Problem 1 :
A 30-liter 10% salt-water solution is mixed with a 20-liter 35% salt-water solution to produce a mixture of 50-liter salt-water solution. What percentage of the mixture is salt?
Problem 2 :
How many liters of 15% acid solution must be mixed with 5% acid solution to produce 20 liters of 10% acid solution?
Problem 3 :
How many ounces of a 90% gold-copper metal must be mixed with a 70% gold-copper metal to make 40 ounces of 80% gold-copper metal?
Problem 4 :
The capacity of a tank is 10 gallons. When it is full, it contains 30% acid and the rest is water. How many gallons be replaced by a 50% salt-water solution to give 10 gallons of 40% acid-water solution?
Problem 5 :
Find the number of pounds of cookies worth $4.20 a pound must be mixed with 10 pounds of cookies worth $1.50 a pound to produce a mixture worth $2.40 a pound.
Problem 6 :
9 lbs. of mixed nuts containing 55% peanuts were mixed with 6 lbs. of another kind of mixed nuts that contain 40% peanuts. What percent of the new mixture is peanuts?
Problem 7 :
5 fl. oz. of a 2% alcohol solution was mixed with 11 fl. oz. of a 66% alcohol solution. Find the concentration of the new mixture.
Problem 8 :
16 lb of Brand M Cinnamon was made by combining 12 lb of Indonesian cinnamon which costs $19/lb with 4 lb of Thai cinnamon which costs $11/lb. Find the cost per lb of the mixture.
1. Answer :
Amount of salt in 10% solution :
= 10% of 30 liters
= 0.1 x 30 liters
= 3 liters
Amount of salt in 35% acid-solution :
= 35% of 20 liters
= 0.35 x 20 liters
= 7 liters
Amount of salt in the mixture 50 liters :
= 3 + 7
= 10 liters
Percentage of salt in the mixture :
= ¹⁰⁄₅₀ x 100%
= 20%
2. Answer :
Let x be the number of liters of 15% acid solution required. Then, the number of liters of 5% acid solution 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 liters.
3. Answer :
Let x be the number of ounces of 90% gold-copper metal required. Then, the number of ounces of 70% gold-copper metal 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 ounces.
4. Answer :
Let x be the number gallons of solution in the tank replaced by 50% acid-water solution.
Then, the remaning number of gallons of 30% salt-water 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.
5. Answer :
Let x be the number of pounds of $4.20 cookies required. Then, the mixture will contain (10 + x) pounds of cookies 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.
6. Answer :
Let x be the percent of peanuts in the mixture.
55% of 9 + 40% of 6 = x% of (9 + 6)
Multiply both sides by 100.
55 ⋅ 9 + 40 ⋅ 6 = x ⋅ 15
495 + 240 = 15x
735 = 15x
Divide both sides by 8.
49 = x
The answer is 49%.
7. Answer :
Let x be the percent of alcohol in the mixture.
2% of 5 + 66% of 11 = x% of (5 + 11)
Multiply both sides by 100.
2 ⋅ 5 + 66 ⋅ 11 = x ⋅ 16
10 + 726 = 16x
736 = 16x
Divide both sides by 16.
46 = x
The concentration of the new mixture is 49%.
8. Answer :
Let x be the cost per lb of the mixture.
12(19) + 4(11) = 16(x)
228 + 44 = 16x
272 = 16x
Divide both sides by 16.
17 = x
The cost of the mixture is $17/lb.
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