An animal shelter takes in an average of 5 animals per day. The shelter must keep its total occupancy below 300. Currently, the shelter has 165 animals.



If none of the animals get adopted, which inequality represents how many more days, $x$, the shelter can continue to take in animals without exceeding its occupancy limit?


A$x < 25$
B$x < 27$
C$x < 33$
D$x < 35$

Verified Answer :

  • Define the total occupancy after \(x\) days as \(165 + 5x\).
  • Set up the inequality \(165 + 5x < 300\) to represent the shelter's capacity limit.
  • Solve the inequality by subtracting \(165\) from both sides and then dividing by \(5\).
  • The resulting inequality is \(x < 27\), which matches the second option.
  • Final answer: \(\boxed{x < 27}\)

Explanation

  1. Analyzing the Problem Data and Setting up the Inequality

To determine the number of days \(x\) the animal shelter can continue to take in animals, we must first model the total occupancy as a function of time. We know the following:

  • The current occupancy is \(165\) animals.
  • The shelter takes in an average of \(5\) animals per day.
  • The total occupancy after \(x\) days will be the initial occupancy plus the number of new animals admitted, which is \(5 \times x\).
  • The total occupancy must remain strictly less than \(300\).

Therefore, the total occupancy after \(x\) days is given by the expression \(165 + 5x\). To ensure the shelter does not exceed its limit, we set this expression to be less than \(300\).

  1. Solving the Inequality for x

We start with the inequality:
\[165 + 5x < 300\]
To solve for \(x\), we first subtract the current occupancy, \(165\), from both sides of the inequality:
\[5x < 300 - 165\]
\[5x < 135\]
Next, we isolate \(x\) by dividing both sides of the inequality by \(5\):
\[x < \frac{135}{5}\]
Performing the division:
\[x < 27\]
This inequality shows that the shelter can continue to take in animals for any number of days strictly less than \(27\).

  1. Selecting the Correct Inequality

Comparing our result, \(x < 27\), with the given options:

  • \(x < 25\)
  • \(x < 27\)
  • \(x < 33\)
  • \(x < 35\)

The correct option is \(x < 27\).

Examples

This type of modeling is very useful in everyday logistics, such as tracking inventory in a warehouse or managing space in a parking garage. By understanding the current capacity and the rate at which items are being added, you can accurately predict how much time you have before reaching your maximum limit. For example, if a garage has \(50\) spaces and adds \(2\) cars per hour, and currently has \(30\) cars, the inequality \(30 + 2x < 50\) helps determine when the garage will be full.