---
title: "NCERT Solutions for Class 7 Maths Chapter 13 Connecting the Dots"
url: https://www.swavid.com/maths/class/7/chapter/connecting-the-dots/ncert-solutions
dateModified: 2026-10-07T15:17:45+00:00
---

# NCERT Solutions for Class 7 Maths Chapter 13 Connecting the Dots

This chapter's questions cover statistical thinking, representative values like mean and median, data visualization methods such as dot plots and clustered bar graphs, and interpreting real-world data sets.

Free PDF (36 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-13-connecting-the-dots-f3712ff86e.pdf

## Of Questions and Statements

### Question 1

*2 marks · Very short answer*

Your teacher tells you that they are meeting two of their childhood friends this evening. One is 5 feet tall and the other is 6 feet tall. What is your guess as to each friend's gender based on this information?

**Solution**

1. Based on everyday experience and statistical thinking, we can guess that the 5-foot-tall person is a woman and the 6-foot-tall person is a man.
2. This is because 5-foot-tall men and 6-foot-tall women are rare, as men are more often taller than women.

**Answer:** The 5-foot-tall person is likely a woman and the 6-foot-tall person is likely a man.

> Common mistake: Assuming the guess is a certain fact rather than a statistical likelihood.

## Which of the following are statistical questions?

### Question 1

*3 marks · Short answer*

Which of the following are statistical questions?

- What is the price of a tennis ball in India?
- How old are the dogs that live on this street?
- What fraction of the students in your class like walking up a hill?
- Do you like reading?
- Approximately how many bricks are in this wall?
- Who was the best bowler in the match yesterday?
- What was the rainfall pattern in Barmer last year?

**Solution**

1. A statistical question is a question that can be answered by collecting data and anticipating variability in that data.
2. Questions (b) 'How old are the dogs that live on this street?', (c) 'What fraction of the students in your class like walking up a hill?', (e) 'Approximately how many bricks are in this wall?', and (g) 'What was the rainfall pattern in Barmer last year?' are statistical questions because answering them requires collecting, analysing, and summarising data where values vary.
3. Questions (a), (d), and (f) have specific single answers or personal opinions not requiring statistical data collection across a group.

**Answer:** (b), (c), (e), and (g)

> Common mistake: Confusing questions that have a single fixed fact or personal preference with questions that require collecting data from a group.

## Representative Values

### Question 1

*3 marks · Short answer*

The runs scored by Shubman and Yashasvi in a cricket series are given in the table below. Who do you think performed better?

**Solution**

1. Shubman's runs across 4 matches: $0, 17, 21, 90$. Total runs = $0 + 17 + 21 + 90 = 128$. Shubman's average = $128 \div 4 = 32$ runs.
2. Yashasvi's runs across 4 matches: $67, 55, 18, 35$. Total runs = $67 + 55 + 18 + 35 = 175$. Yashasvi's average = $175 \div 4 = 43.75$ runs.
3. Comparing their averages, Yashasvi has a higher average score than Shubman, so Yashasvi performed better.

**Answer:** Yashasvi performed better with an average of 43.75 runs compared to Shubman's average of 32 runs.

> Common mistake: Comparing only the total runs or only the maximum score without finding the average.

### Question 2

*2 marks · Very short answer*

What do you think of Vaishnavi's statement?

**Solution**

1. Vaishnavi's statement considers only the total runs scored, but does not account for the different number of matches played by each player.
2. Comparing groups with different sizes using only totals can be misleading, so finding the average per match gives a fairer comparison.

**Answer:** Vaishnavi's statement is incomplete because comparing totals is not appropriate when the number of matches played is different.

> Common mistake: Stating that Vaishnavi is simply wrong without explaining why totals are insufficient for different group sizes.

### Question 3

*2 marks · Very short answer*

Can a single number act as a representative of a group of numbers? For example, can we represent Shubman's or Yashasvi's batting in this series with one number? Discuss.

**Solution**

1. Yes, a single number such as the arithmetic mean or average can act as a representative of a group of numbers by balancing out the highs and lows.
2. For example, we can represent each player's batting performance in a series by calculating their average runs scored per match.

**Answer:** Yes, a single number like the average can represent a group of numbers by balancing out the highs and lows.

> Common mistake: Believing a single number cannot summarize a group of data.

### Question 4

*3 marks · Short answer*

Shreyas and 4 of his friends have collected the following numbers of guavas: 3, 8, 10, 5, and 4. Parag and 5 of his friends have collected the following numbers of guavas: 5, 4, 6, 3, 4, and 8. Each group will share their guavas equally amongst themselves. In which group will each member get a bigger share of guavas?

**Solution**

1. Shreyas's group total = $3 + 8 + 10 + 5 + 4 = 30$ guavas, and the number of members is $5$. Each member's share = $30 \div 5 = 6$ guavas.
2. Parag's group total = $5 + 4 + 6 + 3 + 4 + 8 = 30$ guavas, and the number of members is $6$. Each member's share = $30 \div 6 = 5$ guavas.
3. Comparing the shares, members of Shreyas's group get a bigger share of guavas.

**Answer:** Each member of Shreyas's group gets a bigger share of 6 guavas.

> Common mistake: Dividing the total by the wrong number of people in the group.

### Question 5

*3 marks · Short answer*

Vaishnavi tracks the number of Hibiscus flowers blooming in her garden each day. The data for the last few days' is 2, 7, 9, 4, 3. What is the average number of Hibiscus flowers blooming per day in Vaishnavi's garden?

**Solution**

1. Total number of Hibiscus flowers bloomed = $2 + 7 + 9 + 4 + 3 = 25$.
2. Number of days = $5$.
3. Average number of flowers per day = $25 \div 5 = 5$.

**Answer:** The average number of Hibiscus flowers blooming per day is 5.

> Common mistake: Dividing by the sum instead of the number of days.

## Figure it Out

### Question 1

*3 marks · Short answer*

Shreyas is playing with a bat and a ball — but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.

**Solution**

1. Given data for 8 attempts: $6, 2, 9, 5, 4, 6, 3, 5$
2. Total number of attempts = $8$
3. Sum of all the values in the data = $6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40$
4. $\text{Mean} = \frac{\text{Sum of all the values in the data}}{\text{Number of values in the data}}$
5. $\text{Mean} = \frac{40}{8} = 5$

**Answer:** 5 bounces

> Common mistake: Dividing by the wrong number of attempts or making a calculation error while summing the data.

### Question 2

*Activity*

Try the activity above on your own. Collect data for 7 or more attempts and find the average.

**Solution**

1. Perform the ball-bouncing activity for 7 or more attempts.
2. Record the number of bounces for each attempt.
3. Find the sum of all recorded values and divide by the number of attempts to calculate the average.

**Answer:** Activity-based practical question.

### Question 3

*Activity*

Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?

**Solution**

1. Choose a flowering plant in the neighbourhood and observe it daily over a week (7 days).
2. Track and note down the number of flowers that bloom on each of the 7 days.
3. Calculate the average by dividing the total number of flowers bloomed by 7.

**Answer:** Activity-based practical question.

### Question 4

*3 marks · Short answer*

Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil: 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?

**Solution**

1. Nikhil's running times: $17, 18, 17, 16, 19, 17, 18$
2. Total time for Nikhil = $17 + 18 + 17 + 16 + 19 + 17 + 18 = 122$ seconds, and number of days = $7$
3. Nikhil's average time = $122 \div 7 \approx 17.43$ seconds
4. Sunil's running times: $20, 18, 18, 17, 16, 16, 17$
5. Total time for Sunil = $20 + 18 + 18 + 17 + 16 + 16 + 17 = 122$ seconds, and number of days = $7$
6. Sunil's average time = $122 \div 7 \approx 17.43$ seconds
7. Both friends ran equally on average.

**Answer:** Both Nikhil and Sunil ran equally on average with a mean time of about $17.43$ seconds.

> Common mistake: Dividing by an incorrect number of days or making addition errors.

### Question 5

*3 marks · Short answer*

The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.

**Solution**

1. Given enrolment figures for six consecutive years: $1555, 1670, 1750, 2013, 2040, 2126$
2. Sum of all values = $1555 + 1670 + 1750 + 2013 + 2040 + 2126 = 11154$
3. Number of years = $6$
4. Mean enrolment = $\text{Sum of values} \div \text{Number of values} = 11154 \div 6 = 1859$

**Answer:** 1859

> Common mistake: Adding the enrolment numbers incorrectly or dividing by the wrong number of years.

## Know Your Onions!

### Question 1

*3 marks · Short answer*

The table shows the monthly price of onions, in rupees per kilogram (kg), at two towns. Where are onions costlier, according to you?

**Solution**

1. Find the total price of onions over the year for each town by adding the monthly prices given in the table.
2. The total price for Yahapur is $25 + 24 + 26 + 28 + 30 + 35 + 39 + 43 + 49 + 56 + 59 + 44 = 458$ rupees per kg.
3. The total price for Wahapur is $19 + 17 + 23 + 30 + 38 + 35 + 42 + 39 + 53 + 60 + 52 + 42 = 450$ rupees per kg.
4. Since the total price in Yahapur is higher than in Wahapur, or by looking at individual month comparisons, onions are generally costlier in Yahapur.

**Answer:** Onions are costlier in Yahapur.

> Common mistake: Comparing only the maximum price instead of considering all months or totals.

### Question 2

*2 marks · Very short answer*

Can you think of any other ways to compare the data?

**Solution**

1. Data can be compared by referring to its minimum value, maximum value, average value, sum total, or the difference between maximum and minimum values.

**Answer:** By comparing minimum values, maximum values, averages, total sums, or ranges.

> Common mistake: Naming only one method instead of various ways discussed in the chapter.

### Question 3

*2 marks · Very short answer*

Does this visualisation capture all the data presented in the tables earlier?

**Solution**

1. The dot plot shows the distribution and frequency of prices, but it does not show the chronological month-wise sequence of the data.

**Answer:** No, it does not capture the month-wise sequence of the data.

> Common mistake: Assuming dot plots retain all temporal or chronological information.

### Question 4

*2 marks · Very short answer*

Looking at it, can we tell the price of onions in Yahapur in the month of January?

**Solution**

1. A dot plot displays frequencies of data values on a number line, so individual month-wise values like January cannot be identified directly.

**Answer:** No, we cannot tell the price for a specific month like January from the dot plot alone.

> Common mistake: Thinking that individual data points can be tracked back to specific months from a dot plot.

### Question 5

*3 marks · Short answer*

Find the average price of onions at Yahapur and Wahapur.

**Solution**

1. The monthly prices of onions in Yahapur are 25, 24, 26, 28, 30, 35, 39, 43, 49, 56, 59, and 44.
2. The sum of the prices in Yahapur is $25 + 24 + 26 + 28 + 30 + 35 + 39 + 43 + 49 + 56 + 59 + 44 = 458$.
3. The average price of onions in Yahapur is $458 \div 12 = 38.17$ rupees per kg.
4. The monthly prices of onions in Wahapur are 19, 17, 23, 30, 38, 35, 42, 39, 53, 60, 52, and 42.
5. The sum of the prices in Wahapur is $19 + 17 + 23 + 30 + 38 + 35 + 42 + 39 + 53 + 60 + 52 + 42 = 450$.
6. The average price of onions in Wahapur is $450 \div 12 = 37.50$ rupees per kg.

**Answer:** The average price of onions is ₹38.17 per kg in Yahapur and ₹37.50 per kg in Wahapur.

> Common mistake: Dividing by a wrong number of months or making addition errors while finding the total sum of prices.

### Question 6

*2 marks · Very short answer*

What else do you wonder about?

**Solution**

1. We may wonder about the factors determining onion prices, how seasonal changes affect them, or how prices vary across different shops in the same area.

**Answer:** We wonder about factors determining prices, seasonal effects, and price variations across shops.

> Common mistake: Giving a closed numerical answer instead of open-ended curiosities.

## Outliers and Medians

### Question 1

*3 marks · Short answer*

The heights of the family members of Yaangba and Poovizhi are as follows: Yaangba's family: 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm. Poovizhi's family: 170 cm, 173 cm, 165 cm, 118 cm, 175 cm. Find the average height of each family. Can we say that Yaangba's family is taller than Poovizhi's family?

**Solution**

1. Yaangba's family heights: $169, 173, 155, 165, 160, 164$.
2. Sum of Yaangba's family heights = $169 + 173 + 155 + 165 + 160 + 164 = 986$ cm, and number of members = $6$.
3. Yaangba's family average height = $986 \div 6 = 164.3$ cm.
4. Poovizhi's family heights: $170, 173, 165, 118, 175$.
5. Sum of Poovizhi's family heights = $170 + 173 + 165 + 118 + 175 = 801$ cm, and number of members = $5$.
6. Poovizhi's family average height = $801 \div 5 = 160.2$ cm.
7. Yes, based on the average heights ($164.3$ cm > $160.2$ cm), Yaangba's family is taller than Poovizhi's family.

**Answer:** Yaangba's family average height is $164.3$ cm and Poovizhi's family average height is $160.2$ cm. Yes, Yaangba's family is taller on average.

> Common mistake: Dividing the sum by the wrong number of family members.

### Question 2

*2 marks · Very short answer*

Can you think of any other number that can represent the data better?

**Solution**

1. Sorting the data and picking the number in the middle gives the median, which is less affected by extreme values.

**Answer:** The median can represent the data better when there are outliers.

> Common mistake: Confusing median with mean.

### Question 3

*2 marks · Very short answer*

In this case, does the median represent the heights of the families better than the average?

**Solution**

1. Yes, the median represents the heights better because the mean is lowered significantly by the young child's height.

**Answer:** Yes, the median represents the heights better.

> Common mistake: Ignoring the effect of the outlier on the mean.

### Question 4

*3 marks · Short answer*

Find the mean and median in Poovizhi's data without the outlier value 118. What change do you notice?

**Solution**

1. Given Poovizhi's family heights with the outlier 118 removed: 165, 170, 173, 175.
2. The mean is calculated as $(165 + 170 + 173 + 175) \div 4 = 683 \div 4 = 170.75 \text{ cm}$.
3. The sorted data has 4 values, so the median is the average of the two middle numbers: $(170 + 173) \div 2 = 343 \div 2 = 171.5 \text{ cm}$.
4. Without the outlier, the mean increases significantly from 160.2 cm to 170.75 cm, and the median increases slightly from 170 cm to 171.5 cm.

**Answer:** Mean = 170.75 cm, Median = 171.5 cm; both values increase when the lower outlier is removed.

> Common mistake: Forgetting to take the average of the two middle numbers when finding the median for an even number of observations.

### Question 5

*3 marks · Short answer*

After the summer vacation, a class teacher asked his class how many short stories they had read. Each student answered the number of stories read on a piece of paper, as shown below. Find the mean and median number of short stories read. Before calculating them, can you guess whether the mean will be less than or greater than the median?

**Solution**

1. Data values from the paper chits: $6, 3, 0, 8, 2, 5, 7, 12, 15, 10, 5, 4, 40, 0, 8, 1$.
2. Total number of students = $16$. Sum of values = $86$. Mean = $86 \div 16 = 5.375$ stories.
3. Sorted data: $0, 0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 10, 12, 15, 40$.
4. Median = average of 8th and 9th values = $(5 + 6) \div 2 = 5.5$ stories.
5. The value $40$ is a high outlier, so we can guess the mean will be greater than the median.

**Answer:** Mean is $5.375$ stories and median is $5.5$ stories.

> Common mistake: Not sorting the data correctly before finding the median.

### Question 6

*2 marks · Very short answer*

Which of the values would you consider an outlier?

**Solution**

1. Values that significantly deviate from the rest of the values in the data, such as $118$ cm in family heights or $40$ in short stories, are considered outliers.

**Answer:** Values that significantly deviate from the rest of the data.

> Common mistake: Choosing normal variation values as outliers.

### Question 7

*3 marks · Short answer*

Find the mean and median in the absence of the outlier. What change do you notice?

**Solution**

1. Poovizhi's family heights without the outlier 118 cm are 165 cm, 170 cm, 173 cm, and 175 cm.
2. The mean of these 4 heights is $(165 + 170 + 173 + 175) \div 4 = 683 \div 4 = 170.75 \text{ cm}$.
3. The sorted heights are 165, 170, 173, 175, so the median is the average of the two middle values $(170 + 173) \div 2 = 171.5 \text{ cm}$.
4. Without the outlier, both the mean and the median increase significantly and become very close to each other.

**Answer:** Mean = 170.75 cm, Median = 171.5 cm. Both values increase and become closer to each other.

> Common mistake: Forgetting to divide by the new number of family members (4 instead of 5) when calculating the mean.

## Are We on the Same Page?

### Question 1

*3 marks · Short answer*

Do you read newspapers? Have you noticed how many pages a newspaper has on different days of the week — is it the same or different? The list below shows the number of pages for a particular newspaper from Monday to Sunday: 16, 18, 20, 22, 26, 16, 10. Mark the data, the mean, and the median on the dot plot below.

**Solution**

1. List the data representing the number of pages: $16, 18, 20, 22, 26, 16, 10$.
2. Calculate the mean by dividing the sum of all values by the number of values: $\text{Mean} = \frac{16 + 18 + 20 + 22 + 26 + 16 + 10}{7} = \frac{128}{7} \approx 18.29$.
3. Sort the data to find the median: $10, 16, 16, 18, 20, 22, 26$; the middle value is $18$, so the median is $18$.

**Answer:** Mean = 18.29 pages, Median = 18 pages

> Common mistake: Forgetting to sort the data before finding the median.

### Question 2

*3 marks · Short answer*

In the three examples we considered — the heights, short-stories, and newspaper pages — observe the variability in data when: (a) the mean and median are close to each other (b) the mean and median are comparatively far apart, with mean < median (c) the mean and median are comparatively far apart, with mean > median

**Part (a)**

1. When the data is more balanced or uniformly spread out, the mean and the median appear to be close to each other.

Answer (a): Data is more balanced or uniformly spread out.

**Part (b)**

1. When there is an outlier on the lower end, it pulls the mean down, making the mean less than the median.

Answer (b): An outlier is present on the lower end.

**Part (c)**

1. When there is an outlier on the higher end, it pulls the mean up, making the mean greater than the median.

Answer (c): An outlier is present on the higher end.

**Answer:** The relationship between mean and median reflects the distribution and presence of outliers in the data.

> Common mistake: Confusing how lower and higher outliers affect the shift of the mean relative to the median.

### Question 3

*3 marks · Short answer*

Discuss the effect on the mean and median when outliers are present on both sides. You may take some example data to examine and explain this.

**Solution**

1. Outliers on both sides (very low and very high values) tend to pull the mean in opposite directions depending on their magnitudes.
2. If the low and high outliers are balanced in magnitude, the mean may remain close to the median.
3. The median remains relatively stable as it depends only on the middle values of the sorted data rather than extreme values.

**Answer:** Balanced outliers on both sides keep the mean close to the median, while the median remains unaffected by extreme values.

> Common mistake: Assuming the median changes significantly when outliers are present on both sides.

## How Tall is Your Class?

### Question 1

*2 marks · Very short answer*

Suppose you are asked the question, "How tall is your class?" What would you say?

**Solution**

1. To describe the height of the class, we can use representative values such as the average height or the median height.
2. From the given data of the Grade 5 class, the whole class average height is $144.4\text{ cm}$ and the median height is $145\text{ cm}$.

**Answer:** The class height can be represented by the average height of $144.4\text{ cm}$ or the median height of $145\text{ cm}$.

> Common mistake: Giving only a single student's height instead of a representative measure for the whole class.

### Question 2

*3 marks · Short answer*

What can we infer from the dot plots and the central tendency measures?

**Solution**

1. The boys' heights are more spread out between $128\text{ cm}$ and $158\text{ cm}$, while the girls' heights lie between $136\text{ cm}$ and $156\text{ cm}$.
2. The girls' average height ($146.9\text{ cm}$) is greater than the boys' average height ($142.94\text{ cm}$), indicating that girls are taller on average in this class.
3. For both boys and girls, $\text{mean} < \text{median}$, showing a small influence of values on the lower side.

**Answer:** Boys' heights are more spread out, girls have a higher average height than boys, and both distributions show $\text{mean} < \text{median}$.

> Common mistake: Assuming that because girls are taller on average, every single girl is taller than every boy.

### Question 3

*3 marks · Short answer*

How many students are taller than the class' average height?

**Solution**

1. Given: The class average height is $144.4\text{ cm}$.
2. List all student heights from the table: Boys: $147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 142, 140, 141, 144, 145, 150$; Girls: $143, 136, 150, 144, 154, 140, 145, 148, 156, 150, 150$.
3. Count the number of students with heights greater than $144.4\text{ cm}$ from the combined data of boys and girls.
4. Result: There are $15$ students taller than the class average height.

**Answer:** 15 students

> Common mistake: Including students who have a height exactly equal to the average or miscounting the values.

### Question 4

*3 marks · Short answer*

How many boys are taller than the class' average height?

**Solution**

1. Given: The class average height is $144.4\text{ cm}$ and the list of boy heights is $147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 142, 140, 141, 144, 145, 150$.
2. Identify the boys whose heights are greater than $144.4\text{ cm}$: $147, 154, 158, 155, 146, 146, 145, 150$.
3. Count these valid heights to find the total number of boys.
4. Result: There are $8$ boys taller than the class average height.

**Answer:** 8 boys

> Common mistake: Comparing with the boys' average instead of the class average height.

## How long is a minute?

### Question 1

*3 marks · Short answer*

Two groups of children were asked to estimate the length of 1 minute. They start by closing their eyes and then open when they think 1 minute has passed. Of course, they are not supposed to count while their eyes are closed. The dot plots below show after how many seconds the children opened their eyes. Discuss how well both the groups fared at this activity. Describe and compare the variability in data and their central tendency.

**Solution**

1. For Group A, the mean is 58.21 seconds and the median is 60 seconds, with estimates ranging from around 40 to 70 seconds.
2. For Group B, the mean is 59.28 seconds and the median is 59.5 seconds, with estimates also ranging from around 40 to 70 seconds.
3. Both groups fared very well and were close to 60 seconds, but Group B had a slightly higher mean and median, showing both groups estimated the duration of 1 minute with high accuracy.

**Answer:** Both groups fared equally well with central values very close to 60 seconds, though Group B's mean is slightly higher.

> Common mistake: Comparing only the mean while ignoring the variability and median values.

## Zero Median Runs Scored!

### Question 1

*3 marks · Short answer*

In a cricket match, can a team's median runs scored by a player be 0 but the team's total score be 407/10?

**Solution**

1. Yes, a team's median runs scored by a player can be 0 even if the team's total score is 407 for the loss of 10 wickets.
2. In a cricket team, 11 players bat, but only 10 wickets fall, meaning 10 players get out and one player remains not out.
3. If 6 or more players score 0 runs (ducks), the middle value (median) when all 11 individual scores are arranged in ascending order will be 0.

**Answer:** Yes, it is possible if 6 or more players score 0 runs, making the median 0 while the total score is 407/10.

> Common mistake: Assuming that a high team score means all players must have scored many runs.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the median of onion prices in Yahapur and Wahapur.

**Solution**

1. First, list the monthly onion prices for Yahapur from the table: 25, 24, 26, 28, 30, 35, 39, 43, 49, 56, 59, 44.
2. Sort the Yahapur prices in ascending order: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59.
3. Since there are 12 values (an even number), the median is the average of the 6th and 7th values, which are 35 and 39.
4. Median for Yahapur = $(35 + 39) \div 2 = 37$
5. Next, list the monthly onion prices for Wahapur: 19, 17, 23, 30, 38, 35, 42, 39, 53, 60, 52, 42.
6. Sort the Wahapur prices in ascending order: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60.
7. The 6th and 7th values are 38 and 39.
8. Median for Wahapur = $(38 + 39) \div 2 = 38.5$

**Answer:** The median price of onions in Yahapur is ₹37 and in Wahapur is ₹38.5 per kg.

> Common mistake: Forgetting to sort the data before finding the middle values.

### Question 2

*3 marks · Short answer*

Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, —, 2, 4. Find the mean and median. How would you describe this data?

**Solution**

1. Given data values with 2 missing values represented by dashes, total 20 valid values: 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4.
2. Sum of all 20 values = $0+1+0+4+8+0+0+2+1+1+5+3+4+0+0+10+25+2+2+4 = 72$.
3. Number of values = 20. Mean = $72 \div 20 = 3.6$.
4. Sort the 20 values in ascending order: 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25.
5. The 10th and 11th values are both 2, so the median is 2.
6. The data is skewed with most students having 0 to 4 pets, and a few higher values like 10 and 25.

**Answer:** Mean = 3.6, Median = 2; the data is heavily skewed towards lower values with a few high outliers.

> Common mistake: Including the dashes as zeros or dividing by 22 instead of 20.

### Question 3

*3 marks · Short answer*

Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height?

**Solution**

1. Total number of trees = 29.
2. Sum of all heights = 1621 feet.
3. Mean = $1621 \div 29 \approx 55.90$ feet and Median = 56 feet.
4. 13 trees are shorter than the average height of 55.90 feet.

**Answer:** Mean = 55.90 feet, Median = 56 feet, and 13 trees are shorter than the average height.

> Common mistake: Sorting errors or miscounting the total number of data points.

### Question 4

*3 marks · Short answer*

The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4. (a) Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median. (b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?

**Solution**

1. Given daily water usages are 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, and 7.4 litres.
2. (a) No, the mean or median daily usage cannot lie between 25 and 30 because both measures must always lie between the minimum value (3.09) and the maximum value (20.5) of the given data.
3. (b) No, the mean or median can never be lesser than the minimum value or greater than the maximum value in a dataset.

**Answer:** No, mean and median cannot lie outside the minimum and maximum values of the data.

> Common mistake: Confusing the range of possible values for mean and median with arbitrary numbers.

### Question 5

*3 marks · Short answer*

The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data.

**Solution**

1. Given: Weights of newborn baby boys and girls in kg as 3.5, 4.1, 2.6, 3.2, 3.4, 3.8 for boys and 4.0, 3.1, 3.4, 3.7, 2.5, 3.4 for girls.
2. The dot plot is filled by placing dots corresponding to each weight value along the horizontal scale from 0 to 4.5.
3. Comparing the two sets of weights, both boys' and girls' weights cluster closely between 2.5 kg and 4.1 kg, showing similar central tendency and variability.

**Answer:** The dot plot shows that the weights of newborn baby boys and girls have a similar distribution and cluster between 2.5 kg and 4.1 kg.

> Common mistake: Misinterpreting the scale on the dot plot or missing data points while plotting.

### Question 6

*3 marks · Short answer*

The dot plots of heights of another section of Grade 5 students of the same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures? Compare the heights of the two sections. Share your observations.

**Solution**

1. The dot plots show that the heights of the whole class range from a minimum of around $125\text{ cm}$ to a maximum of around $155\text{ cm}$.
2. For the whole class, the mean height is $141.21\text{ cm}$ and the median height is $142.5\text{ cm}$, showing that the central height lies around $142\text{ cm}$.
3. Comparing the boys and girls, the mean height for boys is $142.05\text{ cm}$ (median $143\text{ cm}$) and for girls is $140.14\text{ cm}$ (median $140\text{ cm}$), indicating that boys are slightly taller on average than girls in this section.

**Answer:** The heights range from $125\text{ cm}$ to $155\text{ cm}$, with the boys having a slightly higher average height ($142.05\text{ cm}$) than the girls ($140.14\text{ cm}$).

> Common mistake: Confusing the values of mean and median or misreading the scale of the dot plot.

### Question 7

*3 marks · Short answer*

The weights of some sumo wrestlers and ballet dancers are: Sumo wrestlers: 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg. Ballet dancers: 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. Approximately how many times heavier is a sumo wrestler compared to a ballet dancer?

**Solution**

1. Given: Sumo wrestlers' weights are 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg and ballet dancers' weights are 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg.
2. Sum of sumo weights = $295.2 + 250.7 + 234.1 + 221.0 + 200.9 = 1201.9\text{ kg}$, so mean weight = $1201.9 \div 5 = 240.38\text{ kg}$.
3. Sum of dancer weights = $40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2 = 254.5\text{ kg}$, so mean weight = $254.5 \div 6 \approx 42.42\text{ kg}$.
4. Ratio of weights = $240.38 \div 42.42 \approx 5.67$.

**Answer:** A sumo wrestler is approximately 5.7 times heavier than a ballet dancer on average.

> Common mistake: Dividing by the wrong number of observations or making arithmetic errors in summation.

## Visualising Data

### Question 1

*2 marks · Very short answer*

What is the scale used in this graph?

**Solution**

1. Observe the markings along the vertical line in the clustered column graph.
2. The vertical axis shows price in rupees, where 1 unit equals 10 rupees.

**Answer:** The scale used in the graph is 1 unit = 10 rupees.

> Common mistake: Confusing the unit length with horizontal axis categories.

### Question 2

*2 marks · Very short answer*

Is it now easier to compare month-wise prices in both places?

**Solution**

1. Examine the side-by-side arrangement of bars for each month in the clustered column graph.
2. The relative heights of the bars for Yahapur and Wahapur placed side by side make it easy to compare month-wise prices.

**Answer:** Yes, it is easier to compare month-wise prices because the side-by-side bars allow direct visual comparison for each month.

> Common mistake: Saying it is harder due to multiple bars.

### Question 3

*3 marks · Short answer*

Share your observations (you may take the teacher's help to identify the countries these organisations belong to).

**Solution**

1. Observe the graph showing the number of worldwide rocket launches by different organisations for the years 2021, 2022, and 2023.
2. Note that the scale used is 1 unit length = 20 rockets, and the data is represented using three adjacent bars for each organisation.
3. The USA (SpaceX), China (CASC), and Russia (Roscosmos) are the leading rocket-launching countries in the given period.

**Answer:** The USA, China, and Russia are the leading countries in rocket launches, with SpaceX showing significant year-on-year increase.

> Common mistake: Misreading the scale of 20 rockets per unit length.

### Question 4

*3 marks · Short answer*

Analyse and interpret each of your observations.

**Solution**

1. Examine the bar lengths for each organisation across the years 2021, 2022, and 2023.
2. SpaceX launched about twice the number of rockets in 2022 compared to 2021, and numbers for Arianespace decreased every year.
3. United Launch Alliance launched more in 2022 than in 2021, but fewer in 2023.

**Answer:** SpaceX doubled its launches in 2022 compared to 2021, while Arianespace showed a continuous decrease over the three years.

> Common mistake: Incorrectly reading the years corresponding to each bar.

### Question 5

*1 mark · MCQ*

Identify which of the following statements can be justified using this data.

- All organisations launched more rockets than the previous years.
- Only an organisation from the USA launched more than 50 rockets in a single year.
- The total number of rockets launched by France in all 3 years is less than 40.
- The average number of rockets launched by CASC in these 3 years is around 40.
- ISRO launched more rockets than Galactic Energy in these 3 years.
- Russia launched more than 60 rockets in these 3 years.

**Solution**

1. Check each statement against the rocket launch graph data.
2. Statement (d) is correct because the average number of rockets launched by CASC in the 3 years is around 40.

**Answer:** (d) The average number of rockets launched by CASC in these 3 years is around 40.

> Common mistake: Choosing a statement that assumes a general trend without checking specific bars.

### Question 6

*3 marks · Short answer*

List the organisations that have consistently launched more rockets every year.

**Solution**

1. Inspect the bars for each organisation from left to right (2021 to 2022 to 2023).
2. Check which organisations have bar heights increasing continuously every year.
3. SpaceX is the organisation that has consistently launched more rockets every year.

**Answer:** SpaceX is the organisation that has consistently launched more rockets every year.

> Common mistake: Listing organisations whose bar lengths fluctuate or decrease in 2023.

### Question 7

*1 mark · MCQ*

Estimate the total number of rockets launched worldwide in 2023.

- less than 200
- 200 to 400
- 400 to 600
- more than 600

**Solution**

1. Observe the heights of the bars for the year 2023 across all organisations in the graph on page 20.
2. Adding up the approximate number of launches for all organisations in 2023 gives a total well above 600 rockets.

**Answer:** (d) more than 600

> Common mistake: Adding only the top few organisations and underestimating the total.

### Question 8

*2 marks · Very short answer*

What are you curious to know after looking at this graph?

**Solution**

1. After looking at the graph of worldwide rocket launches, one might wonder why certain countries like the USA launch significantly more rockets than other countries.

**Answer:** We might wonder why the USA launches so many more rockets than other countries.

> Common mistake: Writing unrelated questions not based on the rocket launch graph.

### Question 9

*3 marks · Short answer*

Analyse and interpret each of your observations. Share appropriate summary and conclusion statements.

**Solution**

1. Observe the clustered bar graph showing average daily sunshine hours in two cities across the months from January to December.
2. Note that the average daylight hours per day in City 1 increase from January to a maximum in June and then decrease to a minimum in December.
3. Note that City 2 shows an inverted pattern, with maximum daylight hours in December and minimum in June, indicating location in opposite hemispheres.

**Answer:** City 1 has maximum daylight in June and minimum in December, while City 2 has the exact opposite pattern.

> Common mistake: Confusing the months of maximum and minimum daylight hours between the two cities.

### Question 10

*2 marks · Very short answer*

Does this give some idea of where these two cities are located?

**Solution**

1. The inverted seasonal daylight patterns between the two cities indicate that they are located in opposite hemispheres away from the Equator.

**Answer:** Yes, it shows that the two cities are located in opposite hemispheres (Northern and Southern hemispheres) away from the Equator.

> Common mistake: Stating that the cities are near the Equator.

### Question 11

*2 marks · Very short answer*

Is there anything more that you wish to explore?

**Solution**

1. Looking at the extreme variation in daylight hours near the poles, one can explore natural phenomena like experiencing the midnight sun during summer.

**Answer:** We can explore how near the poles one can see the sun even at midnight during summer.

> Common mistake: Giving vague answers without connecting to the data or text.

### Question 12

*3 marks · Short answer*

Answer the following questions based on the graph: 1. Can we tell who batted first? Who won the match? 2. How many runs did the blue team score in over 12? 3. In which over did the red team score the least number of runs? 4. Is it easy to tell the target set by the team batting first?

**Part (i)**

1. Look at the double bar graph for the first over.
2. The graph shows runs per over for both teams simultaneously without indicating which team batted in the first innings or who ultimately won the match.

Answer (i): We cannot tell who batted first or who won the match from this runs-per-over graph alone.

**Part (ii)**

1. Locate over 12 on the horizontal axis.
2. Observe the height of the blue bar corresponding to over 12, which reaches the 15-run mark on the vertical scale.

Answer (ii): The blue team scored 15 runs in over 12.

**Part (iii)**

1. Examine the red bars across all overs from 1 to 20.
2. Find the over where the red bar has the smallest height, which corresponds to the least number of runs scored.

Answer (iii): The red team scored the least number of runs in over 4.

**Answer:** Based on the graph, we cannot tell who batted first, the blue team scored 15 runs in over 12, and the red team scored the least in over 4.

> Common mistake: Assuming match winners or batting order without complete scorecard details.

## Figure it Out

### Question 1

*4 marks · Case-based*

The following infographic shows the speeds of a few animals in air, on land, and in water. Can we call this graph a bar graph? (a) What is the scale used in this graph? (b) What did you find interesting in this infographic? What do you want to explore further? (c) Identify a pair of creatures where one's speed is about twice that of the other. (d) Can we say that a sailfish is about 4 times faster than a humpback whale? Can we say that a sailfish is the fastest aquatic animal in the world?

**Part (a)**

1. Observe the scale given at the bottom left of the infographic.
2. The scale is 1 unit = $16\text{ km/h}$.

Answer (a): 1 unit = 16 km/h

**Part (b)**

1. Observe the speeds of various animals across air, land, and water.
2. The Australian tiger beetle moves at about 120 body lengths per second, which is an interesting fact as it runs blind at top speed.

Answer (b): The Australian tiger beetle moves at about 120 body lengths per second and runs blind at top speed.

**Part (c)**

1. Check the speeds of the listed creatures from the infographic.
2. A cheetah runs at $103\text{ km/h}$ and a pronghorn antelope runs at $88\text{ km/h}$, or a green darner dragonfly at $64\text{ km/h}$ and a flying fish at $56\text{ km/h}$.

Answer (c): A pair of creatures with speeds around a 2:1 ratio can be identified from the infographic, such as the cheetah and the ostrich.

**Part (d)**

1. Find the speed of a sailfish ($109\text{ km/h}$) and a humpback whale ($26\text{ km/h}$).
2. Divide $109$ by $26$ to get approximately $4$ times, but a sailfish is not the fastest aquatic animal as it is slower than some birds and land animals.

Answer (d): Yes, a sailfish is about 4 times faster than a humpback whale, but it is not the fastest animal in the world overall.

**Answer:** Infographic analysis of animal speeds in air, land, and water.

> Common mistake: Misreading the scale or confusing aquatic animals with land animals.

### Question 2

*3 marks · Short answer*

Preyashi asked her students 'If you were to get a super power to become aquatic (water-borne), aerial (air-borne), or spaceborne which one would you choose?'. The responses are shown below. Some chose none. Draw a double-bar graph comparing how both grades chose each option. Choose an appropriate scale.

**Solution**

1. Count the frequency of each choice (aquatic 'w', aerial 'a', spaceborne 's', none 'n') for Grade 5 and Grade 9 from the given text.
2. Choose an appropriate scale for the vertical axis, such as 1 unit = 2 students.
3. Draw the double-bar graph comparing the responses of both grades for each category.

**Answer:** Double-bar graph drawn with categories for aquatic, aerial, spaceborne, and none.

> Common mistake: Mixing up the frequency counts of Grade 5 and Grade 9.

### Question 3

*3 marks · Short answer*

The temperature variation over two days in different months in Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use the scale 1 unit = 4°C. Can you guess which two months these days might belong to?

**Solution**

1. Set up the horizontal axis with time intervals from 12 am to 9 pm and use the scale 1 unit = 4°C on the vertical axis.
2. Plot the double bars for Day 1 and Day 2 corresponding to each time slot.
3. Observe that Day 1 has moderate temperatures typical of winter or spring, while Day 2 has very high temperatures (up to 43°C) typical of summer in Jodhpur.

**Answer:** Double-bar graph drawn; Day 1 likely belongs to a winter/cooler month and Day 2 to a summer month like May or June.

> Common mistake: Incorrectly applying the temperature scale on the vertical axis.

### Question 4

*4 marks · Case-based*

The following clustered-bar graph shows the number of electric vehicles registered in some states every year from 2022 to 2024. (a) The data (rounded-off to thousands) for the states of Gujarat and Delhi are given in the table below. Mark the corresponding bars on the bar graph. (It is enough if you place the top of the bars between the two appropriate vertical guidelines.) (b) Notice how the graph is organised, what scale is used, and what patterns the data shows. (c) How would you describe the change for various states between 2022 and 2024? (d) Approximately how many more registrations did Assam get in 2023 compared to 2022? (e) How many times more did the registrations in West Bengal increase from 2022 to 2024? (f) Is this statement correct — 'There were very few new registrations in Uttarakhand in 2023 and 2024, as the increase in the bar lengths is minimal'?

**Part (a)**

1. Locate Gujarat and Delhi on the horizontal axis of the graph in the textbook (Fig. on page 28).
2. Mark the bars corresponding to the given rounded-off values for 2022, 2023, and 2024 between the appropriate vertical guidelines.

Answer (a): Bars marked on the bar graph for Gujarat and Delhi.

**Part (b)**

1. Observe that the vertical axis represents the number of electric vehicles with a scale of 1 unit = 25000 vehicles.
2. The horizontal axis lists the states, with three clustered bars for the years 2022, 2023, and 2024.

Answer (b): Scale is 1 unit = 25000 vehicles, organised by states with yearly clusters.

**Part (c)**

1. Compare the bar heights for each state from 2022 to 2024.
2. Most states show an overall increase in electric vehicle registrations over the three years.

Answer (c): Registrations generally increased across the states from 2022 to 2024.

**Part (d)**

1. Read the bar heights for Assam in 2022 and 2023 from the graph.
2. Estimate the difference between the two bar heights to find the extra registrations.

Answer (d): Approximately 25000 more registrations.

**Part (e)**

1. Observe the bar heights for West Bengal in 2022 and 2024.
2. Calculate the approximate ratio of the 2024 value to the 2022 value.

Answer (e): Registrations increased by about 2 times.

**Part (f)**

1. Examine the bars for Uttarakhand for the years 2022, 2023, and 2024.
2. Assess whether the visual increase in bar lengths is small compared to other states.

Answer (f): Yes, the statement is correct as the bar lengths show very little increase.

**Answer:** All parts of the clustered-bar graph analysis completed.

> Common mistake: Misreading the scale values on the vertical axis when estimating differences.

## Telling Tall Tales

### Question 1

*3 marks · Short answer*

Following are the dot plots of heights of boys (in blue) and girls (in orange) of Grades 6, 7 and 8 (in that order) of two different schools. What do you notice? Share your observations.

**Solution**

1. Observe the dot plots for School A and School B across Grades 6 to 8 shown in the figure in the textbook (Fig. 5.5).
2. Notice that the students in School B are generally taller than those in School A across all grades, as seen from the higher mean and median values.
3. Observe the spread and variability of heights within each grade and compare the average heights of boys and girls across the two schools.

**Answer:** Students in School B are generally taller than students in School A across the corresponding grades, and heights increase progressively from Grade 6 to Grade 8.

> Common mistake: Generalising the height difference of a single school to all schools in the country.

### Question 2

*3 marks · Short answer*

Spend sufficient time observing the data presented in this table. Share your findings with the class. These are some prompts for you to probe — • Changes in the heights of boys or girls of a certain age from 1989 to 2019. • The heights of boys vs. girls at different ages in a particular year. • Changes in height between successive ages in boys and girls in 2019.

**Solution**

1. Observe the average height table for ages 5 to 19 across the years 1989, 1999, 2009, and 2019.
2. Compare the heights of boys and girls of a specific age over the decades to see a general increase in average height over time.
3. Compare the heights between boys and girls at different ages in a particular year, noting that girls are temporarily taller around ages 11 to 13.

**Answer:** Average heights of children at every age have increased from 1989 to 2019, and girls tend to be slightly taller than boys during early adolescence (ages 11 to 13).

> Common mistake: Ignoring the year-wise comparison and only looking at a single column.

### Question 3

*1 mark · MCQ*

Which of the following statements can be justified using the data?

- The average heights of both boys and girls at every age increased from 1989 to 2019.
- The average height of 13-year-old girls in 1989 is more than the average height of 14-year-old girls in 2009.
- The average height of 15-year-old boys in 2019 is more than the average height of 16-year-old boys in 1989.
- All girls aged 13 are taller than all girls aged 11.
- Throughout the age period 5 to 19, the average boy's height is more than the average girl's height.
- Boys keep growing even beyond age 19.

**Solution**

1. Check each statement against the values given in the table of average heights from 1989 to 2019.
2. Statement 1 is correct because comparing the values for every age from 1989 to 2019 shows a consistent increase in average heights.

**Answer:** (1) The average heights of both boys and girls at every age increased from 1989 to 2019.

> Common mistake: Confusing individual data points with group averages and universal statements.

### Question 4

*3 marks · Short answer*

In 2019, between which two successive ages from 5 to 19 did boys grow the most? Between which two successive ages from 5 to 19 did girls grow most?

**Solution**

1. Look at the 2019 column in the height table to find the difference in average height between successive ages.
2. For boys in 2019, the largest increase in average height occurs between age 11 and age 12 ($142.2 - 137 = 5.2\text{ cm}$) or age 12 to 13 ($148.4 - 142.2 = 6.2\text{ cm}$).
3. For girls in 2019, the largest increase occurs between age 10 and age 11 ($138.6 - 132.8 = 5.8\text{ cm}$) or age 11 to 12 ($143.8 - 138.6 = 5.2\text{ cm}$)., confirming the adolescent growth spurt.

**Answer:** Boys grew the most between ages 12 and 13, and girls grew the most between ages 10 and 11.

> Common mistake: Subtracting non-successive ages or looking at the wrong year column.

### Question 5

*3 marks · Short answer*

Suppose the average height of a newborn is 50 cm. Estimate the average height of young children of ages 1 to 4.

**Solution**

1. Given that the average height of a newborn is $50 \text{ cm}$ and observing the trend of rapid growth in early childhood from the height tables in the chapter.
2. Children grow very fast in their first few years, with average heights increasing significantly each year from age 1 to age 4.
3. Based on typical growth data, the estimated average heights are approximately $75 \text{ cm}$ for age 1, $85 \text{ cm}$ for age 2, $95 \text{ cm}$ for age 3, and $105 \text{ cm}$ for age 4.

**Answer:** Estimated average heights: Age 1: $75 \text{ cm}$, Age 2: $85 \text{ cm}$, Age 3: $95 \text{ cm}$, Age 4: $105 \text{ cm}$

> Common mistake: Assuming linear growth by adding a fixed constant every month instead of accounting for the growth curve.

### Question 6

*3 marks · Short answer*

Based on the trend observed in the table, write your estimates of the heights of boys and girls for ages 5 to 19 in the year 2029.

**Solution**

1. Observe the decadal increase in average heights from 2009 to 2019 for each age group in the table.
2. Note that the average heights have been increasing by approximately $1\text{ cm}$ to $3\text{ cm}$ every ten years.
3. Apply a similar upward trend to estimate the heights for ages 5 to 19 in the year 2029 by adding about $1\text{ cm}$ to $2\text{ cm}$ to the 2019 values.

**Answer:** The estimated heights for 2029 will be higher than the 2019 values by roughly $1\text{ cm}$ to $2\text{ cm}$ across the respective ages, continuing the observed historical upward trend.

> Common mistake: Adding a constant arbitrary number without looking at the actual decadal differences.

### Question 7

*2 marks · Very short answer*

How is the graph organised? What information is presented?

**Solution**

1. The horizontal line lists different countries and the vertical line shows the height in centimeters starting from 145 cm.
2. It presents the average heights of 19-year-old boys and girls across various countries for the years 1989 and 2019.

**Answer:** The graph shows country-wise average heights of 19-year-old boys and girls for 1989 and 2019, with the vertical axis starting from 145 cm.

> Common mistake: Failing to mention that the vertical axis starts from 145 cm to zoom in on the height differences.

### Question 8

*2 marks · Very short answer*

What do you find interesting?

**Solution**

1. We notice that across almost all countries, the height markers for 2019 are generally positioned higher than those for 1989.
2. This indicates an overall increase in the average height of 19-year-olds over the thirty-year period.

**Answer:** The heights for 2019 are mostly higher than those for 1989, showing that people have generally grown taller over time across countries.

> Common mistake: Giving a personal opinion instead of an observation based on the visual trend in the graph.

## Figure it Out

### Question 1

*4 marks · Case-based*

The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls. Based on the dot plots, which of the following statements are true? (a) The data varies more for the boys than for the girls. (b) The median number of pockets for the boys is more than that for the girls. (c) The mean number of pockets for the girls is more than that for the boys. (d) The maximum number of pockets for boys is greater than that for the girls.

**Part (a)**

1. Observe the spread of dots in the dot plot for boys and girls.
2. The boys' dot plot covers a wider range of pocket counts compared to the girls.
3. Therefore, the data varies more for the boys than for the girls, making this statement true.

Answer (a): True

**Part (b)**

1. Find the median for both groups from the dot plots.
2. The median number of pockets for boys is 4 and for girls is 4.
3. Thus, the median for boys is not more than that for girls, making this statement false.

Answer (b): False

**Part (c)**

1. Calculate or observe the balance of values for both dot plots.
2. The distribution of dots shows that the mean number of pockets for girls is higher than that for boys.
3. Therefore, this statement is true.

Answer (c): True

**Part (d)**

1. Identify the maximum number of pockets for each group from the dot plots.
2. The maximum number of pockets for boys is 6, while for girls it is 6.
3. Thus, the maximum for boys is equal to, not greater than, that for girls, making this statement false.

Answer (d): False

**Answer:** Statements (a) and (c) are true.

> Common mistake: Confusing the spread (variability) of the data with the central tendency measures.

### Question 2

*3 marks · Case-based*

The following table shows the points scored by each player in four games: Now answer the following questions: (a) Find the average number of points scored per game by A. (b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B? (c) Who is the best performer?

**Part (a)**

1. Sum of points scored by player A = $14 + 16 + 10 + 10 = 50$.
2. Number of games played = 4.
3. Average number of points per game = $50 \div 4 = 12.5$.

Answer (a): 12.5 points

**Part (b)**

1. For player C, divide by 3 because C did not play Game 3 (no value is considered, similar to the method on page 111).
2. For player B, divide by 4 because B played all 4 games, even though zero was scored in one game.

Answer (b): Divide C by 3 and B by 4

**Part (c)**

1. Calculate average for B: $(0 + 8 + 6 + 4) \div 4 = 18 \div 4 = 4.5$ points.
2. Calculate average for C: $(8 + 11 + 13) \div 3 = 32 \div 3 = 10.67$ points.
3. Comparing averages (A: 12.5, B: 4.5, C: 10.67), player A has the highest average performance.

Answer (c): Player A

**Answer:** (a) 12.5 points, (b) Divide C by 3 and B by 4, (c) Player A

> Common mistake: Dividing player C's total by 4 instead of 3, ignoring the 'Did not play' rule.

### Question 3

*3 marks · Short answer*

The marks (out of 100) obtained by a group of students in a General Knowledge quiz are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Another group's scores in the same quiz are 68, 59, 73, 86, 47, 79, 90, 93 and 86. Compare and describe both the groups performance using, mean and median.

**Solution**

1. Group 1 scores: 85, 76, 90, 85, 39, 48, 56, 95, 81, 75 (Total values = 10).
2. Group 1 sorted: 39, 48, 56, 75, 76, 81, 85, 85, 90, 95.
3. Group 1 mean = $730 \div 10 = 73$ and median = $(76 + 81) \div 2 = 78.5$.
4. Group 2 scores: 68, 59, 73, 86, 47, 79, 90, 93, 86 (Total values = 9).
5. Group 2 sorted: 47, 59, 68, 73, 79, 86, 86, 90, 93.
6. Group 2 mean = $679 \div 9 = 75.44$ and median = 79.
7. Comparison: Group 2 has a slightly higher mean and median than Group 1, indicating overall better performance.

**Answer:** Group 1: Mean = 73, Median = 78.5; Group 2: Mean = 75.44, Median = 79; Group 2 performed slightly better.

> Common mistake: Forgetting to sort the data before finding the median.

### Question 4

*3 marks · Short answer*

Consider this data collected from a survey of a colony. Choose an appropriate scale and draw a double-bar graph. Write down your observations.

**Solution**

1. Choose an appropriate scale along the vertical axis, such as 1 unit = 100 people.
2. Draw pairs of adjacent bars for watching and participating for each sport: Cricket, Basketball, Swimming, Hockey, and Athletics.
3. Observe that Cricket is the most watched and played sport, while Athletics has the lowest numbers in both categories.

**Answer:** Double-bar graph constructed with appropriate scale showing Cricket as the most popular sport.

> Common mistake: Failing to include a clear key or legend to distinguish between watching and participating bars.

### Question 5

*3 marks · Short answer*

Consider a group of 17 students with the following heights (in cm): 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101. The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with height less than a particular height and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the 'Telling Tall Tales' section?

**Solution**

1. Sort the 17 heights in ascending order to find the middle value.
2. The sorted heights are: 101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125.
3. The median (9th value) is 115 cm, which splits the group into two equal halves of 8 students each.
4. Based on the tabular data in the 'Telling Tall Tales' section for Grade 5 or 6, the age of these students is approximately 10 to 11 years.

**Answer:** Use the median height of 115 cm to divide the groups; estimated age is 10-11 years.

> Common mistake: Trying to use the mean instead of the median to divide the data into two equal numerical counts.

### Question 6

*Activity*

Describe the mean and median of heights of your class. You can visualise the heights on a dot plot.

**Solution**

1. Collect the height measurements of all students in the class.
2. Represent the data visually on a dot plot to observe clustering and spread.
3. Calculate the arithmetic mean and find the median to describe the central tendency of the class heights.

**Answer:** Activity completed by collecting data, plotting a dot plot, and calculating the mean and median.

### Question 7

*1 mark · MCQ*

There are two 7th grade sections at a school. Each section has 15 boys and 15 girls. In one section, the mean height of students is 154.2 cm. From this information, what must be true about the mean height of students in the other section?

- The mean height of students in the other section is 154.2 cm.
- The mean height of students in the other section is less than 154.2 cm.
- The mean height of students in the other section is more than 154.2 cm.
- The mean height of students in the other section cannot be determined.

**Solution**

1. The mean height depends on the individual values of the students in the section.
2. Knowing the number of students is not sufficient to determine the exact mean height of the other section.

**Answer:** (d) The mean height of students in the other section cannot be determined.

> Common mistake: Assuming that sections with equal numbers of students must have the same mean height.

### Question 8

*3 marks · Short answer*

Standing tall in the storm. (a) Write estimated values for the number of skyscrapers in New York, Tokyo, and London. (b) Are the following statements valid? (i) Only 12 cities have more skyscrapers than Mumbai. (ii) Only 7 cities have fewer skyscrapers than Mumbai. (iii) The tallest building in the world is in Hong Kong.

**Solution**

1. 1. From the bar graph of skyscrapers, read the approximate values for New York (around 251), Tokyo (around 154), and London (around 46).
2. 2. Statement (i) is valid as there are 12 cities with more skyscrapers than Mumbai (which has 86).
3. 3. Statement (ii) is invalid because there are more than 7 cities with fewer skyscrapers than Mumbai, and statement (iii) is false because the tallest building is not necessarily in Hong Kong (Hong Kong has the most skyscrapers).

**Answer:** (a) New York $\approx 251$, Tokyo $\approx 154$, London $\approx 46$. (b) Only statement (i) is valid.

> Common mistake: Misinterpreting the rank of cities and number of skyscrapers from the bar heights.

### Question 9

*Activity*

Estimate and then measure the objects listed in the following table. Draw a double bar graph based on the data. How accurate were your estimates? Find the average difference between the estimated and measured values.

**Solution**

1. This is a practical activity involving estimation, measurement, and drawing a double bar graph.
2. Students record estimated and measured lengths of given objects, calculate differences, and find the average difference.

**Answer:** Perform the activity by estimating, measuring, and plotting the data.

### Question 10

*3 marks · Short answer*

Aditi likes solving puzzles. She recently started attempting the 'Easy' level Sudoku puzzles. The time she took (in seconds) to solve these puzzles are — 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, 280, 270, 230, 220, 240. The first nine values correspond to Week 1 and the rest to Week 2. (a) Construct a dot plot below showing the data for both weeks. (b) Describe the mean, median, and any observations you may have about the data.

**Solution**

1. 1. Week 1 times (first 9 values): 410, 400, 370, 340, 360, 400, 320, 330, 310. Week 2 times: 320, 290, 380, 280, 270, 230, 220, 240.
2. 2. Construct the dot plot with a suitable horizontal scale from 200 to 410, marking data points for both weeks.
3. 3. Calculate the mean and median for each week to compare puzzle-solving times.

**Answer:** Dot plot constructed and summary values described.

> Common mistake: Mixing up the values for Week 1 and Week 2.

### Question 11

*Activity*

Individual Project: Pick at least one of the following: (a) How Long is a Sentence? Pick any two textbooks from different subjects. Choose any page with a lot of text from each book. (i) Use a dot plot to describe how many words the sentences have on each page. (ii) Compare the data of both the pages using mean and median. (b) What is in a Name? Write down the names of all of your classmates. The following are some interesting things you can do with this data! (i) Find the mean and median name length (number of letters in a name). (ii) Visualise the data and describe its variability and central tendency. (iii) Which starting letters are more popular? Which are less popular? (iv) What is the median starting letter? What does this say about the number of names starting with the letters A–M and N–Z? (v) Plot a double-bar graph showing the number of boys' names and girls' names that: • start and end with vowels, • start with vowels and end with consonants, • start with consonants and end with vowels, • start and end with consonants.

**Solution**

1. This is an individual project involving text analysis or name lengths.
2. Data is collected, visualised using dot plots or double-bar graphs, and analysed using mean and median.

**Answer:** Complete the project steps based on personal data collection.

### Question 12

*Activity*

Individual project (long term): This requires collecting data over 2 weeks or more. In and Out: Track how many times you step out of your house in a day. Do this for a month. (i) Describe the variability and central tendency of this data. Make a dot plot. (ii) Do you find anything interesting about this data? Share your observations. (iii) You can ask any of your family members or friends to do this as well.

**Solution**

1. This is a long-term project requiring daily tracking over a month.
2. Data is recorded, represented on a dot plot, and analysed for variability and central tendency.

**Answer:** Track the data over a month and construct the required dot plot and observations.

### Question 13

*Activity*

Small-group project: Pick at least one of the following. Make groups of 8 to 10. Collect data individually as needed. Put together everyone's data and do the appropriate analysis and visualisation. (a) Our heights vs. our family's heights: Collect the heights of your family members. (i) Make a dot plot showing heights of just your family members. Describe its variability and central tendency. (ii) Make a double-bar graph showing each student's height next to their family's mean height. (iii) Look at everyone's data and share your observations. (b) Estimating time: Check the time and close your eyes. Open them when you think 1 minute has passed (no counting). Note down after how m-any seconds you opened your eyes. Collect this data for yourself and for your family members. Repeat this activity to estimate 3 minutes. (i) Make two dot plots (for 1 minute and 3 minutes) showing estimates of just your family members. (ii) Mark these on the respective dot plots. Describe its variability and central tendency. (iii) Make a double bar graph showing each family's mean 1 minute estimate and mean 3 minute estimate. (iv) Look at everyone's data and share your observations.

**Part (a)(i)**

1. Collect the heights of all members in your family.
2. Draw a number line representing heights and mark each height with a dot to create a dot plot.
3. Describe the minimum, maximum, range, mean, and median to understand the variability and central tendency.

Answer (a)(i): A dot plot showing family heights along with a description of their variability (minimum, maximum, range) and central tendency (mean and median).

**Part (a)(ii)**

1. Calculate the mean height of the family members for each student in the group.
2. Draw a double-bar graph with two adjacent bars for each student representing their own height and their family's mean height.

Answer (a)(ii): A double-bar graph comparing each student's height with their family's mean height.

**Part (a)(iii)**

1. Examine the combined data of all students in the group.
2. Share observations comparing student heights and family mean heights across the group.

Answer (a)(iii): Observations based on the group's compiled height data.

**Part (b)(i)**

1. Record the time in seconds estimated for 1 minute and 3 minutes by family members without counting.
2. Construct two separate dot plots for the 1-minute and 3-minute estimates.

Answer (b)(i): Two dot plots showing family estimates for 1 minute and 3 minutes.

**Part (b)(ii)**

1. Mark the individual estimates on the respective dot plots.
2. Calculate and describe the mean, median, and spread of the estimates.

Answer (b)(ii): Marked dot plots with a description of variability and central tendency.

**Part (b)(iii)**

1. Find the mean 1-minute estimate and mean 3-minute estimate for each family.
2. Construct a double-bar graph showing these two means for each family.

Answer (b)(iii): A double-bar graph comparing each family's mean 1-minute estimate with their mean 3-minute estimate.

**Part (b)(iv)**

1. Compare the results across all families in the group.
2. Discuss observations on how closely families estimated 1 minute versus 3 minutes.

Answer (b)(iv): Shared observations on time estimation across the group.

**Answer:** This is a small-group project involving data collection, calculating arithmetic mean, constructing dot plots and double-bar graphs, and making observations about variability and central tendency.

## Frequently asked questions

### How many total questions are there in NCERT Solutions for Class 7 Maths Chapter 13 Connecting the Dots?

This chapter contains a total of 72 questions spread across various sections like Of Questions and Statements, Representative Values, and Visualising Data. You can find step-by-step solutions for all these questions in the free PDF available on this page only.

### Which topics and concepts are covered in Class 7 Maths Chapter 13?

The chapter covers key statistical concepts such as Arithmetic Mean, Median, Outliers, dot plots, and clustered bar graphs. Students also learn about data collection, interpretation, and comparing data sets using measures of central tendency.

### What are the hardest question types in this chapter and how do I approach them?

The case-based and multi-step data interpretation questions involving dot plots and double-bar graphs can be challenging. To approach them, carefully read the given graphical data, identify trends or outliers first, and then apply formulas for mean and median systematically.

### How should I write answers to score full marks in Class 7 Maths Chapter 13 assessments?

To secure full marks, always write down the given data clearly, show every step of the calculation for arithmetic mean or median, and include proper units where necessary. Referring to SwaVid's detailed solutions on this page helps you understand the ideal presentation format.

### Is the free PDF for Class 7 Maths Chapter 13 Connecting the Dots available for the 2026-27 session?

Yes, the complete chapter solutions aligned with the new NCERT book for the 2026-27 session are provided here. You can easily download or view the free PDF on this page to practice all textbook questions.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student