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NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 5 Tales by Dots and Lines
In the fifth chapter of "Tales by Dots and Line," you will learn a new way of looking at data with the concept of a 'balancing act' of statistics. In this chapter, different ways to calculate data will be illustrated through several methods such as; using means as physical 'centers of gravity' for data points, where the 'Mediana' represents an 'anchor' to which all data points can be compared against each other.
Using the analytical tools provided by NCERT Solutions Class 8 is a way of setting the user up for success when it comes to using data (as described above). It enables them to become knowledgeable data consumers with an ability to identify trends and predict how a singular value affects the overall average.
1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 5 Solutions
The chapter focuses on the balancing act of statistics. These NCERT Solutions features detailed explanations of mean-balancing, median shift related questions form the chapter in a clear step by step format. Download them from the link below.
2.0Key Concepts covered in Class 8 Maths : Tales by Dots and Lines
This chapter builds a conceptual bridge between numerical calculation and visual data interpretation.
- The Balancing Act (Mean): Visualizing the arithmetic mean as the "balance point" where the sum of distances to the left equals the sum of distances to the right.
- Mean vs. Median Behavior: Observing how adding extreme values (outliers) causes the mean to increase or decrease significantly, while the median remains more stable.
- Data Representation with Dot Plots: Using dots to represent frequency and distribution, making it easier to see clusters and gaps in data.
- Line Graphs: Learning to plot data points over intervals of time to visualize change, such as temperature fluctuations or growth rates.
- Data Probing: Learning to ask critical questions about data, such as "What do you notice?" and "What do you wonder?" to drive further investigation.
3.0NCERT Solutions for Class 8 Maths Chapter 5 : All Exercises
Exercise 5.1 This section explores the mean as a physical balance point. Students use dot plots to visualize data distribution and calculate the arithmetic mean by finding the center of gravity between various data points.
Exercise 5.2 Students investigate how outliers affect the various measures used in statistics. The students will complete questions that either add or subtract extreme values from their data, observing how both the mean and median are shifted as a result, and demonstrating the different levels of stability associated with the various measures of central tendency.
Exercise 5.3 The final exercise focuses on line graphs and temporal trends. Students create a line graph of a series of data points related to some type of time interval to graphically display change (example: variations in temperature, or growth trends) and provide a narrative of what that graph may indicate.
4.0Detailed Class 8 Maths Chapter 5 Tales by Dots and Lines - NCERT Solutions
Figure it out-01
1. 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.
Sol. (i) The mean of the first 50 natural numbers. The first 50 natural numbers 1, 2, 3, ......, 50 The sum of natural number is So, for The sum Mean Observations: The mean of the first n natural numbers is always
For (ii) Mean of the first 50 odd number.
The first 50 odd numbers 1, 3, 5, 7, 9, ......, 99 Sum of the first n odd numbers So for the sum Mean Observations: The mean of the first n odd numbers is always n . For 50 -odd numbers, the mean is 50. (iii) Mean of the first 50 multiples of 4 .
The first 50 multiples of , 16, ......, 200 Sum of multiples of 4 Mean Observation: The mean of the first n multiples of 4 is 4 times the mean of the first n natural numbers. Mean
2. 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.)
Sol. From the dot plot, the data values4, 7, 8, , xNumber of observations = 11Mean = 9Sum of data values Total sum Missing number
Sol. (i) The teacher does not need to remeasure all the heights. Since the shoes add 1 cm to every student's height, the average height with shoes is 1 cm more than the actual average height. So, the correct average height is 1 cm less than the measured average height. (ii) Correct average height = Option (d) is correct
4. The three dot plots below show the length, in minutes, of songs of different albums. Which of these has a mean of5.57 minutes? Explain how you arrived at the answer.
B
C
Sol. We can determine it by examining the dot plots. In plot A , most song lengths are between 5 and 6.5 minutes, so the mean is likely around 5.57 minutes. In dot plots B and C , all song lengths are below 5.57 minutes, so their means cannot exceed 5.57. Therefore, is the one with a mean of 5.57 minutes.
Check:
For Album A
Data values Mean minutes
For Album B
Data values Mean minutes
For album C
Data values Mean minutes Album (A) has a mean of 5.57 minutes
5. Find the median of , 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?
Sol. The given data set is , 40, 41, 41, 48, 51, 55, 70, 84, 91, 92
The number of observations ( n ) (even) Median = average of and term average of and terms (i) To keep the median 41 with an odd number of values ( ), the new value must be placed at the median itself in the order list for making the middle value the 9th value. So, the value could be 41. (ii) With (even) the median is the average of the 9th and 10th values. To keep the median 41 , we need to add two numbers whose sum is 82 . For this, one value should be less than 41 and the other greater than 41.For example, the two values could be 40 and 42 . (iii) With (odd) the median is the value that is 41 . So, we can remove another 41 from the ordered list.
6. 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.
Sol. (i) Sometimes true: Removing a value less than the median can decrease the median. But if the data set has an even number of elements and the removed value is below the lower of the two middle values, the median may stay the same. (ii) Always true: The mean is the sum of all values divided by the number of values. Adding a value less than the mean decreases the total sum less than proportionally to the increase in the number of values, thus decreasing the mean. (iii) Sometimes true: Including any 4 values could shift the median depending on whether those values are above or below the median and on the original number of observations. (iv) Never true: Including values less than the median will either keep the median the same or decrease it, but it will never increase it.
7. The mean of the numbers , is 10.375 . Find the value of .
Sol. Numbers Total numbers Mean Total Sum Sum of numbers So,
8. The mean of a set of data with 15 value is 134. Find the sum of the data.
Sol. Mean Total values Sum of data
9. Consider the data: , 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
Sol. Arranging the data (excluding p) Total observations = 11 (odd) Median term term So, p must be to keep the value 29 in the middle.
So possible values of p are (iii) 40, (iv) 100, (v) 29, (vi) 47, and (vii) 30.
10. 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.
Sol. From the dot plot Data
(i) Sum
Total numbers of students Average times (ii) Since total number of students Since number of observations is even, we use the average of the two middle numbers
Now, Median =
Thus, median is the average of the and values
Counting the dots from left to right, both the and dots fall in the "4" column.
The median is 4. (iii)(a)Invalid: Three students rode 0 times. (b)Valid: The heavy clustering in the middle shows almost everyone rode a few times. (c) Valid: Three are only 7 days in a week. Some students recorded 8 and 10 rides, meaning they had to ride multiple times in a single day to reach that number. (d)Not valid: More than 5 students have used their cycles more than once on some days. (e) If everyone cycles 1 more time next week, the entire data set simply shifts up by 1 .
Thus, New average old average +1 (approx.) and New median old median
11. 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.
Sol. Minimum: the lowest number of throws to hit the bullseye was 1 .
Maximum: the height number of throws was 10 .
Here, Total number of students
and Number of trails
Finding mean
Mean (approx.)
Since total number of students Since number of observations is even, We use the average of two middle numbers
Now, Median = Thus, median is the average of the and values.
If we keep a running total of the frequencies from left to right, the and values both fall inside the column for 8 throws.
Thus, median is 8
Figure it out-02
1. The average number of customers visiting a shop and the average number of customers actually purchasing item over different days of the week is shown in the table below. Visualise this data on a line graph.
Sol. While drawing a line graph, we follow these steps. First, we decide what to put in x -axis and -axis. We always use time in x-axis - so Day of the week is in x -axis.
So, Number of Customers are in y -axis. Then, we find out our Scale. Since numbers go from 7 to 35. We choose Scale 1-unit length = 4 units. Lastly, we decide our legend - which color to denote.
We choose Blue for Visiting, Red for Purchasing.
(i) What could be the possible method to compile this data? (ii) Mark the data for Mangaluru, Port Blair and Rameshwaram in the line graph shown below. You can round off the values to the nearest integer.
Sol. (i) Just like temperature, meteorologists collect daily rainfall data over several years.
They then add up all the rain that fell in a specific month (say, all the Junes over a 10-year period) and average it out to find the "typical" amount of rain for June. To represent it a line graph is used. (ii) Since the graph is already given, we just plot points and join them by straight lines.
Thus, our graph looks like
(iv) Mangaluru receives the most number of days of rainfall per year, and Rameswaram receives the least. (v) For New Delhi, the numbers peak heavily in July and August (around 10 days of rain each), so that is the rainy season. For Rameswaram, the numbers peak late in the year: October (8.1), November (10.4), and December (7.8), making its rainy season late autumn/early winter.
3. The following line graph shows the number of births in every month in India over a time period.
Sol. (i) The graph shows a distinct, repeating wave pattern. Birth dip low at the beginning of the year and hit a peak in the later months of the year, every single year. (ii)
The dot is between 1.5 M and 2 M . Thus, we can say it is approximately 1.75 Million. (iii) The graph starts slightly before July 2017 and ends slightly after January 2020. So, it roughly covers mid-2017 to early 2020 (iv)
We notice that
- Dot of Jan 2019 is higher than Dot of Jan 2018
- Dot of Jan 2020 is same as than Dot of Jan 2019 (v) We can calculate the approximate values for each month For each month, value is
- Jan
- Feb
- Mar
- Apr
- May
- Jun
- July
- Aug 2019-2M
- Sep 2019-1.95 M
- Oct 2019-2 M
- Nov 2019-1.9 M
- Dec
Thus, Number of births in 2019
Figure it out-03
1. Mean Grids: (i) Fill the grid with 9 distinct numbers such that the average along each row, column and diagonal is 10.
Here is one way to fill the grid
- Row 1: 13, 6, 11
- Row 2: 8, 10, 12
- Row 3: 9, 14, 7
(ii) Yes, but you would have to change all the numbers to a completely new sequence that is still perfectly balanced around 10.
For example: We could use even numbers: . As long as the numbers jump by a consistent amount and center on 10, you can build a new magic square to We use only even numbers Our new 9 numbers will be: , 10, 12, 14, 16, 18
And, 10 our target average is at the center So, our grid will be
- 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.
Sol. (i) 3 numbers whose mean is 8 are (1) 6,8 , and 10 (2) 7,8, and 9 . (ii) 4 numbers whose median is 15.5 are (1) , and 20 (2) , and 18 . (iii) 5 numbers whose mean is 13.6 are (1) , and 18 . (2) , and 18. (iv) 6 numbers whose Mean = Median (1) (Mean Median ) (2) (Mean Median ) (v) 6 numbers whose Mean > Median (1) (Mean=7.5, Median=3.5) (2) (Mean = 6.67, Median = 4.5)
3. Fill in the blanks such that the median of the collection is , . How many possibilities exist if only counting numbers are allowed? Sol. Our data is Number of observations Since number of observations is even Median is the average of middle two terms Since we are given 14 as a term, but median is lesser than 14
So, 14 can be our term Thus, Median Putting term , Median term +14 term +14 term term term Given terms were 5, 21, 14 and term = 12
Let term Arranging data in ascending order, our data becomes
5, , 12, 14, 21, Thus, term can be any number from 5 to 12 i.e. term can be and term can be any number greater than 21 . But there are infinitely many terms greater than 21 , like , 99000000.
So, we can say that infinitely many possibilities exist.
4. Fill in the blanks such that the mean of the collection is , , , 15, 6 . How many possibilities exist if only counting numbers are allowed?
Sol. The collection is Mean , let the other numbers be x and y . For 6 numbers With counting numbers, the pairs ( ) satisfying are ( 1,3 ) and ( 2,2 ).So, the number of possibilities is 2 .
5. 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 .
Sol. (i) False Let the two even numbers be 2 a and 2b. Average
Since 2 a and 2 b is even but we cannot say about parity(odd/even) of ' a ' and ' b '
So, can be odd or even. We take an example Let the two even numbers be 2 & 4 Average Since the average is odd, the given statement is false
(ii) False
Let the two multiples of 5 be 5 m and 5n.
Average can be any number other than integer We take an example Let the two multiples of 5 be 5 & 10 . Average Since 7.5 is not a multiple of 5, The given statement is false
(iii)True
Let the five multiples of 5 be 5a, 5b, 5c, 5d, 5e.
Average Since can be or cannot be multiple of 5 .
We take an example Let five multiples of 5 be , 20, & 25 Average Since 15 is not a multiple of 5, the given statement is true.
6. 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. (2) The average height of the class will remain the same. (3) The heights of the new students have to be measured to find out the new average height. (4) 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.
Sol. (i) To find the average, we need Sum of all heights.
Thus, we cannot find average without knowing the heights of the new students.
So, option (c) is correct (ii) Old mean
Let the number of students be x New mean Average of new two heights As the average of 2 new added heights is greater than old average, the class average increases.
Option (b) is correct. (iii) We do not know the individual heights of the original 24 students. Without seeing the list of numbers, we can't determine how adding two new numbers will shift the middle position.
So, option (d) is correct
7. Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question.
Sol. Method use: Count and Calculate
Sum
Total numbers Mean The average of the data is 17.72 or on a dot plot 17.
8. 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? Sol. Original average weight Let there be n people in the group. Total weight After one person loses 2 kg and two people each gain 1 kg , the net change in total weight is Total weight New mean
The new mean weight will remain unchanged.
The effect on the median depends on the original position of these three people, whose weight gains or lose.
So new median weight cannot be determined exactly without knowing more data.
9. 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?
Sol. (i) Let's choose Andaman & Nicobar, Assam and Gujarat While drawing a line graph, we follow these steps.
- First, we decide what to put in x-axis and -axis
- We always use time in x -axis so Year is in x -axis
- So, Price of Salt is in -axis
- Then, we find out our Scale
- Since numbers go from 6 to 20.99
- We choose Scale 1 -unit length unit
- Lastly, we decide our legend - which colour to denote
- We choose Blue for Andaman, Red for Assam and Green for Gujrat
(ii) Price generally increases over the years for most states. Mizoram shows the highest price jump and the highest overall prices in later years. Gujarat has relatively stable prices compared to other states. (iii) Gujarat's price is incredibly stable, staying right around 14 to 16 for an entire decade.
Uttar Pradesh shows a steep, relentless upward climb from 16 all the way to nearly 27 .
Uttar Pradesh has a large price increase compared to Gujarat. (iv) Calculating the price increase for each state.
Andaman and Nicobar Islands Assam Gujarat Mizoram Uttar Pradesh 8.66West Bengal 14.52The price in West Bengal increases the most (₹ 14.42). (v) Why is Mizoram's salt so expensive? Is it because it is a remote, landlocked state where transportation costs heavily impact retail prices?
What happened in West Bengal around 2021? The price was stable around 11-12 for years, and then suddenly doubled over the next four years. Was there a local supply chain issue or a change in state taxes?
10. Referring to the graph below, which of the following statements are valid? Why?
Sol. (i) The graph clearly shows Kerosene > Electricity for rural and Electricity > Kerosene for urban in 1983 Thus, given statement is valid (ii) This is true as Both yellow lines trend downward toward zero.
Thus, given statement is valid (iii) In the year 2000, the Urban electricity line is way up at roughly , not . Thus, given statement is invalid (iv) This graph tracks which energy source is used, not whether that source is reliable. It says nothing about power cuts. Thus, given statement is invalid
11. 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? (a) 8 years (b) 10 years (c) 12 years (d) 14 years (e) 18 years
Average Daily Time spent on Hobbies and Games
(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.
Sol. (i)
Looking where the blue line hits the 10 years mark
Urban children aged 10 spend little more than 2 hours. each day on hobbies and games. (ii)
The rural (red) line hits the 1.5 - hour mark at (d) 14 years. (iii) Statement (a)
At age 15, the average is roughly 1 hour. At age 10, it is 2 hours. 1 is half of 2, not twice!
Thus, given statement is incorrect. Statement (b) The graph shows the average is 1 hour. Averages are made of highs and lowssome kids probably spend 0 hours, while others spend 3.
Thus, given statement is incorrect.
12. Individual project: Make your own activity strip for different days of the week. (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.
Sol. Do it yourself
13. Small group project: Make a group of 3-4 members. Do at least one of the following: (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. (1) 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.
Sol. Do it yourself
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?
Answer the following questions based on the graphs:
Sol. The sun rises in the early morning and sets in the evening.
Therefore, the bottom lines (which hover between 04:00 and 08:00) represent sunrise, and the top lines (which hover between 16:00 and 20:00) represent sunset. (i)
- Earliest Sunrise: We need to look at the bottom line for the month of January across all four graphs. Kibithu has the lowest starting point in January, sitting exactly on the 06 : : 00 AM mark.
- Day Length: To find the day length, we subtract the sunrise time from the sunset time. In January at Kibithu, the sun rises at roughly 06:00 and the top line (sunset) is around 16 : .
- Calculation: From 06:00 to 16:30 is a gap of 10.5 hours (or 10 hours and 30 minutes). We are looking for the graph with the widest vertical gap between its top line and bottom line. That happens in Srinagar during the summer month of June. The sun rises very early (around 05:00) and sets very late (around 19:45), creating a massive daylight window of nearly 14.75 hours.
Observations: Kanyakumari: Its lines are almost completely flat! The length of the day barely changes from winter to summer. Kibithu vs. Ghuar Moti: Kibithu's entire day is shifted "early" (early to rise, early to set), while Ghuar Moti's day is shifted "late" (late to rise, late to set). Curiosities (The "Why"): As your teacher, I would hope this makes you wonder why Kanyakumari is so stable while Srinagar varies so much! It is entirely based on geography. Kanyakumari is very close to the equator, where day length stays a consistent 12 hours.
Srinagar is much further north, meaning it experiences extreme summer daylight and shorter winter daylight due to the tilt of the Earth.
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: (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?
- Moonrise - Moonset
Sol. (i) We need to understand the relationship between the Sun and Moon during these two phases:
- Amavasya (New Moon): The Moon rises and sets at approximately the same time as the Sun (around 06:00 and 18:00).
- Purnima (Full Moon): The Moon rises when the Sun sets (around 18:00) and sets when the Sun rises (around 06:00). Finding Purnima: Look at the yellow line (moonrise). When does it hit the 18:00 mark? It crosses 18:00 exactly on Day 15.
Finding Amavasya: Look at the yellow line again. When does it hit the 06:00 morning mark? It crosses 06:00 exactly on Day 30. (You can also verify this by checking the blue moonset line for Day 30, which hits exactly at 18:00!) (ii) Notice: The lines form perfect diagonal staircases that repeat themselves. The moon does not rise at the same time every day. Instead, the yellow line creeps higher and higher up the graph. Wonder: By looking at the grid, you can wonder (and mathematically prove!) exactly how much the moon is delayed each day. Because the line travels a full 24 hours (from 00:00 to 24:00) over roughly 29.5 days, it gets delayed by about 50 minutes every single day. That is why the lines slope upward.
5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 5
- Conceptual "Balance" Approach: Instead of just memorizing the formula for the mean, students learn to see it as a "fair share" or center of gravity, which aids in deeper intuitive understanding.
- Dynamic Data Analysis: The solutions demonstrate how inserting or removing specific numbers shifts the mean and median, preparing students for real-world statistical analysis.
- Time-Series Visualization: By mastering line graphs, students gain the ability to track and predict trends, a skill highly valued in science, economics, and geography.
- Encourages Critical Thinking: The "Tales" aspect of the chapter teaches students that data isn't just numbers—it’s a way to tell a story and answer complex questions about society and nature.
- Interactive Learning Foundation: With the inclusion of strategy games like "Hex" and visual dot plots, the solutions make learning statistics engaging and less intimidating.
- High Academic Performance: These concepts are central to the CBSE curriculum and are frequently tested in Olympiads, where data interpretation and logical reasoning are key components.