Before you take a scored quiz for this lesson, try this set of practice questions. Your score on this self-check will be similar to that of your scored quiz. If you do not score well on this self-check, please review this lesson and try again.
If a 5-pound bag of mixed nuts with 45 percent peanuts (by weight) is combined with a 12-pound bag of mixed nuts that is 40 percent peanuts (by weight), what percentage of the mixture is peanuts? (by weight) Which equation would represent this scenario?
- \(\mathsf{ \frac{5(0.45)+ 12(a)}{17} }\) = \(\mathsf{ \frac{40}{100} }\)
- \(\mathsf{ \frac{5(0.40)+ 12(0.45)}{a} }\) = \(\mathsf{ \frac{17}{100} }\)
- \(\mathsf{ \frac{5(0.45)+ 12(0.40)}{17} }\) = \(\mathsf{ \frac{a}{100} }\)
- \(\mathsf{ \frac{5(0.45)+ 12(0.40)}{a} }\) = \(\mathsf{ \frac{17}{100} }\)
Let a = the percentage of the mixture that is peanuts by weight.
The total mixture is 17 pounds.
The equation would be
\(\mathsf{ \frac{5(0.45)+ 12(0.40)}{17} }\) =
\(\mathsf{ \frac{a}{100} }\).
Let a = the percentage of the mixture that is peanuts by weight.
The total mixture is 17 pounds.
The equation would be
\(\mathsf{ \frac{5(0.45)+ 12(0.40)}{17} }\) =
\(\mathsf{ \frac{a}{100} }\).
Let a = the percentage of the mixture that is peanuts by weight.
The total mixture is 17 pounds.
The equation would be
\(\mathsf{ \frac{5(0.45)+ 12(0.40)}{17} }\) =
\(\mathsf{ \frac{a}{100} }\).
Let a = the percentage of the mixture that is peanuts by weight.
The total mixture is 17 pounds.
The equation would be
\(\mathsf{ \frac{5(0.45)+ 12(0.40)}{17} }\) =
\(\mathsf{ \frac{a}{100} }\).
A 12-ounce cup of juice contains 70 percent fruit and 30 percent water.
If combined with 16 ounces of juice that contain 10 percent fruit and 90 percent water, what portion of the mixture is fruit?
Write the answer to the nearest hundredth.
- 24.29%
- 23.45%
- 62.53%
- 35.71%
Let x = the percentage of the mixture that is fruit.
The total mixture is 12 + 16 = 28-ounces.
The equation to use is
\(\mathsf{ \frac{12(0.70)+ 16(0.10)}{28} }\) =
\(\mathsf{ \frac{x}{100} }\);
840 + 160 = 28x; 1000 = 28x; 35.71 = x.
Let x = the percentage of the mixture that is fruit.
The total mixture is 12 + 16 = 28-ounces.
The equation to use is
\(\mathsf{ \frac{12(0.70)+ 16(0.10)}{28} }\) =
\(\mathsf{ \frac{x}{100} }\);
840 + 160 = 28x; 1000 = 28x; 35.71 = x.
Let x = the percentage of the mixture that is fruit.
The total mixture is 12 + 16 = 28-ounces.
The equation to use is
\(\mathsf{ \frac{12(0.70)+ 16(0.10)}{28} }\) =
\(\mathsf{ \frac{x}{100} }\);
840 + 160 = 28x; 1000 = 28x; 35.71 = x.
Let x = the percentage of the mixture that is fruit.
The total mixture is 12 + 16 = 28-ounces.
The equation to use is
\(\mathsf{ \frac{12(0.70)+ 16(0.10)}{28} }\) =
\(\mathsf{ \frac{x}{100} }\);
840 + 160 = 28x; 1000 = 28x; 35.71 = x.
A jar with a 7-pound mixture of dimes and nickels was mixed with 5-pound mixture that was 90 percent dimes to make a 73 percent dimes mixture.
Find the percent of dimes that was in the first mixture.
- 61%
- 79%
- 87%
- 80%
Let d = the percent of dimes in the first mixture.
There is a total of 7 + 5 = 12 pounds.
\(\mathsf{ \frac{7(d)+ 5(0.90)}{12} }\) =
\(\mathsf{ \frac{73}{100} }\).
Cross multiply to get 700d + 450 = 876, 700d = 426, p = 0.61.
The first mixture had 61 percent dimes.
Let d = the percent of dimes in the first mixture.
There is a total of 7 + 5 = 12 pounds.
\(\mathsf{ \frac{7(d)+ 5(0.90)}{12} }\) =
\(\mathsf{ \frac{73}{100} }\).
Cross multiply to get 700d + 450 = 876, 700d = 426, p = 0.61.
The first mixture had 61 percent dimes.
Let d = the percent of dimes in the first mixture.
There is a total of 7 + 5 = 12 pounds.
\(\mathsf{ \frac{7(d)+ 5(0.90)}{12} }\) =
\(\mathsf{ \frac{73}{100} }\).
Cross multiply to get 700d + 450 = 876, 700d = 426, p = 0.61.
The first mixture had 61 percent dimes.
Let d = the percent of dimes in the first mixture.
There is a total of 7 + 5 = 12 pounds.
\(\mathsf{ \frac{7(d)+ 5(0.90)}{12} }\) =
\(\mathsf{ \frac{73}{100} }\).
Cross multiply to get 700d + 450 = 876, 700d = 426, p = 0.61.
The first mixture had 61 percent dimes.
How many gallons of olive oil, containing 0.8 percent acidity, must be added to 4 gallons of oil with acidity of 2.5 percent so that the mixture has an acidity of 2 percent?
Round to the nearest hundredth.
- 2 gallons
- 1 gallon
- 8 gallons
- 1.67 gallons
Let a = the amount of olive oil to be added.
\(\mathsf{ \frac{a(0.008)+ 4(0.025)}{a + 4} }\) =
\(\mathsf{ \frac{2}{100} }\).
Cross multiply to get .8a + 10 = 2a + 8; -1.20a = -2; a = 1.67.
So, you need to add 1.67 gallons to produce a mixture of 2 percent acidity.
Let a = the amount of olive oil to be added.
\(\mathsf{ \frac{a(0.008)+ 4(0.025)}{a + 4} }\) =
\(\mathsf{ \frac{2}{100} }\).
Cross multiply to get .8a + 10 = 2a + 8; -1.20a = -2; a = 1.67.
So, you need to add 1.67 gallons to produce a mixture of 2 percent acidity.
Let a = the amount of olive oil to be added.
\(\mathsf{ \frac{a(0.008)+ 4(0.025)}{a + 4} }\) =
\(\mathsf{ \frac{2}{100} }\).
Cross multiply to get .8a + 10 = 2a + 8; -1.20a = -2; a = 1.67.
So, you need to add 1.67 gallons to produce a mixture of 2 percent acidity.
Let a = the amount of olive oil to be added.
\(\mathsf{ \frac{a(0.008)+ 4(0.025)}{a + 4} }\) =
\(\mathsf{ \frac{2}{100} }\).
Cross multiply to get .8a + 10 = 2a + 8; -1.20a = -2; a = 1.67.
So, you need to add 1.67 gallons to produce a mixture of 2 percent acidity.
Summary
Questions answered correctly:
Questions answered incorrectly:
Mixture Problems
Mixture problems are another example of rational equations at work. When mixing chemicals, food, or even paint, you can model these kinds of problems using rational equations. Let's look at an example.
Suppose you have a 30-ounce snack mix that is half almonds and half raisins. How many ounces of almonds should be added to yield a 70 percent almond mix?
Let a be the amount of ounces of almonds to be added to the mix.
Half of the original mixture was almonds, which translates to 15 ounces. In the mixture there will be 15 + a almonds in the mixture and there will be 30 +a total amount of mixture.
You also know that you need a final mixture with 70 percent almonds, which is written as 0.70 as a decimal. We can write the following rational equation.
\(\mathsf{ \frac{15 + a}{30 + a} }\) = .70
Solve this rational equation using the LCD.
| [\(\mathsf{ \frac{15 + a}{30 + a} }\) = .70]30 + a |
| \(\mathsf{ \frac{(15 + a)(30 + a)}{30 + a} }\) = .70(30 + a) |
| 15 + a = 21 + .70a |
| 0.30a = 6 |
| a = 20 |
In order to yield a snack mixture with 70 percent almonds you must add 20 ounces of almonds.
Mixture Problem Practice
Now it's your turn. Solve the following problem then click on the Answer button to check your work.
Suppose you have a 10 pint mixture of paint that is equal parts blue and yellow paint. In order to create the perfect shade of green you need a paint mixture that is 80 percent blue.
How many pints of blue paint do you need to add to the mixture?
If the mixture is equal parts blue and yellow paint, there are 5 pints of blue paint and 5 pints of yellow paint.
Let b be number of pints of blue paint that needs to be added in order to yield a mixture that is 80 percent blue paint.
Remember that 80 percent is written as .80 in decimal form.
Write the equation.
\(\mathsf{ \frac{5 + b}{10 + b} }\) = .80
| [\(\mathsf{ \frac{5 + b}{10 + b} }\) = .80]10 + b |
| 5 + b = .8(10 + b) |
| 5 + b = 8 + .80b |
| .20b = 3 |
| b = 15 |
In order to have a mixture that is 80 percent blue paint you would need to add 15 pints of blue paint to the existing mixture.
