# Henry paid $.80 for each bag of peanuts. He sold all but 20 of them for$1.50 and made a profit...

Henry paid $.80 for each bag of peanuts. He sold all but 20 of them for$1.50 and made a profit of $54. How many bags did he buy? ## Word problems with algebra This problem may seem confusing since all the expressions we can construct rely on how many bags he bought. Instead of panicking, what we can do is assign the unknown with some sort of letter variable of our choosing. This letter variable acts just like a number and we can use it like one in all of our operations, and eventually we should end up in a situation where we can simply solve for the variable. ## Answer and Explanation: Let's call the number of bags he bought 'n'. Firstly, we establish that he paid$0.80 for each bag. Therefore, his initial cost was 0.80n. Profit is equal to revenue minus expenses, so we know that {eq}\\ Revenue - 0.80n = 54 \\ Revenue = 54 + 0.80n \\ {/eq} We know that he sold all but 20 of them. So, he sold (n - 20) bags. The revenue earned is simply the number he sold times the price he charged for them. So, his revenue is equal to {eq}1.50 \times (n - 20) \\ = 1.50n - 30 \\ {/eq} Now, we have two expressions for revenue. We can equate them and solve for n. {eq}54 + 0.80n = 1.50n - 30 \\ 84 = 0.7n \\ 120 = n \\ {/eq} Therefore, he bought 120 bags and sold 100.

How to Solve Multi-Step Algebra Equations in Word Problems

from Algebra I: High School

Chapter 8 / Lesson 20
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