NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 5: Tales by Dots and Lines

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• Figure it Out: Page 113
• Figure it Out: Page 122
• Figure it Out: Page 127
• Test Papers with Solutions (PDF)

You find missing values, use frequency tables and try spreadsheet formulas like =AVERAGE. Section 5.2, Visualising and Interpreting Data, then turns to line graphs and infographics using real data, from Kerala and Punjab temperatures and coastal rainfall to space launches and sleep across ages.
Get here the Solutions and Answers of Figure it Out at Page 113.

(i) The first 50 natural numbers.
(ii) The first 50 odd numbers.
(iii) The first 50 multiples of 4.
Answer:
(i) The first 50 natural numbers are 1, 2, 3, …, 50.
Pairing the first and last numbers, we get
1 + 50 = 51, 2 + 49 = 51, and so on.
There are 25 such pairs.
Sum = 25 × 51 = 1275
Mean = 1275 ÷ 50 = 25.5

(ii) The first 50 odd numbers are 1, 3, 5, …, 99.
Each pair from opposite ends has a sum of 100.
Sum = 25 × 100 = 2500
Mean = 2500 ÷ 50 = 50

(iii) The first 50 multiples of 4 are 4, 8, 12, …, 200.
Each pair from opposite ends has a sum of 204.
Sum = 25 × 204 = 5100
Mean = 5100 ÷ 50 = 102

In each case, the numbers are equally spaced.
Their mean is halfway between the first and last numbers.
Also, multiplying every natural number by 4 multiplies the mean by 4, so 25.5 × 4 = 102.

Answer:
The 10 values shown are 4, 7, 8, 8, 9, 9, 9, 9, 9 and 11.
Their sum = 4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 = 83
After adding the missing dot, there will be 11 values.
Required sum = 9 × 11 = 99
Missing value = 99 − 83 = 16
So, I mark one dot above 16.

(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
Answer:
(i) The teacher need not measure all the heights again. Each height is 1 cm more because of the shoes. Subtracting 1 cm from every height also reduces the mean by 1 cm.

(ii) Correct average height = 150.2 − 1 = 149.2 cm
Therefore, the correct option is (d) 149.2 cm.

Answer:
The song lengths in album A are 5, 5, 5.25, 5.5, 5.75, 6 and 6.5 minutes.
Total length = 5 + 5 + 5.25 + 5.5 + 5.75 + 6 + 6.5 = 39 minutes
Number of songs = 7
Mean = 39 ÷ 7 ≈ 5.57 minutes

In albums B and C, every song is shorter than 5.57 minutes, so their means cannot be 5.57 minutes.
Therefore, album A has a mean length of about 5.57 minutes.

(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?
Answer:
The 16 values are already in increasing order.
The 8th and 9th values are both 41.
Median = (41 + 41) ÷ 2 = 41
(i) We can include 41. There will be 17 values and the 9th value will be 41.

(ii) We can include 20 and 60, one below 41 and one above 41. There will be 18 values and the 9th and 10th values will both be 41. So, the median remains 41.

(iii) We can remove 8. There will be 15 values and the 8th value will be 41. So, the median remains 41.

(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.
Answer:
(i) Never true. Removing a value below the median shifts the middle towards the larger values. The median can increase or remain the same, but it cannot decrease.
For example, the median of 2, 4, 6 is 4. On removing 2, the median becomes (4 + 6) ÷ 2 = 5.

(ii) Always true. A new value below the mean brings the mean down.
For example, the mean of 6 and 10 is 8. On including 5, the mean becomes (6 + 10 + 5) ÷ 3 = 7.

(iii) Sometimes true. It depends on the four values added.
The median of 1, 2, 3 is 2.
On adding 0, 1, 3 and 4, the ordered data is 0, 1, 1, 2, 3, 3, 4. The median remains 2.
But on adding 4, 5, 6 and 7 to the original data, the median becomes 4.

(iv) Never true. Adding four values below the median shifts the middle towards the smaller values. The median can decrease or remain the same, but it cannot increase.
For example, the median of 1, 5, 9 is 5. On adding 1, 2, 3 and 4, the ordered data is 1, 1, 2, 3, 4, 5, 9. Its median is 3.

Answer:
Number of values = 8
Sum of all the values = 10.375 × 8 = 83
⇒ 8 + 13 + 10 + 4 + 5 + 20 + y + 10 = 83
⇒ 70 + y = 83
On simplifying, y = 83 − 70 = 13.
Therefore, y = 13.

Answer:
Sum of the data = Mean × Number of values
= 134 × 15
= 2010

(i) 10 (ii) 25 (iii) 40 (iv) 100
(v) 29 (vi) 47 (vii) 30
Answer:
Arranging the ten known values in increasing order, we get
8, 8, 12, 18, 25, 29, 35, 39, 47, 73.

There will be 11 values after including p. So, the median will be the 6th value.
There are already five values below 29. To keep 29 in the 6th position, p must be 29 or greater.

Therefore, the possible values are 40, 100, 29, 47 and 30.
The correct options are (iii), (iv), (v), (vi) and (vii).

(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?
Answer:
Counting the dots, we get:

(i) Total number of students = 3 + 1 + 4 + 7 + 7 + 5 + 4 + 6 + 3 + 0 + 2 = 42
Total number of cycle rides
= (0 × 3) + (1 × 1) + (2 × 4) + (3 × 7) + (4 × 7) + (5 × 5) + (6 × 4) + (7 × 6) + (8 × 3) + (9 × 0) + (10 × 2)
= 193
Average = 193 ÷ 42 ≈ 4.60 times

(ii) There are 42 values, so the median is the average of the 21st and 22nd values.
The first 15 values are 0, 1, 2 or 3. The next seven values, from the 16th to the 22nd, are all 4.
Median = (4 + 4) ÷ 2 = 4 times

(iii) (a) Not valid. Three students did not use their cycles at all.

(b) Valid. Of the 42 students, 38 used their cycles at least twice. So, almost everyone used their cycle a few times.

(c) Valid. Three students cycled 8 times and two students cycled 10 times in a week. Since a week has only 7 days, these students must have cycled more than once on at least one day.

(d) Not necessarily valid. At least 5 students must have cycled more than once on some days. Some students with 7 or fewer rides may also have cycled twice on one day and not cycled on another day. So, we cannot say that exactly 5 students did so.

(e) Adding 1 to every value increases both the mean and the median by 1.
New total number of rides = 193 + 42 = 235
New average = 235 ÷ 42 ≈ 5.60 times
New median = 4 + 1 = 5 times

Answer:
The smallest number of trials is 1 and the largest is 10.
Minimum = 1 trial
Maximum = 10 trials
Total number of students = 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62
Total number of trials
= (1 × 1) + (2 × 0) + (3 × 0) + (4 × 1) + (5 × 4) + (6 × 9) + (7 × 12) + (8 × 15) + (9 × 10) + (10 × 10)
= 473
Mean = 473 ÷ 62 ≈ 7.63 trials

There are 62 values, so the median is the average of the 31st and 32nd values.
The number of students who took up to 7 trials is 1 + 0 + 0 + 1 + 4 + 9 + 12 = 27.
The next 15 students took 8 trials, so the values from the 28th to the 42nd position are all 8.
Median = (8 + 8) ÷ 2 = 8 trials

Get here the Solutions and Answers of Figure it Out at Page 122.

Answer:
I take the days of the week on the horizontal axis and the average number of customers on the vertical axis.
Scale on the vertical axis: 1 division = 5 customers.
I plot the two sets of values and join consecutive points separately. The two lines show customers visiting the shop and customers purchasing items.

(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?
Answer:
(i) The weather station in each city can record the number of rainy days in every month for several years. For each month, the total number of rainy days is divided by the number of years. This gives the average number of rainy days for that month.

(ii) Rounding the values to the nearest whole number, we get:

I take the months on the horizontal axis and the average number of rainy days on the vertical axis.
Scale on the vertical axis: 1 division = 5 days.
The completed line graph, with New Delhi also shown to the nearest whole day, is:

(iii) Reading the graph approximately and rounding to whole days, we get:

(iv) Adding the monthly averages, we get:
Mangaluru: 112.1 days
Port Blair: 125.8 days
Rameswaram: 42.3 days
New Delhi: approximately 38 days, using the rounded readings.
Therefore, Port Blair receives the most rainy days in a year and New Delhi receives the least.

(v) In New Delhi, the rainy season is mainly from June to September, with the most rainy days in July and August.
In Rameswaram, the rainy season is mainly from October to December, with the most rainy days in November.

(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.
Answer:
(i) I notice that the number of births changes from month to month. It is roughly between 1.4 million and 2 million in a month. The number generally rises around the middle of the year, reaches a high value around August–October and then falls. A similar pattern appears in all three years.

(ii) In July 2017, there were approximately 1.8 million births.
1.8 million = 18 lakh.
Therefore, there were approximately 18 lakh births.

(iii) Counting the monthly points before July 2017 and after January 2020, the graph covers approximately April 2017 to March 2020, a period of 3 years.

(iv) The approximate numbers of births in January are:

January 2018 has the fewest births. The values for January 2019 and January 2020 are close, with January 2020 slightly higher.

(v) Reading the monthly values for 2019 approximately, in millions, we get:

Total births ≈ (1.8 + 1.6 + 1.8 + 1.5 + 1.6 + 1.6 + 1.8 + 2.0 + 2.0 + 2.0 + 1.9 + 1.9) million
= 21.5 million
= 2.15 crore.
Therefore, approximately 2.15 crore births took place in 2019.

Get here the Solutions and Answers of Figure it Out at Page 127.

(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?

Answer:
(i) Each row, column and diagonal has 3 numbers.
Required sum in each direction = 10 × 3 = 30
I fill the grid as follows:

Row sums: 13 + 6 + 11 = 30; 8 + 10 + 12 = 30; 9 + 14 + 7 = 30.
Column sums: 13 + 8 + 9 = 30; 6 + 10 + 14 = 30; 11 + 12 + 7 = 30.
Diagonal sums: 13 + 10 + 7 = 30; 11 + 10 + 9 = 30.
So, the average in every direction is 30 ÷ 3 = 10. All nine numbers are distinct.

(ii) Yes. I interchange the first and last columns:

The sums along all rows, columns and diagonals are still 30. So, their averages remain 10.

(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.
Answer:
(i) 6, 8, 10: Mean = (6 + 8 + 10) ÷ 3 = 8.
4, 8, 12: Mean = (4 + 8 + 12) ÷ 3 = 8.

(ii) 12, 15, 16, 20: Median = (15 + 16) ÷ 2 = 15.5.
10, 14, 17, 25: Median = (14 + 17) ÷ 2 = 15.5.

(iii) 10, 12, 14, 15, 17: Mean = 68 ÷ 5 = 13.6.
8, 11, 13, 16, 20: Mean = 68 ÷ 5 = 13.6.

(iv) 1, 2, 3, 4, 5, 6:
Mean = 21 ÷ 6 = 3.5
Median = (3 + 4) ÷ 2 = 3.5

2, 4, 6, 8, 10, 12:
Mean = 42 ÷ 6 = 7
Median = (6 + 8) ÷ 2 = 7

(v) 1, 2, 3, 4, 5, 15:
Mean = 30 ÷ 6 = 5
Median = (3 + 4) ÷ 2 = 3.5
So, mean > median.

2, 3, 4, 5, 6, 16:
Mean = 36 ÷ 6 = 6
Median = (4 + 5) ÷ 2 = 4.5
So, mean > median.

Answer:
There are 6 values. So, the average of the 3rd and 4th values in increasing order must be 13.
Their sum must be 13 × 2 = 26.
I fill the blanks with 1, 12 and 22.
In increasing order, the collection is 1, 5, 12, 14, 21, 22.
Median = (12 + 14) ÷ 2 = 13
We can replace 22 by 23, 24, 25, and so on. The middle two values will still be 12 and 14.
Since counting numbers have no largest value, infinitely many possibilities exist.

Answer:
Required sum of the 6 numbers = 6.5 × 6 = 39
Sum of the known numbers = 3 + 11 + 15 + 6 = 35
Sum of the two missing numbers = 39 − 35 = 4

The blanks can be filled as:

Thus, there are 3 ways to fill the two blanks using counting numbers.
If we count only different collections and ignore the order of the blanks, there are 2 possibilities: the pair 1, 3 and the pair 2, 2.

(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.
Answer:
(i) This is not always true.
For example, 2 and 4 are even numbers, but their average is (2 + 4) ÷ 2 = 3, which is odd.

(ii) This is not always true.
For example, 5 and 10 are multiples of 5, but their average is (5 + 10) ÷ 2 = 7.5, which is not a multiple of 5.

(iii) This is not always true.
For example, 5, 10, 15, 20 and 30 are five multiples of 5.
Their average = (5 + 10 + 15 + 20 + 30) ÷ 5 = 80 ÷ 5 = 16.
Since 16 is not a multiple of 5, the statement is false.

(i) Which of the following statements are correct? Why?
(a) The average height of the class will increase as there are 2 new values.
(b) The average height of the class will remain the same.
(c) The heights of the new students have to be measured to find out the new average height.
(d) The heights of everyone in the class has to be measured again to calculate the new average height.

(ii) The heights of the two new joinees are 149 cm and 152 cm. Which of the following statements about the class’ average height are correct? Why?
(a) The average will remain the same.
(b) The average will increase.
(c) The average will decrease.
(d) The information is not sufficient to make a claim about the average.

(iii) Which of the following statements about the new class average height are correct? Why?
(a) The median will remain the same.
(b) The median will increase.
(c) The median will decrease.
(d) The information is not sufficient to make a claim about median.
Answer:
(i) Only (c) is correct. We need the heights of the two new students to find the new average. Adding two students alone does not tell us whether the average will increase, decrease or remain the same. The earlier measurements can be used for the other students.

(ii) The correct option is (b): The average will increase.
Average height of the two new students = (149 + 152) ÷ 2 = 150.5 cm
Since 150.5 cm is greater than the earlier average of 150.2 cm, the class average increases slightly.

(iii) The correct option is (d). The mean does not tell us the middle heights in the ordered data. Without those heights, we cannot decide whether the median increases, decreases or remains the same.

Answer:
I compare the total distances of the dots on either side of 17.
To the left of 17, there are two dots at 14, two at 15 and three at 16.
Total distance on the left = (2 × 3) + (2 × 2) + (3 × 1) = 13

To the right of 17, there are four dots at 18, four at 19, three at 20, one at 21 and one at 23.
Total distance on the right = (4 × 1) + (4 × 2) + (3 × 3) + (1 × 4) + (1 × 6) = 31

The distances are unequal, so 17 is not the average.
The larger distance on the right shows that the average is greater than 17.

There are 25 dots in all.
Mean = 17 + (31 − 13) ÷ 25
= 17 + 0.72
= 17.72

Answer:
Change in total weight = −2 + 1 + 1 = 0 kg
The total weight and the number of people remain the same.
Therefore, the mean weight remains 65.3 kg.

The median depends on the middle weights after arranging them in order.
We do not know which people’s weights changed or how the middle weights were affected.
So, the median may increase, decrease or remain 67 kg.

(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?
Answer:
(i) I choose Gujarat, Uttar Pradesh, and West Bengal.
Rounding their prices to the nearest counting number, we get:

I take the years on the horizontal axis and the price on the vertical axis.

Scale on the horizontal axis: 1 division = 1 year.
Scale on the vertical axis: 1 division = ₹5.

(ii) I notice that the price in 2025 is higher than in 2016 in all six places. Mizoram has the highest price in every year. Gujarat’s price remains ₹14.75 for three years, from 2017 to 2019. West Bengal’s price rises every year after 2020.

(iii) In Gujarat, the price falls from ₹16.50 in 2016 to ₹13 in 2020.
After a few small changes, it rises to ₹19.20 in 2025.
Overall increase in Gujarat = ₹19.20 − ₹16.50 = ₹2.70.

In Uttar Pradesh, the price generally rises from ₹16.15 in 2016 to ₹26.90 in 2024, though it falls slightly in 2018. It then falls to ₹24.81 in 2025.
Overall increase in Uttar Pradesh = ₹24.81 − ₹16.15 = ₹8.66.
The price changes more in Uttar Pradesh than in Gujarat.

(iv) Increase in price from 2016 to 2025:
Andaman and Nicobar Islands: ₹20.99 − ₹16 = ₹4.99
Assam: ₹12.35 − ₹6 = ₹6.35
Gujarat: ₹19.20 − ₹16.50 = ₹2.70
Mizoram: ₹29.80 − ₹20 = ₹9.80
Uttar Pradesh: ₹24.81 − ₹16.15 = ₹8.66
West Bengal: ₹23.99 − ₹9.47 = ₹14.52
Therefore, the price increased the most in West Bengal, by ₹14.52.

(v) I am curious to find out why the price of the same item differs so much between places. I would also like to compare the prices in other months and find out whether transport costs affect them.

(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.
Answer:
(i) Valid. In 1983, more than 80% of rural households used kerosene and more than 60% of urban households used electricity. Both are more than 50%, so they form a majority.

(ii) Valid. The lines for kerosene move downwards over time in both rural and urban areas. Therefore, the percentage of households using kerosene as their primary lighting source decreased.

(iii) Not valid. In 2000, about 90% of urban households used electricity as their primary lighting source. About 10% used kerosene.

(iv) We cannot conclude this. The graph shows the primary lighting source used by households. It does not show whether electricity was available all the time or whether power cuts occurred.

(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
(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.

Answer:
(i) Children aged 10 in urban areas spend approximately 2 hours a day on hobbies and games, on average.

(ii) The rural line is at 1.5 hours at about 14 years of age.
Therefore, the correct option is (d) 14 years.

(iii) (a) Incorrect. Children aged 15 spend less time on hobbies and games than children aged 10. The average for 15-year-olds is roughly half that for 10-year-olds, not twice as much.
(b) We cannot conclude this. The graph gives the average time for rural children aged 15. Some children may spend less than 1 hour and others may spend more than 1 hour. An average does not give the time spent by every child.

(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.
Answer:
I recorded my activities from Monday to Sunday. The time spent on each activity is shown below. All the times are in hours.

Free time includes playing, meeting friends, hobbies and time with my family. Care/exercise includes bathing, getting dressed and exercise. Each day’s total is 24 hours.

(i) I have lunch at 12 noon and dinner at 7 pm every day. I have breakfast at 6:30 am on school days and at 7:30 am on Saturday and Sunday. So, my meal times are mostly regular.
From Monday to Thursday, I go to bed at 10 pm. On Friday, Saturday and Sunday, I go to bed at 9:30 pm. I wake up at 6 am on school days and at 7 am on Saturday and Sunday.
I usually play outdoors for 1 hour on school days and for 2 hours on Saturday and Sunday. This is part of my free time.

(ii) Average time spent on an activity = Total time spent in the week ÷ 7.

Total average time = 8.5 + 1.5 + 7 + 4.5 + 1 + 1.5 = 24 hours.
I divided each strip into 48 equal boxes. Each box represents half an hour. My average-day strip has 17 boxes for sleep, 3 for meals, 14 for school/study, 9 for free time, 2 for care/exercise and 3 for travel.

(iii) I also recorded my mother’s activities for the same week. All the times below are in hours.

Dividing each weekly total by 7, we get:

I sleep 1.5 hours more than my mother each day on average. She spends 1 hour more on work and housework than I spend on school and study. Our average time for meals, free time and travel is the same.
The first seven strips show my activities in time order. The last two strips compare the average time spent on each activity.

(i) Track daily sleep time of all your family members for a week. Daily sleep time includes night sleep, naps, and any sleep during the day.
(a) Represent this on strips.
(b) Put together the data of all your group members. Calculate the average and median sleep time of children, adults, elderly.
(c) Share your findings and observations.
(ii) When do schools start and end? On a weekday, Manoj’s school starts at 9:30 am and ends at 4:30 pm, i.e., 7 hours which include class time and breaks. Collect information on the daily timings of different schools for Grade 8, including class time and break time (the schools can be anywhere in the country. You can ask your neighbours, relatives, parents and friends to find out). Analyse and present the data collected.
Answer:
I formed a group with Riya and Arjun. We chose activity (i). We recorded the daily sleep time of everyone in our three families for one week. We included night sleep and daytime naps.
There are 3 children, 6 adults and 3 elderly people in our records.
Riya, Arjun and I are the children, our parents are the adults, and our grandparents are the elderly people.

All the sleep times below are in hours:

(a) We divided each daily strip into 48 equal boxes. Each box represents half an hour. We coloured the boxes for the total sleep time. The other boxes show the remaining hours of the day.

(b) We combined all seven daily records for every person in each group.
Children:
Number of daily records = 3 × 7 = 21.
Total sleep time = 59.5 + 63 + 59.5 = 182 hours.
Average daily sleep time = 182 ÷ 21 = 8⅔ hours = 8 hours 40 minutes.
On arranging the 21 daily sleep times in ascending order, the middle value is the 11th value.
Median sleep time = 8.5 hours = 8 hours 30 minutes.

Adults:
Number of daily records = 6 × 7 = 42.
Total sleep time = 49 + 49 + 49 + 52.5 + 52.5 + 52.5 = 304.5 hours.
Average daily sleep time = 304.5 ÷ 42 = 7.25 hours = 7 hours 15 minutes.
On arranging the 42 daily sleep times in ascending order, the two middle values are the 21st and 22nd values. Both are 7 hours.
Median sleep time = (7 + 7) ÷ 2 = 7 hours.

Elderly people:
Number of daily records = 3 × 7 = 21.
Total sleep time = 59.5 + 56 + 63 = 178.5 hours.
Average daily sleep time = 178.5 ÷ 21 = 8.5 hours = 8 hours 30 minutes.
The middle value of the 21 daily sleep times in ascending order is the 11th value.
Median sleep time = 8.5 hours = 8 hours 30 minutes.

(c) Our observations:

• In our three families, the children slept the longest on average, followed by the elderly people and then the adults.
• The children slept 1 hour 25 minutes more per day than the adults on average.
• Riya and I slept longer on Saturday and Sunday than on school days. Arjun’s sleep time stayed the same throughout the week.
• Some family members had the same sleep time every day, while others had different sleep times on different days.
• These observations describe the people in our three families during this week.

Answer the following questions based on the graphs:
(i) At which place does the sun rise the earliest in January? What is the approximate day length at this place in January?
(ii) Which place has the longest day length over the year?
(iii) Share your observations – what do you find interesting? What are you curious to find out?
Answer:
The lower lines show sunrise times, and the upper lines show sunset times. The difference between the sunset and sunrise times gives the length of the day.
(i) In January, the sun rises the earliest at Kibithu.
Sunrise ≈ 6:00 am
Sunset ≈ 4:30 pm
Day length ≈ 4:30 pm − 6:00 am
= 10 hours 30 minutes.
Therefore, the approximate day length is 10½ hours.

(ii) Srinagar has the longest day during the year. Around June, the gap between its sunrise and sunset lines is the greatest, giving a day length of about 14½ hours.

(iii) I notice that sunrise and sunset generally occur earlier in Kibithu than in Ghuar Moti. The length of the day changes a lot during the year in Srinagar, but changes much less in Kanyakumari. In most of these places, days are longer around June than around December.
I am curious to find out why places in the same country have different sunrise times and why the change in day length is so small in Kanyakumari.

(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?
Answer:
(i) Around amavasya, the moon rises near sunrise and sets near sunset. Around the 28th, the graph shows moonrise near 6:00 am and moonset near 5:00 pm. So, amavasya was around the 28th.
Around purnima, the moon rises near sunset and sets near sunrise. Around the 13th–14th, moonrise is near 6:00 pm and moonset is near 6:00 am. So, purnima was around the 13th–14th.
These are approximate dates read from the graph.

(ii) I notice that the moon generally rises and sets later on successive days. It can rise during the day as well as at night. The lines restart at the bottom when the time passes midnight; the moon does not suddenly change its motion.
I wonder why the moon rises later each day and whether this pattern is the same in every month and at every place.

Practise interpreting data with 10 Tales by Dots and Lines test papers for Class 8 Maths, Ganita Prakash Part 2 Chapter 5. Revise mean and median, read dot plots and line graphs and explain what changes in a data set tell you. Download the PDF papers with worked solutions to check your calculations and your interpretation of the information shown.

Four slips cost marks in this data handling chapter. With frequency tables, multiply before adding; skipping that step gives the book’s trap answer of 6.5 instead of 5.22. Sort the data before picking the median. When every value goes up or down by the same amount, the mean shifts by the same amount; no need to recalculate. On line graphs, read the scale first and claim only what the graph shows: a lighting graph says nothing about power cuts.
Each of these checks appears in the Tales by Dots and Lines Class 8 solutions above. Next comes Chapter 6, Algebra Play and the rest of the book is covered in our NCERT Solutions for Class 8 Maths.