Question
solve mixture prolems whith proportion practice
Answers
GPT-4o mini
Sure! Mixture problems often involve combining different elements with known properties to achieve a desired outcome. These problems can typically be solved using proportions. Here are a few examples along with their solutions.
### Example 1: Mixing Two Solutions
**Problem:**
You have a 30% salt solution and a 50% salt solution. How many liters of each solution do you need to mix together to obtain 12 liters of a 40% salt solution?
**Solution:**
Let \( x \) be the liters of 30% salt solution and \( 12 - x \) be the liters of 50% salt solution.
Set up the equation based on the total amount of salt in the final mixture:
\[
0.30x + 0.50(12 - x) = 0.40(12)
\]
Expand and simplify:
\[
0.30x + 6 - 0.50x = 4.8
\]
Combine like terms:
\[
-0.20x + 6 = 4.8
\]
Subtract 6 from both sides:
\[
-0.20x = -1.2
\]
Divide by -0.20:
\[
x = 6
\]
So, you need 6 liters of the 30% solution. Now, the amount of the 50% solution is:
\[
12 - x = 12 - 6 = 6 \text{ liters}
\]
### Answer:
You need 6 liters of the 30% solution and 6 liters of the 50% solution.
---
### Example 2: Mixing Two Grades of Cement
**Problem:**
You want to create 100 kg of a mixture that is 70% cement. You have a cement that is 50% cement and another that is 90% cement. How much of each type do you need?
**Solution:**
Let \( x \) be the kg of the 50% cement, and \( 100 - x \) be the kg of the 90% cement.
Set up the equation based on the total amount of cement in the final mixture:
\[
0.50x + 0.90(100 - x) = 0.70(100)
\]
Expand and simplify:
\[
0.50x + 90 - 0.90x = 70
\]
Combine like terms:
\[
-0.40x + 90 = 70
\]
Subtract 90 from both sides:
\[
-0.40x = -20
\]
Divide by -0.40:
\[
x = 50
\]
So, you need 50 kg of the 50% cement. Now, the amount of the 90% cement is:
\[
100 - x = 100 - 50 = 50 \text{ kg}
\]
### Answer:
You need 50 kg of the 50% cement and 50 kg of the 90% cement.
---
### Example 3: Mixing Two Drinks
**Problem:**
A bartender has a drink that is 20% alcohol and another that is 60% alcohol. How many liters of each should he mix to create 10 liters of a drink that is 40% alcohol?
**Solution:**
Let \( x \) be the liters of the 20% alcohol drink, and \( 10 - x \) be the liters of the 60% drink.
Set up the equation based on the total amount of alcohol in the final mixture:
\[
0.20x + 0.60(10 - x) = 0.40(10)
\]
Expand and simplify:
\[
0.20x + 6 - 0.60x = 4
\]
Combine like terms:
\[
-0.40x + 6 = 4
\]
Subtract 6 from both sides:
\[
-0.40x = -2
\]
Divide by -0.40:
\[
x = 5
\]
So, you need 5 liters of the 20% alcohol drink. Now, the amount of the 60% drink is:
\[
10 - x = 10 - 5 = 5 \text{ liters}
\]
### Answer:
You need 5 liters of the 20% alcohol drink and 5 liters of the 60% alcohol drink.
---
These examples illustrate how to set up and solve mixture problems using proportions. If you have a specific problem in mind or need more practice examples, feel free to ask!
### Example 1: Mixing Two Solutions
**Problem:**
You have a 30% salt solution and a 50% salt solution. How many liters of each solution do you need to mix together to obtain 12 liters of a 40% salt solution?
**Solution:**
Let \( x \) be the liters of 30% salt solution and \( 12 - x \) be the liters of 50% salt solution.
Set up the equation based on the total amount of salt in the final mixture:
\[
0.30x + 0.50(12 - x) = 0.40(12)
\]
Expand and simplify:
\[
0.30x + 6 - 0.50x = 4.8
\]
Combine like terms:
\[
-0.20x + 6 = 4.8
\]
Subtract 6 from both sides:
\[
-0.20x = -1.2
\]
Divide by -0.20:
\[
x = 6
\]
So, you need 6 liters of the 30% solution. Now, the amount of the 50% solution is:
\[
12 - x = 12 - 6 = 6 \text{ liters}
\]
### Answer:
You need 6 liters of the 30% solution and 6 liters of the 50% solution.
---
### Example 2: Mixing Two Grades of Cement
**Problem:**
You want to create 100 kg of a mixture that is 70% cement. You have a cement that is 50% cement and another that is 90% cement. How much of each type do you need?
**Solution:**
Let \( x \) be the kg of the 50% cement, and \( 100 - x \) be the kg of the 90% cement.
Set up the equation based on the total amount of cement in the final mixture:
\[
0.50x + 0.90(100 - x) = 0.70(100)
\]
Expand and simplify:
\[
0.50x + 90 - 0.90x = 70
\]
Combine like terms:
\[
-0.40x + 90 = 70
\]
Subtract 90 from both sides:
\[
-0.40x = -20
\]
Divide by -0.40:
\[
x = 50
\]
So, you need 50 kg of the 50% cement. Now, the amount of the 90% cement is:
\[
100 - x = 100 - 50 = 50 \text{ kg}
\]
### Answer:
You need 50 kg of the 50% cement and 50 kg of the 90% cement.
---
### Example 3: Mixing Two Drinks
**Problem:**
A bartender has a drink that is 20% alcohol and another that is 60% alcohol. How many liters of each should he mix to create 10 liters of a drink that is 40% alcohol?
**Solution:**
Let \( x \) be the liters of the 20% alcohol drink, and \( 10 - x \) be the liters of the 60% drink.
Set up the equation based on the total amount of alcohol in the final mixture:
\[
0.20x + 0.60(10 - x) = 0.40(10)
\]
Expand and simplify:
\[
0.20x + 6 - 0.60x = 4
\]
Combine like terms:
\[
-0.40x + 6 = 4
\]
Subtract 6 from both sides:
\[
-0.40x = -2
\]
Divide by -0.40:
\[
x = 5
\]
So, you need 5 liters of the 20% alcohol drink. Now, the amount of the 60% drink is:
\[
10 - x = 10 - 5 = 5 \text{ liters}
\]
### Answer:
You need 5 liters of the 20% alcohol drink and 5 liters of the 60% alcohol drink.
---
These examples illustrate how to set up and solve mixture problems using proportions. If you have a specific problem in mind or need more practice examples, feel free to ask!