---
title: "NCERT Solutions Class 8 Maths Chapter 12 Tales by Dots and Lines"
url: https://www.swavid.com/maths/class/8/chapter/tales-by-dots-and-lines/ncert-solutions
dateModified: 2026-10-07T15:38:21+00:00
---

# NCERT Solutions Class 8 Maths Chapter 12 Tales by Dots and Lines

This chapter's questions cover concepts related to statistical measures like mean, median, frequencies, and data interpretation using line graphs and infographics.

Free PDF (23 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-12-tales-by-dots-and-lines-a664df3d7c.pdf

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the mean of the following data and share your observations: 
(i) The first 50 natural numbers.
(ii) The first 50 odd numbers.
(iii) The first 50 multiples of 4.

**Part (i)**

1. The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.
2. For $n = 50$, the sum is $\frac{50 \times 51}{2} = 1275$.
3. The mean is $\frac{1275}{50} = 25.5$.

Answer (i): 25.5

**Part (ii)**

1. The sum of the first $n$ odd numbers is $n^2$.
2. For $n = 50$, the sum is $50^2 = 2500$.
3. The mean is $\frac{2500}{50} = 50$.

Answer (ii): 50

**Part (iii)**

1. The first 50 multiples of 4 form an arithmetic progression or can be written as $4 \times (1 + 2 + \dots + 50)$.
2. The sum is $4 \times 1275 = 5100$.
3. The mean is $\frac{5100}{50} = 102$.

Answer (iii): 102

**Answer:** The means are (i) 25.5, (ii) 50, and (iii) 102.

> Common mistake: Dividing by the wrong number of terms or mixing up formulas for sum of natural and odd numbers.

### Question 2

*3 marks · Short answer*

The dot plot below shows a collection of data and its average; but one dot is missing. Mark the missing value so that the mean is 9 (as shown below).

**Solution**

1. Let the missing value be $x$.
2. From the given dot plot, the total number of data points is 10, and their sum can be calculated using the given points and the mean $9$.
3. The total sum of 10 values with a mean of $9$ is $10 \times 9 = 90$.
4. Summing the visible data points: $3 + 4 + 5 + 6 + 7 + 8 + 8 + 12 + 13 + x = 66 + x$.
5. Equating the total sum gives $66 + x = 90$, so $x = 90 - 66 = 24$.
6. Thus, the missing value is $24$.

**Answer:** 24

> Common mistake: Forgetting to multiply the mean by the total count of numbers to find the total sum.

### Question 3

*3 marks · Case-based*

Sudhakar, the class teacher, asks Shreyas to measure the heights of all 24 students in his class and calculate the average height. Shreyas informs the teacher that the average height is 150.2 cm. Sudhakar discovers that the students were wearing uniform shoes when the measurements were taken and the shoes add 1 cm to the height.
(i) Should the teacher get all the heights measured again without the shoes to find the correct average height? Or is there a simpler way?
(ii) What is the correct average height of the class?

- 174.2 cm
- 126.2 cm
- 150.2 cm
- 149.2 cm
- 151.2 cm
- None of the above
- Insufficient information

**Part (i)**

1. No, the teacher does not need to measure all heights again.
2. Since each student's height increases by a fixed amount of 1 cm due to shoes, subtracting 1 cm from the average gives the correct average.

Answer (i): No, a simpler way is to subtract 1 cm from the measured average height.

**Part (ii)**

1. The measured average height is $150.2$ cm.
2. Since every height includes 1 cm for shoes, the correct average height is $150.2 - 1 = 149.2$ cm.

Answer (ii): (d) 149.2 cm

**Answer:** (d) 149.2 cm

> Common mistake: Measuring all heights again instead of using the property that adding a fixed number to every data value increases the mean by that same number.

### Question 4

*3 marks · Short answer*

The three dot plots below show the lengths, in minutes, of songs of different albums. Which of these has a mean of 5.57 minutes? Explain how you arrived at the answer.

**Solution**

1. Calculate the sum of values for each dot plot by multiplying each value by its frequency.
2. Plot A has values centered around 5.5 with a mean of 5.57 minutes.
3. Plots B and C have different distributions with means not equal to 5.57 minutes.

**Answer:** Plot A has a mean of 5.57 minutes.

> Common mistake: Guessing the plot without calculating the exact sum and number of data points.

### Question 5

*3 marks · Short answer*

Find the median of $8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92$.
(i) If we include one value to the data (in the given list) without affecting the median, what could that value be?
(ii) If we include two values to the data without affecting the median what could the two values be?
(iii) If we remove one value from the data without affecting the median what could the value be?

**Part (i)**

1. The given data has 16 values (even number), so the median is the average of the 8th and 9th values.
2. The 8th value is 40 and the 9th value is 41, giving a median of $\frac{40+41}{2} = 40.5$.
3. To keep the median unchanged when including one value, that value must be equal to the current median, 40.5.

Answer (i): 40.5

**Part (ii)**

1. When including two values to an even-sized dataset without changing the median, the two new values must be placed one below and one above the middle, or both equal to the median.
2. Any two numbers whose average is 40.5 will keep the median unchanged.

Answer (ii): Any two numbers whose average is 40.5

**Part (iii)**

1. When removing one value from an even-sized dataset to keep the median of the remaining odd-sized dataset equal to 40.5, we must remove a value that does not shift the middle position.
2. Removing either 40 or 41 leaves the remaining middle value as 41 or 40, but to keep the median strictly 40.5, removing a value outside the two middle numbers or choosing appropriately keeps the median robust.

Answer (iii): Any value from 40 to 41

**Answer:** Median is 40.5. (i) 40.5, (ii) Any two values whose average is 40.5, (iii) Any value greater than or equal to 40 and less than or equal to 41.

> Common mistake: Confusing the median position formula for odd and even number of observations.

### Question 6

*3 marks · Short answer*

Examine the statements below and justify if the statement is always true, sometimes true, or never true.
(i) Removing a value less than the median will decrease the median.
(ii) Including a value less than the mean will decrease the mean.
(iii) Including any 4 values will not affect the median.
(iv) Including 4 values less than the median will increase the median.

**Part (i)**

1. Removing a value smaller than the median changes the order, but depending on whether it shifts the middle value, the median may or may not decrease.

Answer (i): Sometimes true

**Part (ii)**

1. Adding a value less than the current mean reduces the total sum relative to the count, thereby decreasing the mean.

Answer (ii): Always true

**Part (iii)**

1. Including 4 arbitrary values changes the total number of observations and shifts the middle position, so it will generally affect the median.

Answer (iii): Never true

**Part (iv)**

1. Including values less than the median adds smaller elements to the lower half, which shifts the sorted order and can decrease or keep the median same rather than always increasing it.

Answer (iv): Sometimes true

**Answer:** (i) Sometimes true, (ii) Always true, (iii) Never true, (iv) Sometimes true.

> Common mistake: Assuming median and mean behave identically when new data points are added or removed.

### Question 7

*3 marks · Short answer*

The mean of the numbers $8, 13, 10, 4, 5, 20, y, 10$ is $10.375$. Find the value of $y$.

**Solution**

1. Given: The numbers are $8, 13, 10, 4, 5, 20, y, 10$ and their mean is $10.375$.
2. Formula: $\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}$
3. Substitution: $10.375 = \frac{8 + 13 + 10 + 4 + 5 + 20 + y + 10}{8}$
4. Working: $10.375 \times 8 = 70 + y$, which gives $83 = 70 + y$
5. Result: $y = 13$

**Answer:** $13$

> Common mistake: Dividing by the wrong number of terms or arithmetic errors in summing the given numbers.

### Question 8

*3 marks · Short answer*

The mean of a set of data with 15 values is 134. Find the sum of the data.

**Solution**

1. Given: Number of values = 15, Mean = 134
2. Formula: $\text{Mean} = \frac{\text{Sum of the data}}{\text{Number of values}}$
3. Substitution: $134 = \frac{\text{Sum}}{15}$
4. Result: $\text{Sum} = 134 \times 15 = 2010$

**Answer:** $2010$

> Common mistake: Multiplying mean by the wrong count or mixing up numerator and denominator.

### Question 9

*1 mark · MCQ*

Consider the data: $12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p$. Which of the following number(s) could be $p$ if the median of this data is 29?
(i) 10 (ii) 25 (iii) 40 (iv) 100
(v) 29 (vi) 47 (vii) 30

- 10
- 25
- 40
- 100
- 29
- 47
- 30

**Solution**

1. Arrange the known values in ascending order: $8, 8, 12, 18, 25, 29, 35, 39, 47, 73$. There are 10 known values.
2. With the inclusion of $p$, there are 11 values in total, so the median will be the 6th value when sorted.
3. For the median to be 29, 29 must be the 6th value in the sorted order, which means $p$ must be greater than or equal to 29.
4. Checking the given options that are $\ge 29$: (iii) 40, (iv) 100, (v) 29, (vi) 47, (vii) 30.

**Answer:** (iii), (iv), (v), (vi), (vii)

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

### Question 10

*3 marks · Short answer*

The number of times students rode their cycles in a week is shown in the dot plot below. Four students rode their cycles twice in that week.
(i) Find the average number of times students rode their cycles.
(ii) Find the median number of times students rode their cycles.
(iii) Which of the following statements are valid? Why?
(a) Everyone used their cycle at least once.
(b) Almost everyone used their cycle a few times.
(c) There are some students who cycled more than once on some days.
(d) Exactly 5 students have used their cycles more than once on some days.
(e) The following week, if all of them cycled 1 more time than they did the previous week, what would be the average and median of the next week’s data?

**Part (i)**

1. Count the total number of students and the sum of all data points from the dot plot.
2. Total number of students = 36, and sum of all values = 144.
3. Average = $\frac{144}{36} = 4$.

Answer (i): $4$

**Part (ii)**

1. With 36 values, the median is the average of the 18th and 19th values.
2. Counting frequencies from the sorted dot plot, both the 18th and 19th values fall at 4.
3. Median = $4$.

Answer (ii): $4$

**Part (iii)**

1. Statement (a) is valid because the dot plot starts at 0 or 1 depending on minimum cycles, but looking at the plot, everyone has at least 1 cycle ride.
2. Statement (c) is valid as multiple students have rides greater than 1.
3. For the next week, if everyone cycles 1 more time, the average increases by 1 to become 5, and the median also increases by 1 to become 5.

Answer (iii): Statements (a) and (c) are valid. Next week's average is 5 and median is 5.

**Answer:** Calculated average, median, valid statements, and next week's values.

> Common mistake: Miscounting the frequencies from the dot plot.

### Question 11

*3 marks · Short answer*

A dart-throwing competition was organised in a school. The number of throws participants took to hit the bull’s eye (the centre circle) is given in the table below. Describe the data using its minimum, maximum, mean and median.

**Solution**

1. Given data from table: Trials 1 (1 student), 4 (1 student), 5 (4 students), 6 (9 students), 7 (12 students), 8 (15 students), 9 (10 students), 10 (10 students). Total students = $1 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62$.
2. Minimum trials = 1, Maximum trials = 10.
3. Sum of all trials = $(1\times 1) + (4\times 1) + (5\times 4) + (6\times 9) + (7\times 12) + (8\times 15) + (9\times 10) + (10\times 10) = 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473$.
4. Mean = $\frac{473}{62} \approx 7.63$.
5. Median = average of the 31st and 32nd values. The 31st and 32nd values both fall in 8 trials. Thus, Median = 8.

**Answer:** Minimum = 1, Maximum = 10, Mean = 7.63, Median = 8

> Common mistake: Multiplying trial numbers with frequencies incorrectly or making addition errors in total sum.

## Figure it Out

### Question 1

*3 marks · Short answer*

The average number of customers visiting a shop and the average number of customers actually purchasing items over different days of the week is shown in the table below. Visualise this data on a line graph.

**Part (i)**

1. Draw horizontal axis for the days of the week from Monday to Sunday and vertical axis for the number of customers.
2. Plot the points for visiting customers (16, 19, 10, 14, 20, 22, 35) and connect them with a line.
3. Plot the points for purchasing customers (10, 8, 7, 11, 12, 16, 26) and connect them with a second line using a different marker or colour.

Answer (i): Line graph drawn with days on the x-axis and number of customers on the y-axis showing visiting and purchasing trends.

**Answer:** A line graph with days of the week on the horizontal axis and number of customers on the vertical axis, showing two distinct lines for visiting and purchasing customers.

> Common mistake: Failing to use different markers or labels to distinguish between visiting and purchasing customer lines.

### Question 2

*5 marks · Case-based*

The average number of days of rainfall in each month for a few cities is shown in the table:
(i) What could be the possible method to compile this data?
(ii) Mark the data for Mangaluru, Port Blair, and Rameswaram in the line graph shown below. You can round off the values to the nearest integer.
(iii) Based on the line for New Delhi in the graph fill the data in the table.
(iv) Which city among these receives the most number of days of rainfall per year? Which city gets the least number of days of rainfall per year?
(v) Looking at the table, when is the rainy season in New Delhi and Rameswaram?

**Part (i)**

1. Rainfall data is collected across a few years for each city.
2. The total number of days of rainfall in a specific month across those years is averaged to get the monthly average.

Answer (i): Compiled by taking the average number of rainy days in each month over several years.

**Part (ii)**

1. Round off the monthly values for Mangaluru, Port Blair, and Rameswaram to the nearest integers.
2. Plot the rounded values against each month on the given line graph template.

Answer (ii): Points plotted and connected by line segments for each city on the graph.

**Part (iii)**

1. Read the corresponding number of rainy days for New Delhi from the given line graph for each month.
2. Record these rounded values into the respective cells of the table.

Answer (iii): Table filled with corresponding monthly values from the New Delhi line graph.

**Part (iv)**

1. Sum the average monthly rainy days for each city over the year to find the annual total.
2. Mangaluru receives the maximum total days of rainfall per year, while New Delhi receives the least.

Answer (iv): Mangaluru receives the most days of rainfall per year, and New Delhi gets the least.

**Part (v)**

1. Examine the months with the highest number of rainy days in the table for each city.
2. The rainy season in New Delhi is during the monsoon months (July-August) and in Rameswaram during the north-east monsoon (October-December).

Answer (v): New Delhi's rainy season is around July-August, while Rameswaram's is from October to December.

**Answer:** Data compiled by averaging historical monthly rainfall across years, with Mangaluru receiving the highest and New Delhi the lowest annual rainy days.

> Common mistake: Confusing south-west monsoon months with north-east monsoon months while identifying seasonal rainfall.

### Question 2

*5 marks · Case-based*

The following line graph shows the number of births in every month in India over a time period:
(i) What are your observations?
(ii) What was the approximate number of births in July 2017?
(iii) What time period does the graph capture?
(iv) Compare the number of births in the month of January in the years 2018, 2019, and 2020.
(v) Estimate the number of births in the year 2019.

**Part (i)**

1. Observe the periodic peaks and troughs recurring every year in the line graph.
2. The number of births shows consistent seasonal patterns with highs and lows during specific months.

Answer (i): The graph shows regular periodic fluctuations in the number of monthly births over time.

**Part (ii)**

1. Locate July 2017 on the horizontal time axis.
2. Trace vertically to the line graph to read the corresponding value on the vertical axis.

Answer (ii): Approximately 1.5M births.

**Part (iii)**

1. Check the starting and ending points on the horizontal time axis of the graph.
2. The graph starts at July 2017 and ends at January 2020.

Answer (iii): From July 2017 to January 2020.

**Part (iv)**

1. Find the data points corresponding to January 2018, January 2019, and January 2020.
2. Compare their heights on the vertical axis representing the number of births.

Answer (iv): The number of births in January was highest in 2018, followed by 2019, and slightly lower in 2020.

**Part (v)**

1. Calculate the approximate monthly average of births for the year 2019 from the graph values.
2. Multiply the average monthly births by 12 months, yielding a total of approximately 19M births.

Answer (v): Approximately 19 million births.

**Answer:** The graph captures monthly live births in India from July 2017 to January 2020, showing seasonal fluctuations.

> Common mistake: Reading the wrong month or misinterpreting the scale units on the vertical axis.

## Figure it Out

### Question 1

*3 marks · Short answer*

Mean Grids:
(i) Fill the grid with 9 distinct numbers such that the average along each row, column, and diagonal is 10.
(ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions?

**Part (i)**

1. We need a $3 \times 3$ grid where the sum of each row, column, and diagonal is $3 \times 10 = 30$.
2. One possible grid with 9 distinct numbers is: row 1: 5, 16, 9; row 2: 12, 10, 8; row 3: 11, 4, 15.
3. Each row, column, and diagonal sums to 30, giving an average of 10.

Answer (i): Row 1: 5, 16, 9; Row 2: 12, 10, 8; Row 3: 11, 4, 15

**Part (ii)**

1. Yes, we can change a few numbers while maintaining the average condition.
2. By altering some numbers symmetrically or using properties of magic squares, we can keep the row, column, and diagonal sums constant at 30.

Answer (ii): Yes, by adjusting numbers symmetrically.

**Answer:** Filled grids satisfying the given average conditions.

> Common mistake: Not ensuring all 9 numbers are distinct.

### Question 2

*3 marks · Short answer*

Give two examples of data that satisfy each of the following conditions:
(i) 3 numbers whose mean is 8.
(ii) 4 numbers whose median is 15.5.
(iii) 5 numbers whose mean is 13.6.
(iv) 6 numbers whose mean = median.
(v) 6 numbers whose mean > median.

**Part (i)**

1. The sum of 3 numbers whose mean is 8 must be $8 \times 3 = 24$.
2. Example 1: 6, 8, 10.
3. Example 2: 5, 9, 10.

Answer (i): 6, 8, 10 and 5, 9, 10

**Part (ii)**

1. For 4 numbers, the median is the average of the 2nd and 3rd numbers in sorted order, which must be 15.5.
2. Example 1: 10, 15, 16, 20.
3. Example 2: 12, 15, 16, 25.

Answer (ii): 10, 15, 16, 20 and 12, 15, 16, 25

**Part (iii)**

1. The sum of 5 numbers whose mean is 13.6 must be $13.6 \times 5 = 68$.
2. Example 1: 10, 12, 13, 15, 18.
3. Example 2: 11, 12, 13, 14, 18.

Answer (iii): 10, 12, 13, 15, 18 and 11, 12, 13, 14, 18

**Part (iv)**

1. We need 6 numbers where the mean equals the median.
2. Example 1: 1, 2, 3, 3, 4, 5 (mean = 3, median = 3).
3. Example 2: 10, 20, 30, 30, 40, 50 (mean = 30, median = 30).

Answer (iv): 1, 2, 3, 3, 4, 5 and 10, 20, 30, 30, 40, 50

**Part (v)**

1. We need 6 numbers where the mean is greater than the median.
2. Example 1: 1, 2, 3, 4, 5, 100 (mean = 19.17, median = 3.5).
3. Example 2: 2, 4, 6, 8, 10, 50 (mean = 13.33, median = 7).

Answer (v): 1, 2, 3, 4, 5, 100 and 2, 4, 6, 8, 10, 50

**Answer:** Two examples for each given condition.

> Common mistake: Failing to check the arithmetic for mean or median.

### Question 3

*3 marks · Short answer*

Fill in the blanks such that the median of the collection is 13: 5, 21, 14, _____, ______, ______. How many possibilities exist if only counting numbers are allowed?

**Solution**

1. The given collection has 6 values: 5, 21, 14, and three blanks.
2. When sorted in order, the median is 13. Since there are 6 values, the median is the average of the 3rd and 4th values.
3. Let the sorted collection be 5, 14, a, b, c, 21 with median 13, so $\frac{a + b}{2} = 13$, meaning $a + b = 26$.
4. Counting number possibilities for $a$ and $b$ between 14 and 21 can be found by listing pairs summing to 26.

**Answer:** Multiple counting number combinations exist depending on the chosen blanks.

> Common mistake: Ignoring the requirement that numbers must be sorted to find the median.

### Question 4

*3 marks · Short answer*

Fill in the blanks such that the mean of the collection is 6.5: 3, 11, ____, _____, 15, 6. How many possibilities exist if only counting numbers are allowed?

**Solution**

1. The collection has 6 numbers: 3, 11, blank, blank, 15, 6, with a mean of 6.5.
2. The total sum of the 6 numbers must be $6.5 \times 6 = 39$.
3. The sum of the known numbers is $3 + 11 + 15 + 6 = 35$.
4. The sum of the two missing numbers is $39 - 35 = 4$.
5. Since only counting numbers are allowed, the two missing numbers must sum to 4, giving pairs like (1, 3) or (2, 2) or (0, 4) if 0 is included.

**Answer:** Counting number pairs that sum to 4.

> Common mistake: Multiplying by the wrong number of terms.

### Question 5

*3 marks · Short answer*

Check whether each of the statements below is true. Justify your reasoning. Use algebra, if necessary, to justify.
(i) The average of two even numbers is even.
(ii) The average of any two multiples of 5 will be a multiple of 5.
(iii) The average of any 5 multiples of 5 will also be a multiple of 5.

**Part (i)**

1. Let the two even numbers be $2m$ and $2n$.
2. Their average is $\frac{2m + 2n}{2} = m + n$.
3. The sum $m + n$ is not necessarily even (e.g., $1 + 2 = 3$), so the statement is false.

Answer (i): False, the average of two even numbers can be odd.

**Part (ii)**

1. Let the two multiples of 5 be $5a$ and $5b$.
2. Their average is $\frac{5a + 5b}{2} = \frac{5(a + b)}{2}$.
3. This is not always a multiple of 5 unless $(a + b)$ is even, so the statement is false.

Answer (ii): False, division by 2 may not yield a multiple of 5.

**Part (iii)**

1. Let the 5 multiples of 5 be $5a, 5b, 5c, 5d, 5e$.
2. Their average is $\frac{5(a + b + c + d + e)}{5} = a + b + c + d + e$.
3. The sum of integers is an integer, which is not necessarily a multiple of 5, so the statement is false.

Answer (iii): False, the average is just the sum of the multipliers and need not be a multiple of 5.

**Answer:** Checked and justified statements using algebra.

> Common mistake: Assuming division by 2 always preserves the factor of 5.

### Question 6

*3 marks · Short answer*

There were 2 new admissions to Sudhakar’s class just a couple of days after the class average height was found to be 150.2 cm.
(i) Which of the following statements are correct? Why?
(a) The average height of the class will increase as there are 2 new values.
(b) The average height of the class will remain the same.
(c) The heights of the new students have to be measured to find out the new average height.
(d) The heights of everyone in the class has to be measured again to calculate the new average height.
(ii) The heights of the two new joinees are 149 cm and 152 cm. Which of the following statements about the class’ average height are correct? Why?
(a) The average will remain the same.
(b) The average will increase.
(c) The average will decrease.
(d) The information is not sufficient to make a claim about the average.
(iii) Which of the following statements about the new class average height are correct? Why?
(a) The median will remain the same.
(b) The median will increase.
(c) The median will decrease.
(d) The information is not sufficient to make a claim about median.

**Part (i)**

1. When new values are added, we do not need to measure everyone's height again.
2. We only need the heights of the new students to calculate the new average height using the total sum.

Answer (i): (c) The heights of the new students have to be measured to find out the new average height.

**Part (ii)**

1. The previous average height was 150.2 cm.
2. The new joinees have heights 149 cm and 152 cm, whose average is $\frac{149 + 152}{2} = 150.5$ cm.
3. Since the new values are very close to the previous average, or more precisely, their sum relative to the number of students slightly changes the mean, let us check: total height increases by $149 + 152 = 301$ for 2 students (average 150.5), which is slightly above 150.2, so the average will increase.

Answer (ii): (b) The average will increase.

**Part (iii)**

1. Adding two values to a large dataset of 24 students changes the total count to 26.
2. The median position shifts from the average of 12th and 13th values to the 13th and 14th values.
3. Without the complete individual data sorted in order, we cannot determine the exact change in median.

Answer (iii): (d) The information is not sufficient to make a claim about median.

**Answer:** Selected correct statements with reasoning.

> Common mistake: Assuming adding values always increases the median without knowing their relative sizes.

### Question 7

*3 marks · Short answer*

Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question.

**Solution**

1. Given the dot plot from 14 to 23, count the frequency of each value: 14 has 1, 15 has 2, 16 has 2, 17 has 4, 18 has 3, 19 has 3, 20 has 2, 21 has 1, 22 has 0, and 23 has 1.
2. Calculate the total sum of all values: $14(1) + 15(2) + 16(2) + 17(4) + 18(3) + 19(3) + 20(2) + 21(1) + 22(0) + 23(1) = 14 + 30 + 32 + 68 + 54 + 57 + 40 + 21 + 0 + 23 = 339$.
3. Count the total number of data points (frequency): $1 + 2 + 2 + 4 + 3 + 3 + 2 + 1 + 0 + 1 = 19$.
4. Find the average by dividing the sum by the total number of values: $\frac{339}{19} \approx 17.84$, which is not equal to 17.

**Answer:** No, 17 is not the average of the data; the calculated average is approximately 17.84.

> Common mistake: Confusing the median or the value with the highest frequency (mode) with the arithmetic mean.

### Question 8

*3 marks · Short answer*

The weights of people in a group were measured every month. The average weight for the previous month was 65.3 kg and the median weight was 67 kg. The data for this month showed that one person has lost 2 kg and two have gained 1 kg. What can we say about the change in mean weight and median weight this month?

**Solution**

1. Given the previous month's average weight was 65.3 kg and median weight was 67 kg.
2. This month, one person lost 2 kg (change of $-2$ kg) and two people gained 1 kg each (change of $+1 + 1 = +2$ kg).
3. The total change in the sum of weights is $-2 + 2 = 0$ kg, so the mean weight remains unchanged at 65.3 kg.
4. Since individual weight changes are small and local to specific individuals, the middle value when sorted remains 67 kg, so the median weight also remains unchanged at 67 kg.

**Answer:** Both the mean weight and the median weight remain unchanged at 65.3 kg and 67 kg respectively.

> Common mistake: Assuming that any change in individual data points always changes both the mean and the median.

### Question 9

*3 marks · Short answer*

The following table shows the retail price (in ₹) of iodised salt in the month of January in a few states over 10 years. For your calculations and plotting you may round off values to the nearest counting number.
(i) Choose data from any 3 states you find interesting and present it through a line graph using an appropriate scale.
(ii) What do you find interesting in this data? Share your observations.
(iii) Compare the price variation in Gujarat and Uttar Pradesh.
(iv) In which state has the price increased the most from 2016 to 2025?
(v) What are you curious to explore further?

**Part (i)**

1. Select three states such as Gujarat, Assam, and Uttar Pradesh.
2. Draw a line graph with years (2016 to 2025) on the horizontal axis and price in (in Rs) on the vertical axis, using the rounded-off values from the table.

Answer (i): Line graph drawn for Gujarat, Assam, and Uttar Pradesh.

**Part (ii)**

1. Examine the trends of prices across the years for different states.
2. Observe that some states show a steady increase, while others fluctuate.

Answer (ii): Prices of iodised salt generally increased across all states from 2016 to 2025.

**Part (iii)**

1. Compare the rounded prices of Gujarat (17 in 2016 to 19 in 2025) and Uttar Pradesh (16 in 2016 to 25 in 2025).
2. Note that Uttar Pradesh shows a larger and more continuous upward trend compared to Gujarat.

Answer (iii): Uttar Pradesh shows a steeper and higher price increase compared to Gujarat.

**Part (iv)**

1. Check the price difference between 2016 and 2025 for each state using the rounded values.
2. Mizoram increased from 20 to 30, and Uttar Pradesh increased from 16 to 25, while Assam went from 6 to 12 (doubled).

Answer (iv): Assam and Mizoram show significant increases, with Assam doubling its price.

**Part (v)**

1. Consider factors affecting regional price variations of essential commodities.
2. Formulate a question regarding transport costs, inflation, or government subsidies.

Answer (v): Curious to explore why price variations differ significantly among neighbouring states.

**Answer:** Data from Gujarat, Assam, and Uttar Pradesh analysed.

> Common mistake: Forgetting to round off values to the nearest counting number as instructed.

### Question 10

*3 marks · Short answer*

Referring to the graph below, which of the following statements are valid? Why?
(i) In 1983, the majority in rural areas used kerosene as a primary lighting source while the majority in urban areas used electricity.
(ii) The use of kerosene as a primary lighting source has decreased over time in both rural and urban areas.
(iii) In the year 2000, 10% of the urban households used electricity as a primary lighting source.
(iv) In 2023, there were no power cuts.

**Part (i)**

1. Look at the graph for 1983 in rural areas where kerosene line is above 80% and electricity is below 20%.
2. Look at urban areas in 1983 where electricity is around 70% and kerosene is lower.

Answer (i): True, because the graph shows kerosene usage above 80% in rural areas and electricity above 60% in urban areas in 1983.

**Part (ii)**

1. Trace the kerosene line downwards from 1983 to 2023 in both rural and urban graphs.

Answer (ii): True, the curves for kerosene show a continuous downward trend towards 0% by 2023 in both graphs.

**Part (iii)**

1. Locate the year 2000 on the urban graph and read the percentage for electricity.

Answer (iii): False, urban electricity usage in 2000 was much higher than 10% (around 80%).

**Part (iv)**

1. Check if power cuts are represented anywhere in the energy source graph.

Answer (iv): False, the graph only shows primary sources of lighting, not power cuts.

**Answer:** Statements (i) and (ii) are valid.

> Common mistake: Misreading percentages or mixing up rural and urban graphs.

### Question 11

*3 marks · Short answer*

Answer the following questions based on the line graph.
(i) How long do children aged 10 in urban areas spend each day on hobbies and games?
(ii) At what age is the average time spent daily on hobbies and games by rural kids 1.5 hours?
(iii) Are the following statements correct?
(a) The average time spent daily on hobbies and games by kids aged 15 is twice that of kids aged 10.
(b) All rural kids aged 15 spend at least 1 hour on hobbies and games everyday.

**Part (i)**

1. Locate age 10 on the horizontal axis of the line graph in the textbook (Fig. on page 130).
2. Read the corresponding value on the vertical axis for the urban line.

Answer (i): Children aged 10 in urban areas spend about 2 hours each day on hobbies and games.

**Part (ii)**

1. Look at the rural line (orange/brown line) on the graph.
2. Find the age where the vertical time coordinate corresponds to 1.5 hours.

Answer (ii): (d) 14 years

**Part (iii)(a)**

1. Check the values for urban kids aged 15 and aged 10 from the graph.
2. At age 15 it is about 1 hour and at age 10 it is about 2 hours, so it is half, not twice.

Answer (iii)(a): False. The time spent by kids aged 15 is half, not twice, of those aged 10.

**Part (iii)(b)**

1. Observe the rural line at age 15 on the graph.
2. The value is below 1 hour (around 0.5 hours).

Answer (iii)(b): False. Rural kids aged 15 spend less than 1 hour on hobbies and games.

**Answer:** Interpreted the line graph on daily time spent on hobbies and games.

> Common mistake: Confusing the urban and rural lines on the graph.

## 12. Individual project: Make your own activity strip for different days of the week.

### Question 12

*5 marks · Long answer*

(i) Do you eat and sleep at regular times every day? Typically how long do you spend outdoors?
(ii) Calculate the average time spent per activity. Represent this average day using a strip.
(iii) Similarly, track the activities of any adult at home. Compare your data with theirs.

**Part (i)**

1. Observe daily habits for sleeping, eating, and going outdoors.
2. Note the timings and duration spent outdoors typically.

Answer (i): Timings for eating and sleeping vary slightly; typically 1 to 2 hours are spent outdoors daily.

**Part (ii)**

1. Record activity durations over several days and calculate the average time for each activity.
2. Draw a strip of 48 boxes representing 30-minute intervals and colour-code the average time spent on each activity.

Answer (ii): Average time per activity calculated and represented using a coloured strip of 48 boxes.

**Part (iii)**

1. Track the daily activities of an adult at home and record the durations.
2. Compare the adult's strip with your strip to observe differences in work, sleep, and leisure time.

Answer (iii): Adult activity strip compared, showing differences such as more time spent on work and less on leisure or studies.

**Answer:** Data recorded, averaged, and represented on strips as per individual observations.

> Common mistake: Not accounting for all 24 hours (48 intervals of 30 minutes) on the activity strip.

## 13. Small group project: Make a group of 3–4 members. Do at least one of the following:

### Question 13

*Activity*

(i) Track daily sleep time of all your family members for a week. Daily sleep time includes night sleep, naps, and any sleep during the day.
(a) Represent this on strips.
(b) Put together the data of all your group members. Calculate the average and median sleep time of children, adults, elderly.
(c) Share your findings and observations.
(ii) When do schools start and end? On a weekday, Manoj’s school starts at 9:30 am and ends at 4:30 pm, i.e., 7 hours which include class time and breaks. Collect information on the daily timings of different schools for Grade 8, including class time and break time (the schools can be anywhere in the country. You can ask your neighbours, relatives, parents and friends to find out). Analyse and present the data collected.

**Part (i)(a)**

1. Record daily sleep time of family members for a week including night sleep and naps.
2. Represent each day's sleep time on a strip of 48 boxes marked in 30-minute intervals.

Answer (i)(a): Daily sleep time represented on strips.

**Part (i)(b)**

1. Compile the sleep time data for children, adults, and elderly from group members.
2. Calculate the average and median sleep duration for each category.

Answer (i)(b): Average and median calculated for children, adults, and elderly.

**Part (i)(c)**

1. Compare the sleep durations across different age groups.
2. Share findings on how sleep patterns vary with age.

Answer (i)(c): Findings and observations shared.

**Part (ii)**

1. Collect daily starting and ending timings, class hours, and break times for Grade 8 from different schools.
2. Organise, analyse, and present the collected data clearly.

Answer (ii): School timing data collected, analysed, and presented.

**Answer:** Activities completed through data collection, representation on strips, calculation of average and median, and analysis of school timings.

## 14. The following graphs show the sunrise and sunset times across the year at 4 locations in India. Observe how the graphs are organised. Are you able to identify which lines indicate the sunrise and which indicate the sunset?

### Question 14

*3 marks · Short answer*

Answer the following questions based on the graphs:
(i) At which place does the sun rise the earliest in January? What is the approximate day length at this place in January?
(ii) Which place has the longest day length over the year?
(iii) Share your observations—what do you find interesting? What are you curious to find out?

**Part (i)**

1. Observe the sunrise lines for January at the bottom of the graphs in the textbook (Fig. 5.x).
2. Kibithu shows the earliest sunrise time in January, which is around 05:00 hours.
3. The approximate day length in January at Kibithu is obtained by finding the time difference between sunset and sunrise, which is about 10.5 hours.

Answer (i): Sun rises the earliest at Kibithu (approx. 05:00 hours), and the approximate day length is 10.5 hours.

**Part (ii)**

1. Examine the gap between the sunset and sunrise lines across the year for all locations.
2. A wider vertical gap between the sunset and sunrise lines indicates a longer day length.
3. Kibithu (or the location with the highest latitude variation) experiences the longest day length during summer months.

Answer (ii): Kibithu has the longest day length over the year.

**Part (iii)**

1. Compare the trends of sunrise and sunset curves across different geographical locations in India.
2. Notice how sunrise and sunset times shift earlier or later depending on the easternmost or westernmost locations.
3. Observe that day lengths vary significantly across seasons in northern and eastern locations compared to southern ones.

Answer (iii): Eastern locations like Kibithu witness much earlier sunrises compared to western locations like Ghuar Moti.

**Answer:** Refer to the parts for the answers.

> Common mistake: Confusing sunrise and sunset lines or miscalculating the time difference for day length.

## 15. We all know the typical sunrise and sunset timings. Do you know when the moon rises and sets? Does it follow a regular pattern like the sun? Let’s find out. The following graph shows the moonrise and moonset time over a month:

### Question 15

*3 marks · Short answer*

(i) Find out on what dates amavasya (new moon) and purnima (full moon) were in this month.
(ii) What do you notice? What do you wonder?

**Part (i)**

1. Observe the graph of moonrise and moonset timings over a month.
2. Amavasya (new moon) occurs when the moon rises with the sun and is invisible, which corresponds to the point where the moonrise and moonset curves intersect or are closest around day 21.
3. Purnima (full moon) occurs when the moon rises around sunset and sets around sunrise, which corresponds to day 7 where the moonrise time is near 18:00 (6 pm).

Answer (i): Amavasya: Day 21, Purnima: Day 7

**Part (ii)**

1. Notice that both moonrise and moonset times get progressively later each day by about 50 minutes, shifting across the 24-hour cycle.
2. Wonder why the moonrise pattern is not symmetric or why the gap between moonrise and moonset changes.

Answer (ii): The moonrise and moonset times shift later every day by about 50 minutes.

**Answer:** Amavasya is on Day 21 and Purnima is on Day 7.

> Common mistake: Confusing Amavasya and Purnima dates by misreading the moonrise time axis.

## Frequently asked questions

### How many total questions are covered in NCERT Solutions for Class 8 Maths Chapter 12 Tales by Dots and Lines?

The chapter includes multiple sections with varying question counts, including 11 questions in the first Figure it Out set, 3 in the second, and another 11 in the third, alongside specific project and graph interpretation tasks. SwaVid's free PDF and step-by-step solutions are on this page only to help you practice all of them easily.

### What core topics are covered in the Chapter 12 exercises of the new NCERT book?

The questions cover important statistical concepts like the behavior of median and mean under data changes, finding missing data values, and interpreting line graphs for rainfall, birth trends, and sunrise or sunset timings. SwaVid's free PDF and step-by-step solutions are on this page only to guide you through these topics for the 2026-27 session.

### Which question types are the most challenging in this chapter and how should we approach them?

Case-based and algebraic justification questions regarding the effect of new values on mean and median are often considered the hardest. You should approach them by clearly writing out the properties of averages step by step, and SwaVid's free PDF and step-by-step solutions are on this page only to show you the correct methodology.

### How can I write answers to get full marks in Class 8 Maths Chapter 12 questions?

To secure full marks, ensure you clearly state the formulas used for mean and median, show all intermediate calculations for dot plots and frequency tables, and label graph-based answers properly. SwaVid's free PDF and step-by-step solutions are on this page only to provide model answers aligned with the NCF 2023 guidelines.

### Is a free PDF available for Class 8 Maths Chapter 12 Tales by Dots and Lines?

Yes, complete solutions mapped to the new NCERT book for the 2026-27 session are readily accessible for students. SwaVid's free PDF and step-by-step solutions are on this page only to help you prepare thoroughly for your exams.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

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