50(6900 + x) ≤ 25000

1 answer

To determine the greatest number of 50-kilogram crates that can be loaded onto the shipping container, we need to establish a relationship between the total weight in the container and its maximum capacity.

  1. Let \( z \) represent the number of 50-kilogram crates that can be added.
  2. The weight of the crates is \( 50z \) kilograms.
  3. The weight already loaded into the container is \( 6900 \) kilograms.
  4. The maximum weight that can be accommodated in the container is \( 25000 \) kilograms.

The total weight in the container, therefore, can be expressed as:

\[ 6900 + 50z \]

We want this total weight to be less than or equal to the maximum capacity of the container (25000 kg). This gives us the inequality:

\[ 6900 + 50z \leq 25000 \]

Now, rearranging the inequality to isolate \( z \):

\[ 50z \leq 25000 - 6900 \]

\[ 50z \leq 18100 \]

Dividing by 50:

\[ z \leq \frac{18100}{50} \]

Thus, the inequality that represents the situation is:

\[ 6900 + 50z \leq 25000 \]

In terms of the options provided, the correct option that expresses this situation is:

50(6900 + z) ≤ 25000

However, if we take into consideration the specific wording around \( z \) (where \( z \) is the number of crates), the clear relationship should rather be expressed directly with respect to \( z \):

25000 ≥ (6900 + 50z)

If we had to choose from the provided options, the correct inequality that directly accounts for the maximum weight with the outlined context is:

25000 ≥ 50z + 6900