Practice Worksheets for Class 6 Mathematics: Chapter 04 Data Handling and Presentation
Access printable practice worksheets for Chapter 04 Data Handling and Presentation designed to align with the 2026-27 academic syllabus for Class 6 Mathematics. These structured exercises help students evaluate their conceptual understanding and improve exam readiness.
Practice Chapter 04 Data Handling and Presentation Worksheets for Class 6 Mathematics
Access the complete worksheet PDF for Chapter 04 Data Handling and Presentation below. Regular practice with these targeted questions builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.
Long Answer Questions
(i) Organise this raw list into a table with three columns: Fruit, Tally Marks and Frequency.
(ii) Name the fruit liked by the greatest number of students and the fruit liked by the smallest number of students.
(iii) How many more students chose mango than papaya? Show that the frequencies in your table add up to the total number of students.
Answer:
(i) Going through the list once and putting one tally mark for each name, we get:
| Fruit | Tally Marks | Frequency |
|---|---|---|
| Mango | 9 | |
| Banana | 7 | |
| Apple | 6 | |
| Guava | 5 | |
| Papaya | 3 |
(ii) Mango (frequency 9) is liked by the most students; Papaya (frequency 3) by the fewest.
(iii) Mango − Papaya = 9 − 3 = 6 more students.
Sum of frequencies = 9 + 7 + 6 + 5 + 3 = 30, exactly the number of students, so no answer is missed or counted twice.
In simple words: Mango 9, Banana 7, Apple 6, Guava 5, Papaya 3; mango is the favourite and the total is 30.
Teacher's Note:
1. Strike off each name while making tally marks.
2. The total check is a must for every frequency table.
3. The table shows in five rows what the long list hid.
| Game | Kabaddi | Cricket | Hockey | Football | Badminton |
|---|---|---|---|---|---|
| Number of students | 12 | 16 | 10 | 8 | 6 |
(ii) Find the height, in unit lengths, of the bar for each game.
(iii) Why must all the bars have the same width and equal gaps between them?
Answer:
(i) Steps:
1. Draw a horizontal line and a vertical line meeting at a corner.
2. Along the horizontal line mark the five games at equal distances; label it “Game”.
3. Along the vertical line write the number of students; label it “Number of students”.
4. Take 1 unit length = 2 students and mark 0, 2, 4, 6, … in equal steps from zero.
5. Above each game draw a bar of equal width up to its frequency.
6. Leave equal gaps between the bars and write a title.
(ii) Kabaddi 12 ÷ 2 = 6, Cricket 16 ÷ 2 = 8, Hockey 10 ÷ 2 = 5, Football 8 ÷ 2 = 4, Badminton 6 ÷ 2 = 3 unit lengths.
(iii) Only the height should carry the number. Wider bars would look bigger and mislead. Equal gaps show the bars are free-standing and the categories are separate and equally important.
In simple words: Draw axes, mark a scale of 2 per unit, draw bars of 6, 8, 5, 4, 3 units with equal widths and gaps.
Teacher's Note:
1. Height in units = number ÷ scale.
2. Insist on labels and a title.
3. Width and gaps never carry numbers.
| Village | P | Q | R | S | T |
|---|---|---|---|---|---|
| Symbols shown | 4 full | 2 full and 1 half | 6 full | 3 full | 5 full and 1 half |
(ii) Find the total milk collected from all five villages.
(iii) Which village gave the most milk and which gave the least? Explain how a half symbol helps the person who draws the pictograph.
Answer:
(i) One full symbol = 10 litres, half a symbol = 5 litres.
Village P: 4 × 10 = 40 litres; Village Q: 2 × 10 + 5 = 25 litres; Village R: 6 × 10 = 60 litres; Village S: 3 × 10 = 30 litres; Village T: 5 × 10 + 5 = 55 litres.
(ii) Total = 40 + 25 + 60 + 30 + 55 = 210 litres.
(iii) Village R gave the most (60 litres) and Village Q the least (25 litres).
Data is not always a multiple of the key. Without a half symbol, 25 and 55 litres could not be shown with a key of 10 litres; the half symbol keeps the pictograph short and still exact.
In simple words: P 40, Q 25, R 60, S 30, T 55 litres; total 210; R most, Q least.
Teacher's Note:
1. Add 5 for every half symbol.
2. Compare the rows: the longest row is the most milk.
3. Half symbols handle values ending in 5.
Question 4. The bar graph below shows the number of books borrowed from a school library on five working days.
(i) On which day were the most books borrowed, and on which day the fewest?
(ii) Find the total number of books borrowed during these five days.
(iii) How many more books were borrowed on Friday than on Thursday?
(iv) What scale has been used on the vertical axis, and how do you know?
Answer:
From the graph: Monday 25, Tuesday 40, Wednesday 30, Thursday 15, Friday 45 books.
(i) The tallest bar is Friday’s: most books (45) on Friday. The shortest is Thursday’s: fewest (15) on Thursday.
(ii) Total = 25 + 40 + 30 + 15 + 45 = 155 books.
(iii) 45 − 15 = 30 more books.
(iv) The vertical axis rises from 0 in equal steps of 5, so 1 unit length = 5 books. We know this because the marks are equally spaced, begin at zero and each gap covers 5 books.
In simple words: Most on Friday (45), fewest on Thursday (15); total 155; scale 5 books per unit.
Teacher's Note:
1. List every bar’s value before answering.
2. The gap between two marks gives the scale.
Question 5. The number of bicycles kept in four shops of a market is: Shop P = 30, Shop Q = 45, Shop R = 25, Shop S = 40.
(i) If a pictograph is made with one cycle symbol for 5 bicycles, how many symbols will each shop need?
(ii) If a bar graph is made with the scale 1 unit length = 5 bicycles, how tall will each bar be?
(iii) Compare the two presentations. Which one is easier to prepare for this data and why? Give one advantage of each.
Answer:
(i) Shop P: 30 ÷ 5 = 6; Shop Q: 45 ÷ 5 = 9; Shop R: 25 ÷ 5 = 5; Shop S: 40 ÷ 5 = 8 symbols.
(ii) The bars are 6, 9, 5 and 8 unit lengths for P, Q, R and S. These match the numbers of symbols because the key and the scale are the same.
(iii) The bar graph is easier: the pictograph needs 6 + 9 + 5 + 8 = 28 neat cycle symbols, while four bars are drawn in a minute.
Advantage of the pictograph: it is colourful and understood at a glance without counting divisions.
Advantage of the bar graph: it shows any number exactly, even when it is not a multiple of the key, and is quicker for large data.
In simple words: Both need 6, 9, 5, 8; the bar graph is quicker, the pictograph more eye-catching.
Teacher's Note:
1. Same key and scale give the same counts.
2. Count total symbols to show the effort.
(i) Make a frequency table with tally marks for this data.
(ii) Which family size occurs most often?
(iii) How many families have more than 4 members?
(iv) Suggest a suitable scale for a bar graph of this data and give the height of each bar in unit lengths.
Answer:
(i) Frequency table:
| Members in the family | Tally Marks | Number of families |
|---|---|---|
| 3 | 6 | |
| 4 | 12 | |
| 5 | 8 | |
| 6 | 4 |
Check: 6 + 12 + 8 + 4 = 30 families.
(ii) The highest frequency, 12, is against 4. So a family of 4 members occurs most often.
(iii) Families of 5 or 6 members: 8 + 4 = 12 families.
(iv) The largest frequency is only 12, so take 1 unit length = 2 families. Heights: 6 ÷ 2 = 3, 12 ÷ 2 = 6, 8 ÷ 2 = 4, 4 ÷ 2 = 2 unit lengths for sizes 3, 4, 5 and 6.
In simple words: Sizes 3, 4, 5, 6 occur 6, 12, 8, 4 times; size 4 is most common.
Teacher's Note:
1. Tick off the data while tallying.
2. More than 4 means 5 and 6 only.
3. 1 unit = 1 family is also fine; heights would be 6, 12, 8, 4.
Question 7. A bar graph was drawn on the board for the number of mango baskets sold by five vendors, but part of it was wiped off. The bars still visible are: Vendor P, 7 unit lengths; Vendor Q, 4 unit lengths; Vendor R, 6 unit lengths; Vendor S, 9 unit lengths. The bar for Vendor T has been wiped off. The record book says that Vendor P sold 35 baskets and Vendor T sold 15 baskets.
(i) Find the scale of this bar graph.
(ii) Find the number of baskets sold by Q, R and S.
(iii) How many unit lengths should the bar for Vendor T be?
(iv) Write the numbers you would mark on the vertical axis and explain why the markings must start from zero.
Answer:
(i) 7 unit lengths stand for 35 baskets, so 1 unit length = 35 ÷ 7 = 5 baskets.
(ii) Q: 4 × 5 = 20; R: 6 × 5 = 30; S: 9 × 5 = 45 baskets.
(iii) 15 ÷ 5 = 3 unit lengths.
(iv) Mark 0, 5, 10, 15, 20, 25, 30, 35, 40, 45 at equal distances.
The markings must start at zero because a bar’s length must be proportional to its number. If the axis began at 15, Vendor T’s bar would vanish and the other bars would show only the part above 15, so the comparison would be wrong.
In simple words: Scale is 5 baskets per unit; Q 20, R 30, S 45; T’s bar is 3 units.
Teacher's Note:
1. Find the scale from the one bar whose value is known.
2. The top mark must reach the largest value, 45.
| Item | House rent | Food | School fees | Travel | Others |
|---|---|---|---|---|---|
| Amount (in ₹) | 2400 | 3000 | 1800 | 900 | 1500 |
(ii) Find the height of each bar in unit lengths.
(iii) Find the total monthly spending of the family.
(iv) Is the amount spent on travel half of the amount spent on school fees? Show your working.
Answer:
(i) Every amount is an exact multiple of 300, so every bar ends exactly on a marking. The largest bar is 3000 ÷ 300 = 10 unit lengths, which fits a page, and the smallest is 3 unit lengths, still easy to see.
(ii) House rent 2400 ÷ 300 = 8; Food 3000 ÷ 300 = 10; School fees 1800 ÷ 300 = 6; Travel 900 ÷ 300 = 3; Others 1500 ÷ 300 = 5 units.
(iii) 2400 + 3000 + 1800 + 900 + 1500 = ₹9600 per month.
(iv) Half of school fees = 1800 ÷ 2 = ₹900, and travel is ₹900. So yes. On the graph the travel bar (3 units) is half the school-fees bar (6 units).
In simple words: ₹300 per unit gives neat bars of 8, 10, 6, 3, 5; total ₹9600; travel is half of school fees.
Teacher's Note:
1. A good scale divides every value exactly.
2. Halving shows up as half the bar height.
Question 9. The bar graph below was put up on a school notice board to compare the strength of four sections. The true numbers are: Section A = 40, Section B = 42, Section C = 44 and Section D = 48 students.
(i) At what number does the vertical axis of this graph begin?
(ii) The bar of Section D looks about five times as long as the bar of Section A. Using the true numbers, show that Section D does not have five times as many students as Section A.
(iii) Explain clearly why this graph gives a wrong impression.
(iv) Describe how the graph should be redrawn so that it shows the data honestly.
Answer:
(i) The vertical axis begins at 38, not at 0.
(ii) Five times Section A would be 5 × 40 = 200 students, but Section D has only 48. It has just 48 − 40 = 8 more students than Section A, nowhere near five times.
(iii) Since the axis starts at 38, each bar shows only the part above 38: A shows 40 − 38 = 2, B shows 4, C shows 6 and D shows 10. The eye compares 2 with 10 and sees 1 to 5, though 40 and 48 are very close. Bar lengths are no longer proportional to the numbers, so small differences are exaggerated.
(iv) Start the vertical axis at 0 with equal steps, e.g. 1 unit length = 5 students (0, 5, 10, …, 50). Keep equal widths and gaps, label the axes “Section” and “Number of students”, and write the scale. The bars become 8, 8.4, 8.8 and 9.6 unit lengths, nearly equal, which is the true picture.
In simple words: The axis starts at 38, so tiny differences look huge; start it at 0.
Teacher's Note:
1. Always check the bottom number of the axis.
2. Compare bar lengths with true numbers.
3. Honest graphs start at zero.
| Activity | Playing | Reading | Drawing | Singing | Gardening |
|---|---|---|---|---|---|
| Number of students | 21 | 15 | 12 | 6 | 6 |
(ii) Choose a suitable scale for a bar graph and give the height of each bar in unit lengths.
(iii) Anil says, “More than twice as many students like playing as like singing.” Check his statement by calculation and say whether he is right.
(iv) For a poster on the school notice board, which of the two presentations would you choose, and why?
Answer:
(i) Key: one symbol = 3 students, because 21, 15, 12, 6 and 6 are all multiples of 3, so no part symbol is needed.
Symbols: Playing 21 ÷ 3 = 7, Reading 15 ÷ 3 = 5, Drawing 12 ÷ 3 = 4, Singing 6 ÷ 3 = 2, Gardening 6 ÷ 3 = 2.
(ii) Scale: 1 unit length = 3 students. Bar heights: 7, 5, 4, 2 and 2 unit lengths. The tallest is only 7 units, so it fits easily.
(iii) Twice the singers = 2 × 6 = 12. Playing = 21, and 21 > 12. So Anil is right.
(iv) The pictograph: a student walking past understands it at a glance without reading numbers, and it looks attractive from a distance. A bar graph would be better if exact numbers had to be read, or if the values were not neat multiples of 3.
In simple words: Use 3 students per symbol or per unit; Anil is right since 21 > 12.
Teacher's Note:
1. Check 21 + 15 + 12 + 6 + 6 = 60.
2. Find a number that divides all the values.
3. Choice of display depends on the audience.
Case Study Based Questions
Meera wanted to know which subject her classmates like the best. She stood near the door during the lunch break and asked every student of her class to name one subject. She wrote each answer on a sheet. Later she said that the long list did not tell her anything at a glance. Her teacher advised her to organise the answers in a table using tally marks. After counting the tally marks, Meera prepared the frequency table given below. Every student of the class answered her question, and nobody named two subjects.
| Subject | Mathematics | Science | English | Hindi | Art |
|---|---|---|---|---|---|
| Number of students | 12 | 9 | 7 | 8 | 4 |
(ii) How many students are there in Meera’s class?
(iii) How many more students chose Mathematics than English?
(iv) If Meera draws a bar graph with the scale 1 unit length = 2 students, how tall will the bar for Hindi be?
Answer:
(i) Mathematics, with the highest frequency, 12.
(ii) 12 + 9 + 7 + 8 + 4 = 40 students.
(iii) 12 − 7 = 5 more students.
(iv) 8 ÷ 2 = 4 unit lengths.
In simple words: Maths is the favourite; 40 students; 5 more chose Maths; Hindi bar is 4 units.
Teacher's Note:
1. Every student named one subject, so the frequencies add to the class size.
2. Bar height = frequency ÷ scale.
The class monitor of Class VI writes the attendance on the board every morning. At the end of the week he copied the six numbers into his notebook so that the class teacher could see how regular the students were. The class has 42 students on its rolls. The monitor said that a table of six numbers is much easier to read than six pages of the attendance register, and the teacher agreed that the table would be clearer still as a bar graph.
| Day | Mon | Tue | Wed | Thu | Fri | Sat |
|---|---|---|---|---|---|---|
| Students present | 38 | 35 | 40 | 32 | 36 | 29 |
(ii) On which day were the fewest students present?
(iii) How many more students were present on Wednesday than on Thursday?
(iv) Find the total of the six numbers in the table.
Answer:
(i) Wednesday, with 40 students present.
(ii) Saturday, with only 29 present.
(iii) 40 − 32 = 8 more students.
(iv) 38 + 35 + 40 + 32 + 36 + 29 = 210. (This is the sum of daily attendances, not the number of different students, since a student is counted on each day he or she comes.)
In simple words: Most on Wednesday, fewest on Saturday, 8 more on Wednesday, total 210.
Teacher's Note:
1. 210 is more than 42 because each student is counted every day.
2. Pair numbers to add quickly.
The Eco Club of a school watched the birds that came to the school pond one morning. They counted the birds of five kinds and then drew a pictograph on a chart. In their pictograph one full bird symbol stands for 2 birds, so half a symbol stands for 1 bird. The table below shows how many symbols they drew for each kind of bird.
| Bird | Crow | Sparrow | Pigeon | Myna | Parrot |
|---|---|---|---|---|---|
| Symbols drawn | 7 full | 5 full | 6 full and 1 half | 3 full | 4 full |
(ii) Which bird was seen the greatest number of times?
(iii) How many pigeons were seen?
(iv) How many birds were counted altogether?
Answer:
(i) One full symbol stands for 2 birds (half a symbol for 1 bird).
(ii) The crow: 7 × 2 = 14 birds, the most of all.
(iii) Pigeons: 6 × 2 + 1 = 13 birds.
(iv) Crow 14, Sparrow 5 × 2 = 10, Pigeon 13, Myna 3 × 2 = 6, Parrot 4 × 2 = 8. Total = 14 + 10 + 13 + 6 + 8 = 51 birds.
In simple words: Each symbol is 2 birds; crows were most (14); 13 pigeons; 51 birds in all.
Teacher's Note:
1. Pigeon has 6½ symbols but crow’s 7 is still more.
2. Convert each row to a number before adding.
Imran walks to school every day along the same road. For one week he kept a small notebook and wrote down the kind of every tree he passed. He then counted how many trees of each kind he had written down and made the table below. He wants to show this table to his class as a bar graph, so he must first choose a scale that fits his chart paper.
| Tree | Neem | Peepal | Mango | Banyan | Gulmohar |
|---|---|---|---|---|---|
| Number of trees | 9 | 12 | 15 | 6 | 3 |
(ii) Which kind of tree did he see the least?
(iii) How many trees did he count in all?
(iv) If he takes 1 unit length = 3 trees, how tall will the bars for Peepal and Banyan be?
Answer:
(i) The mango tree, seen 15 times.
(ii) The gulmohar tree, seen only 3 times.
(iii) 9 + 12 + 15 + 6 + 3 = 45 trees.
(iv) Peepal: 12 ÷ 3 = 4 unit lengths; Banyan: 6 ÷ 3 = 2 unit lengths.
In simple words: Mango most, gulmohar least, 45 trees, bars of 4 and 2 units.
Teacher's Note:
1. All values are multiples of 3, so the scale works neatly.
2. Height = number ÷ 3.
The clerk of a village grain market writes down how many bags of wheat the farmers bring in each month. After the harvest the number of bags rises sharply and then falls again. The figures of five months of one year are given in the table. The teacher of Class VI asked her students to show this data as a bar graph on chart paper. She told them that the bars must all have the same width, must have equal gaps between them, and that the vertical axis must begin at zero and rise in equal steps.
| Month | March | April | May | June | July |
|---|---|---|---|---|---|
| Bags of wheat | 150 | 300 | 250 | 200 | 50 |
(ii) How many more bags were brought in April than in July?
(iii) If the scale is 1 unit length = 50 bags, what will be the height of the bar for May?
(iv) Ravi says the number of bags in April is six times the number in July. Check whether he is right.
Answer:
(i) April, with 300 bags.
(ii) 300 − 50 = 250 more bags.
(iii) 250 ÷ 50 = 5 unit lengths.
(iv) 6 × 50 = 300, which is exactly the April figure. So Ravi is right.
In simple words: April had the most; 250 more than July; May bar 5 units; Ravi is right.
Teacher's Note:
1. ‘Six times’ means multiply July by 6.
2. With scale 50, April’s bar (6) is six times July’s (1).
Question 16. A school fair.
Class VI put up five stalls at the school fair. At the end of the day the students counted how many items each stall had sold and made the bar graph given below. The bars are of equal width with equal gaps, and the numbers on the vertical axis go up in equal steps from zero.
(i) Which stall sold the greatest number of items?
(ii) How many cards were sold?
(iii) How many more sweets were sold than plants?
(iv) Find the total number of items sold at the fair.
Answer:
From the graph: Books 30, Toys 45, Sweets 60, Plants 25, Cards 40.
(i) The sweets stall; its bar is the tallest and reaches 60.
(ii) The cards bar reaches 40, so 40 cards were sold.
(iii) 60 − 25 = 35 more sweets.
(iv) 30 + 45 + 60 + 25 + 40 = 200 items.
In simple words: Sweets sold most (60); 40 cards; 35 more sweets than plants; 200 items in all.
Teacher's Note:
1. Bars ending between labelled marks need care: 25 and 45 lie on the 5-step lines.
2. Write all five values first.
The Health Club of a middle school asked all the students of the senior classes how they travel to school. There were too many students to draw one symbol for each, so the club drew a pictograph in which one full symbol stands for 10 students. Where the number was not a whole multiple of ten, they drew half a symbol, which stands for 5 students. The pictograph they made is described in the table below.
| Mode of travel | Walking | Bicycle | School bus | Public bus | Private car |
|---|---|---|---|---|---|
| Symbols drawn | 4 full | 2 full and 1 half | 3 full | 1 full and 1 half | 1 full |
(ii) Which mode of travel is used by the greatest number of students?
(iii) How many students come by bicycle?
(iv) How many students were asked in all?
Answer:
(i) 10 students (half a symbol stands for 5).
(ii) Walking: 4 × 10 = 40 students, more than any other mode.
(iii) Bicycle: 2 × 10 + 5 = 25 students.
(iv) Walking 40, Bicycle 25, School bus 3 × 10 = 30, Public bus 10 + 5 = 15, Private car 10. Total = 40 + 25 + 30 + 15 + 10 = 120 students.
In simple words: Each symbol is 10 students; walking is most common; 25 by bicycle; 120 in all.
Teacher's Note:
1. Half symbols add 5 each.
2. Convert each row before adding.
A teacher gave a five-question test to her class. Each carried one mark, so every student scored a whole number of marks from 0 to 5. After checking the papers she made the frequency table below, in which the first row gives the marks and the second row gives the number of students who scored those marks. No student scored zero.
| Marks scored | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Number of students | 2 | 5 | 8 | 6 | 4 |
(ii) Which score was obtained by the greatest number of students?
(iii) How many students scored more than 3 marks?
(iv) Find the total of the marks scored by the whole class, and explain why adding 1 + 2 + 3 + 4 + 5 does not give this total.
Answer:
(i) 8 students scored 3 marks.
(ii) The score 3, with the highest frequency, 8.
(iii) Scores 4 and 5: 6 + 4 = 10 students.
(iv) 1 × 2 + 2 × 5 + 3 × 8 + 4 × 6 + 5 × 4 = 2 + 10 + 24 + 24 + 20 = 80 marks.
Adding 1 + 2 + 3 + 4 + 5 counts each score only once and ignores how many students got it, so it cannot give the class total.
In simple words: 8 scored 3; 3 is the most common score; 10 scored above 3; the class total is 80.
Teacher's Note:
1. Do not confuse the score with the number of students.
2. Multiply each score by its frequency, then add.
A book fair was held in a town for four days. The organisers counted the visitors each day and put up a bar graph outside the gate. The vertical axis began at 0 and was marked correctly, but the artist drew the Sunday bar three times as wide as the other bars and coloured it bright red, so that it would catch the eye. The gaps between the bars were also left unequal. A person looking at the graph said that Sunday had brought in far more people than all the other days. The true numbers counted by the organisers are given in the table.
| Day | Thursday | Friday | Saturday | Sunday |
|---|---|---|---|---|
| Visitors | 420 | 440 | 480 | 500 |
(ii) How many more visitors came on Sunday than on Thursday?
(iii) Find the total number of visitors during the four days.
(iv) Explain why the graph misled that person, and state two things the artist should change.
Answer:
(i) Sunday, with 500 visitors.
(ii) 500 − 420 = 80 more visitors.
(iii) 420 + 440 + 480 + 500 = 1840 visitors.
(iv) Only the height of a bar should carry the number. The Sunday bar is three times as wide, so it covers about three times too much area, and the bright red colour pulls the eye to it. So Sunday looks far ahead though it had only 80 more visitors than Thursday (about one-fifth more). Unequal gaps add to the confusion.
The artist should (1) draw all four bars with the same width, and (2) leave equal gaps between them, using colour only for decoration, not to make one bar stand out.
In simple words: Sunday had the most (500), but the wide red bar exaggerated it; use equal widths and gaps.
Teacher's Note:
1. Area grabs attention more than height.
2. Equal width, equal gaps, start at zero.
A school cricket team played five matches in an inter-school tournament. The scorer noted the team total in each match in the table below. The sports teacher wants this data put on the sports board as a bar graph, large enough to be read from a distance but small enough to fit on a chart that allows bars of at most 20 unit lengths. Aditi says a pictograph with one bat symbol for one run would be clearer, but her friend does not agree.
| Match | Match 1 | Match 2 | Match 3 | Match 4 | Match 5 |
|---|---|---|---|---|---|
| Runs scored | 45 | 80 | 15 | 60 | 100 |
(ii) Find the total number of runs scored in the five matches.
(iii) Taking 1 unit length = 5 runs, find the height of the bar for each match, and say whether every bar fits within 20 unit lengths.
(iv) A newspaper wrote that in Match 5 the team scored more than double its Match 4 score. Check this statement, and give one reason why the pictograph idea of Aditi is not practical here.
Answer:
(i) Match 5, with 100 runs.
(ii) 45 + 80 + 15 + 60 + 100 = 300 runs.
(iii) Match 1: 45 ÷ 5 = 9; Match 2: 80 ÷ 5 = 16; Match 3: 15 ÷ 5 = 3; Match 4: 60 ÷ 5 = 12; Match 5: 100 ÷ 5 = 20 unit lengths. The tallest bar is exactly 20, so every bar fits.
(iv) Double of Match 4 = 2 × 60 = 120 runs, but Match 5 had only 100, and 100 < 120. So the newspaper is wrong.
Aditi’s idea needs 300 bat symbols in all, which would take very long to draw and be impossible to count quickly; a bigger key (e.g. one symbol = 5 runs) or a bar graph is far better.
In simple words: Match 5 was best; 300 runs in all; bars of 9, 16, 3, 12, 20 fit; the newspaper is wrong.
Teacher's Note:
1. ‘Double’ means multiply by 2 before comparing.
2. One symbol per run is far too many for large data.
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Download CBSE Practice Material: Class 6 Mathematics Chapter 04 Data Handling and Presentation
CBSE Practice Material: Class 6 Mathematics Chapter 04 Data Handling and Presentation
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