# MIXTURE PROBLEMS WORKSHEET

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.

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%

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

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

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

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

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

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.

⋅ 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%.

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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