Q. A chemist has two alloys of gold and copper. The first alloy contains 40% gold and the second alloy contains 60% gold. How many grams of each alloy should the chemist combine to create a new alloy that contains 50% gold?
A) 20 and 20
B) 25 and 25
Let’s say we use x grams of the first alloy and y grams of the second alloy. Then we can set up two equations based on the amount of gold and the total weight of the new alloy:0.4x + 0.6y = 0.5(x + y), x + y = 50. Solving for x and y, we get:x = y = 25.Therefore, we need to use 25 grams of each alloy to create a new alloy that contains 50% gold.
C) 30 and 30
D) 35 and 35
Q. A solution of 12% salt is mixed with a solution of 18% salt to obtain a 15-liter solution of 15% salt. How many liters of each solution were used?
A) 3 and 12
B) 4.5 and 10.5
Let’s say we use x liters of the 12% salt solution and y liters of the 18% salt solution. Then we can set up two equations based on the amount of salt and the total volume of the mixture:0.12x + 0.18y = 0.15(15)x + y = 15 .Solving for x and y, we get:x = 4.5 y = 10.5. Therefore, we need to use 4.5 liters of the 12% salt solution and 10.5 liters of the 18% salt solution.
C) 6 and 9
D) 7.5 and 7.5
Q. A jar contains a mixture of 40% almonds and 60% cashews. How many grams of cashews must be added to the jar to make the mixture 50% almonds and 50% cashews? (Assume that the volume of the mixture is proportional to the weight of the mixture.)
A) 10
B) 20
C) 30
Let’s say there are x grams of cashews in the jar. Then there are 0.4(1000 - x) grams of almonds in the jar, where 1000 is the total weight of the mixture. After adding y grams of cashews to the jar, there are 0.5(1000 + y) grams of almonds and 0.5(1000 + y) grams of cashews in the jar. We can set up an equation to solve for y:
0.4(1000 - x) = 0.5(1000 + y) - 0.4y
Solving for y, we get:
y = 30. Therefore, we need to add 30 grams of cashews to the jar to make the mixture 50% almonds and 50% cashews.
D) 40
Q. A farmer has two types of milk, one that is 2% fat and another that is 4% fat. He wants to make a mixture of milk that is 3% fat. How many liters of each type of milk should he use if he wants to make a total of 60 liters of the mixture?
A) 15 and 45
B) 20 and 40
C) 25 and 35
Let’s say we use x liters of the 2% fat milk and y liters of the 4% fat milk. Then we can set up two equations based on the amount of fat and the total volume of the mixture:
.
. Solving for x and y, we get:
. Therefore, we need to use 25 liters of the 2% fat milk and 35 liters of the 4% fat milk.
D) 30 and 30
Q. A painter has two cans of paint, one that is red and another that is blue. He mixes some of each paint to create a new color that is purple. If he uses twice as much red paint as blue paint, what is the ratio of red paint to purple paint in the mixture?
A) 1:2
B) 1:3
C) 2:3
Let’s say we use x units of red paint and y units of blue paint to create a new color that is purple. Since he uses twice as much red paint as blue paint, we can set up an equation:
. The ratio of red paint to purple paint is:
. Substituting x = 2y, we get:
. Therefore, the ratio of red paint to purple paint is 2:3.
D) 2:5
Q. A chemist has two solutions of acid. The first solution contains 20% acid and the second solution contains 50% acid. How many liters of each solution should the chemist mix to create a new solution that contains 40% acid?
A) 1 and 2
B) 2 and 1
C) 3 and 2
D) 2 and 3
Let’s say we use x liters of the first solution and y liters of the second solution. Then we can set up two equations based on the amount of acid and the total volume of the new solution:
Solving for x and y, we get:
Therefore, we need to mix 2 liters of the first solution with 3 liters of the second solution to create a new solution that contains 40% acid.
Q. A farmer has two types of feed, one that is 10% protein and another that is 20% protein. He wants to make a mixture of feed that is 15% protein. How many pounds of each type of feed should he use if he wants to make a total of 100 pounds of the mixture?
A) 30 and 70
30 and 70 Let’s say we use x pounds of the first feed and y pounds of the second feed. Then we can set up two equations based on the amount of protein and the total weight of the new mixture:
0.1x + 0.2y = 0.15(100)
x + y = 100
Solving for x and y, we get:
x = 30
y = 70
Therefore, we need to use 30 pounds of the first feed and 70 pounds of the second feed to make a mixture that is 15% protein.
B) 40 and 60
C) 50 and 50
D) 60 and 40
