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Chapter 4: Problem 24

How many liters of a \(10 \%\) salt solution must be mixed with 15 liters of a \(40 \%\) salt solution to obtain a \(20 \%\) salt solution?

Short Answer

Expert verified
30 liters of the 10% salt solution must be added.

Step by step solution

01

Understanding the Problem

The problem involves mixing two salt solutions of different concentrations to achieve a desired concentration. Specifically, we have a 10% salt solution and a 40% salt solution, and we want the final mixture to be a 20% salt solution with an unknown quantity from the 10% solution.
02

Define Variables

Let \( x \) be the amount of 10% salt solution (in liters) we need to add. We already know that we have 15 liters of a 40% salt solution.
03

Equations Setup

Set up an equation based on the salt content: the salt from the 10% solution plus the salt from the 40% solution must equal the salt in the combined 20% solution. The equation is: \( 0.10x + 0.40(15) = 0.20(x + 15) \).
04

Simplify and Solve the Equation

Start solving the equation by distributing and combining like terms: 1. Expand the equation: \( 0.10x + 6 = 0.20x + 3 \).2. Rearrange terms to isolate \( x \): \( 6 - 3 = 0.20x - 0.10x \)3. This simplifies to \( 3 = 0.10x \).4. Solve for \( x \): \( x = \frac{3}{0.10} = 30 \).
05

Verify Solution

Check the solution by inserting \( x = 30 \) back into the context. You should have 30 liters of a 10% salt solution and 15 liters of a 40% salt solution. The total salt is \( 0.10(30) + 0.40(15) = 3 + 6 = 9 \) liters in 45 liters, and \( 9/45 = 0.20 \) or 20%, confirming the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixture Problems
Mixture problems in algebra involve combining different substances to achieve a new mixture with specific properties. These problems often deal with solutions, where one component, like salt, is mixed with a liquid, like water, to form a solution of a particular concentration. The goal is usually to find out how much of each component is needed to reach a desired concentration. In these problems:
  • We typically know the concentrations and quantities of some components.
  • We aim to find the unknown quantity of one component necessary to achieve a desired final concentration.
  • Setting up an equation based on the conservation of the core substance (like salt) is key to solving these problems.
Percent Concentration
Percent concentration is a way of expressing how much of one component is present in a solution compared to the total mixture, expressed as a percentage. It's used to easily convey the strength of a solution. For example, saying a solution is 10% salt means 10% of the total weight (or volume, depending on the context) is salt. Understanding percent concentration is crucial when dealing with mixture problems. Here's why:
  • It provides a clear and standardized way to understand the proportion of components in a mixture.
  • In algebra problems, concentrations are often converted into decimals to facilitate calculations, such as using 0.10 for a 10% solution.
  • It helps to ensure that when combining substances, the overall concentration aligns with the desired target concentration.
Equations Setup
Setting up equations is a critical step in solving mixture problems. The core idea is to write an equation that represents the conservation of the substance of interest (e.g., salt). Here鈥檚 how it鈥檚 generally done:
  • Identify the unknown quantity and represent it with a variable, like using \( x \) for the unknown in our problem.
  • Write out the expression that represents the total amount of the substance before and after the mixture is combined.
  • For our example, you add the salt content from each solution: \( 0.10x \) from the 10% solution and \( 0.40 \times 15 \) from the 40% solution.
  • Then, set this equal to the desired concentration of the final mixture, showing how amounts combine to give the target concentration: \( 0.20(x + 15) \).
By aligning the total initial amount with the final desired amount, solving this equation gives you the quantity needed.

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Most popular questions from this chapter

Jeff leaves home on his bicycle and rides out into the country for 3 hours. On his return trip, along the same route, it takes him three-quarters of an hour longer. If his rate on the return trip was 2 miles per hour slower than on the trip out into the country, find the total roundtrip distance.

A sum of \(\$ 6000\) is invested, part of it at \(5 \%\) interest and the remainder at \(7 \%\). If the interest earned by the \(5 \%\) investment is \(\$ 160\) less than the interest earned by the \(7 \%\) investment, find the amount invested at each rate.

The circumference of a circle is \(2.24\) centimeters more than six times the length of a radius. Find the radius of the circle. (Use \(3.14\) as an approximation for \(\pi\).)

For Problems 1-12, solve each equation. You will be using these types of equations in Problems \(13-41\). $$ 0.2(20)+x=0.3(20+x) $$

For Problems \(1-12\), solve each of the equations. These equations are the types you will be using in Problems 13-40. $$ 500(0.08) t=1000 $$
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