A mixture problem involves mixing two different solutions of a certain ingredient to get a desired concentration of the ingredient.
For example, consider two solutions A and B, where A contains 30% acid and B contains 40% acid. I mix 2 liters of solution A and 3 liters of solution B. Then the quantity of the mixture or resultant solution is 5 liters. What percentage of the mixture 5 liters is acid?. Here, you will learn, how to solve this kind of problems.
The amount of an ingredient in a solution can be given as a percent of the total solution. For example, a 25% acid-water solution means that 25% of the total solution is acid and the remaining 75% is water.
Consider a 10-liter 35% salt-water solution.
Amount of salt :
= 35% of 10 liters
= 0.35 x 10 liters
= 3.5 liters
Amount of water :
= 65% of 10 liters
= 0.65 x 10 liters
= 6.5 liters
Problem 1 :
A 20-liter 35% acid-solution is mixed with a 30-liter 10% acid solution to produce a mixture of 50-liter acid solution. Fiond the percentage of acid in the mixture.
Solution :
Amount of acid in 35% acid-solution :
= 35% of 20 liters
= 0.35 x 20 liters
= 7 liters
Amount of acid in 10% acid-solution :
= 10% of 30 liters
= 0.1 x 30 liters
= 3 liters
Amount of acid in the mixture 50 liters :
= 7 + 3
= 10 liters
Percentage of acid in the mixture :
= ¹⁰⁄₅₀ x 100%
= 20%
Problem 2 :
How many gallons of cream that is 15% fat must be mixed with milk that is 5% fat to produce 20 gallons of cream that is 10% fat?
Solution :
Let x be the number of gallons of cream required. Then, the number of gallons of milks required is (20 - x).
15% of x + 5% of (20 - x) = 10% of 20
0.15x + 0.05(20 - x) = 0.1(20)
0.15x + 1 - 0.05x = 2
0.1x + 1 = 2
Subtract 1 from both sides.
0.1x = 1
Divide both sides by 0.1.
x = 10
The answer is 10 gallons.
Problem 3 :
How many kilograms of a 90% nickel alloy must be mixed with a 70% nickel alloy to make 40 kilograms of 80% nickel alloy?
Solution :
Let x be the number kilograms of 90% nickel alloy required. Then, the number of kilograms of 70% nickel alloy required is (40 - x).
90% of x + 70% of (40 - x) = 80% of 40
0.9x + 0.7(40 - x) = 0.8(40)
0.9x + 28 - 0.7x = 32
0.2x + 28 = 32
Subtract 28 from both sides.
0.2x = 4
Divide both sides by 0.2.
x = 20
The answer is 20 kilograms.
Problem 4 :
A tank has a capacity of 10 gallons. When it is full, it contains 30% alcohol. How many gallons, to the nearest tenth, must be replaced by a 50% alcohol solution to give 10 gallons of 40% alcohol solution?
Solution :
Let x be the number gallons of solution in the tank relaced by 50% alchohol solution.
Then, the remaning number of gallons of 30% alcohol solution in the tank is (10 - x).
50% of x + 30% of (10 - x) = 40% of 10
0.5x + 0.3(10 - x) = 0.4(10)
0.5x + 3 - 0.3x = 4
0.2x + 3 = 4
Subtract 3 from both sides.
0.2x = 1
Divide both sides by 0.2.
x = 5
The answer is 5 gallons.
Problem 5 :
How many pounds of chocolate worth $4.20 a pound must be mixed with 10 pounds of chocolate worth $1.50 a pound to produce a mixture worth $2.40 a pound?
Solution :
Let x be the number of pounds of $4.20 chocolate required. Then, the mixture will contain (10 + x) pounds of chocolate worth $2.40 a pound.
4.2x + 1.5(10) = 2.4(10 + x)
4.2x + 15 = 24 + 2.4x
1.8x = 9
x = 5
The answer is 5 pounds.
Problem 6 :
2 m^{3} of soil containing 35% sand was mixed into 6 m^{3} of soil containing 15% sand. What percentage of the mixture is sand?
Solution :
Let x be the percentage of sand in the mixture.
35% of 2 + 15% of 6 = x% of (2 + 6)
Multiply both sides by 100.
35 ⋅ 2 + 15 ⋅ 6 = x ⋅ 8
70 + 90 = 8x
160 = 8x
Divide both sides by 8.
20 = 8
The answer is 20%.
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