Chapter 4 Data Handling and Presentation

To find the most popular game among Naresh's and Navya's classmates from the list, I would need to determine which game was chosen by the highest number of students.

The steps to do this would be:

1. Go through the list provided by Navya, reading each student's favourite game.

2. Identify all the distinct games mentioned in the list.

3. For each distinct game, count how many times it appears in the list. This is called finding the frequency of each game.

Methods for counting frequency could include:

• Making a new list of the unique games and putting a tally mark next to the game name every time it appears in the original list.
• Creating a frequency table with two columns: one for the Game Name and one for the Count (or Frequency).

4. After counting the occurrences of every game, compare the counts.

5. The game with the largest count is the most popular game among the classmates.

To find the most popular game, we need to count how many students chose each game from the list provided in the image (Page 75).

Let's list the games and tally their frequencies:

• Cricket: Ajay, Vijay, Ramu ($|||$ - 3 students)
• Kho-Kho: Aarti, Kavita, Satish, Vimal ($||||$ - 4 students)
• Badminton: Navya, Naresh, David ($|||$ - 3 students)
• Football: Nawab ($|$ - 1 student)
• Hockey: Sarita ($|$ - 1 student)
• Table Tennis: Jaspal ($|$ - 1 student)

Now let's make a frequency count:

Comparing the number of students for each game, we see that Kho-Kho has the highest number of students (4).

Therefore, the most popular game in their class is Kho-Kho.

As an AI, I am unable to physically interact with students in a classroom to collect data on their favourite games.

To find the most popular game among your classmates, you would need to conduct a survey similar to what Naresh and Navya did. Here are the steps you would follow:

1. Prepare a question: Decide exactly what you will ask, e.g., "What is your favourite game?"

2. Collect data: Go to each of your classmates and ask them the question. Record their answers on a list or in a notebook.

3. Organize the data: Go through your list of answers. Identify all the different games that were mentioned.

4. Tally the results: Count how many times each game was chosen. You can use tally marks to help keep track.

Example of a tally table structure:

5. Find the maximum count: Look at the "Count" column in your table. Identify the game with the highest number of students who chose it.

6. State the most popular game: The game with the highest count is the most popular game among your classmates.

The specific answer will depend entirely on the preferences of the students in your class.

Let's analyze each question to determine if data collection is required.

a. What is the most popular TV show among her classmates?

To find the most popular TV show among a specific group of people (her classmates), Pari would need to ask each classmate about their favourite TV show and record their answers. This involves gathering information directly from the source, which is data collection.

Answer: ✔

b. When did India get independence?

This is a historical fact with a specific date. The answer is already known and can be found in history books, online resources, or from knowledgeable people. It does not require conducting a survey or collecting new measurements or observations.

Answer: ✗

c. How much water is getting wasted in her locality?

To determine the amount of water wasted in a locality, Pari would need to gather information specific to that area. This could involve observing water leaks, measuring water usage (if possible), or collecting data from local authorities or residents about water consumption patterns and issues. This requires collecting data about a specific situation or area.

Answer: ✔

d. What is the capital of India?

This is a geographical fact. The capital of India is a piece of standard information that can be found in geographical texts, maps, encyclopedias, or online databases. It does not require collecting new data.

Answer: ✗

To complete the table and answer the questions, we need to count the number of students who chose each type of sweet from the list provided in the image.

Let's list the sweets and count the students who chose each:

• Jalebi: Rohit, Lalit, Reena, Kiran ($||||$ - 4 students)
• Barfi: Manjari, Bhupinder, Javed, Priya ($||||$ - 4 students)
• Gujiya: Ramesh, Dinesh, Ashok, Santosh ($||||$ - 4 students)
• Rasgulla: Swati, Rakesh, Geeta, Anju ($||||$ - 4 students)
• Gulab Jamun: Kavita, Shalini, Rajeev ($|||$ - 3 students)

Now, let's complete the table (implicitly, a frequency table) and answer the questions based on these counts:

Frequency Table:

Based on the counts:

• How many students chose jalebi? 4
• Barfi was chosen by 4 students?
• How many students chose gujiya? 4
• Rasgulla was chosen by 4 students?
• How many students chose gulab jamun? 3

The frequency table that lists the number of students who chose each sweet is not sufficient to distribute each type of sweet to the correct student.

Explanation:

The frequency table only provides the total count for each sweet. For example, it tells Shri Nilesh that 4 students chose Jalebi, but it does not tell him which specific 4 students chose Jalebi. Similarly, for Barfi, Gujiya, Rasgulla, and Gulab Jamun, he knows how many are needed, but not who requested them.

To give the correct sweet to the correct student, Shri Nilesh needs to know the preference of each individual student.

Alternative:

The alternative is to use the original list prepared by Naresh and Navya.

This list contains the name of each student and their corresponding favourite sweet (e.g., "Rohit - Jalebi", "Manjari - Barfi"). By referring to this list, Shri Nilesh can look up each student's name and see which sweet they chose, ensuring that each student receives the sweet they requested.

So, the original, unsummarised list of student names and their chosen sweets is required for accurate distribution, not just the summary table.

We are given the shoe sizes of students, arranged in ascending order:

3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7

Let's answer each part of the question based on this list:

The largest shoe size in the class:

This is the largest number in the sorted list, which is the last number.

The largest shoe size is 7.

The smallest shoe size in the class:

This is the smallest number in the sorted list, which is the first number.

The smallest shoe size is 3.

There are _________ students who wear shoe size 5:

We need to count how many times the number 5 appears in the list.

Counting the '5's in the list: 5, 5, 5, 5, 5, 5, 5, 5, 5, 5.

There are 10 occurrences of the number 5.

There are 10 students who wear shoe size 5.

There are _________ students who wear shoe sizes larger than 4:

We need to count the number of students whose shoe size is greater than 4. From the list, these are the students who wear size 5, 6, or 7.

Number of students wearing size 5 = 10

Number of students wearing size 6 = 4 (counting the '6's in the list: 6, 6, 6, 6)

Number of students wearing size 7 = 1 (counting the '7's in the list: 7)

Total students wearing shoe sizes larger than 4 = (Number of size 5) + (Number of size 6) + (Number of size 7)

Total students = $10 + 4 + 1 = 15$.

There are 15 students who wear shoe sizes larger than 4.

Arranging the data (shoe sizes) in ascending order helped to answer the questions in several ways:

1. Finding the Smallest and Largest values: When data is arranged in ascending order, the smallest value is always the first entry in the list, and the largest value is always the last entry in the list. This makes it very easy to quickly identify the minimum and maximum values without having to scan through the entire unsorted list.

2. Counting Frequencies: When data is sorted, identical values are grouped together. This makes it much easier and less error-prone to count how many times a specific value appears. For example, to count the number of students wearing size 5 shoes, we can just look at the block of '5's in the sorted list and count them consecutively.

3. Finding values within a range or greater/less than a value: When the data is sorted, it is easy to find how many entries satisfy a condition like being "larger than 4". We just need to look at where the numbers transition from 4 to 5 and count all the numbers from that point onwards (the 5s, 6s, and 7s in this case). In an unsorted list, we would have to check every single entry.

In summary, arranging data in ascending (or descending) order makes it easier to find the range of the data, count the occurrences of specific values, and identify values that meet certain criteria by grouping similar values together and ordering them sequentially.

Yes, there are other ways to arrange or represent the data besides arranging it in ascending order.

Here are a few common ways:

1. Descending Order: The data can be arranged from the largest value to the smallest value (e.g., 7, 6, 6, 6, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3). This is the reverse of ascending order and also helps in identifying the largest and smallest values easily and counting frequencies.

2. Frequency Distribution Table: This involves creating a table that lists each unique shoe size and the number of students who wear that size (frequency). This is a very useful way to summarize data and directly shows how often each value occurs.

Example:

3. Using Tally Marks: Similar to a frequency table, but using tally marks as an intermediate step to count the frequency of each shoe size.

Example:

4. Graphical Representation: The data can be represented visually using graphs, such as:

• Pictograph: Using a symbol (like a shoe) to represent a certain number of students for each shoe size.
• Bar Graph: Using bars of different heights to show the number of students for each shoe size.

These methods arrange and present the data in different formats, each useful for different purposes, like summarization, easy counting, or visual comparison.

As an AI, I cannot physically observe trees or travel from home to school to collect this data. This task requires real-world observation.

However, I can provide the names of a few common trees often seen around:

Mango Tree, Neem Tree, Banyan Tree, Peepal Tree, Gulmohar Tree, Ashoka Tree, Coconut Tree.

To answer the question fully, you would need to perform the observation yourself. Here is how you would do it:

1. Choose your route (e.g., home to school).

2. Walk along the route and identify the different types of trees you see.

3. Keep a count of each type of tree encountered. You can use tally marks as shown in the table structure.

4. After completing your observation, count the tally marks for each type of tree and write the total number in the 'Number of Trees' column.

Let's fill the table with some hypothetical example data to demonstrate the process and answer the questions (a, b, c).

Example Data Collection:

Now, using this hypothetical example data, let's answer the questions:

a. Which tree was found in the greatest number?

Looking at the 'Number of Trees' column (8, 12, 4, 5, 8), the greatest number is 12.

The tree found in the greatest number is the Neem Tree (with 12 occurrences).

b. Which tree was found in the smallest number?

Looking at the 'Number of Trees' column (8, 12, 4, 5, 8), the smallest number is 4.

The tree found in the smallest number is the Peepal Tree (with 4 occurrences).

c. Were there any two trees found in the same numbers?

Looking at the 'Number of Trees' column (8, 12, 4, 5, 8), we see that the number 8 appears twice.

Yes, the Mango Tree and the Ashoka Tree were found in the same number (8 each).

Remember, these answers are based on the example data provided above. Your answers will likely be different when you collect the data from your own observation.

As an AI, I cannot physically perform the task of cutting and pasting a news article or counting letters on a piece of paper. This task requires manual data collection from a physical source.

However, I can explain the process you would follow to complete the task and discuss the likely outcomes based on typical letter frequencies in the English language used in news articles.

To answer the questions, you need to follow the steps outlined in the question: paste a news article, count the occurrences of the letters 'c', 'e', 'i', 'r', and 'x', and fill the table.

Let's assume, for demonstration purposes, a hypothetical count based on the typical frequency of these letters in English text. The letter 'e' is generally the most common, 'x' is very uncommon, while 'c', 'i', and 'r' are moderately common, with 'r' and 'i' often more frequent than 'c'.

Hypothetical Example Table:

Based on this hypothetical example data, let's answer parts a and b:

a. The letter found the most number of times is ________

Looking at the counts (45, 120, 75, 88, 3), the highest count is 120.

The letter found the most number of times is e.

b. The letter found the least number of times is ________

Looking at the counts (45, 120, 75, 88, 3), the lowest count is 3.

The letter found the least number of times is x.

c. List the five letters ‘c’, ‘e’, ‘i’, ‘r’, ‘x’ in ascending order of frequency. Now, compare the order of your list with that of your classmates. Is your order the same or nearly the same as theirs? (Almost everyone is likely to get the order ‘x, c, r, i, e’.) Why do you think this is the case?

Listing the letters based on the hypothetical counts in ascending order of frequency (from least frequent to most frequent):

• x (3)
• c (45)
• i (75)
• r (88)
• e (120)

The order of frequency is x, c, i, r, e based on the hypothetical data.

When comparing your list with classmates using different news articles, the order is likely to be nearly the same. This is because the frequency of letters is a property of the English language itself. While small samples (like a single news item) can show some variation, larger texts tend to show consistent patterns of letter frequencies (e.g., 'e' is almost always the most frequent, 'x' is one of the least frequent). The order 'x, c, r, i, e' is a commonly observed approximate order of frequency for these specific letters in standard English text.

d. Write the process you followed to complete this task.

The process followed to complete this task is:

1. Obtain a blank piece of paper and a small news item from a newspaper.

2. Paste the news item onto the paper.

3. Draw a table with two columns: 'Letter' and 'Number of times the letter occurred'. Add rows for the letters 'c', 'e', 'i', 'r', and 'x'.

4. Read through the news article carefully, word by word, letter by letter.

5. Every time one of the specified letters ('c', 'e', 'i', 'r', 'x') is encountered, make a tally mark next to that letter in the table. It is helpful to count each letter as you read to avoid missing any. One way is to go through the text once for each letter, counting all occurrences of 'c', then all 'e's, and so on.

6. After scanning the entire article for all five letters, count the total number of tally marks for each letter.

7. Write the total count for each letter in the 'Number of times the letter occurred' column of the table.

8. Use the completed table to answer questions a, b, and c.

e. Discuss with your friends the processes they followed.

This part requires a classroom activity where students share how they counted the letters (e.g., reading word by word, scanning for one letter at a time, using different marking methods) and compare their results.

f. If you do this task with another news item, what process would you follow?

If I were to do this task with another news item, I would follow the same process as described in part d. The method of collecting data (counting letter frequencies) remains the same regardless of the specific text of the news article. The only difference would be the news article itself and potentially the resulting counts for each letter.

The image shows a pictograph representing the different modes of travel used by students to go to school.

The key of the pictograph states that 1 picture symbol represents 5 students.

A half picture symbol represents half the number of students represented by a full symbol. So, a half symbol represents $\frac{1}{2} \times 5 = 2.5$ students.

Let's calculate the number of students for each mode of travel:

• Cycle: There are 3 full symbols.

Number of students = $3 \times 5 = 15$

• Bus: There are 4 full symbols.

Number of students = $4 \times 5 = 20$

• Scooter: There is 1 full symbol and 1 half symbol.

Number of students = $1 \times 5 + 1 \times 2.5 = 5 + 2.5 = 7.5$

• Walking: There are 2 full symbols and 1 half symbol.

Number of students = $2 \times 5 + 1 \times 2.5 = 10 + 2.5 = 12.5$

• Car: There is 1 full symbol.

Number of students = $1 \times 5 = 5$

The number of students using each mode of travel is:

Cycle: 15

Bus: 20

Scooter: 7.5

Walking: 12.5

Car: 5

Comparing these numbers, the largest number is 20, which corresponds to the mode of travel "Bus".

Therefore, the mode of travel used by the most number of students is Bus.

Based on the pictograph and the key where 1 picture symbol represents 5 students and a half symbol represents 2.5 students, we determined the number of students using each mode of travel:

• Cycle: $3 \times 5 = 15$ students
• Bus: $4 \times 5 = 20$ students
• Scooter: $1 \times 5 + 1 \times 2.5 = 7.5$ students
• Walking: $2 \times 5 + 1 \times 2.5 = 12.5$ students
• Car: $1 \times 5 = 5$ students

The number of students for each mode is:

Cycle: 15

Bus: 20

Scooter: 7.5

Walking: 12.5

Car: 5

Comparing these numbers, the smallest number is 5, which corresponds to the mode of travel "Car".

Therefore, the mode of travel used by the least number of students is Car.

The pictograph shows the number of books borrowed on different days of a week. The key states that 1 book symbol represents 10 books.

Let's find the number of books borrowed on each day:

• Monday: 5 book symbols.

Number of books = $5 \times 10 = 50$

• Tuesday: 4 book symbols.

Number of books = $4 \times 10 = 40$

• Wednesday: 6 book symbols.

Number of books = $6 \times 10 = 60$

• Thursday: 2 book symbols.

Number of books = $2 \times 10 = 20$

• Friday: 5 book symbols.

Number of books = $5 \times 10 = 50$

• Saturday: 7 book symbols.

Number of books = $7 \times 10 = 70$

Now we can answer the questions:

a. On which day were the minimum number of books borrowed?

Comparing the number of books borrowed each day (50, 40, 60, 20, 50, 70), the minimum number is 20.

The minimum number of books was borrowed on Thursday.

b. What was the total number of books borrowed during the week?

To find the total number of books, we add the number of books borrowed each day:

Total books = Books on Mon + Tue + Wed + Thu + Fri + Sat

Total books = $50 + 40 + 60 + 20 + 50 + 70$

Total books = $90 + 60 + 20 + 50 + 70$

Total books = $150 + 20 + 50 + 70$

Total books = $170 + 50 + 70$

Total books = $220 + 70$

The total number of books borrowed during the week was 290.

c. On which day were the maximum number of books borrowed? What may be the possible reason?

Comparing the number of books borrowed each day (50, 40, 60, 20, 50, 70), the maximum number is 70.

The maximum number of books was borrowed on Saturday.

Possible Reason:

Saturday is typically the last day of the school week before the weekend. Students might borrow more books on Saturday to have reading material for their two days off when the library might be closed or they have more free time.

The number of kites purchased by each shopkeeper is given:

• Jasmeet: 150
• Chaman: 200
• Rani: 100
• Poonam Ben: 300
• Rukhsana: 250
• Deepak: 150

We are asked to prepare a pictograph using the symbol 嬨 to represent 100 kites.

Since some numbers are not exact multiples of 100, we will need a half symbol. Let the half symbol represent $\frac{1}{2} \times 100 = 50$ kites.

Key: 嬨 = 100 kites, Half 嬨 = 50 kites.

Number of symbols for each shopkeeper:

• Jasmeet: $150 = 100 + 50 \implies 1$ full 嬨 and 1 half 嬨
• Chaman: $200 = 2 \times 100 \implies 2$ full 嬨 symbols
• Rani: $100 = 1 \times 100 \implies 1$ full 嬨 symbol
• Poonam Ben: $300 = 3 \times 100 \implies 3$ full 嬨 symbols
• Rukhsana: $250 = 2 \times 100 + 50 \implies 2$ full 嬨 symbols and 1 half 嬨
• Deepak: $150 = 1 \times 100 + 50 \implies 1$ full 嬨 and 1 half 嬨

Pictograph:

Key: 嬨 = 100 kites, 嬨_1/2 = 50 kites

Now, let's answer the questions:

a. How many symbols represent the kites that Rani purchased?

Rani purchased 100 kites. Since each symbol represents 100 kites, the number of symbols for Rani is $100 \div 100 = 1$.

One symbol represents the kites that Rani purchased.

b. Who purchased the maximum number of kites?

Comparing the number of kites purchased: Jasmeet (150), Chaman (200), Rani (100), Poonam Ben (300), Rukhsana (250), Deepak (150).

The maximum number is 300.

Poonam Ben purchased the maximum number of kites.

c. Who purchased more kites, Jasmeet or Chaman?

Jasmeet purchased 150 kites.

Chaman purchased 200 kites.

Since $200 > 150$, Chaman purchased more kites than Jasmeet.

d. Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why?

Number of kites Rani purchased = 100.

Double the number of kites Rani purchased = $2 \times 100 = 200$ kites.

Number of kites Poonam Ben purchased = 300.

We need to check if Poonam Ben purchased more than double Rani's amount, i.e., if $300 > 200$.

Yes, $300$ is indeed greater than $200$.

So, Rukhsana is correct because Poonam Ben purchased 300 kites, which is more than double the 100 kites purchased by Rani.

The bar graph shows the number of students absent in different classes on a particular day.

The horizontal axis represents the Class (Class 1, Class 2, Class 3, Class 4, Class 5).

The vertical axis represents the Number of Absent Students. The scale on the vertical axis increases in increments of 2 (0, 2, 4, 6, 8, 10).

Let's read the number of absent students for each class from the height of the bars:

• Class 1: The bar reaches the level of 4 on the vertical axis.

Number of absent students in Class 1 = 4

• Class 2: The bar reaches the level of 6 on the vertical axis.

Number of absent students in Class 2 = 6

• Class 3: The bar reaches the level of 2 on the vertical axis.

Number of absent students in Class 3 = 2

• Class 4: The bar reaches the level of 0 on the vertical axis (the bar has no height).

Number of absent students in Class 4 = 0

• Class 5: The bar reaches the level of 10 on the vertical axis.

Number of absent students in Class 5 = 10

Now we can answer the questions:

1. In Class 2, ___________ students were absent that day.

From the bar graph, the height of the bar for Class 2 corresponds to 6 on the vertical axis.

In Class 2, 6 students were absent that day.

2. In which class were the maximum number of students absent?___________

Comparing the number of absent students in each class (4, 6, 2, 0, 10), the maximum number is 10.

The maximum number of students were absent in Class 5.

3. Which class had full attendance that day? ___________

Full attendance means the number of absent students is 0.

From the bar graph, the number of absent students in Class 4 is 0.

Class 4 had full attendance that day.

The bar graph shows the number of vehicles passing through a crossing each hour from 6 am to 12:00 noon.

The horizontal axis represents the Time (in hourly intervals from 6-7 am to 11 am - 12 noon).

The vertical axis represents the Number of Vehicles. The scale on the vertical axis indicates that one unit of length stands for 100 vehicles.

We need to read the number of vehicles for each hourly interval from 6 am to 12:00 noon:

• 6 am - 7 am: The bar reaches the level between 200 and 300. It aligns with 250.

Number of vehicles = 250

• 7 am - 8 am: The bar reaches the level of 400.

Number of vehicles = 400

• 8 am - 9 am: The bar reaches the level of 550 (halfway between 500 and 600).

Number of vehicles = 550

• 9 am - 10 am: The bar reaches the level of 450 (halfway between 400 and 500).

Number of vehicles = 450

• 10 am - 11 am: The bar reaches the level of 300.

Number of vehicles = 300

• 11 am - 12 noon: The bar reaches the level of 200.

Number of vehicles = 200

To find the total number of vehicles (the question mistakenly asks for "cars" but the graph title and vertical axis label refer to "vehicles"), we add the number of vehicles from each hourly interval:

Total vehicles = (6-7 am) + (7-8 am) + (8-9 am) + (9-10 am) + (10-11 am) + (11 am-12 noon)

Total vehicles = $250 + 400 + 550 + 450 + 300 + 200$

Total vehicles = $(250 + 400) + (550 + 450) + (300 + 200)$

Total vehicles = $650 + 1000 + 500$

Total vehicles = $1650 + 500$

Assuming "cars" in the question refers to "vehicles" as shown in the graph, the total number of vehicles passed through the crossing between 6 am and noon is 2150.

Looking at the bar graph, the number of vehicles passing through the crossing between 6 am and 7 am is 250, which is significantly lower than the traffic volume during the subsequent hours from 7 am to 12 noon (400, 550, 450, 300, 200).

There are several possible reasons why the traffic is much lower in the 6 am - 7 am hour compared to 7 am - 12 noon:

1. Start of the Day/Commute: The time between 7 am and noon, especially from 7 am to 10 am, typically corresponds to the peak hours when people are traveling to work, school, and other activities. The 6 am - 7 am hour is earlier, and fewer people have started their daily commute or activities by this time.

2. Business and School Timings: Most offices, schools, and businesses start operations around 8 am or 9 am. The heaviest traffic is usually observed in the hour or two leading up to these start times as people rush to reach their destinations. The 6 am - 7 am slot is simply too early for most people's standard schedules.

3. Gradual Wake-up and Preparation: People typically wake up and prepare for their day before leaving home. Very few individuals start their commute or activities as early as 6 am unless their job requires it or they prefer to avoid peak traffic.

In summary, the period from 6 am to 7 am precedes the main rush hour when daily activities (like going to work and school) begin for the majority of the population, leading to considerably lower traffic volume compared to later hours in the morning.

Looking at the bar graph, the traffic volume for the hour 7 am - 8 am is 400 vehicles. While the traffic is also heavy in other hours (8 am - 9 am with 550 vehicles, and 9 am - 10 am with 450 vehicles), the question states "heaviest between 7 am and 8 am", which seems inconsistent with the graph as 8 am - 9 am shows higher traffic (550). Let's assume there might be a slight misunderstanding in the question phrasing and analyze why traffic is high from 7 am onwards, particularly peaking between 8 am and 9 am (or possibly the question refers to the *start* of the heavy traffic period).

Considering the period 7 am to 8 am has 400 vehicles, making it the second highest count after 8 am - 9 am (550) and 9 am - 10 am (450), here are possible reasons why traffic increases significantly during this hour:

1. Start of the Morning Rush Hour: The 7 am - 8 am hour marks the beginning of the main morning commute period in many places. People who need to reach their workplaces or schools by 8:30 am or 9:00 am typically start leaving their homes around this time.

2. Preparation Time: Most people would have woken up, prepared, and are ready to leave by 7 am, making this the logical starting point for their travel to reach destinations within the city or locality.

3. Shift Changes: For some professions or industries, shift changes might occur around 7 am or 8 am, leading to increased traffic as one shift ends and another begins.

4. School Start Times: Many schools, especially for younger students, might start early, leading parents and buses to be on the roads during the 7 am - 8 am slot.

The graph shows the traffic is heaviest between 8 am and 9 am (550 vehicles), which aligns even better with typical school and office start times (usually 9 am or slightly earlier), with travel time accounting for the peak occurring in the preceding hour.

Therefore, the traffic significantly increases from 7 am and becomes heavy because it coincides with the start of the peak morning commute as people begin their daily journeys to work, school, and other destinations.

Looking at the bar graph, the traffic volume decreases consistently after the hour of 8 am - 9 am (which is the actual peak hour in the graph) until 11 am - 12 noon.

Here's why this trend likely occurs:

1. Completion of Morning Commute: The period from roughly 7 am to 9 am is the typical morning rush hour when most people travel to their workplaces, schools, and other destinations. By 9 am or shortly after, the majority of commuters have reached their destinations.

2. Reduced Travel Activity: Once people have arrived at work or school, there is a significant reduction in the number of people traveling. Mid-morning hours (like 9 am to 12 noon) are generally periods of relatively lower travel activity compared to the peak commute times.

3. Specific Destination Arrivals: Traffic volume drops as the main waves of people arriving for specific start times (like 9 am for offices and schools) subside.

Therefore, the traffic becomes progressively lesser each hour after the morning peak because the primary reason for the heavy morning traffic – the daily commute – has largely been completed, and fewer people are traveling during the mid-morning hours before the lunch period or afternoon activities might increase traffic again.

The bar graph shows the monthly expenditure of Imran's family on various items.

The horizontal axis represents the Items (House Rent, Food, Education, Electricity, Transport, Miscellaneous).

The vertical axis represents the Monthly Expenditure (in Rupees). The scale on the vertical axis increases in increments of 100 (0, 100, 200, 300, ...).

Let's read the monthly expenditure for each item from the height of the bars:

• House Rent: The bar reaches the level of 900.

Expenditure on House Rent = $\textsf{₹} 900$

• Food: The bar reaches the level of 1200.

Expenditure on Food = $\textsf{₹} 1200$

• Education: The bar reaches the level of 700.

Expenditure on Education = $\textsf{₹} 700$

• Electricity: The bar reaches the level of 400.

Expenditure on Electricity = $\textsf{₹} 400$

• Transport: The bar reaches the level of 600.

Expenditure on Transport = $\textsf{₹} 600$

• Miscellaneous: The bar reaches the level of 400.

Expenditure on Miscellaneous = $\textsf{₹} 400$

Now let's identify the most and second most expensive items:

Expenditures are: House Rent ($\textsf{₹} 900$), Food ($\textsf{₹} 1200$), Education ($\textsf{₹} 700$), Electricity ($\textsf{₹} 400$), Transport ($\textsf{₹} 600$), Miscellaneous ($\textsf{₹} 400$).

Arranging these in descending order:

$\textsf{₹} 1200$ (Food)

$\textsf{₹} 900$ (House Rent)

$\textsf{₹} 700$ (Education)

$\textsf{₹} 600$ (Transport)

$\textsf{₹} 400$ (Electricity and Miscellaneous)

The item on which Imran's family spends the most is the one with the highest expenditure, which is $\textsf{₹} 1200$.

Item with most expenditure: Food.

The item on which Imran's family spends the second most is the one with the second highest expenditure, which is $\textsf{₹} 900$.

Item with second most expenditure: House Rent.

From the bar graph, we can find the monthly expenditure on Electricity and Education.

Expenditure on Electricity = $\textsf{₹} 400$

Expenditure on Education = $\textsf{₹} 700$

Now, let's calculate one-half the cost of education:

We need to compare the cost of Electricity ($\textsf{₹} 400$) with one-half the cost of Education ($\textsf{₹} 350$).

Is $\textsf{₹} 400$ about $\textsf{₹} 350$?

The difference is $\textsf{₹} 400 - \textsf{₹} 350 = \textsf{₹} 50$.

While $\textsf{₹} 400$ is somewhat close to $\textsf{₹} 350$, calling it "about one-half" might be debatable depending on the required precision. However, $\textsf{₹} 400$ is closer to $\textsf{₹} 350$ than it is to the full cost of education ($\textsf{₹} 700$) or other significant fractions.

In many contexts, $\textsf{₹} 400$ would be considered "about" $\textsf{₹} 350$.

Yes, the cost of electricity ($\textsf{₹} 400$) is about one-half the cost of education ($\textsf{₹} 700 \div 2 = \textsf{₹} 350$).

From the bar graph, we can find the monthly expenditure on Education and Food.

Expenditure on Education = $\textsf{₹} 700$

Expenditure on Food = $\textsf{₹} 1200$

Now, let's calculate one-fourth the cost of food:

We need to check if the cost of education is less than one-fourth the cost of food.

Is $\textsf{₹} 700 < \textsf{₹} 300$?

Comparing the values, $\textsf{₹} 700$ is greater than $\textsf{₹} 300$.

$700 \not< 300$

No, the cost of education ($\textsf{₹} 700$) is not less than one-fourth the cost of food ($\textsf{₹} 300$). In fact, it is more than double one-fourth the cost of food.

To prepare a bar graph representing the data collected by Samantha, we need to follow these steps:

Data Collected:

• Butterfly: 8
• Dragonfly: 5
• Bee: 12
• Spider: 6
• Beetle: 4

Steps to Prepare a Bar Graph:

1. Draw two perpendicular lines, one horizontal (x-axis) and one vertical (y-axis).

2. Label the horizontal axis. It will represent the 'Type of Insect/Critter'. Mark equal spaces along this axis for each type: Butterfly, Dragonfly, Bee, Spider, Beetle.

3. Label the vertical axis. It will represent the 'Number Seen'.

4. Choose a suitable scale for the vertical axis. The numbers range from 4 to 12. An appropriate scale would be to represent a certain number of insects/critters by a unit length. Let's use a scale where 1 unit length represents 2 insects/critters. Mark the vertical axis with points at 0, 2, 4, 6, 8, 10, 12, 14.

5. Draw rectangular bars for each type of insect/critter. The height of each bar should correspond to the number seen, based on the chosen scale on the vertical axis.

• For Butterfly (8): Draw a bar above 'Butterfly' extending up to the level of 8 on the vertical axis.
• For Dragonfly (5): Draw a bar above 'Dragonfly' extending up to the level of 5 on the vertical axis (halfway between 4 and 6).
• For Bee (12): Draw a bar above 'Bee' extending up to the level of 12 on the vertical axis.
• For Spider (6): Draw a bar above 'Spider' extending up to the level of 6 on the vertical axis.
• For Beetle (4): Draw a bar above 'Beetle' extending up to the level of 4 on the vertical axis.

6. Ensure the bars are of equal width and the spacing between consecutive bars is uniform.

7. Give the bar graph a suitable title, such as "Insects and Critters Seen in a Tea Garden".

Below is a textual description of the resulting bar graph, as a visual image cannot be displayed here directly in the answer format.

Description of the Bar Graph:

Title: Insects and Critters Seen in a Tea Garden

Horizontal Axis (x-axis): Type of Insect/Critter

Labels: Butterfly, Dragonfly, Bee, Spider, Beetle (evenly spaced)

Vertical Axis (y-axis): Number Seen

Scale: 1 unit = 2 insects/critters

Labels: 0, 2, 4, 6, 8, 10, 12, 14

Bars:

• Bar for Butterfly: Height reaches 8.
• Bar for Dragonfly: Height reaches 5.
• Bar for Bee: Height reaches 12.
• Bar for Spider: Height reaches 6.
• Bar for Beetle: Height reaches 4.

This bar graph visually represents the frequency of each type of insect/critter Samantha observed in the tea garden.

We are given the data on the number of tickets sold for different cities and a partially erased bar graph.

Data:

• Vidisha: 300 tickets
• Sagar: 400 tickets
• Jabalpur: 250 tickets
• Seoni: 150 tickets
• Indore: 500 tickets

Let's address each part of the question:

a. Write the number of tickets sold for Vidisha above the bar.

From the data, the number of tickets sold for Vidisha is 300. We should write this number above the bar labeled "Vidisha" in the bar graph.

b. Write the number of tickets sold for Jabalpur above the bar.

From the data, the number of tickets sold for Jabalpur is 250. We should write this number above the bar labeled "Jabalpur" in the bar graph.

c. The bar for Vidisha is 6 unit lengths and the bar for Jabalpur is 5 unit lengths. What is the scale for this graph?

We know the number of tickets for Vidisha is 300 and its bar length is 6 units.

Let's verify with Jabalpur: The number of tickets is 250 and the bar length is 5 units.

The scale is consistent.

The scale for this graph is 1 unit length represents 50 tickets.

e. Add the scale of the bar graph placing the correct numbers on the vertical axis.

Based on the scale (1 unit = 50 tickets), we can mark the vertical axis starting from 0 and increasing by 50 for each unit length.

0, 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, ...

These numbers should be placed along the vertical axis corresponding to the appropriate height (number of unit lengths from the horizontal axis).

d. Draw the correct bar for Sagar.

Number of tickets sold for Sagar = 400.

Using the scale, the length of the bar for Sagar should be:

The correct bar for Sagar should have a height corresponding to 8 unit lengths on the vertical axis (which represents 400 tickets).

f. Are the bars for Seoni and Indore correct in this graph? If not, draw the correct bar(s).

We need to determine the correct bar lengths for Seoni and Indore based on the data and the scale.

Number of tickets for Seoni = 150.

Looking at the incomplete graph image, the bar for Seoni appears to go up to the level of 3 unit lengths (which would correspond to 150 on the vertical axis). So, the bar for Seoni appears correct.

Number of tickets for Indore = 500.

Looking at the incomplete graph image, the bar for Indore appears to go up to the level of 10 unit lengths (which would correspond to 500 on the vertical axis). So, the bar for Indore appears correct.

Therefore, the bars for Seoni and Indore appear to be correct in the given graph.

Chinu listed the following means of transport that passed in front of his house:

Car, Bus, Car, Cycle, Car, Bus, Scooter, Scooter, Car, Bus, Cycle, Car, Scooter, Bus, Car, Cycle, Scooter, Car, Bus, Bus, Car, Cycle, Bus, Scooter, Scooter, Cycle, Bus, Bus, Car, Scooter

a. Prepare a frequency distribution table for the data.

To prepare a frequency distribution table, we need to list each unique means of transport and count how many times it appears in the list. We can use tally marks to assist in counting.

Unique means of transport are: Car, Bus, Cycle, Scooter.

Let's tally the counts:

• Car: Car, Car, Car, Car, Car, Car, Car, Car ($|||| \text{ } ||||$ - 8 times)
• Bus: Bus, Bus, Bus, Bus, Bus, Bus, Bus ($|||| \text{ } ||$ - 7 times)
• Cycle: Cycle, Cycle, Cycle, Cycle ($||||$ - 4 times)
• Scooter: Scooter, Scooter, Scooter, Scooter, Scooter, Scooter ($|||| \text{ } |$ - 6 times)

Now, let's create the frequency distribution table:

b. Which means of transport was used the most?

Looking at the 'Frequency' column in the table (8, 7, 4, 6), the highest frequency is 8.

The means of transport used the most is Car.

c. If you were there to collect this data, how could you do it? Write the steps or process.

If I were there to collect this data, the process would be as follows:

1. Identify the time period for observation (9 am to 10 am, which is 1 hour).

2. Prepare a sheet of paper or a notebook and make columns for each type of transport I expect to see (Car, Bus, Cycle, Scooter, etc.). Or, prepare a list of the specific types given in the question.

3. Position myself in a place where I can clearly see the traffic passing on the road in front of the house.

4. Start the timer or note the start time (9 am).

5. For every vehicle of the listed types that passes, make a tally mark in the corresponding column on the paper.

6. Continue observing and marking for the entire duration of the observation period (until 10 am).

7. Stop counting when the observation period is over (at 10 am).

8. Count the total number of tally marks for each means of transport to get the frequency.

This process allows for systematic collection of data by recording each occurrence as it happens within the specified time frame.

As an AI, I cannot physically roll a die. This task requires conducting an experiment and collecting real data.

To answer the question, you need to perform the experiment yourself: roll a standard six-sided die 30 times and record the outcome of each roll. The possible outcomes for a standard die are the numbers {1, 2, 3, 4, 5, 6}.

After collecting the data, you will prepare a frequency distribution table with tally marks.

Steps to follow:

1. Get a standard die.

2. Prepare a table with columns for 'Outcome (Number on Die)', 'Tally Marks', and 'Frequency (Number of Times)'. Include rows for each possible outcome (1, 2, 3, 4, 5, 6).

3. Roll the die 30 times.

4. After each roll, look at the number that appears on the top face of the die and make a tally mark in the corresponding row of your table.

5. After 30 rolls, count the tally marks for each number and write the total frequency in the last column.

6. Use the completed table to answer questions a, b, and c.

Let's create a hypothetical frequency distribution table based on simulated results to demonstrate how you would answer the questions. Your actual results will vary due to randomness.

Hypothetical Data (Example after 30 rolls):

Total frequency = $6 + 5 + 4 + 8 + 3 + 7 = 33$ (Error in total, should be 30) - Let's adjust counts to sum to 30.

Let's use a valid hypothetical example where the sum is 30:

Total frequency = $5 + 4 + 7 + 6 + 3 + 5 = 30$. This is a valid hypothetical example.

Now, using this hypothetical example data, let's answer the questions:

a. Find the number that appeared the minimum number of times.

Looking at the 'Frequency' column (5, 4, 7, 6, 3, 5), the minimum frequency is 3.

The number that appeared the minimum number of times is 5.

b. Find the number that appeared the maximum number of times.

Looking at the 'Frequency' column (5, 4, 7, 6, 3, 5), the maximum frequency is 7.

The number that appeared the maximum number of times is 3.

c. Find numbers that appeared an equal number of times.

Looking at the 'Frequency' column (5, 4, 7, 6, 3, 5), we see that the frequency 5 appears twice.

The numbers that appeared an equal number of times (with frequency 5) are 1 and 6.

Remember that the answers based on your actual experiment will likely be different from this hypothetical example.

The table shows the number of wickets taken by Jaspreet Bumrah in his last 30 matches, categorized by the number of wickets taken in a single match and the frequency (how many matches) for each category.

a. What information is this table giving?

This table provides information about the frequency distribution of the number of wickets taken by Jaspreet Bumrah per match over his last 30 matches. Specifically, it tells us how many matches he played where he took 0 wickets, how many matches where he took 1 wicket, 2 wickets, and so on, up to 7 wickets.

b. What may be the title of this table?

A suitable title for this table could be:

• Number of Wickets Taken by Jaspreet Bumrah in 30 Matches
• Frequency Distribution of Wickets Taken by Jaspreet Bumrah
• Bumrah's Wicket Tally in Last 30 Matches

c. What caught your attention in this table?

Several things might catch attention, depending on what one is looking for. Based on the table data:

• The fact that Bumrah took 0 wickets in 5 out of 30 matches might be notable.
• The most frequent number of wickets taken per match is 2 (occurred in 9 matches).
• He took 5, 6, or 7 wickets in a match relatively less frequently (1, 2, and 1 matches respectively).
• He took 4 wickets in 4 matches.

What specifically catches attention is subjective.

d. In how many matches has Bumrah taken 4 wickets?

Looking at the row where the 'Number of Wickets' is 4, the 'Number of Matches (Frequency)' column shows the value 4.

Bumrah has taken 4 wickets in 4 matches.

e. Mayank says “If we want to know the total number of wickets he has taken in his last 30 matches, we have to add the numbers 0, 1, 2, 3 …, up to 7.” Can Mayank get the total number of wickets taken in this way? Why?

No, Mayank cannot get the total number of wickets taken by simply adding the numbers 0, 1, 2, ..., up to 7.

Reason: The numbers 0, 1, 2, ..., 7 represent the *different possible counts* of wickets taken in a match, not the *total number of wickets* taken across all matches. The table shows how many *times* (frequency) each of those wicket counts occurred. To find the total number of wickets, we need to consider how many matches had 0 wickets, how many had 1 wicket, etc., and multiply the number of wickets by the number of matches it occurred in for each row, then sum these products.

f. How would you correctly figure out the total number of wickets taken by Bumrah in his last 30 matches, using this table?

To correctly figure out the total number of wickets, we need to multiply the number of wickets taken in each category by the frequency (number of matches) for that category, and then sum up these products.

Total Wickets = (0 wickets/match $\times$ 5 matches) + (1 wicket/match $\times$ 7 matches) + (2 wickets/match $\times$ 9 matches) + (3 wickets/match $\times$ 1 match) + (4 wickets/match $\times$ 4 matches) + (5 wickets/match $\times$ 1 match) + (6 wickets/match $\times$ 2 matches) + (7 wickets/match $\times$ 1 match)

So, the total number of wickets taken by Bumrah in his last 30 matches is 68.

The pictograph shows the number of tractors in five different villages (Village A, B, C, D, E). The key states that 1 tractor symbol represents 1 tractor.

Since each symbol represents exactly one tractor, the number of tractors in each village is equal to the number of symbols shown for that village.

Let's find the number of tractors in each village by counting the symbols:

• Village A: There are 6 symbols.

Number of tractors = 6

• Village B: There are 5 symbols.

Number of tractors = 5

• Village C: There are 8 symbols.

Number of tractors = 8

• Village D: There are 3 symbols.

Number of tractors = 3

• Village E: There are 6 symbols.

Number of tractors = 6

Now we can answer the questions:

a. Which village has the smallest number of tractors?

Comparing the number of tractors in each village (6, 5, 8, 3, 6), the smallest number is 3.

The village with the smallest number of tractors is Village D.

b. Which village has the most tractors?

Comparing the number of tractors in each village (6, 5, 8, 3, 6), the largest number is 8.

The village with the most tractors is Village C.

c. How many more tractors does Village C have than Village B?

Number of tractors in Village C = 8

Number of tractors in Village B = 5

Difference = Number of tractors in Village C - Number of tractors in Village B

Village C has 3 more tractors than Village B.

d. Komal says, “Village D has half the number of tractors as Village E.” Is she right?

Number of tractors in Village D = 3.

Number of tractors in Village E = 6.

Half the number of tractors in Village E = $\frac{1}{2} \times \text{Number of tractors in Village E}$

The number of tractors in Village D (3) is equal to half the number of tractors in Village E (3).

So, Komal is right. Village D has half the number of tractors as Village E.

The pictograph shows the number of girl students in each class (Class 1 to Class 8). The key states that 1 girl symbol represents 4 girl students.

A half symbol represents half the number of students represented by a full symbol. So, a half symbol represents $\frac{1}{2} \times 4 = 2$ girl students.

Let's find the number of girl students in each class by counting the symbols and using the key:

• Class 1: 6 full symbols. Number of girls = $6 \times 4 = 24$.
• Class 2: 4 full symbols and 1 half symbol. Number of girls = $4 \times 4 + 1 \times 2 = 16 + 2 = 18$.
• Class 3: 5 full symbols. Number of girls = $5 \times 4 = 20$.
• Class 4: 3 full symbols and 1 half symbol. Number of girls = $3 \times 4 + 1 \times 2 = 12 + 2 = 14$.
• Class 5: 2 full symbols and 1 half symbol. Number of girls = $2 \times 4 + 1 \times 2 = 8 + 2 = 10$.
• Class 6: 4 full symbols. Number of girls = $4 \times 4 = 16$.
• Class 7: 3 full symbols. Number of girls = $3 \times 4 = 12$.
• Class 8: 1 full symbol and 1 half symbol. Number of girls = $1 \times 4 + 1 \times 2 = 4 + 2 = 6$.

Now we can answer the questions:

a. Which class has the least number of girl students?

Comparing the number of girls in each class (24, 18, 20, 14, 10, 16, 12, 6), the smallest number is 6.

The class with the least number of girl students is Class 8.

b. What is the difference between the number of girls in Class 5 and 6?

Number of girls in Class 5 = 10.

Number of girls in Class 6 = 16.

Difference = Number of girls in Class 6 - Number of girls in Class 5

The difference between the number of girls in Class 5 and 6 is 6.

c. If 2 more girls were admitted in Class 2, how would the graph change?

Currently, Class 2 has 18 girl students, represented by 4 full symbols and 1 half symbol ($4 \times 4 + 2 = 18$).

If 2 more girls were admitted, the new number of girls in Class 2 would be $18 + 2 = 20$.

To represent 20 girls using the key (1 symbol = 4 girls, half symbol = 2 girls):

The graph for Class 2 would change from showing 4 full symbols and 1 half symbol to showing 5 full symbols.

d. How many girls are there in Class 7?

Looking at the row for Class 7 in the pictograph, there are 3 full symbols.

Using the key (1 symbol = 4 girls), the number of girls in Class 7 is $3 \times 4 = 12$.

There are 12 girls in Class 7.

The number of Mudhol dogs in six villages are:

Village A : 18

Village B : 36

Village C : 12

Village D : 48

Village E : 18

Village F : 24

a. What will be a useful scale or key to draw this pictograph?

We need to choose a symbol to represent a certain number of dogs. Look at the given numbers (18, 36, 12, 48, 18, 24). All these numbers are multiples of 6. They are also multiples of 3, 4, 2.

A useful scale would be a number that divides all or most of the values evenly. Representing 6 dogs with one symbol seems appropriate as all numbers are multiples of 6, which will result in whole numbers of symbols.

Let's use a dog symbol 🐶.

A useful scale/key would be: 1 🐶 represents 6 dogs.

b. How many symbols will you use to represent the dogs in Village B?

Number of dogs in Village B = 36.

Using the key (1 🐶 = 6 dogs), the number of symbols needed is:

You will use 6 symbols to represent the dogs in Village B.

Pictograph Construction (based on the chosen key):

Key: 🐶 = 6 dogs

• Village A (18): $18 \div 6 = 3$ symbols 🐶 🐶 🐶
• Village B (36): $36 \div 6 = 6$ symbols 🐶 🐶 🐶 🐶 🐶 🐶
• Village C (12): $12 \div 6 = 2$ symbols 🐶 🐶
• Village D (48): $48 \div 6 = 8$ symbols 🐶 🐶 🐶 🐶 🐶 🐶 🐶 🐶
• Village E (18): $18 \div 6 = 3$ symbols 🐶 🐶 🐶
• Village F (24): $24 \div 6 = 4$ symbols 🐶 🐶 🐶 🐶

Pictograph Table:

c. Kamini said that the number of dogs in Village B and Village D together will be more than the number of dogs in the other 4 villages. Is she right? Give reasons for your response.

Number of dogs in Village B = 36.

Number of dogs in Village D = 48.

Number of dogs in Village B and Village D together = $36 + 48 = 84$.

The other 4 villages are Village A, C, E, and F.

Number of dogs in Village A = 18.

Number of dogs in Village C = 12.

Number of dogs in Village E = 18.

Number of dogs in Village F = 24.

Total number of dogs in the other 4 villages = $18 + 12 + 18 + 24 = 72$.

Now, let's compare the two sums:

Number of dogs in B and D together = 84.

Number of dogs in the other 4 villages = 72.

Is $84 > 72$?

Yes, $84$ is greater than $72$.

So, Kamini is right.

Reason: The total number of dogs in Village B (36) and Village D (48) is $36 + 48 = 84$. The total number of dogs in the other four villages (A, C, E, F) is $18 + 12 + 18 + 24 = 72$. Since $84 > 72$, the number of dogs in Village B and Village D together is more than the number of dogs in the other 4 villages.

The data collected from the survey of 120 students about their preferred activity in free time is:

• Playing: 45 students
• Reading Story Books: 30 students
• Watching TV: 20 students
• Listening to Music: 10 students
• Painting: 15 students

Total students: $45 + 30 + 20 + 10 + 15 = 120$. The total matches the survey size.

Prepare a Bar Graph:

We are asked to draw a bar graph with a scale of 1 unit length = 5 students.

We need to calculate the height (in unit lengths) for each bar by dividing the number of students by the scale value (5):

• Playing: $45 \div 5 = 9$ unit lengths
• Reading Story Books: $30 \div 5 = 6$ unit lengths
• Watching TV: $20 \div 5 = 4$ unit lengths
• Listening to Music: $10 \div 5 = 2$ unit lengths
• Painting: $15 \div 5 = 3$ unit lengths

Steps to Draw the Bar Graph:

1. Draw horizontal and vertical axes. Label the horizontal axis as 'Preferred Activity' and the vertical axis as 'Number of Students'.

2. Mark equal intervals on the horizontal axis for each activity: Playing, Reading Story Books, Watching TV, Listening to Music, Painting.

3. On the vertical axis, mark the scale. Starting from 0, mark points representing 5, 10, 15, 20, 25, 30, ..., up to a value slightly greater than the highest frequency (45), or simply mark up to 9 unit lengths (representing 45 students) if drawing by hand with unit lengths.

4. Draw bars of equal width above each activity label. The height of each bar should correspond to the calculated number of unit lengths or the actual number of students based on the vertical scale.

• Playing: Bar height is 9 units (or reaches 45 on the vertical axis).
• Reading Story Books: Bar height is 6 units (or reaches 30 on the vertical axis).
• Watching TV: Bar height is 4 units (or reaches 20 on the vertical axis).
• Listening to Music: Bar height is 2 units (or reaches 10 on the vertical axis).
• Painting: Bar height is 3 units (or reaches 15 on the vertical axis).

5. Give the graph a title, e.g., "Preferred Activities of 120 School Students in Free Time".

Below is a textual description of the bar graph:

Title: Preferred Activities of 120 School Students in Free Time

Horizontal Axis (x-axis): Preferred Activity

Labels: Playing, Reading Story Books, Watching TV, Listening to Music, Painting

Vertical Axis (y-axis): Number of Students

Scale: 1 unit length = 5 students (Vertical axis marks: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45)

Bars:

• Playing: Bar reaches 45.
• Reading Story Books: Bar reaches 30.
• Watching TV: Bar reaches 20.
• Listening to Music: Bar reaches 10.
• Painting: Bar reaches 15.

Which activity is preferred by most students other than playing?

We need to find the activity with the highest number of students among all activities *except* Playing.

The number of students for each activity are: Playing (45), Reading Story Books (30), Watching TV (20), Listening to Music (10), Painting (15).

Excluding 'Playing' (45), we look at the remaining counts: 30, 20, 10, 15.

The maximum among these is 30.

The activity preferred by most students other than playing is Reading Story Books (with 30 students).

The table shows the number of saplings planted each day of the first week of July.

a. The total number of saplings planted on Wednesday and Thursday is __________.

Number of saplings planted on Wednesday = 38

Number of saplings planted on Thursday = 40

Total saplings on Wednesday and Thursday = Saplings on Wednesday + Saplings on Thursday

The total number of saplings planted on Wednesday and Thursday is 78.

b. The total number of saplings planted during the whole week is __________.

To find the total number of saplings planted during the whole week, we add the number of saplings planted each day:

Total saplings = Mon + Tue + Wed + Thu + Fri + Sat + Sun

Total saplings = $20 + 35 + 38 + 40 + 48 + 30 + 25$

Let's add them:

The total number of saplings planted during the whole week is 236.

c. The greatest number of saplings were planted on ___________, and the least number of saplings were planted on ___________. Why do you think that is the case? Why were more saplings planted on certain days of the week and less on others? Can you think of possible explanations or reasons? How could you try and figure out whether your explanations are correct?

Comparing the number of saplings planted each day (20, 35, 38, 40, 48, 30, 25):

The greatest number is 48, which occurred on Friday.

The least number is 20, which occurred on Monday.

The greatest number of saplings were planted on Friday, and the least number of saplings were planted on Monday.

Possible explanations for why more saplings were planted on certain days and less on others:

• Planning and Preparation: Planting might have started slow on Monday (lowest number) as the team organized and prepared the saplings, tools, and planting sites. By Friday, they might have been in full swing, leading to the highest number planted.
• Availability of Participants: More students and teachers might have been available or dedicated time for planting on certain days (like Friday) compared to others (like Monday).
• Logistics: Delivery of saplings, availability of water for irrigation, or other logistical factors might have influenced the number of saplings planted each day.
• Weather Conditions: Favorable weather conditions on some days might have allowed for more planting compared to less favorable conditions on others.
• Specific Events or Targets: There might have been a specific target for planting by the end of the week, leading to increased efforts on later weekdays. Or, a special event related to planting was scheduled on Friday.

How to figure out if your explanations are correct:

1. Ask the Organizers/Participants: The most direct way is to talk to the students and teachers who participated in the planting activity. They can provide insights into the daily schedule, attendance, challenges faced, and any specific plans that influenced the number of saplings planted each day.

2. Check Records: Look for any records related to the tree planting week, such as attendance registers for volunteers, notes on daily activities, weather reports for that week, or details about sapling deliveries.

3. Observe Future Activities: If similar planting drives are held in the future, observing the pattern and reasons given then might provide clues.

By gathering more information from those involved or related records, you can verify if your hypothetical explanations align with the actual events and reasons behind the daily variations in the number of saplings planted.

The frequency table provides the following data on the number of tigers in India:

The bar graph has the 'Year' on the horizontal axis and the 'Number of Tigers' on the vertical axis. The vertical scale increments are shown at 0, 500, 1000, 1500, 2000, 2500, 3000, 3500.

Let's compare the number of tigers from the table with the height of the bars in the graph based on the vertical scale:

• Year 2006: Data is 1411. The bar should reach a level just below 1500. The graph shows the bar correctly at this approximate level.
• Year 2010: Data is 1706. The bar should reach a level noticeably above 1500, approximately one-third of the way between 1500 and 2000. The graph shows the bar reaching a level that looks closer to 1800 or 1900, which is incorrect.
• Year 2014: Data is 2226. The bar should reach a level above 2000, approximately two-fifths of the way between 2000 and 2500. The graph shows the bar reaching a level that looks only slightly above 2000, which is incorrect.
• Year 2018: Data is 2967. The bar should reach a level just below 3000. The graph shows the bar correctly at this approximate level.
• Year 2022: Data is 3167. The bar should reach a level above 3000, approximately one-third of the way between 3000 and 3500. The graph shows the bar reaching exactly the 3000 mark, which is incorrect.

Mistakes in the Bar Graph:

The mistakes are in the heights of the bars for the years 2010, 2014, and 2022. Their heights do not accurately represent the number of tigers for those years according to the given scale.

How to Fix the Mistakes:

To fix the graph, the bars need to be redrawn to the correct heights corresponding to the number of tigers on the vertical scale:

• The bar for 2010 should be lowered slightly to accurately represent 1706 (approximately one-third of the way from 1500 to 2000).
• The bar for 2014 should be raised significantly to accurately represent 2226 (approximately two-fifths of the way from 2000 to 2500).
• The bar for 2022 should be raised to accurately represent 3167 (approximately one-third of the way from 3000 to 3500).

If I wanted to visually represent the data of the heights of the tallest persons in each class, I would typically use a graph with vertical bars (a vertical bar graph).

Reason:

Heights are a measure of vertical distance. Representing heights using vertical bars provides a more intuitive and direct visual comparison. The height of the bar naturally corresponds to the actual height being measured. This makes it easier for the viewer to immediately grasp that a taller bar signifies a greater height.

While a horizontal bar graph could also represent this data, the correspondence between the horizontal length of the bar and vertical height being measured is less natural or immediately understandable.

Therefore, a vertical bar graph is generally preferred for representing data related to vertical measurements like height or length, as it provides a more visually congruent representation.

If I were making a table of the longest rivers on each continent and their lengths, I would prefer to use a bar graph with horizontal bars.

Reason:

River names can often be long (e.g., Nile, Amazon, Yangtze, Mississippi-Missouri-Jefferson). Using a horizontal bar graph allows for the labels (the names of the rivers or continents) to be placed clearly along the vertical axis without overlapping or requiring excessive rotation, which can happen with long labels on the horizontal axis of a vertical bar graph. The length of the horizontal bar then represents the length of the river.

Data Collection:

Finding the absolute "longest river on each continent" can be tricky due to different definitions of river length (source to mouth, inclusion of tributaries, etc.). However, here is generally accepted information for the longest river system on each continent:

• Africa: Nile River (approx. 6,650 km)
• Asia: Yangtze River (approx. 6,300 km)
• North America: Mississippi-Missouri-Jefferson River System (approx. 6,275 km)
• South America: Amazon River System (approx. 6,992 km, although exact length is debated)
• Europe: Volga River (approx. 3,530 km)
• Australia: Murray-Darling River System (approx. 3,672 km)

Note: Antarctica has no major rivers due to ice cover.

Table of Longest Rivers by Continent:

Bar Graph (Horizontal Bars):

To draw a horizontal bar graph, the vertical axis lists the Continents (or River Names), and the horizontal axis represents the Length in kilometers. We need a scale for the horizontal axis.

The lengths range from 3530 to 6992 km. A suitable scale could be 1 unit length = 500 km or 1000 km.

Let's use a scale of 1 unit length = 500 km.

Bar lengths (approximately):

• Africa (Nile): $6650 \div 500 = 13.3$ units
• Asia (Yangtze): $6300 \div 500 = 12.6$ units
• North America (Miss.-Mo.-Jeff.): $6275 \div 500 = 12.55$ units
• South America (Amazon): $6992 \div 500 = 13.984$ units
• Europe (Volga): $3530 \div 500 = 7.06$ units
• Australia (Murray-Darling): $3672 \div 500 = 7.344$ units

Description of the Horizontal Bar Graph:

Title: Longest River System on Each Continent

Vertical Axis (y-axis): Continent

Labels: Africa, Asia, North America, South America, Europe, Australia (evenly spaced)

Horizontal Axis (x-axis): Approximate Length (km)

Scale: 1 unit length = 500 km (Horizontal axis marks: 0, 500, 1000, ..., up to 7000 or 7500)

Bars: (Horizontal bars starting from the vertical axis)

• Bar for Africa: Length reaches approximately 6650 on the horizontal scale.
• Bar for Asia: Length reaches approximately 6300 on the horizontal scale.
• Bar for North America: Length reaches approximately 6275 on the horizontal scale.
• Bar for South America: Length reaches approximately 6992 on the horizontal scale.
• Bar for Europe: Length reaches approximately 3530 on the horizontal scale.
• Bar for Australia: Length reaches approximately 3672 on the horizontal scale.

Which continents have the longest rivers?

Looking at the table or the conceptual bar lengths, the continents with the longest rivers (lengths approximately 6000 km or more) are:

• South America (Amazon - 6992 km)
• Africa (Nile - 6650 km)
• Asia (Yangtze - 6300 km)
• North America (Mississippi-Missouri-Jefferson - 6275 km)

These four continents have significantly longer longest rivers compared to Europe and Australia.