Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Cumulative Standards Review

Exercise 32 Page 230

What is the discounted price of one box of paper?

Compound inequality: 1400 ≤ 28.80x ≤ 1500
Number of boxes: Between 49 and 52 boxes.

Practice makes perfect

First, let's form the inequality and then solve it.

Forming a compound inequality

It is given that the maximum amount of money we can spend on paper is 1500 dollars. Also, if we spend at least $1400, we receive will shipping. Thus, the minimum amount we can spend (if we want the discount) is 1400 dollars. Now we can express the boundaries of how much we can spend as: 1400 ≤ ... ≤ 1500. To determine the expression in the middle of our inequality, we will use the following pieces of given information:

  • Each box of paper is $32.
  • We can receive 10 % off the cost of each box.To determine the discounted price for a box, we can use the percent proportion and subtract that amount from 32.
    a/b=p/100
    a/32=10/100
    a/32=0.1
    a=3.2
    By subtracting the discount from the total amount, we get the price per box to be: 32 - 3.2 = 28.8. Thus, the discounted price of each box is $28.80. If we let x represent the number of boxes purchased, this gives the expression: 28.80x which represents the discounted cost of the boxes we buy. We can now complete our inequality as: 1400 ≤ 28.80x ≤ 1500.

    Solving the compound inequality

    Solving our inequality will give us a range for the number of boxes we can buy and receive free shipping but also not go over our maximum budget. We can first separate the compound inequality into two inequalities. First Inequality: 1400 ≤ 28.80x Second Inequality: 28.80x ≤ 1500 Let's solve these inequalities one at a time.
    1400 ≤ 28.80x
    48.61 ≤ x
    x ≥ 48.61
    As we cannot order a portion of a box, and we need to spend at least 1400 to get the free shipping, we round this solution up to the nearest whole number to x ≥ 49. Let's also solve the second inequality as well.
    28.80x ≤ 1500
    x ≤ 52.08
    Again, we cannot buy a portion of a box and we cannot exceed our budget, so we will round our answer to the nearest whole number to x ≤ 52. Combining both solutions we have: 49 ≤ x ≤ 52. Thus, we can buy between 49 and 52 boxes.