You need a 50% alcohol solution. On hand, you have a 110 mL of a 10% alcohol mixture. You also have 60% alcohol mixture. How much of the 60% mixture will you need to add to obtain the desired solution? You will need mL of the 60% solution
You need a 50% alcohol solution. On hand, you have a 110 mL of a 10% alcohol mixture. You also have 60% alcohol mixture. How much of the 60% mixture will you need to add to obtain the desired solution? You will need mL of the 60% solution
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Mixing Solutions to Obtain a Desired Concentration
**Problem Statement:**
You need a 50% alcohol solution. On hand, you have 110 mL of a 10% alcohol mixture. You also have a 60% alcohol mixture. How much of the 60% mixture will you need to add to obtain the desired solution?
**Solution:**
To solve this problem, you will need to determine the amount of the 60% alcohol solution required to mix with the 110 mL of the 10% alcohol mixture to achieve a new solution with a 50% alcohol concentration. A mathematical approach usually involving systems of equations is used to solve this.
**Question Help:**
- You can refer to a video tutorial to understand the steps involved in solving this type of problem.
**Answer Box:**
- There is a blank space where you can enter the amount of the 60% alcohol solution required.
**Equation:**
To solve the problem mathematically, you can set up the following equation:
Let \( x \) be the amount (in mL) of the 60% solution needed.
The total amount of alcohol from the 110 mL of a 10% solution:
\[ 0.1 \times 110 = 11 \text{ mL of alcohol} \]
The total amount of alcohol from \( x \) mL of the 60% solution:
\[ 0.6 \times x = 0.6x \text{ mL of alcohol} \]
The total volume of the resulting mixture:
\[ 110 + x \text{ mL of solution} \]
To get the desired 50% alcohol solution, the total mL of alcohol should be:
\[ \frac{11 + 0.6x}{110 + x} = 0.5 \]
Solving this equation will give you the value of \( x \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53b99f33-5097-48ca-9202-113ad6b052b9%2Fc504dc80-0f52-43b3-b2af-b3b8937cd132%2Fquqwzgp_processed.png&w=3840&q=75)
Transcribed Image Text:### Mixing Solutions to Obtain a Desired Concentration
**Problem Statement:**
You need a 50% alcohol solution. On hand, you have 110 mL of a 10% alcohol mixture. You also have a 60% alcohol mixture. How much of the 60% mixture will you need to add to obtain the desired solution?
**Solution:**
To solve this problem, you will need to determine the amount of the 60% alcohol solution required to mix with the 110 mL of the 10% alcohol mixture to achieve a new solution with a 50% alcohol concentration. A mathematical approach usually involving systems of equations is used to solve this.
**Question Help:**
- You can refer to a video tutorial to understand the steps involved in solving this type of problem.
**Answer Box:**
- There is a blank space where you can enter the amount of the 60% alcohol solution required.
**Equation:**
To solve the problem mathematically, you can set up the following equation:
Let \( x \) be the amount (in mL) of the 60% solution needed.
The total amount of alcohol from the 110 mL of a 10% solution:
\[ 0.1 \times 110 = 11 \text{ mL of alcohol} \]
The total amount of alcohol from \( x \) mL of the 60% solution:
\[ 0.6 \times x = 0.6x \text{ mL of alcohol} \]
The total volume of the resulting mixture:
\[ 110 + x \text{ mL of solution} \]
To get the desired 50% alcohol solution, the total mL of alcohol should be:
\[ \frac{11 + 0.6x}{110 + x} = 0.5 \]
Solving this equation will give you the value of \( x \).
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