Practise Class 9 Maths Part 2 Chapter 10 – How Quantities Combine: Understanding Data with this NCERT Ganita Manjari worksheet. Prepare more with important questions and revision exercises designed to strengthen data-handling concepts, improve mathematical practice and support effective preparation for class tests and examinations.
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Class 9 Maths – Chapter 10
How Quantities Combine: Understanding Data
Fresh Worksheet
This worksheet contains newly framed questions based on the concepts of averages, weighted averages, mixtures, custom weights, stacked bar charts and 100% stacked bar charts. The questions are different from the textbook questions. All solutions are kept hidden initially.
This worksheet contains newly framed questions based on the concepts of averages, weighted averages, mixtures, custom weights, stacked bar charts and 100% stacked bar charts. The questions are different from the textbook questions. All solutions are kept hidden initially.
Section A – Multiple Choice Questions
Choose the correct option.
1. A group of 12 students has an average score of 48. Another group of 8 students has an average score of 60. What is the average score of all 20 students?
Total score of first group = 12 × 48 = 576.
Total score of second group = 8 × 60 = 480.
Total score = 576 + 480 = 1056.
Number of students = 20.
Total score of second group = 8 × 60 = 480.
Total score = 576 + 480 = 1056.
Number of students = 20.
Average = 1056 ÷ 20 = 52.8
Answer: (B) 52.8
2. Two litres of a 10% juice mixture are mixed with three litres of a 20% juice mixture. What is the percentage of juice in the final mixture?
Juice in first mixture = 10% of 2 L = 0.20 L.
Juice in second mixture = 20% of 3 L = 0.60 L.
Total juice = 0.80 L.
Total mixture = 5 L.
Juice in second mixture = 20% of 3 L = 0.60 L.
Total juice = 0.80 L.
Total mixture = 5 L.
Percentage = (0.80 ÷ 5) × 100 = 16%
Answer: (C) 16%
3. Find the weighted mean of 40, 50 and 70 when their weights are 2, 3 and 5 respectively.
Weighted mean = (40×2 + 50×3 + 70×5) ÷ (2+3+5)
= (80 + 150 + 350) ÷ 10= 580 ÷ 10
= 58.
Answer: (C) 58
4. If all the weights in a weighted mean are multiplied by 4, what happens to the weighted mean?
Multiplying every weight by the same number changes both the numerator and denominator by that number. Therefore, the value of the weighted mean does not change.
Answer: (C) It remains unchanged
5. Which chart is especially useful for comparing how a whole is divided into different parts when every bar represents 100%?
A 100% stacked bar chart makes every whole equal to 100%, so the proportions of its parts can be compared easily.
Answer: (A) 100% stacked bar chart
6. What is the main difference between a normal stacked bar chart and a 100% stacked bar chart?
In a 100% stacked bar chart, every bar represents 100%. Therefore, it focuses on proportions rather than the actual totals.
Answer: (B)
7. A 400 g mixture contains 25% salt. Another 600 g mixture contains 10% salt. What is the percentage of salt in the final mixture?
Salt in first mixture = 25% of 400 = 100 g.
Salt in second mixture = 10% of 600 = 60 g.
Total salt = 160 g.
Total mixture = 1000 g.
Salt in second mixture = 10% of 600 = 60 g.
Total salt = 160 g.
Total mixture = 1000 g.
Percentage = (160 ÷ 1000) × 100 = 16%
Answer: (B) 16%
8. Marks 72, 80 and 90 have weights 2, 3 and 5 respectively. What is the weighted mean?
(72×2 + 80×3 + 90×5) ÷ 10
= (144 + 240 + 450) ÷ 10= 834 ÷ 10
= 83.4.
Answer: (D) 83.4
9. When can the average of the group averages be used directly to find the combined average?
If all groups have the same number of observations, each group average has equal importance.
Answer: (A) When the groups have equal sizes
10. In a combined average, which group has greater influence?
A larger group contributes more observations and therefore has greater influence on the combined average.
Answer: (B) The group with more observations
Section B – Fill in the Blanks and True/False
11. The formula for weighted mean is:
Weighted mean = __________ ÷ __________
Weighted mean = Sum of (value × weight) ÷ Sum of weights.
Answer: Sum of (value × weight) ÷ Sum of weights
12. If all the weights are equal, the weighted mean becomes the ordinary __________.
If all weights are equal, every value gets equal importance.
Answer: arithmetic mean
13. Two equal quantities of mixtures having concentrations 12% and 18% are mixed. The final concentration is 15%. True or False?
Since the quantities are equal, the final concentration is the average of 12% and 18%.
(12 + 18) ÷ 2 = 15%.
(12 + 18) ÷ 2 = 15%.
Answer: True
14. In a 100% stacked bar chart, every complete bar represents 100%. True or False?
Every complete bar in a 100% stacked bar chart represents the whole, or 100%.
Answer: True
15. A 100% stacked bar chart always tells us the exact total number represented by each bar. True or False?
All bars have the same total length, so the chart shows proportions but not the actual totals.
Answer: False
16. If every value in a data set is the same, its weighted mean will also be that same value. True or False?
Giving different weights to the same value does not change that value.
Answer: True
17. When larger quantities are mixed, the final concentration is generally pulled closer to the concentration of the larger quantity. True or False?
The larger quantity has greater influence on the final mixture.
Answer: True
18. A normal stacked bar chart can show both the total and the individual parts. True or False?
The full length of the bar can show the total, while each section shows a component.
Answer: True
19. Multiplying every weight by the same number changes the weighted mean. True or False?
The common factor appears in both the numerator and denominator and cancels.
Answer: False
20. Adding the same number to every weight always leaves the weighted mean unchanged. True or False?
Adding the same number changes the relative importance of the values. Therefore, the weighted mean can change.
Answer: False
Section C – Combined Averages and Weighted Means
21. There are 15 students in Group A with an average score of 42 and 25 students in Group B with an average score of 54. Find the combined average.
Total score of Group A = 15 × 42 = 630.
Total score of Group B = 25 × 54 = 1350.
Combined score = 1980.
Total students = 40.
Total score of Group B = 25 × 54 = 1350.
Combined score = 1980.
Total students = 40.
Combined average = 1980 ÷ 40 = 49.5
Answer: 49.5
22. Three teams have 10, 15 and 25 players. Their average scores are 18, 22 and 20 respectively. Find the average score of all players.
Total score = (10×18) + (15×22) + (25×20)
= 180 + 330 + 500
= 1010.
Total players = 50.
= 180 + 330 + 500
= 1010.
Total players = 50.
Average = 1010 ÷ 50 = 20.2
Answer: 20.2
23. Class A has 18 students with an average mark of 72, Class B has 22 students with an average of 68, and Class C has 10 students with an average of 75. Find the combined average.
Total marks = (18×72) + (22×68) + (10×75)
= 1296 + 1496 + 750
= 3542.
Total students = 50.
= 1296 + 1496 + 750
= 3542.
Total students = 50.
Average = 3542 ÷ 50 = 70.84
Answer: 70.84
24. The average of 20 readings is 35. Three more readings, 30, 40 and 50, are added. Find the new average.
Total of original readings = 20 × 35 = 700.
Total of new readings = 30 + 40 + 50 = 120.
New total = 820.
New number of readings = 23.
Total of new readings = 30 + 40 + 50 = 120.
New total = 820.
New number of readings = 23.
New average = 820 ÷ 23 ≈ 35.65
Answer: Approximately 35.65
25. The average price of 30 books is ₹240. Ten more books having an average price of ₹180 are added. Find the new average price.
Cost of first 30 books = 30 × 240 = ₹7200.
Cost of next 10 books = 10 × 180 = ₹1800.
Total cost = ₹9000.
Total books = 40.
Cost of next 10 books = 10 × 180 = ₹1800.
Total cost = ₹9000.
Total books = 40.
New average = 9000 ÷ 40 = ₹225
Answer: ₹225
26. A student’s average score in four quizzes is 16. In the fifth quiz, the student scores 19. Find the new average.
Total score in four quizzes = 4 × 16 = 64.
New total = 64 + 19 = 83.
New total = 64 + 19 = 83.
New average = 83 ÷ 5 = 16.6
Answer: 16.6
27. One group of 12 students has an average score of 55. Another group of 18 students has an average score of 63. By how much is the total score of the second group greater than that of the first group?
First group total = 12 × 55 = 660.
Second group total = 18 × 63 = 1134.
Second group total = 18 × 63 = 1134.
Difference = 1134 − 660 = 474
Answer: 474
28. Two groups have the same number of members. Their averages are 64 and 78. Find the combined average.
Since both groups have equal sizes, we can directly average their averages.
(64 + 78) ÷ 2 = 71
Answer: 71
29. Find the weighted mean of 12, 18 and 25 with weights 1, 2 and 3 respectively.
Weighted mean = (12×1 + 18×2 + 25×3) ÷ 6
= (12 + 36 + 75) ÷ 6= 123 ÷ 6
= 20.5.
Answer: 20.5
30. Find the weighted mean of 80, 70 and 90 with weights 4, 3 and 3 respectively.
(80×4 + 70×3 + 90×3) ÷ 10
= (320 + 210 + 270) ÷ 10= 800 ÷ 10
= 80.
Answer: 80
Section D – Mixtures and Concentrations
31. Mix 500 mL of an 8% solution with 500 mL of a 14% solution. Find the concentration of the final solution.
Since both quantities are equal:
(8 + 14) ÷ 2 = 11%
Answer: 11%
32. Mix 300 mL of a 5% solution with 700 mL of a 15% solution. Find the final concentration.
Solute in first solution = 300 × 5/100 = 15 mL.
Solute in second solution = 700 × 15/100 = 105 mL.
Total solute = 120 mL.
Total solution = 1000 mL.
Solute in second solution = 700 × 15/100 = 105 mL.
Total solute = 120 mL.
Total solution = 1000 mL.
Concentration = 120 ÷ 1000 × 100 = 12%
Answer: 12%
33. A farmer mixes 2 kg of 20% fertilizer with 3 kg of 8% fertilizer. Find the percentage of fertilizer in the final mixture.
Fertilizer in first mixture = 2 × 20/100 = 0.40 kg.
Fertilizer in second mixture = 3 × 8/100 = 0.24 kg.
Total fertilizer = 0.64 kg.
Total mixture = 5 kg.
Fertilizer in second mixture = 3 × 8/100 = 0.24 kg.
Total fertilizer = 0.64 kg.
Total mixture = 5 kg.
Percentage = 0.64 ÷ 5 × 100 = 12.8%
Answer: 12.8%
34. A painter mixes 400 g of paint containing 30% pigment with 600 g of paint containing 10% pigment. Find the percentage of pigment in the final mixture.
Pigment in first paint = 400 × 30/100 = 120 g.
Pigment in second paint = 600 × 10/100 = 60 g.
Total pigment = 180 g.
Total paint = 1000 g.
Pigment in second paint = 600 × 10/100 = 60 g.
Total pigment = 180 g.
Total paint = 1000 g.
Percentage = 180 ÷ 1000 × 100 = 18%
Answer: 18%
35. Four litres of syrup contain 25% sugar. How much water must be added to make the concentration 10%?
Sugar present = 25% of 4 L = 1 L.
For a 10% solution, 1 L sugar must be 10% of the total mixture.
For a 10% solution, 1 L sugar must be 10% of the total mixture.
Required total mixture = 1 ÷ 0.10 = 10 L
Water to add = 10 − 4 = 6 L.Answer: 6 litres of water
36. Two litres of a 30% solution are mixed with some amount of a 10% solution to obtain a 20% solution. How much of the 10% solution is needed?
Let the required amount be x litres.
0.20 = 0.10x
x = 2.
(0.30×2 + 0.10x) ÷ (2+x) = 0.20
0.60 + 0.10x = 0.40 + 0.20x0.20 = 0.10x
x = 2.
Answer: 2 litres
37. Find the concentration obtained by mixing 800 mL of a 6% solution with 200 mL of a 16% solution.
Solute in first = 800 × 6/100 = 48 mL.
Solute in second = 200 × 16/100 = 32 mL.
Total solute = 80 mL.
Total mixture = 1000 mL.
Solute in second = 200 × 16/100 = 32 mL.
Total solute = 80 mL.
Total mixture = 1000 mL.
Concentration = 80 ÷ 1000 × 100 = 8%
Answer: 8%
38. A 1.5 L mixture contains 12% sugar and a 2.5 L mixture contains 20% sugar. Find the percentage of sugar in the combined mixture.
Sugar in first mixture = 1.5 × 12/100 = 0.18 L.
Sugar in second mixture = 2.5 × 20/100 = 0.50 L.
Total sugar = 0.68 L.
Total mixture = 4 L.
Sugar in second mixture = 2.5 × 20/100 = 0.50 L.
Total sugar = 0.68 L.
Total mixture = 4 L.
Percentage = 0.68 ÷ 4 × 100 = 17%
Answer: 17%
Section E – Custom Weights and Ratings
39. A student’s marks in three tests are 65, 75 and 85. Their weights are 1, 2 and 3. Find the weighted mean.
(65×1 + 75×2 + 85×3) ÷ 6
= (65 + 150 + 255) ÷ 6= 470 ÷ 6
≈ 78.33.
Answer: Approximately 78.33
40. A product receives ratings 4, 3 and 5 for three features. Their weights are 2, 3 and 1 respectively. Find the weighted rating.
(4×2 + 3×3 + 5×1) ÷ 6
= (8 + 9 + 5) ÷ 6= 22 ÷ 6
≈ 3.67.
Answer: Approximately 3.67
41. A travel service receives scores of 8 for transport, 7 for comfort and 9 for punctuality. Their weights are 3, 2 and 5 respectively. Find the weighted score.
(8×3 + 7×2 + 9×5) ÷ 10
= (24 + 14 + 45) ÷ 10= 83 ÷ 10
= 8.3.
Answer: 8.3
42. A final score is made using 20% from Assignment, 30% from Project and 50% from Examination. A student scores 70, 80 and 90 respectively. Find the final score.
Final score = 70×0.20 + 80×0.30 + 90×0.50
= 14 + 24 + 45.
= 14 + 24 + 45.
Final score = 83
Answer: 83
43. A student’s test score is 78 with weight 2, project score is 88 with weight 3, and examination score is 92 with weight 5. Find the weighted mean.
(78×2 + 88×3 + 92×5) ÷ 10
= (156 + 264 + 460) ÷ 10= 880 ÷ 10
= 88.
Answer: 88
44. A product gets ratings of 4 for durability, 5 for design and 3 for price. Their weights are 4, 2 and 4 respectively. Find the weighted rating.
(4×4 + 5×2 + 3×4) ÷ 10
= (16 + 10 + 12) ÷ 10= 38 ÷ 10
= 3.8.
Answer: 3.8
45. Explain why a score of 90 with weight 1 can have less influence than a score of 60 with weight 5.
The weight tells us how much importance a value has. A weight of 5 gives five times as much influence as a weight of 1.
Answer: The score 60 has greater influence because its weight is 5, while the score 90 has weight 1.
46. Find the weighted mean of 20, 30 and 40 with weights 2, 2 and 6 respectively.
(20×2 + 30×2 + 40×6) ÷ 10
= (40 + 60 + 240) ÷ 10= 340 ÷ 10.
Answer: 34
Section F – Stacked Bar Charts and 100% Stacked Bar Charts
47. A family’s monthly expenses are given below. Which type of chart would be suitable for showing both the total expenditure and its different parts?
| Category | Amount |
|---|---|
| Housing | ₹8000 |
| Food | ₹5000 |
| Travel | ₹2000 |
| Education | ₹3000 |
| Health | ₹2000 |
A normal stacked bar chart can show the total expenditure through the full length of the bar and each expense through a separate section.
Answer: A stacked bar chart
48. Using the data in Question 47, find the percentage spent on each category.
Total expenditure = 8000 + 5000 + 2000 + 3000 + 2000 = ₹20,000.
Housing = 8000/20000 × 100 = 40%.
Food = 5000/20000 × 100 = 25%.
Travel = 2000/20000 × 100 = 10%.
Education = 3000/20000 × 100 = 15%.
Health = 2000/20000 × 100 = 10%.
Housing = 8000/20000 × 100 = 40%.
Food = 5000/20000 × 100 = 25%.
Travel = 2000/20000 × 100 = 10%.
Education = 3000/20000 × 100 = 15%.
Health = 2000/20000 × 100 = 10%.
Answer: Housing 40%, Food 25%, Travel 10%, Education 15%, Health 10%.
49. A family spends ₹30,000 per month. Its expenses are 20% on housing, 30% on food, 25% on travel, 15% on education and 10% on health. Find the amount spent on food and health.
Food = 30% of ₹30,000 = ₹9,000.
Health = 10% of ₹30,000 = ₹3,000.
Health = 10% of ₹30,000 = ₹3,000.
Answer: Food = ₹9,000; Health = ₹3,000
50. Three shops have total monthly sales of ₹40,000, ₹60,000 and ₹50,000. Which shop has the greatest total sales? Which chart can show the total and the parts of each shop’s sales?
The second shop has the greatest total because ₹60,000 is the largest amount.
A stacked bar chart can show both totals and components.
A stacked bar chart can show both totals and components.
Answer: Second shop; stacked bar chart
51. A 100% stacked bar chart shows the following sports participation shares:
| School | Sports | Arts | Music |
|---|---|---|---|
| A | 50% | 30% | 20% |
| B | 40% | 35% | 25% |
Which school has the greater share of students in sports? Can we conclude that School A has more students in sports?
School A has the greater share in sports: 50% compared with 40%.
However, we cannot say that School A has more actual students in sports unless we know the total number of students in each school.
However, we cannot say that School A has more actual students in sports unless we know the total number of students in each school.
Answer: School A has the greater share, but the actual number cannot be determined from the percentages alone.
52. Class X has 120 students, of whom 40 enjoy reading. Class Y has 200 students, of whom 50 enjoy reading. Which class has more reading students? Which class has the greater percentage?
Class X reading percentage = 40/120 × 100 = 33.33%.
Class Y reading percentage = 50/200 × 100 = 25%.
Class Y has more reading students because 50 > 40.
Class X has the greater percentage.
Class Y reading percentage = 50/200 × 100 = 25%.
Class Y has more reading students because 50 > 40.
Class X has the greater percentage.
Answer: More students: Class Y. Greater percentage: Class X.
53. Class X has 150 students, of whom 45 participate in science activities. Class Y has 100 students, of whom 40 participate. Find the percentage for each class. Which class has more participating students?
Class X = 45/150 × 100 = 30%.
Class Y = 40/100 × 100 = 40%.
Class X has 45 participating students and Class Y has 40.
Class Y = 40/100 × 100 = 40%.
Class X has 45 participating students and Class Y has 40.
Answer: Class X = 30%, Class Y = 40%; Class X has more participating students.
54. A survey gives four categories with values 15, 25, 10 and 50. Find the percentage represented by each category.
Total = 15 + 25 + 10 + 50 = 100.
Since the total is already 100, the values themselves are the percentages.
Since the total is already 100, the values themselves are the percentages.
Answer: 15%, 25%, 10% and 50%.
55. Why can a 100% stacked bar chart fail to show the exact total number of objects in each group?
Every complete bar is scaled to 100%, regardless of the actual size of the group. Therefore, two groups with very different totals can have bars of the same length.
Answer: Because all bars are normalised to the same total of 100%.
56. A stacked bar represents a total of 800 objects. One segment represents 200 objects. What percentage of the total does this segment represent?
Percentage = 200 ÷ 800 × 100
= 25%.Answer: 25%
Section G – Reasoning and Challenge Questions
57. The average of five numbers is 24. Each number is increased by 6. What is the new average?
If every number increases by 6, the average also increases by 6.
New average = 24 + 6 = 30
Answer: 30
58. The average of a data set is 40. Every value is multiplied by 1.5. What will be the new average?
Multiplying every value by 1.5 also multiplies the average by 1.5.
New average = 40 × 1.5 = 60
Answer: 60
59. A weighted mean is 50. Every value is increased by 10 while the weights remain unchanged. What is the new weighted mean?
Adding the same number to every value increases the weighted mean by that number.
New weighted mean = 50 + 10 = 60
Answer: 60
60. A weighted mean is 50. All the weights are multiplied by 7. What is the new weighted mean?
Multiplying every weight by the same number does not change the weighted mean.
Answer: 50
61. Values 10, 20 and 30 have weights 1, 2 and 3. Find the weighted mean. Then multiply all weights by 2 and check whether the answer changes.
Original weighted mean:
= 140 ÷ 6
≈ 23.33.
New weights are 2, 4 and 6. These are all twice the original weights, so the weighted mean remains the same.
(10×1 + 20×2 + 30×3) ÷ 6
= (10 + 40 + 90) ÷ 6= 140 ÷ 6
≈ 23.33.
New weights are 2, 4 and 6. These are all twice the original weights, so the weighted mean remains the same.
Answer: Approximately 23.33 in both cases.
62. Consider values 10 and 30. Their weights are 1 and 3. Find the weighted mean. Then add 1 to both weights and find the new weighted mean. What do you notice?
Original weighted mean:
(10×1 + 30×3) ÷ 4 = 100 ÷ 4 = 25
New weights are 2 and 4.
(10×2 + 30×4) ÷ 6 = 140 ÷ 6 ≈ 23.33
The answer changes.
Answer: Original = 25; New = approximately 23.33. Adding the same number to all weights does not generally preserve the weighted mean.
63. Mix 1 litre of a 5% solution with 4 litres of a 20% solution. Find the concentration of the final mixture.
Solute in first = 1 × 5/100 = 0.05 L.
Solute in second = 4 × 20/100 = 0.80 L.
Total solute = 0.85 L.
Total mixture = 5 L.
Solute in second = 4 × 20/100 = 0.80 L.
Total solute = 0.85 L.
Total mixture = 5 L.
Percentage = 0.85 ÷ 5 × 100 = 17%
Answer: 17%
64. A 100% stacked bar shows Category A = 20%, Category B = 50% and Category C = 30%. If the total number is 240, find the actual number in each category.
Category A = 20% of 240 = 48.
Category B = 50% of 240 = 120.
Category C = 30% of 240 = 72.
Category B = 50% of 240 = 120.
Category C = 30% of 240 = 72.
Answer: A = 48, B = 120, C = 72.
65. Two shops have the same percentage distribution of customers among three categories. Shop 1 has 100 customers and Shop 2 has 400 customers. Can we say that both shops have the same number of customers in each category?
No. Equal percentages do not mean equal actual numbers when the totals are different.
Shop 2 has four times as many customers as Shop 1. Therefore, if the percentage distribution is the same, the number in each category in Shop 2 will be four times that of Shop 1.
Shop 2 has four times as many customers as Shop 1. Therefore, if the percentage distribution is the same, the number in each category in Shop 2 will be four times that of Shop 1.
Answer: No.
66. Three classes have the following data. Find the combined average age.
| Class | Number of Students | Average Age |
|---|---|---|
| A | 30 | 72 |
| B | 20 | 78 |
| C | 10 | 81 |
Total = (30×72) + (20×78) + (10×81)
= 2160 + 1560 + 810
= 4530.
Total students = 60.
= 2160 + 1560 + 810
= 4530.
Total students = 60.
Combined average = 4530 ÷ 60 = 75.5
Answer: 75.5
67. A mixture contains 2 kg of a 15% substance, 3 kg of a 25% substance and 5 kg of a 10% substance. Find the percentage of the substance in the final mixture.
Substance in first = 2 × 15/100 = 0.30 kg.
Substance in second = 3 × 25/100 = 0.75 kg.
Substance in third = 5 × 10/100 = 0.50 kg.
Total substance = 1.55 kg.
Total mixture = 10 kg.
Substance in second = 3 × 25/100 = 0.75 kg.
Substance in third = 5 × 10/100 = 0.50 kg.
Total substance = 1.55 kg.
Total mixture = 10 kg.
Percentage = 1.55 ÷ 10 × 100 = 15.5%
Answer: 15.5%
68. A restaurant gives ratings of 4 for food, 5 for service and 4 for cleanliness. Their weights are 3, 4 and 3. Find the weighted rating.
(4×3 + 5×4 + 4×3) ÷ 10
= (12 + 20 + 12) ÷ 10= 44 ÷ 10
= 4.4.
Answer: 4.4
69. Four families have monthly incomes of ₹40,000, ₹35,000, ₹50,000 and ₹45,000. If you want to show both the total income and different sources of income for each family, which chart would be more suitable: a stacked bar chart or a 100% stacked bar chart? Explain.
A normal stacked bar chart is more suitable if the aim is to compare actual totals as well as the individual sources.
A 100% stacked bar chart would be better if the main purpose were to compare the percentage shares of the different income sources.
A 100% stacked bar chart would be better if the main purpose were to compare the percentage shares of the different income sources.
Answer: Use a stacked bar chart for actual totals and parts; use a 100% stacked bar chart for proportions.
70. Design your own weighted rating system for a school activity. Choose three criteria, give each a score and assign suitable weights. Calculate the final weighted rating.
One possible example:
= 83 ÷ 10
= 8.3.
| Criterion | Score | Weight |
|---|---|---|
| Creativity | 8 | 3 |
| Presentation | 7 | 2 |
| Teamwork | 9 | 5 |
Weighted rating = (8×3 + 7×2 + 9×5) ÷ 10
= (24 + 14 + 45) ÷ 10= 83 ÷ 10
= 8.3.
Answer: One possible weighted rating is 8.3.
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