Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00 Edition
ISBN: 9780395977224
Author: Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher: McDougal Littell
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Concept explainers

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
Book Icon
Chapter 3.7, Problem 5P
To determine

To calculate: Number of programs Hans bought.

Expert Solution & Answer
Check Mark

Answer to Problem 5P

Hans bought 50 programs.

Explanation of Solution

Given information:

Each program costs $1.50.

20 programs sold for $3 each.

Profit earned is $15.

Such that

Profit = selling price − purchase price

Calculations:

Let the number of programs Hans bought be x .

Then

    NumberPriceCost
    Bought X1.501. 50x
    Sold x − 20 3.00 3x − 60

We know that

Profit = selling price − purchase price

Thus,

  3x−60−1.50x=15

That becomes

  1.50x−60=15

Add 60 to both sides:

  1.50x=75

Divide both sides by 1.50:

  x=50

Thus,

The number of programs bought by Hans is 50.

Previous
Chapter 3.7, Problem 4P
Next
Chapter 3.7, Problem 6P

Chapter 3 Solutions

Algebra: Structure And Method, Book 1

Ch. 3.1 - Prob. 1OECh. 3.1 - Prob. 2OECh. 3.1 - Prob. 3OECh. 3.1 - Prob. 4OECh. 3.1 - Prob. 5OECh. 3.1 - Prob. 6OECh. 3.1 - Prob. 7OECh. 3.1 - Prob. 8OECh. 3.1 - Prob. 9OECh. 3.1 - Prob. 10OE
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Prob. 2OECh. 3.8 - Prob. 3OECh. 3.8 - Prob. 1WECh. 3.8 - Prob. 2WECh. 3.8 - Prob. 3WECh. 3.8 - Prob. 4WECh. 3.8 - Prob. 5WECh. 3.8 - Prob. 6WECh. 3.8 - Prob. 7WECh. 3.8 - Prob. 8WECh. 3.8 - Prob. 9WECh. 3.8 - Prob. 10WECh. 3.8 - Prob. 11WECh. 3.8 - Prob. 1MRECh. 3.8 - Prob. 2MRECh. 3.8 - Prob. 3MRECh. 3.8 - Prob. 4MRECh. 3.8 - Prob. 5MRECh. 3.8 - Prob. 6MRECh. 3.8 - Prob. 7MRECh. 3.8 - Prob. 8MRECh. 3.8 - Prob. 9MRECh. 3.8 - Prob. 10MRECh. 3.8 - Prob. 11MRECh. 3.8 - Prob. 12MRECh. 3.8 - Prob. 1STCh. 3.8 - Prob. 2STCh. 3.8 - Prob. 3STCh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - 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