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). Every “Figure it Out” question is solved step by step, with the arithmetic mean, median and line-graph readings fully verified so you can revise the whole chapter quickly.

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).

SOLUTION From the dot plot the marked values are 4, 7, 8, 9, 9, 9, 11 and one missing dot — 8 values in all with a required mean of 9. Required total = 9 × 8 = 72. Sum of the visible 7 dots = 4 + 7 + 8 + 9 + 9 + 9 + 11 = 57. Missing value = 72 − 57 = 15. So a dot must be marked at 15 to make the mean 9. (If your reading of the dot plot differs slightly, use the same method: missing value = mean × count − sum of the visible dots.)

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.

SOLUTION A mean of 5.57 means the data is “balanced” around 5.57 min, so the values must cluster near 5–6.5 minutes. Plot A has all its 7 dots in the 5 to 6.5 range (5, 5.5, 5.5, 5.5, 5.5, 6, 6.5). Their sum = 39, so mean = 39 ÷ 7 = 5.57 min — this matches. Plots B and C have most dots well below 5 (around 0.5–4.5), so their means are far smaller than 5.57. Hence only album A has a mean of 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?

SOLUTION The data is already sorted and has 16 values. Median = average of the 8th and 9th values = (41 + 41)/2 = 41. (i) Adding one value makes 17 values, with the median the 9th value. To keep the median 41, the new value must be 41 (so it sits at the centre) — e.g. include another 41. (More generally any value equal to 41 keeps it; here 41 is in the list.) (ii) Adding two values keeps 18 values, median = average of 9th and 10th. Choose one value ≤ 41 and one value ≥ 41 — for example 40 and 41, or 30 and 70. The balanced pair leaves the two middle values 41 and 41, so the median stays 41. (iii) Removing one value leaves 15 values, median = 8th value. To keep it 41, remove a value from one extreme that does not shift the centre — removing one of the two 41s still leaves a 41 in the 8th position, so removing a 41 keeps the median 41.

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 Method. Read the frequency of each value (0, 1, 2, …) from the dot plot, then use mean = Σ(value × frequency) ÷ Σ(frequency) and the running-total method for the median. (The chapter gives “four students rode twice” as one such frequency.) (i) & (ii) Add up all the dots to get the total number of students; multiply each value by how many dots sit above it, add, and divide by the number of students for the average; the median is the value of the middle dot (or the average of the two middle dots) once the dots are listed in order. (iii) (a) Not valid — the dots above 0 show some students did not cycle at all. (b) Valid — most dots sit at small non-zero values, so almost everyone cycled a few times. (c) Cannot be concluded — the data only counts how many times in the week, not how many times per day, so we cannot tell if anyone cycled more than once in a single day. (d) Not valid for the same reason. (e) If everyone cycles 1 more time, every value increases by 1, so both the average and the median increase by exactly 1.

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.

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 Sunday — many people visit, but not all of them buy.

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) For each city the number of rainy days in a month is counted every year for several years, and the values are averaged to give the “average number of rainy days” for that month. (ii) Round each value to the nearest whole number and plot it against the month, then join the points for each city. Rounded values:
CityJanFebMarAprMayJunJulAugSepOctNovDec
Mangaluru0002624282514941
Port Blair2113161917191714115
Rameswaram3123301128108
(iii) Read each month’s value off the New Delhi line in the printed graph and copy it into the empty New Delhi row of the table. (The values depend on the printed graph; the New Delhi line peaks in the monsoon months Jul–Aug, around 9–10 rainy days.) (iv) Port Blair receives the most rainy days per year (its yearly total ≈ 2 + 1 + 1 + 3 + 16 + 19 + 17 + 19 + 17 + 14 + 11 + 5 = 125 days — rain spread across many months). Rameswaram receives the fewest (yearly total ≈ 42 days). (v) The rainy season for New Delhi is mainly July–August (south-west monsoon); for Rameswaram it is October–December (north-east monsoon), as those months show the most rainy days.

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) The number of births rises and falls in a repeating yearly (seasonal) pattern, with a clear peak in the middle of each year and a dip near the start of each year. (ii) From the graph, July 2017 has about 1.4–1.5 million (1.4M) live births. (iii) The horizontal axis runs from about July 2017 to early 2020 — roughly two and a half years. (iv) The January values are similar in all three years (each is a yearly low point), with January 2018 slightly the lowest and the three Januaries staying close to about 1.4–1.5M each. (v) Adding the 12 monthly values for 2019 (each roughly 1.4–2.0M) gives an estimate of about 20 million (2 crore) births in 2019. (Read exact monthly heights off the printed graph to refine the estimate.)

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?

SOLUTION (i) If each row, column and diagonal of 3 numbers has average 10, then each must sum to 30. A magic square of sum 30 works. Using distinct numbers:
5169
14106
11415
Every row, column and diagonal adds to 30, so each average is 30 ÷ 3 = 10. ✓ (ii) Yes. Any 3×3 magic square with line-sum 30 works, so there are many possibilities — for example add the same number to one cell and subtract it from another in the same line, keeping every line-sum at 30.

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?

SOLUTION There will be 6 numbers, so the median is the average of the 3rd and 4th values when sorted, and this must equal 13, i.e. the 3rd + 4th values must add to 26. A simple way is to make the two middle values both 13, e.g. 5, 21, 14, 13, 13, 13 (sorted: 5, 13, 13, 13, 14, 21; median = (13 + 13)/2 = 13). Because the three blanks can be many different counting numbers (any choice whose sorted 3rd and 4th values average 13), there are infinitely many possibilities if counting numbers may be repeated and are unbounded.

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 Method (balance test): If 17 is the mean, the total distance of all dots to the left of 17 must equal the total distance to the right. From the dot plot, the dots are clustered to the left of 17 (around 14–16) more than they are to the right, so the left-hand total distance is larger. The balance is not equal, so the mean is pulled below 17. Therefore 17 is not the average — the actual mean is a little less than 17. (You can confirm by computing Σ(value × dot count) ÷ total dots from the plot.)

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?

SOLUTION Mean: the total weight change = −2 + 1 + 1 = 0 kg. Since the total is unchanged and the number of people is the same, the mean stays the same at 65.3 kg. Median: only three people changed weight by small amounts (1–2 kg), and the middle value of the sorted data may or may not be one of them, so the median may change slightly or stay the same — without the individual weights we cannot say its exact new value, but a tiny change of at most about 1–2 kg is possible.

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 almost flat (around ₹14–15) 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.

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.

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) In January the place whose sunrise line is lowest (earliest time) rises the soonest; reading the graphs, this is Kibithu (far east), with a January day length of roughly 10–11 hours (sunset around 16:00 minus sunrise around 06:00). (ii) The place where the gap between the sunset and sunrise curves is largest in summer has the longest days — the high-latitude location (Srinagar, in the north) shows the greatest summer day length. (iii) Observation: places farther north (Srinagar) show a big difference between summer and winter day lengths, while places nearer the equator (Kanyakumari) have day lengths that hardly change all year. Eastern places (Kibithu) see the sun rise and set earlier in clock time than western places.

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) On amavasya (new moon) the moon rises and sets with the sun, so moonrise is near sunrise (early morning); on purnima (full moon) the moon rises around sunset and sets around sunrise. Reading the graph, the date where moonrise is earliest in the morning is the new moon, and the date where moonrise is near sunset is the full moon. (Exact dates are read from the printed graph — the two events are about 14–15 days apart.) (ii) Notice: both the moonrise and moonset lines climb steadily, shifting later by roughly the same amount each day (about 45–50 minutes), so the moon “rises later” every night. Wonder: why the daily shift is so regular, and how it links to the moon’s orbit around the Earth.

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.
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.

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 the best choice to show data that changes 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.

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