Problem 9 - Prealgebra

Question

Solve each of the following word problems by translating the statement into a proportion. Be sure to show the proportion used in each case. [Examples \(1-4]\) Farming A farmer knows that of every 50 eggs his chickens lay, only 45 will be marketable. If his chickens lay \(1,000\) eggs in a week, how many of them will be marketable?

Step-by-Step Solution

Answer

Out of 1,000 eggs, 900 will be marketable.

Step 1: Understand the Problem

We are given a total amount of eggs laid (1,000 eggs), and we know that out of every 50 eggs, 45 are marketable. We need to find out how many eggs will be marketable out of the total 1000 laid.

Step 2: Set Up the Proportion

The relationship given is 45 marketable eggs out of 50 laid eggs. We set up a proportion to find the marketable eggs, denoted as \(x\), out of 1000 laid eggs: \(\frac{45}{50} = \frac{x}{1000}\).

Step 3: Solve the Proportion

Cross-multiply to solve for \(x\). We have \(45 \times 1000 = 50 \times x\). This simplifies to \(45000 = 50x\).

Step 4: Calculate x

Divide both sides by 50 to solve for \(x\): \(x = \frac{45000}{50}\). Calculating gives \(x = 900\).

Step 5: Interpret the Result

Therefore, out of 1,000 eggs laid, 900 eggs will be marketable.

Key Concepts

Solving Proportions

Proportions are a fundamental concept in mathematics that involve two ratios being equal to each other. In the problem where a farmer needs to find out how many of his chickens' eggs are marketable, this concept is crucial. A proportion allows us to set up an equation based on the relationship provided in the word problem.

To solve a proportion like the one given in this exercise, we first identify the known ratio and the unknown ratio. In this case, we know that 45 out of 50 eggs are marketable. The unknown is how many out of 1,000 eggs are marketable. This sets up the proportion:

• \(\frac{45}{50} = \frac{x}{1000}\)

To solve the proportion, we use the cross-multiplication method. We multiply diagonal elements and equate them, which simplifies finding the unknown value. Here, the equation becomes \(45 \times 1000 = 50 \times x\). Solving for \(x\) gives us the number of marketable eggs out of the total.

Basic Algebra

Basic algebra is the cornerstone for solving many math problems, including the egg marketing problem faced by our farmer. Understanding algebra means knowing how to manipulate equations to find unknown variables.

In this scenario, basic algebra helps us solve the proportion. Once the proportion is set up, cross-multiplying gives us an equation in one variable:

• \(45000 = 50x\)

Here, the goal is to isolate the variable \(x\). We achieve this by performing the same operation on both sides of the equation. In this case, dividing both sides by 50 simplifies to:

• \(x = \frac{45000}{50}\)

Calculating this gives us \(x = 900\). This means 900 eggs are marketable, showing how algebraic manipulation helps us solve practical problems.

Word Problem Solving Steps

Solving word problems can sometimes seem daunting, but following structured steps can simplify the process greatly. For our farmer's egg problem, breaking it down step-by-step ensures that we understand and solve correctly.

Here's how we approached it:

• **Understand the Problem**: Identifying what is known (e.g., the total eggs and how many per batch are marketable) and what is asked (number of marketable eggs out of 1,000).


• **Set Up the Proportion**: Formulate a proportion based on the given relationship \(\frac{45}{50} = \frac{x}{1000}\).


• **Solve the Proportion**: Use cross-multiplication to create an equation and solve for the unknown \(x\).


• **Calculate and Interpret the Result**: Perform the necessary arithmetic to solve \(x\) and relate it back to the problem, determining how many eggs are marketable.

By following these clear steps, solving word problems becomes more manageable and less intimidating.