Chapter 4: Data Handling and Presentation

Step 1 – Decide exactly what is to be found
Navya and Naresh want to know which game is liked by the largest number of their classmates. In data-handling language, they must find the mode of the data set “favourite game”.

Step 2 – Collect the raw data
Let the total number of pupils in the class be $$n$$. They should ask each of the $$n$$ pupils the question:

“What is your favourite game?”

Write down every answer exactly as it is given. This list of answers is the raw data.

Step 3 – Choose suitable labels
From the raw list, note all different games that appear. Suppose the replies contain only these four games:

• Cricket
• Kho-Kho
• Football
• Badminton

(If more games occur, simply add them to the list.) These become the categories of the data.

Step 4 – Organise the data with tally marks
Prepare a table like the one below and go through the raw list once. For every pupil, put one tally ( | ) in the row of the game he or she names. Every fifth tally is crossed as \\ to make counting easier:

(The tallies above are only an illustration; the actual tallies will depend on the real answers.)

Step 5 – Count the tallies to get frequencies
Convert each group of tallies into an ordinary numeral. In symbol form, if a game gets $$f$$ tallies, then $$f$$ is its frequency.

Step 6 – Find the largest frequency
Locate the row that has the greatest frequency. Let that maximum frequency be $$f_{\max}$$. The corresponding game is the most popular one in the class.

Step 7 – State the conclusion clearly
Suppose after counting they obtain $$f_{\max}=6$$ for Football (as in the sample table). Then they must conclude:

“Football is the most popular game in our class because it is liked by $$6$$ pupils, which is more than any other game.”

Therefore, by collecting the data, organising it with tally marks, and comparing the frequencies, Navya and Naresh can unambiguously find the most popular game.

Step 1 – Decide what information you need
To know which game is liked by the greatest number of children, we must first collect one piece of data from every classmate: “Which game do you like the most?”

Step 2 – Prepare a recording sheet
Make a small table in your notebook with three columns.

(If you are not sure which games may appear, leave a few blank rows so you can write new games when they are mentioned.)

Step 3 – Collect the data (survey)
Ask each classmate exactly the same question. Each time a classmate names a game, put one tally mark | in that game’s row. Remember the tally-mark rule: after four vertical lines, the fifth line crosses the previous four, e.g. |||| becomes ||||\.

Step 4 – Convert tally marks to numbers
When everyone has answered, count the groups of five and single sticks in every row and write the total in the “Number (Frequency)” column. For example, if Kabaddi has ||||\ || the number is $$5 + 2 = 7$$.

Step 5 – Compare the frequencies
Look at the numbers you have just written. Find the greatest number. Suppose the largest frequency is $$15$$ for Cricket, $$12$$ for Football, $$7$$ for Kabaddi, etc. The game with the highest frequency ($$15$$ in this example) is the most popular game.

Step 6 – State the result
Write a sentence such as: “Cricket is the most popular game among Naresh’s and Navya’s classmates because $$15$$ out of $$N$$ students chose it.” (Replace $$N$$ by the class strength.)

Summary. To find the most popular game you: (i) survey every classmate, (ii) record answers with tally marks, (iii) count the tallies, and (iv) pick the game with the highest frequency.

Step 1 – Read the raw data
Each child in the class was asked to write down his or her favourite game. The list collected was:

Cricket, Basketball, Football, Cricket, Badminton, Cricket, Basketball, Badminton, Cricket, Hockey,
Football, Cricket, Basketball, Hockey, Cricket, Badminton, Cricket, Basketball, Football, Cricket

There are 20 entries in all, one for each student.

Step 2 – Prepare a tally-mark table

Going through the list once and putting a tally mark in the right row each time, we get. A group of five tallies is written as four vertical lines with a slash through them: ||||/.

Check: $$8+4+3+3+2=20$$, which matches the total number of students, so the counting is correct.

Step 3 – Compare the frequencies

• Cricket: $$8$$
• Basketball: $$4$$
• Football: $$3$$
• Badminton: $$3$$
• Hockey: $$2$$

The greatest frequency is $$8$$, which belongs to Cricket.

Step 4 – Conclusion

Since Cricket has the highest frequency, it is the most popular game in the class.

Step 1 : Decide what to ask
We want to know which game each classmate likes the most, so we prepare the single question:
“Which of the following is your favourite game? — Cricket, Football, Badminton, Kabaddi, Kho-Kho, Chess.”

Step 2 : Collect the raw data
Suppose we have 40 students in the class. After asking everybody we note their answers one after another (this is the raw data):
Cricket, Football, Cricket, Kabaddi, Chess, Football, Cricket, Cricket, Kho-Kho, Badminton, Football, Kabaddi, Cricket, Cricket, Chess, Kho-Kho, Football, Cricket, Badminton, Kabaddi, Cricket, Football, Football, Cricket, Kho-Kho, Badminton, Kabaddi, Cricket, Football, Cricket, Kabaddi, Cricket, Cricket, Football, Badminton, Kabaddi, Cricket, Football, Cricket, Badminton, Cricket.

Step 3 : Make a tally-mark table
Group every five tallies by crossing the previous four ($$\bcancel{||||}$$). Then the table is:

Check: $$14+9+5+6+3+3 = 40$$, which equals the class strength, so the counting is correct.

Step 4 : Represent the information (optional but useful)
You may draw a bar graph. On the horizontal axis mark the six games, on the vertical axis mark numbers from 0 to 15. Draw a bar for each game whose height equals its frequency. (The tallest bar will clearly stand out for Cricket.)

Step 5 : Find the most popular game
The game with the greatest frequency is Cricket with 14 votes. In mathematical language:
$$\text{Maximum frequency} = 14 \implies \text{Most popular game} = \text{Cricket}.$$

Hence, among our 40 classmates, Cricket turns out to be the most popular game.

Step 1 – Read the information that the table really contains

The table placed just before this question lists only two things for every variety of sweet:

• the name of the sweet, and
• the number of students who want that sweet (often written as tally-marks or numerals).

For example, one row of the table may look like :

Such a table is called a frequency distribution because it tells us only the frequency (how many) of each choice.

Step 2 – Ask what is required for distributing the sweets

To actually hand over a sweet to every child, the teacher must know which particular child wants which sweet. In other words, we need a direct link:

$$\text{Child’s name} \longrightarrow \text{Chosen sweet}$$

Without the names, the teacher can buy the right total quantity, but cannot decide whom to give which packet.

Step 3 – Check whether the table is sufficient

The present table does not provide any names, roll-numbers, or identifying symbols for the children. Therefore the table by itself is not sufficient to distribute the sweets correctly.

Step 4 – Suggest an alternative

The remedy is to prepare a raw-data list (sometimes called an unordered list or a response sheet) that records every student separately:

With such a two-column (or three-column) table, the teacher can walk to each child and hand over the exact sweet he or she asked for.

Conclusion

The frequency table summarises the class preference but is insufficient for individual distribution. A name-wise (raw data) list is the correct alternative.

Given in the textbook The raw observations were written one after another, for example

$$34,\;17,\;28,\;45,\;39,\;17,\;50,\;42,\;28,\;34$$

The questions that followed were of the type:

• Which observation is the smallest?
• Which observation is the greatest?
• How many observations are greater than a certain number?
• What is the range of the data?  etc.

Step 1 – Arrange the data
We first write the same numbers from the smallest to the largest (ascending order):
$$17,\;17,\;28,\;28,\;34,\;34,\;39,\;42,\;45,\;50$$

Step 2 – See how every answer becomes obvious

1. Smallest observation: the first entry $$17$$.
2. Greatest observation: the last entry $$50$$.
3. Range: $$50-17 = 33$$ (just subtract the two ends).
4. How many observations are greater than 40?
   Locate the first number >$$40$$ (here it is $$42$$, the 8^th term).
   There are $$10-7=3$$ numbers to its right, so there are $$3$$ such observations.

Why sorting helped

• The minimum and maximum automatically occupy the two ends, so no searching is needed.
• Equal observations sit together; counting them is quick and error-free.
• To find “how many above/below a value” you locate one position and then just count straight to the end.
• Measures like range and median (middle value) can be read or calculated directly.

Thus, arranging the data in ascending order converted a jumbled list into a neat, ordered line where every required fact could be read off at a glance, saving both time and effort.

Yes. In the example given in the book we first arranged the raw list of numbers in ascending order. That makes it easy to see the smallest and the largest value at a glance. But ascending order is only one of several ways to organise the same data so that the information becomes clearer.

Other equally good ways, each serving a particular purpose, are listed below:

• Descending order  – write the numbers from the largest to the smallest. This is just the reverse of ascending order and is useful whenever we wish to focus quickly on higher values first.
• Tally-mark table  – instead of writing the same number again and again, we prepare a two-column table: one column shows each distinct number and the other column records its frequency with tally marks ||||. After the fifth tally we cross the previous four, which helps us count fives at a glance.
• Frequency table (with numerals)  – after completing the tally marks we replace every group of tallies by the corresponding numeral (5, 7, 12, …). This table is more compact and is the usual starting point for drawing graphs.
• Pictograph  – each picture (a face, a book, a stick figure, etc.) represents a fixed number of items. Equal pictures standing in a row give a simple visual comparison.
• Bar graph  – on squared paper we draw equal-width bars whose heights are proportional to the frequencies. The bar graph is one of the most common graphical presentations in newspapers and on television.
• Grouped (class-interval) table  – when the number of observations is very large, instead of listing every value we place them in class intervals such as 0 – 10, 10 – 20, 20 – 30, and so on. Then we count how many observations fall inside each class.

Thus, depending on what we want to highlight (smallest value, highest value, comparisons, overall pattern, etc.) we can choose any of these methods to arrange the same set of data.

The pictograph converts the number of smileys into exact numbers of students. Writing these numbers from the picture:

To find the mode used by the most students, we compare the numbers:

$$11 > 8 > 5 > 4 > 3$$

The greatest number is $$11$$, which corresponds to the school bus.

Hence, the school bus is the mode of travel chosen by the maximum number of students.

The pictograph given in the textbook shows different modes by which children come to school. In that pictograph one picture = 5 students.

To find which mode is used by the least number of students, we compare the calculated numbers:

• School Bus : $$45$$ students
• Cycle    : $$35$$ students
• Auto-rickshaw : $$25$$ students
• Walking : $$20$$ students
• Car       : $$10$$ students

$$10$$ is the smallest number. Therefore, the mode of travel used by the least number of students is travelling by car.

Step 1 – Recall what Lakhanpal did
Lakhanpal’s data for absent students were only the numbers
$$3,\;5,\;4,\;2,\;0,\;1,\;5,\;7$$. The largest of these is $$7$$, so he had to draw at most seven smileys (☺) in any one row of his pictograph. That is quick to draw, needs little space, and is easy for a reader to count.

Step 2 – Jarina & Sangita’s data
For present students they noted

To copy Lakhanpal’s rule “one ☺ represents one student”, they would have to draw exactly the same number of smileys as children present. Let us count how many those are altogether:

So the pictograph would need 215 individual smileys.

Step 3 – Why this becomes a problem

• Too much drawing – Sketching 215 neat, recognisable symbols is tiring and time-consuming.
• Space required – Each row (one class) would need 20 – 35 symbols. Even if a symbol were only 1 cm wide, the row for Class II (35 symbols) would already be about 35 cm long, wider than an ordinary notebook page.
• Hard to read – When many small, identical pictures are packed together, the reader may miscount or lose place while counting.
• Neatness/alignment – Keeping hundreds of hand-drawn symbols the same size and in straight lines is difficult; a messy picture loses the advantage of quick understanding.
• Printing or photocopying – Even if drawn on a computer, shrinking the picture to fit a page can make individual smileys blur into patches.
• No room for explanation – The legend “1 ☺ = 1 student” and the class labels themselves must also fit; with very long rows there may be no convenient space left.

Step 4 – Typical remedy
Because of these difficulties, we usually change the scale, for example “1 ☺ = 5 students”. Then the largest number, 35, needs only $$35\div5 = 7$$ symbols, almost the same as Lakhanpal’s picture, and all the problems listed above disappear.

In the textbook example a single picture represents 5 students. When the actual attendance is not a multiple of 5, the arithmetic immediately shows the difficulty.

1. For 33 students:
   Pictures required $$\frac{33}{5}=6.6$$.
2. For 27 students:
   Pictures required $$\frac{27}{5}=5.4$$.

Because $$6.6$$ and $$5.4$$ are not whole numbers, one would have to draw partial pictures (for example, 6 full symbols and three-fifths of another symbol). Such fractional symbols look untidy and may confuse readers. The only alternatives are

• to draw and label broken symbols, or
• to change the scale (say, one picture = 1 student or 3 students) so that the totals divide exactly.

Thus, totals like 33 or 27 make it troublesome to prepare a neat and easily understandable pictograph at the scale "1 picture = 5 students".

We are told that Lakhanpal has already converted the information on absentees into a bar graph. In the bar graph one small square (unit length) on the vertical scale represents one student.

From the description of the bars we read the exact height for each class:

The bar for Class 2 reaches up to the fifth small square on the vertical axis.

Since each unit square represents one student, the height $$5$$ corresponds to $$5$$ absent students.

Therefore, the number of students absent in Class 2 that day is $$5$$.

Step 1 : Read the numbers from the bar-graph

Step 2 : Compare the numbers

The largest among $$5,\,12,\,8,\,10$$ is $$12$$.

Step 3 : Identify the corresponding class

The value $$12$$ occurs for Class VI.

Conclusion

Therefore, the maximum number of students were absent in Class VI.

Step 1 : Read the pictograph

The pictograph in the textbook shows rows for Classes I to VI. Next to every class there are small pictures of a child.
Key printed below the chart: one picture represents $$2$$ absent students.

Step 2 : Count the pictures for every class

Step 3 : Identify the class with full attendance

A class with full attendance means no one is absent, i.e. $$0$$ absentees. From the last column of the table only Class VI has $$0$$ absentees.

Conclusion

Therefore, Class VI had full attendance on that day.

Step 1 · Understand the scale
The bar graph says that one unit of bar length represents $$100$$ vehicles.

Step 2 · Read each bar value

Step 3 · Add the vehicles

\[ 150+1200+1000+800+700+600 \]

First add the two smallest numbers:
$$150+600=750$$

Add to $$700$$:
$$750+700=1450$$

Add $$800$$:
$$1450+800=2250$$

Add $$1000$$:
$$2250+1000=3250$$

Finally add $$1200$$:
$$3250+1200=4450$$

Step 4 · State the result
Therefore, the total number of vehicles that passed through the crossing between 6 a.m. and 12 noon is $$4450$$.

Step 1 – Read the bar graph correctly
In the graph given in the textbook, one bar represents every one-hour interval. The height of the first bar (6 – 7 a.m.) shows only about 200 vehicles, while each of the next five bars (7 – 8, 8 – 9, 9 – 10, 10 – 11 and 11 – 12) shows between 550 and 700 vehicles.

Step 2 – Compare the numbers
Taking the average of the later five readings, let us say the usual traffic is roughly 600 vehicles per hour. Then

$$\text{difference}=600-200=400\text{ vehicles}. $$

So the 6 – 7 a.m. traffic is less by about 400 vehicles.

Step 3 – Give a realistic reason

• Most offices, factories and schools start after 8 a.m. Therefore, very few people need to be on the road as early as 6 a.m.
• Buses and other public transport also begin their regular peak-hour services only after about 7 a.m.
• Because sunrise itself is usually after 6 a.m. for a good part of the year, many people prefer not to travel before daylight.

Conclusion
As a result, much less traffic is recorded from 6 – 7 a.m., whereas the period from 7 a.m. to noon covers the morning rush hour, producing much heavier traffic.

Step 1 – Read the pictograph.
The pictograph in the textbook shows the number of vehicles that passed a busy crossing during four one-hour intervals. For the time-slot 7–8 a.m. the row has the greatest number of symbols.

Step 2 – Convert symbols into numbers.
If one symbol represents, say, $$100$$ vehicles, and the 7–8 a.m. row has, for example, $$5$$ symbols, then the number of vehicles in that hour is
\[100 \times 5 = 500\]
For every other interval the product is smaller than $$500$$, so mathematically the 7–8 a.m. count is the highest.

Step 3 – Give the real-life reason.
Most offices, factories and schools in cities start between 8 a.m. and 9 a.m. People therefore leave home between 7 a.m. and 8 a.m. This creates a “rush hour”, so naturally the volume of traffic—cars, buses, scooters and cycles—is largest in that period.

Conclusion.
Both the pictograph (largest numerical count) and everyday experience (office- and school-going rush) show that traffic is heaviest between 7–8 a.m.

Step 1 : Observe what the bar/line graph is telling us
Assume the graph supplied in the textbook lists the number of vehicles that passed a particular traffic-signal in each one-hour slot. A careful reading gives – for instance – 

Step 2 : Check the trend after 8 a.m.
Starting from 8 a.m. the counts go down step-by-step:

• Between 8 and 9 a.m. the drop is $$120-90 = 30$$ vehicles.
• Between 9 and 10 a.m. the drop is $$90-55 = 35$$ vehicles.
• Between 10 and 11 a.m. the drop is $$55-30 = 25$$ vehicles.
• Between 11 a.m. and 12 noon the drop is $$30-15 = 15$$ vehicles.

Thus the numerical evidence clearly shows a steady decrease every succeeding hour after 8 a.m.

Step 3 : Connect the data with real-life context
Morning traffic is usually caused by people going to schools, colleges and offices. In most Indian cities:

• Schools begin around 7 – 8 a.m.
• Offices begin around 9 a.m.
• Those who must reach by 9 a.m. leave home before 8 a.m., so the rush hour culminates at 8 a.m.

Once students and office-goers have reached their destinations, very few people need to be on the road until the afternoon. Because the main cause of heavy traffic (the rush to reach on time) disappears, the supply of vehicles on the road diminishes.

Step 4 : State the concluding reason logically
Therefore, the traffic becomes lesser and lesser each hour after 8 a.m. simply because the morning peak rush has ended; most commuters have already arrived at their workplaces or schools, leaving the roads comparatively empty until noon.

Step 1 – Read what the bar graph tells us

• The title says the graph shows India’s population (in crores) once every ten years from 1951 to 2001.
• The scale printed on the graph is: 1 unit ≈ 10 crores.
• From the bars we read the exact numbers that have already been provided in the question:
  1951 = $$36$$ crores, 1961 = $$44$$ crores, 1971 = $$54$$ crores,
  1981 = $$68$$ crores, 1991 = $$84$$ crores, 2001 = $$102$$ crores.

Step 2 – Recall what makes a good "question on a bar graph"

• The question must be answerable only by looking at the graph.
• Typical things we can ask are:
  • Read a single value ("What was the population in 1981?").
  • Find the least or greatest value ("In which decade was the population smallest?").
  • Find the increase or decrease between two decades ("How much did the population rise from 1991 to 2001?").
  • Compare two non-adjacent decades ("Which decade shows a larger population: 1961 or 1981?").
  • Add or subtract several bars ("What is the total population shown for the first three decades?").

Step 3 – Write a few sample questions

You may ask your friends questions such as:

1. What was the population of India in 1981?
2. In which decade shown was the population the least?
3. By how many crores did the population increase between 1951 and 1961?
4. During which ten-year period did the population rise by the greatest amount?
5. What is the difference between the populations of 1991 and 2001?
6. What is the total population for the first three decades (1951, 1961 and 1971) taken together?
7. Has the population ever decreased in any decade shown? (If yes, name the decade; if no, say so.)
8. Express the increase from 1971 to 2001 in crores.

All the above questions can be answered directly by observing the heights of the bars and using the scale $$1\text{ unit}=10\text{ crores}$$.

Step 1 : Read the population figures from the bar graph/table

Step 2 : Find the total increase over 50 years (1951 to 2001)

Total increase = Population in 2001 − Population in 1951

$$\text{Total increase}=102-36$$

Step 3 : Find the increase in each decade

• 1951 to 1961 : $$43-36 = 7$$ crores
• 1961 to 1971 : $$55-43 = 12$$ crores
• 1971 to 1981 : $$68-55 = 13$$ crores
• 1981 to 1991 : $$85-68 = 17$$ crores
• 1991 to 2001 : $$102-85 = 17$$ crores

Step 4 : State the results clearly

The population of India increased by $$66$$ crores in the 50 years from 1951 to 2001.

The increases in each decade were $$7$$ crores, $$12$$ crores, $$13$$ crores, $$17$$ crores and $$17$$ crores respectively.

We want to find the tallest and the second-tallest bars in the bar graph.
Because the graph was drawn to the scale

\[ 1 \text{ unit length } = \text{Rs. }200 \]

the height (in units) of every bar is obtained by dividing the expenditure (in rupees) by 200.

Comparing the heights:

• The tallest bar is for Food (17 units).
• The second tallest bar is for House rent (15 units).

Hence, Imran’s family spends the most on food and the second most on house rent.

Given information (read from the bar graph in the textbook)

• Expenditure on Education = $$4000\text{ rupees}$$
• Expenditure on Electricity = $$2000\text{ rupees}$$

Step 1 – Form a ratio

$$\dfrac{\text{Electricity cost}}{\text{Education cost}} = \dfrac{2000}{4000}$$

Step 2 – Simplify the ratio

$$\dfrac{2000}{4000} = \dfrac{2000 \div 2000}{4000 \div 2000} = \dfrac{1}{2}$$

Step 3 – Interpret the result

The simplified fraction $$\dfrac{1}{2}$$ means “one-half.” Therefore the money spent on electricity is exactly one-half of the money spent on education.

Conclusion: Yes, the cost of electricity is about (in fact, exactly) one-half the cost of education.

Step 1 – Read the two amounts from the bar graph.

• Cost of food $$=\;\text{Rs }3400$$
• Cost of education $$=\;\text{Rs }800$$

Step 2 – Work out one-fourth of the cost of food.

\[\dfrac{1}{4}\times 3400 = 850\]

Step 3 – Compare the two amounts.

The cost of education is Rs $$800$$ and one-fourth of the cost of food is Rs $$850$$.

Since $$800 \lt 850$$, the cost of education is less than one-fourth of the cost of food.

Step 1 : Read the data

Samantha noted the following numbers of tiny creatures in the tea garden:

Step 2 : Choose a suitable scale

• The biggest number is $$10$$ (caterpillars).
• For easy counting we take one small square on the graph paper to represent $$1$$ insect.

This way the tallest bar will be $$10$$ small squares high, which fits comfortably on ordinary squared paper.

Step 3 : Draw the axes

1. On the left-hand side draw a vertical line and mark it “Number of insects”. This is the y-axis.
2. Along the bottom draw a horizontal line and mark it “Type of insect / critter”. This is the x-axis.
3. Starting from the origin (where the two lines meet), mark equal divisions of one small square up the vertical axis and write the numbers $$1, 2, 3, \ldots , 10$$ beside them.

Step 4 : Mark equal spaces for the five categories

Along the horizontal axis leave one small square as a gap, then draw a rectangle (bar) for “Mites”, leave another square as a gap, then another rectangle for “Caterpillars”, and so on. All bars must be of the same width and all gaps of the same width.

Step 5 : Draw each bar to the correct height

• Mites – height $$=6$$ squares.
• Caterpillars – height $$=10$$ squares.
• Beetles – height $$=5$$ squares.
• Butterflies – height $$=3$$ squares.
• Grasshoppers – height $$=2$$ squares.

Shade or colour each bar lightly (optional) and write the number on top of every bar for extra clarity.

Step 6 : Give the graph a title

Write a title such as “Insects and Critters Seen in the Tea Garden”.

What the finished bar graph should look like (description):

• Five vertical bars resting on the x-axis.
• From left to right they are labelled: Mites, Caterpillars, Beetles, Butterflies, Grasshoppers.
• Their heights, measured on the y-axis, are 6, 10, 5, 3, and 2 squares respectively.
• The scale key reads “1 small square = 1 insect”.

With these steps Samantha (or you!) can neatly present the data in a clear bar graph.

Step 1 : Rewrite the data clearly

Step 2 : Choose a scale for the bar graph

The question tells us to take the scale

$$1\text{ unit length } = 5\text{ students}$$

That means for every 5 students, the bar will rise by 1 square (or 1 cm) on the graph paper.

Step 3 : Convert each number of students into "units of height"

Step 4 : Draw the axes

• Draw two perpendicular lines. The horizontal line (x-axis) will show the activities.
• The vertical line (y-axis) will show the number of students. Mark 0 at the origin.
• Going up the y-axis, mark equal divisions of 1 unit, writing the scale beside it: 1 unit = 5 students. Hence label the y-axis 5, 10, 15, 20, …, 45.

Step 5 : Draw one bar for each activity

• Above "Playing" draw a bar 9 units high (reaches up to 45).
• Above "Reading story books" draw a bar 6 units high (up to 30).
• Above "Watching TV" draw a bar 4 units high (up to 20).
• Above "Listening to music" draw a bar 2 units high (up to 10).
• Above "Painting" draw a bar 3 units high (up to 15).

All bars should have equal width and equal spacing between them.

Step 6 : Answer the question

The tallest bar is for "Playing" (45 students). Excluding this, the next tallest bar is for "Reading story books" (30 students).

Therefore, apart from playing, the activity preferred by the maximum number of students is Reading story books.

Step 1 – Rewrite the collected data

This frequency table is correct; we only have to check the bar graph that Shagufta and Divya prepared for it.

Step 2 – Recall the rules for a correct bar graph (Class 6 syllabus)

• The horizontal axis (x-axis) must carry qualitative data – here the “Year”.
• The vertical axis (y-axis) must carry quantitative data – here “Number of Tigers”.
• All bars must have the same width and must be equally spaced.
• The numerical scale on the y-axis must start from $$0$$ and go up in equal steps – e.g. $$200,\;500\text{ or }1000$$.
• The top of every bar must be exactly level with the required value from the table.

Step 3 – Locate the mistakes in the given (wrong) graph

1. Missing axis titles. Neither the x-axis nor the y-axis is labelled, so a reader cannot tell what the bars represent.
2. Unequal scale on the y-axis. The graph jumps from “1000” straight to “2000”, then to “2600”, then to “3000”, so the divisions are not equal. This makes the bars look taller or shorter than they should be.
3. Wrong height for the year 2014. In the graph the 2014 bar stops at about $$2100$$ instead of the correct $$2200$$.
4. Bars touch each other and are of different widths. In a bar graph the bars must be of equal width with small, equal gaps between them.
5. The years are not written in equal intervals on the x-axis. 2006 and 2010 are very close, while 2018 and 2022 are far apart.

Step 4 – Draw the corrected bar graph

Follow these instructions on graph paper:

• Draw two perpendicular axes. Mark the horizontal axis “Year” and the vertical axis “Number of Tigers”.
• Choose a convenient, equal scale on the y-axis. Example: let one small square represent $$100\text{ tigers}$$. Then the highest value $$3700$$ will be at $$37$$ small squares, which easily fits on ordinary graph paper.
• Write the five years equally spaced on the x-axis: 2006, 2010, 2014, 2018, 2022.
• Draw five bars of equal width (say 2 cm each) with a gap of 1 cm between any two bars.
• For each year, take the bar straight up to the correct height calculated from the table:

• Shade or colour all the bars lightly and give the graph a title such as “Number of Tigers in India (2006–2022)”.

Now the bar graph satisfies every rule: equal scale, equal bar width, correctly plotted heights, labelled axes and title.

Step 5 – Summary of corrections

• Add axis titles and an overall title.
• Use an equal scale (e.g. 100 tigers per small square) on the y-axis; start at $$0$$.
• Plot 2014 at $$2200$$, not $$2100$$.
• Make all bars the same width with equal spacing.
• Place the years at equal distances on the x-axis.

After these five corrections the bar graph accurately represents the tiger-population data from 2006 to 2022.

Step 1 – Recall what a bar graph is
A bar graph represents categories on one axis and the corresponding numerical values (called frequencies or measurements) on the other. Every category gets its own bar, and all bars must have equal width and equal spacing.

Step 2 – Identify the data we have
Here the categories are the classes in the school (for example Class 6A, 6B, 7A, …). The numerical values are the heights (say in cm) of the tallest student in each class.

Step 3 – Decide which axis should carry which item
It is easier to read a graph if the categories are written from left to right along the horizontal axis. Then we can let the vertical axis carry the scale for height (0 cm, 20 cm, 40 cm, …).

Step 4 – Choose the bar orientation
Because the categories are placed horizontally and the numerical scale is vertical, the bars will automatically rise vertically. This is the usual, most familiar way of drawing a bar graph for such data.

Step 5 – State the choice with reason
Therefore, we should draw vertical bars. They make it simple to compare the different heights and match the common classroom convention where the y-axis shows the measurement.

(Note: Horizontal bars would also work if we placed the classes on the vertical axis and the scale on the horizontal axis, but that is less common for this type of data.)

Step 1 · Deciding the orientation of the bars

When we make a bar graph we must be able to read two things easily:

• the names of the items, and
• the length of the bars (the numbers).

The names of rivers such as Mississippi–Missouri or Murray–Darling are rather long. If we draw vertical bars we would have to write these long names below the $x$-axis, one under another, and they would overlap or have to be written very slantingly. A horizontal bar graph leaves plenty of room on the left of every bar, so the labels can be written neatly. Therefore we prefer a graph with horizontal bars.

Step 2 · Collecting the data (one longest river from each continent)

(River lengths are rounded to the nearest $25\text{–}50\,\text{km}$ so that they are easy to plot.)

Step 3 · Choosing a scale

The longest river in the list is about $$6\,650\,\text{km}$$. We want a graph that fits comfortably on an ordinary sheet of paper. Let us decide on the scale

$$\text{1\ cm} \;\longrightarrow\; 500\,\text{km}$$

With this scale the longest bar will be

$$6\,650 \text{ km}\;\div\;500\,\text{km per cm}=13.3\,\text{cm}.$$

That fits nicely on a page.

Step 4 · Drawing the horizontal bar graph (description)

1. Draw a long horizontal $x$-axis, mark equal divisions of $1\,\text{cm}$ (or $0.5\,\text{cm}$ if you like) and label them $0,500,1000,\dots ,7000\,\text{km}$.
2. Draw a vertical $y$-axis on the left. On this axis leave equal gaps and write the river names with their continents: Nile (Africa), Amazon (S. America), Yangtze (Asia), Mississippi–Missouri (N. America), Murray–Darling (Australia), Volga (Europe), Onyx (Antarctica).
3. For each river draw a bar of the appropriate length, measured from the $y$-axis according to the scale.
4. Shade or colour all the bars lightly and write the exact value at the end of each bar if you wish.
5. Add a title: "Longest River on each Continent" and write the scale under the graph.

(Because the answer must be text only, no picture is shown; follow the steps to get a neat graph.)

Step 5 · Reading the finished bar graph

Comparing the bar lengths, the four continents whose longest rivers are over $$6\,000\,\text{km}$$ are, in decreasing order:

• Africa — Nile (about $$6\,650\,\text{km}$$),
• South America — Amazon (about $$6\,400\,\text{km}$$),
• Asia — Yangtze (about $$6\,300\,\text{km}$$), and
• North America — Mississippi–Missouri (about $$6\,275\,\text{km}$$).

Australia’s Murray–Darling and Europe’s Volga are roughly half this length, while Antarctica’s Onyx is very small in comparison.

Conclusion

A horizontal bar graph is more convenient here because the labels are long. From the data we can see that Africa (Nile) has the single longest river, followed very closely by South America (Amazon), then Asia (Yangtze) and North America (Mississippi–Missouri).

The problem asks us to multiply the 4-digit number $$5642$$ by $$2$$.

1. Write the numbers one below the other in the standard multiplication layout.

   	Thousands	Hundreds	Tens	Ones
   Multiplicand	5	6	4	2
   Multiplier	$$\times\;2$$

2. Multiply each digit, starting from the ones place, and carry over whenever the product is 10 or more.

   • Ones place: $$2 \times 2 = 4$$ (no carry). Write $$4$$ in the ones column.
   • Tens place: $$2 \times 4 = 8$$ (no carry). Write $$8$$ in the tens column.
   • Hundreds place: $$2 \times 6 = 12$$. Write $$2$$ in the hundreds column and carry $$1$$ thousand.
   • Thousands place: $$2 \times 5 = 10$$, then add the carried $$1$$ to get $$11$$. Write $$11$$ in the thousands (and ten-thousands) columns.

   The partial products and carries can be shown in a short column-method table:

   	Ten-thousands	Thousands	Hundreds	Tens	Ones
   Result	1	1	2	8	4

3. Read off the final product.

   \[11284\]

Therefore, $$5642 \times 2 = 11284$$.