9. Mixture Problems as Part of Arithmetic Problems

Below are 10 Mixture Problems with step-by-step solutions and detailed explanations. Mixture problems involve combining substances with different properties (e.g., concentrations, costs, or quantities) to achieve a desired result, typically requiring equations based on the conserved property. Each problem includes the problem statement, solution steps, and an explanation.

Problem 1

How many liters of a 15% sugar solution must be mixed with 12 liters of a 40% sugar solution to get a 25% sugar solution?

Step-by-Step Solution:

Define variables: Let ( x ) be the liters of 15% sugar solution.
Set up the equation: Total sugar in the mixture equals the sum from each solution.
- Sugar from 15% solution: ( 0.15x )
- Sugar from 40% solution:
  0.4 \cdot 12 = 4.8
- Total volume:
  x + 12
- Sugar in final mixture:
  0.25 (x + 12)
- Equation:
  0.15x + 4.8 = 0.25 (x + 12)
Solve the equation:
0.15x + 4.8 = 0.25x + 3
4.8 - 3 = 0.25x - 0.15x
1.8 = 0.1x \implies x = \frac{1.8}{0.1} = 18
Verify: Total volume =
18 + 12 = 30
liters. Total sugar =
0.15 \cdot 18 + 0.4 \cdot 12 = 2.7 + 4.8 = 7.5
. Concentration =
\frac{7.5}{30} = 0.25 = 25\%
.
Answer: 18 liters of 15% sugar solution are needed.

Explanation: The 15% solution dilutes the 40% solution to reach 25%. The equation balances the sugar contributions to achieve the target concentration.

Problem 2

A grocer mixes 8 kg of rice worth $6 per kg with rice worth $9 per kg to make a blend worth $7.50 per kg. How much of the $9 rice is needed?

Step-by-Step Solution:

Define variables: Let ( x ) be the kg of $9 rice.
Set up the equation: Total cost equals the sum of costs.
- Cost of $6 rice:
  6 \cdot 8 = 48
- Cost of $9 rice: ( 9x )
- Total weight:
  8 + x
- Cost of blend:
  7.5 (8 + x)
- Equation:
  48 + 9x = 7.5 (8 + x)
Solve the equation:
48 + 9x = 60 + 7.5x
9x - 7.5x = 60 - 48
1.5x = 12 \implies x = \frac{12}{1.5} = 8
Verify: Total weight =
8 + 8 = 16
kg. Total cost =
48 + 9 \cdot 8 = 48 + 72 = 120
. Cost per kg =
\frac{120}{16} = 7.5
.
Answer: 8 kg of $9 rice is needed.

Explanation: The blend’s cost ($7.50) is between $6 and $9, so equal quantities balance the costs to achieve the target.

Problem 3

How much water must be added to 10 liters of a 50% salt solution to reduce it to a 20% salt solution?

Step-by-Step Solution:

Define variables: Let ( x ) be the liters of water added.
Set up the equation: The salt amount remains constant.
- Initial salt:
  0.5 \cdot 10 = 5
  liters
- Total volume after adding water:
  10 + x
- Final salt concentration:
  0.2 (10 + x)
- Equation:
  5 = 0.2 (10 + x)
Solve the equation:
5 = 0.2 (10 + x)
5 = 2 + 0.2x
5 - 2 = 0.2x \implies 3 = 0.2x \implies x = \frac{3}{0.2} = 15
Verify: Total volume =
10 + 15 = 25
liters. Salt = 5 liters. Concentration =
\frac{5}{25} = 0.2 = 20\%
.
Answer: 15 liters of water must be added.

Explanation: Water dilutes the solution by increasing volume while keeping salt constant. The equation ensures the salt concentration reaches 20%.

Problem 4

A jeweler mixes 5 grams of silver that is 90% pure with silver that is 60% pure to get 10 grams of silver that is 75% pure. How much 60% pure silver is needed?

Step-by-Step Solution:

Define variables: Let ( x ) be the grams of 60% pure silver.
Set up the equation: Total pure silver equals the sum from each type.
- Pure silver from 90% silver:
  0.9 \cdot 5 = 4.5
- Pure silver from 60% silver: ( 0.6x )
- Total weight:
  5 + x = 10 \implies x = 5
- Pure silver in final mixture:
  0.75 \cdot 10 = 7.5
- Equation:
  4.5 + 0.6 \cdot 5 = 7.5
Solve (confirm):
4.5 + 3 = 7.5
The equation holds, confirming
x = 5
.
Verify: Total weight =
5 + 5 = 10
grams. Pure silver =
4.5 + 0.6 \cdot 5 = 7.5
. Purity =
\frac{7.5}{10} = 0.75 = 75\%
.
Answer: 5 grams of 60% pure silver are needed.

Explanation: The total weight constraint determines ( x ). The equation confirms the pure silver content achieves the desired purity.

Problem 5

A store mixes tea worth $4 per kg with tea worth $7 per kg to make 15 kg of a blend worth $5 per kg. How much of each type is used?

Step-by-Step Solution:

Define variables: Let ( x ) be the kg of $4 tea, then
15 - x
is the kg of $7 tea.
Set up the equation: Total cost equals the sum of costs.
- Cost of $4 tea: ( 4x )
- Cost of $7 tea:
  7 (15 - x)
- Total cost:
  5 \cdot 15 = 75
- Equation:
  4x + 7 (15 - x) = 75
Solve the equation:
4x + 105 - 7x = 75
-3x + 105 = 75
-3x = -30 \implies x = 10
15 - x = 15 - 10 = 5
Verify: Total weight =
10 + 5 = 15
kg. Total cost =
4 \cdot 10 + 7 \cdot 5 = 40 + 35 = 75
. Cost per kg =
\frac{75}{15} = 5
.
Answer: 10 kg of $4 tea and 5 kg of $7 tea.

Explanation: The blend’s cost is closer to $4, so more of the cheaper tea is used. The equation balances the costs to achieve $5 per kg.

Problem 6

How much of a 70% fertilizer solution must be mixed with a 30% fertilizer solution to make 40 liters of a 50% fertilizer solution?

Step-by-Step Solution:

Define variables: Let ( x ) be the liters of 70% fertilizer solution, then
40 - x
is the liters of 30% solution.
Set up the equation: Total fertilizer equals the sum from each solution.
- Fertilizer from 70% solution: ( 0.7x )
- Fertilizer from 30% solution:
  0.3 (40 - x)
- Total fertilizer:
  0.5 \cdot 40 = 20
- Equation:
  0.7x + 0.3 (40 - x) = 20
Solve the equation:
0.7x + 12 - 0.3x = 20
0.4x + 12 = 20
0.4x = 8 \implies x = \frac{8}{0.4} = 20
40 - x = 40 - 20 = 20
Verify: Total volume =
20 + 20 = 40
liters. Total fertilizer =
0.7 \cdot 20 + 0.3 \cdot 20 = 14 + 6 = 20
. Concentration =
\frac{20}{40} = 0.5 = 50\%
.
Answer: 20 liters of 70% solution and 20 liters of 30% solution.

Explanation: The 50% target is the average of 70% and 30%, so equal volumes balance the fertilizer content.

Problem 7

A baker mixes flour worth $2.50 per kg with flour worth $4 per kg to make 25 kg of a mixture worth $3.10 per kg. How much of each is used?

Step-by-Step Solution:

Define variables: Let ( x ) be the kg of $2.50 flour, then
25 - x
is the kg of $4 flour.
Set up crucifix equation: Total cost equals the sum of costs.
- Cost of $2.50 flour: ( 2.5x )
- Cost of $4 flour:
  4 (25 - x)
- Total cost:
  3.1 \cdot 25 = 77.5
- Equation:
  2.5x + 4 (25 - x) = 77.5
Solve the equation:
2.5x + 100 - 4x = 77.5
-1.5x + 100 = 77.5
-1.5x = -22.5 \implies x = \frac{22.5}{1.5} = 15
25 - x = 25 - 15 = 10
Verify: Total weight =
15 + 10 = 25
kg. Total cost =
2.5 \cdot 15 + 4 \cdot 10 = 37.5 + 40 = 77.5
. Cost per kg =
\frac{77.5}{25} = 3.1
.
Answer: 15 kg of $2.50 flour and 10 kg of $4 flour.

Explanation: The mixture’s cost is closer to $2.50, requiring more of the cheaper flour. The equation balances the costs.

Problem 8

How much pure water must be added to 6 liters of a 25% vinegar solution to make a 15% vinegar solution?

Step-by-Step Solution:

Define variables: Let ( x ) be the liters of pure water (0% vinegar).
Set up the equation: The vinegar amount remains constant.
- Initial vinegar:
  0.25 \cdot 6 = 1.5
  liters
- Vinegar from water:
  0 \cdot x = 0
- Total volume:
  6 + x
- Total vinegar:
  0.15 (6 + x)
- Equation:
  1.5 = 0.15 (6 + x)
Solve the equation:
1.5 = 0.15 (6 + x)
1.5 = 0.9 + 0.15x
1.5 - 0.9 = 0.15x \implies 0.6 = 0.15x \implies x = \frac{0.6}{0.15} = 4
Verify: Total volume =
6 + 4 = 10
liters. Vinegar = 1.5 liters. Concentration =
\frac{1.5}{10} = 0.15 = 15\%
.
Answer: 4 liters of pure water must be added.

Explanation: Water dilutes the vinegar solution. The equation ensures the vinegar concentration reaches 15%.

Problem 9

A chemist mixes 20 liters of a 5% acid solution with a 35% acid solution to make a 15% acid solution. How much of the 35% solution is needed?

Step-by-Step Solution:

Define variables: Let ( x ) be the liters of 35% acid solution.
Set up the equation: Total acid equals the sum from each solution.
- Acid from 5% solution:
  0.05 \cdot 20 = 1
- Acid from 35% solution: ( 0.35x )
- Total volume:
  20 + x
- Total acid:
  0.15 (20 + x)
- Equation:
  1 + 0.35x = 0.15 (20 + x)
Solve the equation:
1 + 0.35x = 3 + 0.15x
0.35x - 0.15x = 3 - 1
0.2x = 2 \implies x = \frac{2}{0.2} = 10
Verify: Total volume =
20 + 10 = 30
liters. Total acid =
0.05 \cdot 20 + 0.35 \cdot 10 = 1 + 3.5 = 4.5
. Concentration =
\frac{4.5}{30} = 0.15 = 15\%
.
Answer: 10 liters of 35% acid solution are needed.

Explanation: The 35% solution increases the acid concentration. The equation balances the acid to achieve 15%.

Problem 10

A farmer mixes two types of grain: one with 10% protein and another with 25% protein, to make 50 kg of grain with 16% protein. How much of each type is used?

Step-by-Step Solution:

Define variables: Let ( x ) be the kg of 10% protein grain, then
50 - x
is the kg of 25% protein grain.
Set up the equation: Total protein equals the sum from each type.
- Protein from 10% grain: ( 0.1x )
- Protein from 25% grain:
  0.25 (50 - x)
- Total protein:
  0.16 \cdot 50 = 8
- Equation:
  0.1x + 0.25 (50 - x) = 8
Solve the equation:
0.1x + 12.5 - 0.25x = 8
-0.15x + 12.5 = 8
-0.15x = -4.5 \implies x = \frac{4.5}{0.15} = 30
50 - x = 50 - 30 = 20
Verify: Total weight =
30 + 20 = 50
kg. Total protein =
0.1 \cdot 30 + 0.25 \cdot 20 = 3 + 5 = 8
. Protein percentage =
\frac{8}{50} = 0.16 = 16\%
.
Answer: 30 kg of 10% protein grain and 20 kg of 25% protein grain.

Explanation: The 16% protein target requires more of the lower-protein grain. The equation balances the protein contributions.

General Notes on Mixture Problems

Key Concept: Mixture problems conserve a property (e.g., sugar, salt, cost, or protein) while balancing quantities. The total amount of the property in the mixture equals the sum from each component.
Formula: Typically,
c_1 x + c_2 y = c_f (x + y)
, where
c_1, c_2
are concentrations/costs, ( x, y ) are quantities, and
c_f
is the final concentration/cost.
Key Steps:
1. Define variables for the quantities.
2. Set up an equation based on the conserved property.
3. Solve and verify.
Alternative Method: Use the alligation rule for ratios. For example, in Problem 6:
\text{Ratio of 70% to 30%} = \frac{50 - 30}{70 - 50} = \frac{20}{20} = 1:1
This confirms equal volumes (20 liters each).
Verification: Check total quantity and final property to ensure consistency.

These problems cover diverse scenarios (sugar, salt, vinegar, etc.) and demonstrate practical applications of mixture problems. If you need more problems or further clarification, let me know!

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9. Mixture Problems as Part of Arithmetic Problems

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