Chapter-wise Worksheets for Class 6 Mathematics: Chapter 04 Data Handling and Presentation
Access comprehensive chapter-wise worksheets for Chapter 04 Data Handling and Presentation using the CBSE Class 6 Mathematics Data Handling and Presentation Practice Paper Worksheet Set 03 with Answers. Designed to align with the 2026-27 academic syllabus for Class 6 Mathematics, these printable practice sets help students reinforce key concepts and improve their overall exam readiness.
Practice Class 6 Mathematics Worksheets: Chapter 04 Data Handling and Presentation
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Class 6 Ganita Prakash Maths Chapter 4 (Data Handling and Presentation) Practice Paper 12 has 20 questions for 40 marks across Sections A to E, and every question comes with a clear, step-by-step answer. Here you practise the highest frequency and the range of a data set, pictographs of bicycles, toys and saplings with half symbols, a tally table of sports clubs, a bar graph of zoo visitors with an average, and reading a ready-made bar graph of traffic at a busy crossing.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation
Answer: (c) 2
Explanation:
1. Count each value: 0, 2, 4 and 5 appear once each, and 3 appears twice.
2. So the greatest number of times any value appears is 2.
3. The highest frequency is 2 (for the value 3).
Teacher's Note:
1. Frequency is a count of how often a value appears, not the value itself.
2. 3 is the value; 2 is its frequency.
(a) 15 books
(b) 500 books
(c) 750 books
(d) 150 books
Show Answer & Explanation
Answer: (c) 750 books
Explanation:
1. First bar: 6 × 50 = 300 books. Second bar: 9 × 50 = 450 books.
2. Total = 300 + 450 = 750.
3. Or (6 + 9) × 50 = 15 × 50 = 750 books.
Teacher's Note:
1. 15 is the total of units, not of books.
2. Multiply by the scale at the end.
(a)
(b)
(c)
(d)
Show Answer & Explanation
Answer: (b)
Explanation:
1. 70 ÷ 20 = 3.5.
2. 3 symbols show 60 bicycles; the remaining 10 is half of 20.
3. So 70 bicycles = 3 symbols and a half symbol.
Teacher's Note:
1. 4 and a half symbols would mean 90 bicycles.
2. Always multiply back to check: 3.5 × 20 = 70.
(a) 5000
(b) 500
(c) 1000
(d) 100
Show Answer & Explanation
Answer: (a) 5000
Explanation:
1. Each unit stands for 500 people.
2. 10 units stand for 10 × 500 = 5000 people.
3. So the population is 5000.
Teacher's Note:
1. Multiply the height by the scale.
2. 500 is just one unit.
(a) 15
(b) 20
(c) 10
(d) 5
Show Answer & Explanation
Answer: (a) 15
Explanation:
1. Cricket: 20 students; Tennis: 5 students.
2. Difference = 20 − 5 = 15.
3. So 15 more students prefer Cricket.
Teacher's Note:
1. “How many more” means subtract.
2. Ignore the data for the other sports.
(a) 5 complete symbols and a partial symbol
(b) 5 complete symbols and a half symbol
(c) 6 complete symbols
(d) 5 complete symbols
Show Answer & Explanation
Answer: (a) 5 complete symbols and a partial symbol
Explanation:
1. 5 complete symbols show 5 × 5 = 25 balloons; 28 − 25 = 3 balloons remain.
2. 3 is more than half of 5 (2.5) but less than a whole symbol, so a part of a symbol is needed.
3. So we need 5 complete symbols and a partial symbol.
Teacher's Note:
1. A half symbol would show 2.5, and a full symbol 5; neither is 3.
2. Partial means some fraction of a symbol.
(a) 8 units
(b) 2 units
(c) 4 units
(d) 16 units
Show Answer & Explanation
Answer: (c) 4 units
Explanation:
1. Each unit stands for 2 movies.
2. Height = 8 ÷ 2 = 4 units.
3. So the bar is 4 units tall.
Teacher's Note:
1. Height = value ÷ scale.
2. 16 comes from multiplying, which is wrong.
(a) 5
(b) 7
(c) 10
(d) 15
Show Answer & Explanation
Answer: (b) 7
Explanation:
1. Range = largest value − smallest value.
2. Largest = 15, smallest = 8.
3. Range = 15 − 8 = 7.
Teacher's Note:
1. The range shows how spread out the data is.
2. Sort the data first: 8, 10, 10, 12, 15.
Assertion (A): Organizing data into a frequency distribution table helps in finding the mode of the data easily.
Reason (R): The frequency table shows the number of times each value appears, making the most frequent value (mode) evident.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. The mode is the value that occurs most often. In a frequency table it is the row with the largest frequency, so A is true.
2. R is true: the table lists how many times each value appears.
3. R explains why the table makes the mode easy to find, so the answer is (a).
Teacher's Note:
1. Mode = most frequent value.
2. Look down the frequency column for the biggest number.
Assertion (A): Bar graphs and pictographs can both be used to represent the same set of data.
Reason (R): A pictograph is a visual representation of data using pictures, while a bar graph uses bars of uniform width.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The same data, such as books read by five children, can be shown as pictures or as bars. A is true.
2. R correctly describes each kind of graph, so R is true.
3. But R only defines the two graphs; it does not say why the same data can be shown both ways. So the answer is (b).
Teacher's Note:
1. A true reason is not always an explanation.
2. Ask: does R tell us why A happens?
Section B: Short Answer Questions (2 Marks Each)
Question 11. A survey was conducted on the number of books read by 10 students in a month:
4, 2, 5, 3, 4, 1, 5, 4, 2, 3.
Arrange the data in ascending order. What is the most common number of books read?
Answer:
Ascending order: 1, 2, 2, 3, 3, 4, 4, 4, 5, 5.
4 appears 3 times; 2, 3 and 5 appear twice and 1 once.
So the most common number of books read is 4.
In simple words: After sorting, 4 shows up three times, more than any other number.
Teacher's Note:
1. Sorting puts equal values next to each other, so they are easy to count.
2. Check that there are still 10 values.
Question 12. A bar graph shows the number of cars passing a point. The bar for Monday is 8 units high, and the bar for Tuesday is 11 units high. If the scale is 1 unit length = 15 cars, what is the total number of cars that passed on both days?
Answer:
Monday: 8 × 15 = 120 cars.
Tuesday: 11 × 15 = 165 cars.
Total = 120 + 165 = 285 cars.
Check: (8 + 11) × 15 = 19 × 15 = 285.
In simple words: Turn each bar into cars (120 and 165) and add them to get 285.
Teacher's Note:
1. Multiply each height by the scale.
2. Both methods must give the same total.
Question 13. A pictograph uses the symbol to represent 25 trees. How many symbols are needed to represent 150 trees? What would be the total number of trees if the pictograph had 5 symbols?
Answer:
Symbols for 150 trees = 150 ÷ 25 = 6 symbols.
5 symbols stand for 5 × 25 = 125 trees.
In simple words: 150 trees need 6 symbols, and 5 symbols mean 125 trees.
Teacher's Note:
1. Divide to find symbols; multiply to find the number of items.
2. Check: 6 × 25 = 150.
Question 14. The following data shows the marks obtained by a student in 5 subjects out of 100:
English: 85, Math: 95, Science: 80, Social Studies: 75, Hindi: 90.
If a bar graph is drawn with a scale of 1 unit length = 5 marks, what would be the height of the bars for Math and Science?
Answer:
Height = marks ÷ 5.
Math: 95 ÷ 5 = 19 units.
Science: 80 ÷ 5 = 16 units.
In simple words: Divide the marks by 5: Math gets 19 units and Science 16 units.
Teacher's Note:
1. Only Math and Science are asked; the rest is extra data.
2. Check: 19 × 5 = 95.
Section C: Short Answer Questions (3 Marks Each)
Football: 25, Cricket: 40, Basketball: 30, Tennis: 15.
(a) Represent this data using tally marks in a frequency distribution table.
(b) Which club has the least number of students?
Answer:
(a) Each crossed group stands for 5 students: Football 5 groups, Cricket 8 groups, Basketball 6 groups, Tennis 3 groups.
| Club | Tally Marks | Number of students |
|---|---|---|
| Football | 25 | |
| Cricket | 40 | |
| Basketball | 30 | |
| Tennis | 15 | |
| Total | 110 |
(b) The Tennis club has the least number of students (15).
In simple words: Draw crossed fives for each club; Tennis has the fewest students.
Teacher's Note:
1. For multiples of 5 the tallies are only full groups.
2. Count the groups: 25 = 5 groups of 5.
Question 16. A pictograph represents the number of toys sold by a shop over 4 days. The key is = 5 toys.
(a) What was the total number of toys sold over the 4 days?
(b) What was the difference between the number of toys sold on Tuesday and Wednesday?
Answer:
Symbols: Monday 5, Tuesday 7, Wednesday 4, Thursday 3.
(a) Total symbols = 5 + 7 + 4 + 3 = 19, so toys = 19 × 5 = 95 toys.
(b) Tuesday: 7 × 5 = 35 toys; Wednesday: 4 × 5 = 20 toys.
Difference = 35 − 20 = 15 toys.
In simple words: Count the teddies and multiply by 5: 95 toys in all, and Tuesday sold 15 more than Wednesday.
Teacher's Note:
1. The difference can be found as (7 − 4) × 5 = 15.
2. Count the symbols carefully in each row.
Question 17. A family's monthly expenditure on food, rent, and savings is ₹5000, ₹8000, and ₹2000 respectively. They want to represent this data on a bar graph with a scale of 1 unit length = ₹1000. Calculate the height of the bars for each category.
Answer:
Height = amount ÷ 1000.
Food: 5000 ÷ 1000 = 5 units.
Rent: 8000 ÷ 1000 = 8 units.
Savings: 2000 ÷ 1000 = 2 units.
In simple words: Divide each amount by ₹1000 to get 5, 8 and 2 units.
Teacher's Note:
1. “Respectively” means in the same order as the items.
2. Rent has the tallest bar.
Section D: Long Answer Question (5 Marks)
(a) Prepare a frequency distribution table for this data.
(b) Draw a bar graph to represent the data, choosing a suitable scale.
(c) What is the average number of visitors per day for these five days?
Answer:
(a)
| Day | Number of visitors |
|---|---|
| Monday | 300 |
| Tuesday | 400 |
| Wednesday | 250 |
| Thursday | 500 |
| Friday | 350 |
| Total | 1800 |
(b) All numbers are multiples of 50, so take 1 unit length = 50 visitors.
Bar heights: Monday 6 units, Tuesday 8 units, Wednesday 5 units, Thursday 10 units, Friday 7 units.
(c) Total visitors = 300 + 400 + 250 + 500 + 350 = 1800.
Average = 1800 ÷ 5 = 360 visitors per day.
In simple words: Make the table, draw bars with 1 unit = 50 visitors, and share 1800 equally over 5 days to get 360.
Teacher's Note:
1. Average = total ÷ number of days.
2. A scale of 50 makes every bar end exactly on a grid line.
3. Remember the title, axis names and scale.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1: The School's Green Initiative
The students of a school organized a “Green Initiative” to plant saplings in the school campus. They collected data on the number of saplings planted by each class. They prepared a pictograph to showcase this effort, using a symbol to represent 10 saplings. A half- symbol represents 5 saplings.Based on this information, answer the following questions:
(a) How would the number of saplings planted by Class 9 be represented?
(b) What is the difference in the number of symbols used for Class 8 and Class 7?
(c) What is the total number of saplings planted by the school?
(d) If the scale was changed to = 5 saplings, what would be the number of symbols needed for Class 10?
Answer:
(a) Class 9 planted 45 saplings and 45 ÷ 10 = 4.5, so it is shown by 4 complete symbols and a half symbol.
(b) Class 8 has 6 symbols and Class 7 has 5 symbols; difference = 1 symbol (that is, 10 saplings).
(c) Total = 40 + 50 + 60 + 45 + 55 = 250 saplings.
(d) With 1 symbol = 5 saplings, Class 10 needs 55 ÷ 5 = 11 symbols.
In simple words: Half a tree means 5 saplings; the school planted 250 in all and Class 10 needs 11 symbols on the new key.
Teacher's Note:
1. Halving the key doubles the number of symbols.
2. In (b) the answer is in symbols, not saplings.
Question 20. Case Study 2: Daily Traffic Count
A traffic police officer recorded the number of vehicles passing through an intersection in a busy city over four different hours. The data is represented in the bar graph below, where the scale is 1 unit length = 200 vehicles.Based on the bar graph, answer the following questions:
(a) What was the total number of vehicles that passed through the intersection between 7 am and 11 am?
(b) In which hour was the traffic the heaviest, and how many vehicles passed during that time?
(c) How many more vehicles passed between 8-9 am compared to 9-10 am?
(d) What is the combined total of vehicles for the two hours with the least traffic?
Answer:
Reading the bars: 7-8 am 2000 (10 units), 8-9 am 2400 (12 units), 9-10 am 1600 (8 units), 10-11 am 1200 (6 units).
(a) Total = 2000 + 2400 + 1600 + 1200 = 7200 vehicles.
(b) Heaviest traffic was in 8-9 am, with 2400 vehicles.
(c) 2400 − 1600 = 800 more vehicles.
(d) The two lowest bars are 10-11 am (1200) and 9-10 am (1600): 1200 + 1600 = 2800 vehicles.
In simple words: Read each bar, then add or subtract: 7200 in all, 8-9 am is busiest, and the two quietest hours make 2800.
Teacher's Note:
1. Read the bar top against the scale carefully, using the grid lines.
2. Units × 200 must match the numbers on the axis.
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Chapter 04 Data Handling and Presentation Printable Worksheets and Exercises for Class 6 Mathematics
Practice Exercises for Class 6 Mathematics Chapter 04 Data Handling and Presentation
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