51. Weighted average problems

 51. Weighted average problems

Weighted average problems involve calculating the average of quantities where each quantity contributes to the average based on its "weight" (e.g., quantity, frequency, or importance). Unlike a simple arithmetic mean, the weighted average accounts for the relative contributions of each component. The formula for a weighted average is:

\text{Weighted Average} = \frac{\sum (\text{Value} \times \text{Weight})}{\sum \text{Weights}}

Below are 10 weighted average problems with step-by-step solutions and detailed explanations. Each problem includes the problem statement, solution procedure, verification, and an explanation of the concept. These problems are designed to be diverse, practical, and distinct from previous mixture or alligation problems, focusing on scenarios like grades, costs, speeds, and compositions. No alcohol-related scenarios are included, as per your preference.

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Problem 1

A student scores 80, 90, and 70 on three tests with weights of 20%, 30%, and 50%, respectively. What is the student’s weighted average score?

Step-by-Step Solution:

1. Identify values and weights:
   • Scores: 80, 90, 70
   • Weights: 20% (0.2), 30% (0.3), 50% (0.5)
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(80 \cdot 0.2) + (90 \cdot 0.3) + (70 \cdot 0.5)}{0.2 + 0.3 + 0.5}
   • Numerator:

     80 \cdot 0.2 = 16

     ,

     90 \cdot 0.3 = 27

     ,

     70 \cdot 0.5 = 35
   • Total:

     16 + 27 + 35 = 78
   • Denominator:

     0.2 + 0.3 + 0.5 = 1
   • Weighted average:

     \frac{78}{1} = 78
3. Verify: The weights sum to 1 (100%), and the calculation aligns with the formula.
4. Answer: The weighted average score is 78.

Explanation: The weighted average reflects the importance of each test. The third test (50% weight) contributes more than the others, pulling the average closer to 70 compared to a simple average (

\frac{80 + 90 + 70}{3} = 80

).

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Problem 2

A store sells 10 kg of apples at $3 per kg and 15 kg of oranges at $5 per kg. What is the weighted average price per kg of the fruit sold?

Step-by-Step Solution:

1. Identify values and weights:
   • Prices: $3/kg (apples), $5/kg (oranges)
   • Weights: 10 kg (apples), 15 kg (oranges)
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(3 \cdot 10) + (5 \cdot 15)}{10 + 15}
   • Numerator:

     3 \cdot 10 = 30

     ,

     5 \cdot 15 = 75
   • Total:

     30 + 75 = 105
   • Denominator:

     10 + 15 = 25
   • Weighted average:

     \frac{105}{25} = 4.2
3. Verify: Total cost =

   30 + 75 = 105

   . Total weight = 25 kg. Average price =

   \frac{105}{25} = 4.2

   .
4. Answer: The weighted average price is $4.20 per kg.

Explanation: The weights are the quantities sold, and the average price reflects the contribution of each fruit type. More oranges (15 kg vs. 10 kg) pull the average closer to $5.

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Problem 3

A car travels 100 km at 60 km/h and 200 km at 80 km/h. What is the weighted average speed for the entire journey?

Step-by-Step Solution:

1. Identify values and weights:
   • Speeds: 60 km/h, 80 km/h
   • Weights: Time taken for each segment (since average speed is weighted by time, not distance).
   • Time for 100 km at 60 km/h:

     \frac{100}{60} = \frac{5}{3}

     hours
   • Time for 200 km at 80 km/h:

     \frac{200}{80} = 2.5

     hours
2. Apply the weighted average formula:
   \text{Weighted Average Speed} = \frac{(60 \cdot \frac{5}{3}) + (80 \cdot 2.5)}{\frac{5}{3} + 2.5}
   • Numerator:

     60 \cdot \frac{5}{3} = 100

     ,

     80 \cdot 2.5 = 200
   • Total:

     100 + 200 = 300
   • Denominator:

     \frac{5}{3} + 2.5 = \frac{5}{3} + \frac{5}{2} = \frac{10 + 15}{6} = \frac{25}{6}
   • Weighted average:

     \frac{300}{\frac{25}{6}} = 300 \cdot \frac{6}{25} = 72
3. Verify: Total distance =

   100 + 200 = 300

   km. Total time =

   \frac{25}{6}

   hours. Average speed =

   \frac{300}{\frac{25}{6}} = 72

   km/h.
4. Answer: The weighted average speed is 72 km/h.

Explanation: Average speed is weighted by time, not distance, because speed is distance per unit time. The second segment (longer time) has more influence, making the average less than a simple mean (

\frac{60 + 80}{2} = 70

).

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Problem 4

A class has 20 students with an average height of 150 cm and 30 students with an average height of 160 cm. What is the weighted average height of the class?

Step-by-Step Solution:

1. Identify values and weights:
   • Heights: 150 cm, 160 cm
   • Weights: 20 students, 30 students
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(150 \cdot 20) + (160 \cdot 30)}{20 + 30}
   • Numerator:

     150 \cdot 20 = 3000

     ,

     160 \cdot 30 = 4800
   • Total:

     3000 + 4800 = 7800
   • Denominator:

     20 + 30 = 50
   • Weighted average:

     \frac{7800}{50} = 156
3. Verify: Total height = 7800 cm. Total students = 50. Average =

   \frac{7800}{50} = 156

   cm.
4. Answer: The weighted average height is 156 cm.

Explanation: The weights are the number of students, and the average height reflects the contribution of each group. More students in the 160 cm group pull the average above the midpoint (155 cm).

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Problem 5

A company produces 200 units of a product at $10 per unit and 300 units at $15 per unit. What is the weighted average cost per unit?

Step-by-Step Solution:

1. Identify values and weights:
   • Costs: $10/unit, $15/unit
   • Weights: 200 units, 300 units
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(10 \cdot 200) + (15 \cdot 300)}{200 + 300}
   • Numerator:

     10 \cdot 200 = 2000

     ,

     15 \cdot 300 = 4500
   • Total:

     2000 + 4500 = 6500
   • Denominator:

     200 + 300 = 500
   • Weighted average:

     \frac{6500}{500} = 13
3. Verify: Total cost =

   2000 + 4500 = 6500

   . Total units = 500. Average cost =

   \frac{6500}{500} = 13

   .
4. Answer: The weighted average cost is $13 per unit.

Explanation: The weights are the number of units, and the average cost reflects the contribution of each batch. More units at $15 increase the average above the simple mean (

\frac{10 + 15}{2} = 12.5

).

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Problem 6

A recipe uses 2 liters of a 20% sugar solution and 3 liters of a 40% sugar solution. What is the weighted average sugar concentration?

Step-by-Step Solution:

1. Identify values and weights:
   • Concentrations: 20%, 40%
   • Weights: 2 liters, 3 liters
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(20 \cdot 2) + (40 \cdot 3)}{2 + 3}
   • Numerator:

     20 \cdot 2 = 40

     ,

     40 \cdot 3 = 120
   • Total:

     40 + 120 = 160
   • Denominator:

     2 + 3 = 5
   • Weighted average:

     \frac{160}{5} = 32
3. Verify: Total volume = 5 liters. Total sugar =

   0.2 \cdot 2 + 0.4 \cdot 3 = 0.4 + 1.2 = 1.6

   liters. Concentration =

   \frac{1.6}{5} = 0.32 = 32\%

   .
4. Answer: The weighted average sugar concentration is 32%.

Explanation: The weights are the volumes, and the average concentration reflects the sugar contribution of each solution. More of the 40% solution increases the average above the simple mean (30%).

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Problem 7

A train travels 150 km in 3 hours and 250 km in 4 hours. What is the weighted average speed?

Step-by-Step Solution:

1. Identify values and weights:
   • Speeds: First segment =

     \frac{150}{3} = 50

     km/h, Second segment =

     \frac{250}{4} = 62.5

     km/h
   • Weights: Time (3 hours, 4 hours)
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(50 \cdot 3) + (62.5 \cdot 4)}{3 + 4}
   • Numerator:

     50 \cdot 3 = 150

     ,

     62.5 \cdot 4 = 250
   • Total:

     150 + 250 = 400
   • Denominator:

     3 + 4 = 7
   • Weighted average:

     \frac{400}{7} \approx 57.14
3. Verify: Total distance =

   150 + 250 = 400

   km. Total time = 7 hours. Average speed =

   \frac{400}{7} \approx 57.14

   km/h.
4. Answer: The weighted average speed is approximately 57.14 km/h.

Explanation: Speed is weighted by time, as the train spends more time at the higher speed (62.5 km/h), increasing the average above the simple mean of speeds.

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Problem 8

A bakery sells 50 loaves of bread at $4 each and 70 loaves at $6 each. What is the weighted average price per loaf?

Step-by-Step Solution:

1. Identify values and weights:
   • Prices: $4/loaf, $6/loaf
   • Weights: 50 loaves, 70 loaves
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(4 \cdot 50) + (6 \cdot 70)}{50 + 70}
   • Numerator:

     4 \cdot 50 = 200

     ,

     6 \cdot 70 = 420
   • Total:

     200 + 420 = 620
   • Denominator:

     50 + 70 = 120
   • Weighted average:

     \frac{620}{120} = 5.1667
3. Verify: Total cost =

   200 + 420 = 620

   . Total loaves = 120. Average price =

   \frac{620}{120} \approx 5.17

   .
4. Answer: The weighted average price is approximately $5.17 per loaf.

Explanation: The weights are the number of loaves, and more loaves at $6 increase the average above the simple mean (

\frac{4 + 6}{2} = 5

).

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Problem 9

A farmer mixes 4 tons of fertilizer with 10% nitrogen content and 6 tons with 20% nitrogen content. What is the weighted average nitrogen content?

Step-by-Step Solution:

1. Identify values and weights:
   • Nitrogen contents: 10%, 20%
   • Weights: 4 tons, 6 tons
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(10 \cdot 4) + (20 \cdot 6)}{4 + 6}
   • Numerator:

     10 \cdot 4 = 40

     ,

     20 \cdot 6 = 120
   • Total:

     40 + 120 = 160
   • Denominator:

     4 + 6 = 10
   • Weighted average:

     \frac{160}{10} = 16
3. Verify: Total weight = 10 tons. Total nitrogen =

   0.1 \cdot 4 + 0.2 \cdot 6 = 0.4 + 1.2 = 1.6

   tons. Nitrogen content =

   \frac{1.6}{10} = 16\%

   .
4. Answer: The weighted average nitrogen content is 16%.

Explanation: The weights are the tons of fertilizer, and more of the 20% nitrogen fertilizer increases the average above the simple mean (15%).

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Problem 10

A company employs 40 workers earning $20 per hour and 60 workers earning $25 per hour. What is the weighted average hourly wage?

Step-by-Step Solution:

1. Identify values and weights:
   • Wages: $20/hour, $25/hour
   • Weights: 40 workers, 60 workers
2. Apply the weighted average formula:
   \text{Weighted Average} = \frac{(20 \cdot 40) + (25 \cdot 60)}{40 + 60}
   • Numerator:

     20 \cdot 40 = 800

     ,

     25 \cdot 60 = 1500
   • Total:

     800 + 1500 = 2300
   • Denominator:

     40 + 60 = 100
   • Weighted average:

     \frac{2300}{100} = 23
3. Verify: Total wages = 2300. Total workers = 100. Average wage =

   \frac{2300}{100} = 23

   .
4. Answer: The weighted average hourly wage is $23.

Explanation: The weights are the number of workers, and more workers earning $25 increase the average above the simple mean (

\frac{20 + 25}{2} = 22.5

).

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General Notes on Weighted Average Problems

• Key Concept: The weighted average accounts for the relative contribution of each value, using weights such as quantities, time, or frequencies. It’s common in scenarios like grades, costs, speeds, and concentrations.
• Formula:
  \text{Weighted Average} = \frac{\sum (\text{Value} \times \text{Weight})}{\sum \text{Weights}}
• Key Steps:
  1. Identify the values (e.g., scores, prices) and their weights (e.g., quantities, time).
  2. Multiply each value by its weight and sum the products.
  3. Divide by the sum of the weights.
  4. Verify by checking the total contribution and weight.
• Comparison to Simple Average: A simple average assumes equal weights, while a weighted average reflects unequal contributions, making it more accurate for real-world scenarios.
• Applications: Weighted averages are used in education (grades), finance (costs), physics (speeds), and agriculture (compositions), as shown in these problems.
• Verification: Always verify by recalculating the total contribution (e.g., total cost, distance, or property) and dividing by the total weight.

These problems illustrate the versatility of weighted averages across different contexts, emphasizing the importance of weights in calculating meaningful averages.