Math  /  Algebra

QuestionA clerk must use the elevator to move boxes of paper. The elevator's maximum weight limit is 1491 pounds. If each box of paper weighs 67 pounds and the clerk weighs 150 pounds, use an inequality to find the number of whole boxe she can move on the elevator at one time. a. Give the answer as an inequality. b. Explain the meaning of the answer to part a

Studdy Solution

STEP 1

What is this asking? How many boxes can the clerk fit in the elevator without going over the weight limit? Watch out! Don't forget the clerk's weight!

STEP 2

1. Set up the inequality
2. Solve for the number of boxes

STEP 3

Alright, let's **define** some variables!
Let nn be the **number of boxes** the clerk can move.
Each box weighs 67 pounds\text{67 pounds}, so the **total weight** of the boxes is 67n67 \cdot n.

STEP 4

The clerk weighs 150 pounds\text{150 pounds}, and the **elevator's limit** is 1491 pounds\text{1491 pounds}.
The **combined weight** of the clerk and the boxes *must be less than or equal to* the elevator's limit.

STEP 5

This gives us the **inequality**: 67n+150149167 \cdot n + 150 \le 1491

STEP 6

Let's **isolate** nn.
First, **subtract** 150 from *both sides* of the inequality: 67n+150150149115067 \cdot n + 150 - 150 \le 1491 - 150 67n134167 \cdot n \le 1341We're subtracting 150 because we're trying to get *n* by itself.
Subtracting 150 from both sides keeps the inequality balanced.

STEP 7

Now, **divide** *both sides* by 67: 67n67134167\frac{67 \cdot n}{67} \le \frac{1341}{67} n20n \le 20Dividing both sides by 67 keeps the inequality balanced and gets us closer to our answer.

STEP 8

Since nn represents the **number of boxes**, and we can't have a fraction of a box, nn must be a **whole number**.
Our solution tells us that nn must be *less than or equal to* 20.

STEP 9

a. n20n \le 20 b. The clerk can move a **maximum** of **20 boxes** on the elevator at one time without exceeding the weight limit.

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