Class 8 Maths Ganita Prakash Chapter 12 Solutions (NCERT 2026–27) – Tales by Dots and Lines

These Class 8 Maths Ganita Prakash Chapter 12 solutions cover Tales by Dots and Lines from the new NCF-2023 textbook (2026–27). This is Chapter 5 of Ganita Prakash Part II (the 12th chapter of the Class 8 course). The worked solutions use original dot plots, line graphs and checked calculations. Values estimated from a printed graph are labelled approximate; they are not exact measurements or current statistics.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 12 (Part II Ch 5) Exercises: Figure it Out (3 sets), Math Talk / Try This Session: 2026–27
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Chapter 12 Overview

Chapter 12, Tales by Dots and Lines, revisits the arithmetic mean (average) and the median from a fresh, visual point of view. Using dot plots, you see that the mean is the “balancing point” of the data — the total distance of the values on its left equals the total distance on its right. The chapter then explores how the mean and median change when values are added, removed, increased by a fixed amount or multiplied, and how to compute the mean and median from a frequency table and from a spreadsheet. The second half, Visualising and Interpreting Data, introduces the line graph as the best tool for showing how a quantity changes over time, and shows how to read trends, peaks and lows from temperature, rainfall, sleep and other real data. The Class 8 Maths Ganita Prakash Chapter 12 solutions below work through every “Figure it Out” set step by step.

Key Concepts & Definitions

Arithmetic mean (average): the sum of all the values divided by the number of values.

Mean as a balance point: the total distance of the data values to the left of the mean equals the total distance to the right. There is exactly one such centre.

Median: the middle value of the data when it is sorted; for an even number of values it is the average of the two middle values.

Effect of changing data: inserting a value greater than the mean raises the mean; a value less than the mean lowers it; a value equal to the mean leaves it unchanged.

Adding / multiplying a constant: if every value increases by k, the mean increases by k; if every value is multiplied by k, the mean is multiplied by k.

Line graph: data points joined by line segments, used to visualise how a quantity changes over time; a steeper segment means a faster change.

Important Formulas (Chapter 12)

Mean = (sum of all the observations) ÷ (number of observations) = (x1 + x2 + … + xn) / n

Sum of observations = Mean × (number of observations)

Mean from a frequency table = Σ(value × frequency) / Σ(frequency)

Add a constant: new mean = old mean + k.  •  Multiply by a constant: new mean = k × old mean.

Median = middle value of sorted data (for n even, the average of the (n/2)th and (n/2 + 1)th values).

Figure it Out — Mean & Median (page 113–116)

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.

SOLUTION (i) Sum of first 50 natural numbers = (50 × 51)/2 = 1275. Mean = 1275 ÷ 50 = 25.5. (ii) Sum of first 50 odd numbers = 502 = 2500. Mean = 2500 ÷ 50 = 50. (iii) Sum of first 50 multiples of 4 = 4 × 1275 = 5100. Mean = 5100 ÷ 50 = 102. Observation: In each list the mean is exactly the average of the first and last term — (1 + 50)/2 = 25.5, (1 + 99)/2 = 50, (4 + 200)/2 = 102. The mean of a set of equally-spaced numbers is the midpoint of its smallest and largest values.

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).

SOLUTIONThe ten visible observations are 4, 7, 8, 8, 9, 9, 9, 9, 9, 11. Their sum is 83. Including the missing dot gives 11 observations.Required total = mean × count = 9 × 11 = 99. Missing value = 99 − 83 = 16. Add one dot at 16.Balance check about 9: left distance = 5 + 2 + 1 + 1 = 9; right distance = 2 + 7 = 9. The five dots already at 9 contribute zero distance.
Ten visible dots total 83. Adding the missing dot at 16 gives eleven values with mean 9.
Ten visible dots total 83. Adding the missing dot at 16 gives eleven values with mean 9.

3. 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? (a) 174.2 cm (b) 126.2 cm (c) 150.2 cm (d) 149.2 cm (e) 151.2 cm (f) None of the above (g) Insufficient information

SOLUTION (i) There is no need to remeasure everyone. Every height was 1 cm too much, so the mean is also exactly 1 cm too much. Just subtract 1 cm from the average. (ii) Correct average = 150.2 − 1 = 149.2 cm. The answer is (d) 149.2 cm.

4. 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.

SOLUTIONAlbum A is the match. Its seven values are 5, 5, 5.25, 5.5, 5.75, 6 and 6.5 minutes. Their sum is 39, so the mean is 39/7 ≈ 5.57 minutes. Dots halfway between half-minute grid marks represent quarter-minutes.Album B has mean 19.25/8 ≈ 2.41 minutes. Album C has mean 31.25/8 ≈ 3.91 minutes. These do not match 5.57.
Album A has lengths 5, 5, 5.25, 5.5, 5.75, 6 and 6.5 minutes, giving mean 39/7, approximately 5.57 minutes.
Album A has lengths 5, 5, 5.25, 5.5, 5.75, 6 and 6.5 minutes, giving mean 39/7, approximately 5.57 minutes.

5. 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?

SOLUTIONThe list contains 16 sorted values. Both the 8th and 9th are 41, so the median is (41 + 41)/2 = 41.(i) Adding any single value keeps the new 9th observation equal to 41. For example, add 10, 41 or 100. It is not necessary to add only 41.(ii) One sufficient choice is a value at or below 41 and a value at or above 41, such as 30 and 70. Then the 9th and 10th values are both 41. Other choices may also work; check the new middle pair.(iii) Removing any one observation also keeps the 8th of the remaining 15 values equal to 41. For example, remove 8, one 41, or 92.
The eighth and ninth of the sixteen sorted values are both 41. Adding any one value or deleting any one observation preserves this median.
The eighth and ninth of the sixteen sorted values are both 41. Adding any one value or deleting any one observation preserves this median.

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.

SOLUTION (i) Never true. Removing a value below the median shifts the centre upward (or leaves it the same), so the median can only increase or stay the same — it never decreases. (ii) Always true. Adding a value smaller than the mean lowers the total relative to the new count, so the mean always decreases. (iii) Sometimes true. Adding 4 values can leave the median unchanged (e.g. two below and two above, suitably placed) but in general it shifts the median, so it is not always true. (iv) Never true. Adding values below the median pulls the centre downward, so the median decreases or stays the same — it never increases.

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

SOLUTION There are 8 numbers, so total = 10.375 × 8 = 83. Sum of the known numbers = 8 + 13 + 10 + 4 + 5 + 20 + 10 = 70. y = 83 − 70 = 13.

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

SOLUTION Sum = Mean × number of values = 134 × 15 = 2010.

9. 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

SOLUTION Without p, sorting the 10 numbers gives: 8, 8, 12, 18, 25, 29, 35, 39, 47, 73. Including p makes 11 values, so the median is the 6th value. There are 5 values below 29 (8, 8, 12, 18, 25). For 29 to land in the 6th position, p must be 29 or greater (so it does not push 29 out of the centre). Checking: p = 10 or 25 (both < 29) would make the 6th value 25, not 29 — rejected. Any p ≥ 29 keeps the 6th value at 29. So p could be (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. (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?

SOLUTION
Weekly rides012345678910
Students31477546302
(i) Total students = 42; total rides = 193. Mean = 193/42 ≈ 4.60 rides per student.(ii) The median is the mean of observations 21 and 22. Cumulative frequencies through 0, 1, 2, 3 and 4 rides are 3, 4, 8, 15 and 22. Both middle observations are 4, so median = 4.(iii)(a) False: three students rode zero times. (b) Reasonable as an informal description: 39 of 42 rode at least once and 38 rode at least twice; “a few” is not an exact mathematical threshold.(c) True: five students rode 8 or 10 times in seven days, so each must have ridden more than once on at least one day. (d) Not established: at least five did so; students with fewer total rides could also have taken multiple rides on a day.(e) Adding one to every observation increases both statistics by one: new mean = 235/42 ≈ 5.60, new median = 5.
Frequencies for zero to ten weekly rides are 3, 1, 4, 7, 7, 5, 4, 6, 3, 0 and 2.
Frequencies for zero to ten weekly rides are 3, 1, 4, 7, 7, 5, 4, 6, 3, 0 and 2.

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.

SOLUTION
No. of trials12345678910
No. of students10014912151010
Total number of students = 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62. Minimum = 1 trial, Maximum = 10 trials. Mean = Σ(trials × students) ÷ 62. Σ = (1×1) + (4×1) + (5×4) + (6×9) + (7×12) + (8×15) + (9×10) + (10×10) = 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473. Mean = 473 ÷ 62 ≈ 7.63 trials. Median = average of the 31st and 32nd values. Running totals: up to 5 → 6 students; up to 6 → 15; up to 7 → 27; up to 8 → 42. So the 31st and 32nd values both fall in “8 trials”. Median = 8 trials. Description: most participants needed 7–8 throws, with a typical (median) of 8 and an average of about 7.63; the easiest hit took 1 throw and the hardest took 10.
Frequency plot for dart throws: 62 students, weighted total 473 trials, mean about 7.63 and median 8.
Frequency plot for dart throws: 62 students, weighted total 473 trials, mean about 7.63 and median 8.

Figure it Out — Line Graphs (page 122–123)

1. 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.

SOLUTION Draw the days Mon–Sun along the horizontal axis and the number of customers along the vertical axis. Plot two lines — one for “Visiting” and one for “Purchasing” — using these points:
DayMonTueWedThuFriSatSun
Visiting16191014202235
Purchasing108711121626
Observation: both lines rise sharply towards the weekend and peak on Sunday (35 visiting, 26 purchasing). The “Purchasing” line always stays below the “Visiting” line, and the gap is widest on Tuesday (19 − 8 = 11, compared with Sunday’s 35 − 26 = 9) — many people visit, but not all of them buy.
Two lines plot visiting and purchasing customers. The largest difference is 11 on Tuesday, not Sunday.
Two lines plot visiting and purchasing customers. The largest difference is 11 on Tuesday, not Sunday.

2. The average number of days of rainfall in each month for a few cities is shown in the table below: (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?

SOLUTION(i) Count rainy days for each month over several years at each location and average the counts month by month, using the same definition of a rainy day.(ii) The graph below plots rounded table values. (iii) Approximate New Delhi readings are included in the last row; because its points are read from a printed graph, small reading differences are expected.
CityJanFebMarAprMayJunJulAugSepOctNovDec
Mangaluru0002624282514941
Port Blair2113161917191714115
Rameswaram3123301128108
New Delhi12212410104101
(iv) Sum the original, unrounded table entries for the first three cities: Mangaluru 112.1, Port Blair 125.8, Rameswaram 42.3 rainy days per year. New Delhi is roughly 36–38 from the plotted readings. Therefore Port Blair has the most and New Delhi the fewest.(v) New Delhi has its main rainy period in June–September, peaking in July–August. Rameswaram peaks in October–December. These are conclusions from this textbook dataset, not a forecast.
Rainfall-day data rounded for plotting. New Delhi values are approximate graph readings; Port Blair has the most yearly rainy days and New Delhi the fewest.
Rainfall-day data rounded for plotting. New Delhi values are approximate graph readings; Port Blair has the most yearly rainy days and New Delhi the fewest.

2 (births graph). 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.

SOLUTION(i) Birth counts show repeated rises and falls, with higher values around late summer/autumn and lower values in several early-year months. The peaks approach 2 million. January is not consistently the lowest month.(ii) July 2017 is approximately 1.8 million (about 18 lakh), not 1.4 million.(iii) Counting the monthly points around the labelled July/January ticks, the graph covers approximately April 2017 to March 2020: three years.(iv) January 2018 is about 1.7 million; January 2019 and January 2020 are each about 1.8 million. The latter two are close and slightly higher.(v) Adding the twelve approximate 2019 readings gives about 21 million births (2.1 crore). A reasonable graph-based estimate is around 20–22 million. These are estimates of the plotted series, not exact counts independently verified against birth records.
Approximate reconstruction of the textbook births graph, April 2017 to March 2020. July 2017 is about 1.8 million; the 2019 total is about 21 million.
Approximate reconstruction of the textbook births graph, April 2017 to March 2020. July 2017 is about 1.8 million; the 2019 total is about 21 million.

Figure it Out — Means, Medians & More Graphs (page 127–132)

1. 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?

SOLUTIONEach three-number line must total 3 × 10 = 30. Both grids below work and use nine distinct numbers.Check every row, column and both diagonals. The second grid retains some entries and changes others while keeping all eight line sums at 30.Changing two cells to preserve one row alone is not enough: it may spoil their columns or diagonals. A useful general form with centre 10 is [10+a, 10−a−b, 10+b; 10−a+b, 10, 10+a−b; 10−b, 10+a+b, 10−a]. Choose a and b so all nine entries are distinct.
Two different 3 by 3 grids contain nine distinct numbers and have mean 10 along each row, column and diagonal.
Two different 3 by 3 grids contain nine distinct numbers and have mean 10 along each row, column and diagonal.

2. 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.

SOLUTION (i) Sum must be 24: e.g. {6, 8, 10} and {2, 8, 14}. (ii) Two middle values must average 15.5: e.g. {10, 15, 16, 20} (middle 15, 16) and {12, 14, 17, 19} (middle 14, 17). (iii) Sum must be 13.6 × 5 = 68: e.g. {10, 12, 14, 16, 16} and {13, 13, 14, 14, 14}. (iv) e.g. {4, 6, 8, 10, 12, 14} — mean = 9, median = (8 + 10)/2 = 9; and {2, 4, 6, 6, 8, 10} — mean = 6, median = 6. (v) The mean is pulled up by a large value: e.g. {1, 2, 3, 4, 5, 30} — median = 3.5, mean = 7.5; and {2, 2, 2, 2, 2, 20} — median = 2, mean = 5.

3. 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?

SOLUTIONFor six sorted values, the 3rd and 4th must sum to 26. One answer is 13, 13, 13: the sorted list becomes 5, 13, 13, 13, 14, 21.There are infinitely many possibilities. For example, put 12, 12, M in the blanks, for any counting number M ≥ 21. The sorted list is 5, 12, 12, 14, 21, M, whose median is (12 + 14)/2 = 13. Varying M gives infinitely many answers.

4. 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 There are 6 numbers, so the total must be 6.5 × 6 = 39. Sum of the four known numbers = 3 + 11 + 15 + 6 = 35. So the two blanks must add to 39 − 35 = 4. Counting-number pairs adding to 4: (1, 3), (2, 2), (3, 1). Treating the two ordered blanks separately, there are 3 possibilities — e.g. 3, 11, 1, 3, 15, 6 or 3, 11, 2, 2, 15, 6.

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.

SOLUTION (i) False. Take 2 and 4: average = 3, which is odd. (Algebra: (2a + 2b)/2 = a + b, which can be odd.) (ii) False. Take 5 and 10: average = 7.5, not a multiple of 5. (Algebra: (5a + 5b)/2 = 5(a + b)/2 is a multiple of 5 only when a + b is even.) (iii) False. Take 5, 5, 5, 5, 10: sum = 30, average = 6, which is not a multiple of 5. (Average = 5 × (sum of the five whole numbers)/5; this is a multiple of 5 only when the five multipliers add to a multiple of 5.)

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. (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) remain the same (b) increase (c) decrease (d) information not sufficient. (iii) Which of the following statements about the new class average height are correct? Why? (about the median: remain / increase / decrease / not sufficient)

SOLUTION (i) Only (c) is correct. Adding two values may raise, lower or leave the average unchanged, so (a) and (b) are not always true; we do not need to remeasure the whole class — just measure the two new students — so (d) is wrong. (ii) The two new heights 149 and 152 have average (149 + 152)/2 = 150.5, which is slightly more than 150.2. Adding values whose average is above the current mean raises the mean, so (b) the average will increase (very slightly). (iii) (d) The information is not sufficient. The median depends on the individual heights of all 24 students (not just their average), which are not given, so we cannot say how it changes.

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

SOLUTION
Value14151617181920212223
Frequency2235443101
There are 25 observations and their sum is 443. Mean = 443/25 = 17.72. Thus 17 is not the mean; the mean is greater than 17.Balance check at 17: total left distance = 2×3 + 2×2 + 3×1 = 13. Total right distance = 4×1 + 4×2 + 3×3 + 1×4 + 1×6 = 31. The excess 18 divided among 25 observations shifts the balance point by 18/25 = 0.72.
The page 128 dot plot contains 25 values totalling 443. Its mean is 17.72, not 17.
The page 128 dot plot contains 25 values totalling 443. Its mean is 17.72, not 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?

SOLUTIONThe total weight changes by −2 + 1 + 1 = 0 kg. The group size is unchanged, so the mean remains 65.3 kg.The median may increase, decrease or stay the same, depending on which people occupied the middle positions and whether the ordering changes. The original median 67 kg and the three weight changes do not determine its new value. Avoid describing an unknown change as a measured amount.

9. The following table shows the retail price (in ₹) of iodised salt in the month of January in a few states over 10 years. (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?

SOLUTION (i) Take, for example, Mizoram, Uttar Pradesh and West Bengal. Put the years 2016–2025 on the horizontal axis and price (₹) on the vertical axis (scale: 1 unit = ₹5), then plot and join each state’s points. (ii) The price of salt has generally risen over the decade in most states, but Gujarat’s price stayed mostly around ₹13–15 in 2017–2024 before rising to ₹19.20 in 2025 while Mizoram and West Bengal climbed steeply. (iii) Gujarat: 2016 ₹16.5 → 2025 ₹19.2, a small change of about ₹2.7 — nearly steady. Uttar Pradesh: 2016 ₹16.15 → 2025 ₹24.81, a rise of about ₹8.66 — much greater variation. (iv) Compare 2016 → 2025 increases: Andaman & Nicobar +4.99, Assam +6.35, Gujarat +2.70, Mizoram +9.80, Uttar Pradesh +8.66, West Bengal +14.52. The largest increase is in West Bengal (about ₹14.5). (West Bengal also has the largest factor increase, from ₹9.47 to ₹23.99.) (v) You might explore why salt is cheaper or steadier in some states, how transport and local taxes affect price, or how these prices compare with inflation over the same period.
Original line graph of the textbook salt-price table for Mizoram, Uttar Pradesh and West Bengal, 2016 to 2025.
Original line graph of the textbook salt-price table for Mizoram, Uttar Pradesh and West Bengal, 2016 to 2025.

10. Referring to the graph below (primary source of energy for household lighting over time), 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.

SOLUTION (i) Valid. In 1983 the rural kerosene line is high (above 80%) while the urban electricity line is already above 60% — so most rural homes used kerosene and most urban homes used electricity. (ii) Valid. In both graphs the kerosene line falls steadily towards almost 0% by 2023. (iii) Not valid. In 2000 the urban electricity line is high (around 90%), not 10% — it is the kerosene share that is low. (iv) Not valid. The graph only shows the primary lighting source, not power cuts, so no conclusion about power cuts can be drawn.

11. Answer the following questions based on the line graph (average daily time spent on hobbies and games). (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 (b) 10 (c) 12 (d) 14 (e) 18 years (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.

SOLUTION (i) Read the urban line at age 10: it is at about 2 hours per day. (ii) Follow the rural line down to the 1.5-hour level; it crosses there at about age 14, so the answer is (d) 14 years. (iii) (a) Incorrect — the time at age 15 is lower than at age 10, not twice as much (the lines fall with age). (b) Incorrect — a graph of the average cannot tell us what every individual child does, so we cannot say all rural 15-year-olds spend at least 1 hour.
Approximate textbook hobby-time curves show about two hours for urban ten-year-olds and 1.5 hours for rural fourteen-year-olds.
Approximate textbook hobby-time curves show about two hours for urban ten-year-olds and 1.5 hours for rural fourteen-year-olds.

12. Individual project: your activity strips

Record your own week in half-hour blocks, using 48 blocks per day. For each activity, add its total hours over seven days and divide by 7. The category means should sum to 24 hours if every interval is counted once. Compare regularity of meals, sleep and outdoor time. The example below is illustrative, not a record of your day.

An illustrative twenty-four-hour activity strip uses 48 half-hour boxes, with separate colours for sleep, meals, leisure, study, personal care and travel.
An illustrative twenty-four-hour activity strip uses 48 half-hour boxes, with separate colours for sleep, meals, leisure, study, personal care and travel.

13. Group projects: collect and compare data

For the sleep project, record night sleep and naps consistently for seven days. For each age group, clearly choose whether your observations are daily person-records or each person’s seven-day average; then calculate the mean and median of that chosen list. Do not treat a small convenience sample as representative of everyone.

For school timings, record start, end, breaks and total daily duration for each school. For Manoj, 09:30–16:30 is 7 hours including breaks. Compare like-for-like daily totals; subtract breaks separately only when reporting teaching time.

14. The following graphs show the sunrise and sunset times across the year at 4 locations in India. (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.

SOLUTION(i) Kibithu has the earliest January sunrise, at approximately 06:00. Its sunset is around 16:30, giving roughly 10.5 hours of daylight.(ii) Srinagar has the greatest maximum day length shown, around June–July: roughly 14–14.5 hours. Compare the vertical gap between sunrise and sunset for the same month, not sunset time alone.(iii) Kibithu rises and sets earlier by the clock than Ghuar Moti. Srinagar shows a large seasonal change in daylight; Kanyakumari stays near 12 hours. These are approximate readings of the textbook curves.
Approximate reconstruction of the textbook sunrise and sunset curves for Kibithu and Ghuar Moti. Day length is sunset time minus sunrise time.
Approximate reconstruction of the textbook sunrise and sunset curves for Kibithu and Ghuar Moti. Day length is sunset time minus sunrise time.
Approximate reconstruction of the textbook sunrise and sunset curves for Srinagar and Kanyakumari. Day length is sunset time minus sunrise time.
Approximate reconstruction of the textbook sunrise and sunset curves for Srinagar and Kanyakumari. Day length is sunset time minus sunrise time.

15. 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?

SOLUTION(i) Look for the full-moon pattern when moonrise is near sunset and moonset near sunrise: in this sketch it is around day 13–14. The new-moon pattern has moonrise near sunrise and moonset near sunset, around day 28–29.These are graph-based estimates. The chart gives no month, year or location, and its rise/set times alone do not specify an exact astronomical phase date. An exact calendar answer would require that missing context.(ii) The times generally move later each day. When a time passes midnight it wraps from near 24:00 back to near 00:00; that is why each curve breaks. The delay is not exactly the same every day. The earliest clock-time moonrise near midnight is not the new-moon criterion.
Approximate textbook moon-time graph. Broken curves mark the midnight wrap; the full-moon pattern is near days 13 to 14 and the new-moon pattern near days 28 to 29.
Approximate textbook moon-time graph. Broken curves mark the midnight wrap; the full-moon pattern is near days 13 to 14 and the new-moon pattern near days 28 to 29.

Math Talk & Try This (in-text questions)

Math Talk — The mean as a centre Q. Can you explain how the mean is the centre of each collection? Is the mean the midpoint of the two endpoints? Can there be more than one such centre? Answer. The mean is the “balance point”: the total distance of the values below it equals the total distance of the values above it. It is not always the midpoint of the smallest and largest values. There is exactly one such centre — if you pick any value larger than the mean, the left-side distances grow and the right-side distances shrink, so the balance is lost; the same happens for a value below the mean.
Dots at 10, 10, 11 and 17 balance at 12 because their total left and right distances are both 5.
Dots at 10, 10, 11 and 17 balance at 12 because their total left and right distances are both 5.
Math Talk — Adding, removing, scaling values Q. What happens to the mean when a value is included or removed? What if every value is increased by a fixed number, or doubled? Answer. Including a value greater than the mean raises the mean; one smaller than the mean lowers it; one equal to the mean leaves it unchanged. Removing a value works the opposite way. If every value increases by a fixed number k, the mean increases by k; if every value is multiplied by k, the mean is multiplied by k (proved with algebra in the chapter: average of (xi + k) = a + k, and average of (k·xi) = k·a).
Try This — Subtracting a constant; fair-share Q. Using algebra, what is the average when a fixed number, e.g. 2, is subtracted from every value? Answer. If the original average is a, then for the values (xi − 2): new average = (Σxi − 2n)/n = a − 2. So subtracting 2 from every value lowers the average by exactly 2 — matching the fair-share idea (everyone simply gives back 2).
In-text worked examples (verify these) Coach Balwan: the smudged weight w satisfies (42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 + w)/10 = 39.2, so 349 + w = 392 and w = 43 kg. Venkayya’s coconuts: total = 25.6 × 15 = 384; the true total is 384 − 3 = 381, so the correct average = 381 ÷ 15 = 25.4. Family size (with frequencies): mean = [(3×3)+(4×11)+(5×9)+(6×7)+(7×3)+(8×1)+(9×1)+(10×1)] / 36 = 188 ÷ 36 = 5.22; the median (average of 18th and 19th values) is 5.

Further in-text activities

Spreadsheet calculations (pages 111–113)

With the textbook headers in row 1 and the 22 students in rows 2–23: Farooq’s Maths mark is E5 = 42; B7 = 27 is Gowri’s Odia mark. Ashwin scored above 30 in Telugu (31), English (33) and Maths (34).

Use =SUM(B2:G2) for Ratna’s total and copy down. Use =AVERAGE(G2:G23) for Science; use the corresponding B–F ranges for the other subjects.

SubjectClass average (rounded)
Odia31.23
Telugu33.59
English32.64
Maths34.14
Social Science31.36
Science33.14

Odia’s mean (31.23) is less than Telugu’s (33.59).

StudentTotal marks
Ratna200
Nagesh250
Ashwin185
Farooq266
Mrinal194
Gowri183
Pankaj112
Jaya248
Ganesh225
Shravan102
Aishwarya273
Hari147
Trupti188
Veeresh148
Vidhya213
Sanskruti242
Shanker232
Vyshnavi197
Govind103
Shiva177
Tarun246
Jyothi183

Keep the mean unchanged (pages 105–108)

To add k observations without changing a current mean m, their sum must be km. For example, at mean 12 add 10 and 14, or 10, 11 and 15, or 13, 14 and 9. Removing a group preserves the mean only if that group also has mean m and some observations remain. For the eleven-number list on page 106, sum = 79 and mean = 79/11 ≈ 7.18; adding 10 to each value makes it 189/11 ≈ 17.18, and doubling each value makes it 158/11 ≈ 14.36.

Rice and wheat infographic (page 124)

Read this as a historical preference index from the textbook’s 2011–12 survey data, not a current percentage. Karnataka’s colour suggests a positive value, roughly +80; the exact hidden number cannot be recovered from the shade alone. The five highest rice indices are Manipur +100, Nagaland +99, Mizoram +97, Tripura +96, Meghalaya +95.

Among states, the five strongest wheat preferences are Rajasthan −93, Haryana −81, Punjab −78, Madhya Pradesh −60, Gujarat −37. Delhi (a union territory) is −45 and would rank ahead of Gujarat if states and union territories are considered together. Bihar +3 is closest to balance; Maharashtra −15, Uttarakhand −18 and Himachal Pradesh −19 are also relatively close. +100 does not mean zero wheat consumption.

Manoj’s activity strips (page 125)

From the long overnight band and the repeated school block, a reasonable colour key is: blue = sleep; green = meals; orange = leisure; yellow = study/classes; purple = personal care/exercise; grey = travel. The middle strip fits Friday, the right strip a possible Saturday, and the left a possible Sunday; swapping the two weekend labels cannot be ruled out from colours alone.

The long leisure block on the left strip, about 13:30–16:30, with travel on either side, plausibly represents the movie outing. The Friday strip indicates lunch around 12:30–13:00. Activity identities and weekend ordering are inferences from the pattern, not uniquely encoded labels.

Common Mistakes to Avoid

Watch out for these

  • Finding the mean of values without their frequencies — multiply each value by how many times it occurs before adding.
  • Confusing the mean (balance point / sum ÷ count) with the midpoint of the smallest and largest values — they are not the same.
  • For an even number of values, forgetting that the median is the average of the two middle values, not just the lower one.
  • When every value changes by a constant, recomputing from scratch instead of using “new mean = old mean ± k” or “× k”.
  • Reading a line graph at the wrong axis — always match a point to both its horizontal (time) and vertical (value) coordinates.
  • Drawing conclusions about individuals from a graph of averages — an average tells you nothing certain about any single person.

Practice MCQs & Assertion–Reason

1. The arithmetic mean of a set of values is:

(a) the largest value    (b) the middle value    (c) the sum ÷ the number of values    (d) the most frequent value

2. The mean of the first 10 natural numbers is:

(a) 5    (b) 5.5    (c) 10    (d) 55

3. If the mean of 6 values is 12, the sum of the values is:

(a) 2    (b) 18    (c) 72    (d) 6

4. If every value of a data set is increased by 5, the mean:

(a) stays the same    (b) increases by 5    (c) is multiplied by 5    (d) decreases by 5

5. If every value of a data set is doubled, the mean is:

(a) halved    (b) unchanged    (c) doubled    (d) increased by 2

6. The median of 4, 7, 9, 11, 15 is:

(a) 7    (b) 9    (c) 11    (d) 9.2

7. The median of 6, 8, 10, 12 is:

(a) 8    (b) 9    (c) 10    (d) 12

8. Including a value equal to the mean in a data set will make the mean:

(a) increase    (b) decrease    (c) stay the same    (d) double

9. A line graph is most suitable for showing:

(a) parts of a whole    (b) how a quantity changes over time    (c) exact frequencies    (d) a single value

10. On a line graph, a steeper line segment between two points indicates:

(a) no change    (b) a smaller change    (c) a greater (faster) change    (d) an error

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(c), 6-(b), 7-(b), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The mean of 2 and 8 is 5.

Reason: The mean of two numbers is their sum divided by 2.

A-R 2. Assertion: If every value of a data set is increased by 10, the mean increases by 10.

Reason: Adding a constant k to every value adds k to the sum for each value, so the mean rises by k.

A-R 3. Assertion: The mean is always one of the values in the data.

Reason: The mean is the balance point of the data.

A-R 4. Assertion: Including a value greater than the mean increases the mean.

Reason: The new value pulls the balance point towards itself.

A-R 5. Assertion: A line graph is a suitable choice for showing a numerical trend over time.

Reason: Joining data points by line segments makes trends easy to see.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Mean = sum of all values ÷ number of values; it is the balance point where left-side and right-side distances are equal.
  • There is exactly one mean (one centre) for a data set.
  • Adding a value above the mean raises it; below the mean lowers it; equal to the mean leaves it unchanged.
  • Add a constant k to every value → mean rises by k; multiply every value by k → mean is multiplied by k.
  • With frequencies, mean = Σ(value × frequency) ÷ Σ(frequency); the median is found using running totals of the frequencies.
  • The median is the middle value (or the average of the two middle values) of the sorted data.
  • Line graphs visualise change over time; steeper segments mean faster change. Always read both axes before concluding.

How to score full marks in this chapter

Show the formula “sum = mean × count” for every “find the missing value” question, and write the running totals when finding a median from a frequency table. For line-graph questions, state the value you read and the axis you read it from, and never claim something about an individual from a graph of averages. Keep fractions exact (e.g. 39/7) before rounding to a decimal.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 12 about?

Chapter 12, Tales by Dots and Lines (Chapter 5 of Ganita Prakash Part II), is about the arithmetic mean and median seen as balance points on dot plots, how these change when data changes, the mean from frequency tables and spreadsheets, and reading and drawing line graphs to interpret data over time.

What is the formula for the arithmetic mean?

The mean (average) equals the sum of all the observations divided by the number of observations: mean = (x1 + x2 + … + xn) / n. You can also write sum = mean times the number of values.

How do you find a missing value when the mean is given?

Multiply the mean by the number of values to get the required total, then subtract the sum of the known values. For example, if the mean of 10 numbers is 39.2, the total is 392, so a missing value is 392 minus the sum of the other nine numbers.

Are these Class 8 Maths Ganita Prakash Chapter 12 solutions free?

Yes. All ClearStudy NCERT solutions for Class 8 Maths Ganita Prakash Chapter 12 are free and follow the official NCERT Ganita Prakash textbook for 2026-27.

Textbook reference: NCERT Ganita Prakash, Chapter 12 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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