NEED HELP PLEASE
Mixture Problems
1.Joe created a metal containing 40% platinum by combining two other metals. One of these other metals weighed 12 lb. and contained 45% platinum. If the other weighed 3 lb. then what percent of it was platinum?
2. 12 gal. of a saline solution was mixed with 3 gal. of a 30% saline solution to make a 38% saline solution. Find the percent concentration of the first solution.
3. 9 ml of a 20% alcohol was mixed with 6 ml of pure water. Find the concentration of the new mixture.
4. How much clay do you need to add to 8m^3 of soil with 30% clay in order to make a soil with 60%?
5. 9 L of an alcohol solution was mixed with 1L of a 80% alcohol solution. Find the percent concentration of the first solution.
1)
The final mixture weighs 12 + 3 = 15 pounds and is 40% platinum.
Thus you need .40 x 15 = 6 pounds of platinum total from the two metals that were combined.
The first metal contained .45 x 12 = 5.4 pounds of platinum.
The second metal must then contain 6 – 5.4 = 0.6 pounds of platinum.
You know the second metal weighs 3 pounds.
You need to determine what percent 0.6 is of 3.
The second metal is then 0.6 / 3 = 0.2 or 20% platinum.
To solve these mixture problems, we need to use the concept of percentages and the formula for calculating the concentration of a mixture.
The formula for calculating the concentration of a mixture is:
(concentration of component A * volume of component A + concentration of component B * volume of component B) / total volume of mixture
Let's apply this formula to the given problems.
1. Joe created a metal containing 40% platinum by combining two other metals. One of these other metals weighed 12 lb. and contained 45% platinum. If the other weighed 3 lb., then what percent of it was platinum?
Let's denote the concentration of the platinum in the unknown metal as x%. Using the formula, we can set up the following equation:
(45% * 12 lb. + x% * 3 lb.) / (12 lb. + 3 lb.) = 40%
Simplifying the equation:
(540 lb. + 3x lb.) / 15 lb. = 40%
To solve for x, we can cross-multiply and solve the resulting equation:
540 lb. + 3x lb. = 600 lb.
3x lb. = 60 lb.
x% = 60 lb. / 3 lb. = 20%
Therefore, the other metal contained 20% platinum.
2. 12 gal. of a saline solution was mixed with 3 gal. of a 30% saline solution to make a 38% saline solution. Find the percent concentration of the first solution.
Let's denote the percent concentration of the first solution as x%. Using the formula, we can set up the following equation:
(x% * 12 gal. + 30% * 3 gal.) / (12 gal. + 3 gal.) = 38%
Simplifying the equation:
(12x gal. + 90 gal.) / 15 gal. = 38%
To solve for x, we can cross-multiply and solve the resulting equation:
12x gal. + 90 gal. = 38% * 15 gal.
12x gal. + 90 gal. = 5.7 gal.
12x gal. = 5.7 gal. - 90 gal.
12x gal. = -84.3 gal.
x% = (-84.3 gal. / 12 gal.) * 100%
x% = -702.5%
It seems that there is an error in the given data or the calculations since we cannot have a negative percent concentration. Please recheck the question.
3. 9 ml of a 20% alcohol was mixed with 6 ml of pure water. Find the concentration of the new mixture.
Here, we are only mixing alcohol and water, so the concentration of the alcohol in the new mixture would be the same as the concentration of the original alcohol.
Therefore, the concentration of the new mixture would be 20% alcohol.
4. How much clay do you need to add to 8m^3 of soil with 30% clay in order to make soil with 60%?
Let's denote the amount of clay to be added in m^3 as x. Using the formula, we can set up the following equation:
(30% * 8m^3 + 100% * x) / (8m^3 + x) = 60%
Simplifying the equation:
(2.4m^3 + x) / (8m^3 + x) = 60%
To solve for x, we can cross-multiply and solve the resulting equation:
2.4m^3 + x = 60% * (8m^3 + x)
2.4m^3 + x = 4.8m^3 + 0.6x
0.4x = 2.4m^3
x = (2.4m^3 / 0.4)
x = 6m^3
Therefore, you need to add 6m^3 of clay to the soil to make a soil with 60% clay.
5. 9 L of an alcohol solution was mixed with 1 L of an 80% alcohol solution. Find the percent concentration of the first solution.
Let's denote the percent concentration of the first solution as x%. Using the formula, we can set up the following equation:
(x% * 9 L + 80% * 1 L) / (9 L + 1 L) = ?%
Simplifying the equation:
(9x L + 80 L) / 10 L = ?%
To solve for x, we can cross-multiply and solve the resulting equation:
9x L + 80 L = ?% * 10 L
9x L + 80 L = ? * 10 L
Since the desired percent concentration of the solution is not provided, we cannot calculate the value of x% or the percent concentration of the first solution.
I hope this helps to understand how to solve mixture problems using the given information and the formula for concentration of a mixture.