CBSE Class 8 Mathematics Part II Chapter 5 Questions and Answers

Class 8 Maths

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

---

Figure it Out (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:
For evenly spaced numbers, the mean equals the sum of the first and last terms divided by 2.
Mean = (First term + Last term) / 2

(i) The first 50 natural numbers are: 1, 2, 3, …, 50
Mean = (1 + 50) / 2 = 51 / 2 = 25.5.

(ii) The first 50 odd numbers are: 1, 3, 5, …, 99
Mean = (1 + 99) / 2 = 100 / 2 = 50.

(iii) The first 50 multiples of 4 are: 4, 8, 12, …, 200
Mean = (4 + 200) / 2 = 204 / 2 = 102.

Observations:
(i) The mean for the first n natural numbers is (n + 1) / 2.
(ii) The mean for the first n odd numbers is n.
(iii) The mean of the first n multiples of 4 is 4 times the mean of the first n natural numbers.

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:
Mean = 9
Let the missing value be x.
Total values are: 4, 7, 8, 8, 9, 9, 9, 9, 9, 11, x.
Mean = (4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 + x) / 11
9 = (83 + x) / 11
99 = 83 + x
x = 99 – 83 = 16.

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) No, the teacher does not need to measure all the heights again. Since each student’s height was increased by 1 cm due to the shoes, we can simply subtract 1 cm from the calculated average height.
(ii) Correct average height = 150.2 − 1 = 149.2 cm.
Therefore, the correct option 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:
By observing the dot plots, the song lengths in Album A are mostly clustered between 5 and 6.5 minutes, so its mean is around 5.57 minutes. The song lengths in Albums B and C are generally smaller, resulting in means less than 5.57 minutes.
Therefore, 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:
Number of observations = 16 (even)
Median = Average of 8^th and 9^th terms = (41 + 41) / 2 = 41.

(i) After including one value, there will be 17 observations. The median becomes the 9^th observation. To keep the median 41, the new value can be 41.
(ii) After including two values, there will be 18 observations. The median will be the average of the 9^th and 10^th observations. To keep the median 41, we can add any two numbers whose average is 41 (e.g., one value less than or equal to 41 and one value greater than or equal to 41, such that the middle two remain 41).
(iii) After removing one value, there will be 15 observations. The median becomes the 8^th observation. Since 41 occurs twice, removing one 41 leaves the median unchanged at 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) Sometimes true: Removing a value less than the median shifts the middle position to the right, which may increase the median or leave it unchanged; it will not decrease it.
(ii) Always true: Adding any value lower than the current average always lowers the overall average.
(iii) Sometimes true: Adding 4 values may or may not shift the middle observation(s).
(iv) Never true: Adding values strictly below the median shifts the center to the left, which can only decrease or keep the median unchanged, never increase it.

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

Solution:
Number of observations = 8
Mean = Sum of observations / 8
10.375 = (8 + 13 + 10 + 4 + 5 + 20 + y + 10) / 8
10.375 = (70 + y) / 8
70 + y = 10.375 × 8 = 83
y = 83 – 70 = 13.

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

Solution:
Mean = Sum of observations / Number of observations
134 = Sum of observations / 15
Sum of observations = 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:
Arranging the known 10 values in ascending order:
8, 8, 12, 18, 25, 29, 35, 39, 47, 73
Total terms with p = 11 (odd). The median is the 6^th term.
Since the 6^th term is already 29, p must be placed at or after 29 so that 29 remains in the 6^th position.
Hence, p ≥ 29.
Possible values of p: 29, 30, 40, 47, 100.

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) Some students 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:

Total students = 3 + 1 + 4 + 7 + 7 + 5 + 4 + 6 + 3 + 2 = 42
(i) Total cycling sessions = (0 × 3) + (1 × 1) + (2 × 4) + (3 × 7) + (4 × 7) + (5 × 5) + (6 × 4) + (7 × 6) + (8 × 3) + (10 × 2)
= 0 + 1 + 8 + 21 + 28 + 25 + 24 + 42 + 24 + 20 = 193
Mean = 193 / 42 ≈ 4.6 times.

(ii) Median is the average of the 21^st and 22^nd observations.

Both 21^st and 22^nd terms are 4. Median = (4 + 4) / 2 = 4.

(iii)
(a) Invalid: 3 students cycled 0 times.
(b) Valid: Most students rode their cycle between 2 and 8 times.
(c) Invalid: The data gives weekly totals, not daily breakdown.
(d) Invalid: Daily frequency cannot be determined from weekly totals.
(e) Adding 1 to every score increases both mean and median by 1:
New Mean = 4.59 + 1 = 5.59
New Median = 4 + 1 = 5

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:
(i) Minimum = 1
(ii) Maximum = 10
(iii) Total participants = 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62
Total throws = (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.

(iv) Median is the average of 31^st and 32^nd observations.

Both 31^st and 32^nd values are 8.
Median = (8 + 8) / 2 = 8.

Figure it Out (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:

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) Line graph.
(ii)

(iii)

(iv) Mangaluru receives the most number of rainy days per year; Rameswaram gets the least.
(v) In New Delhi: June to August. In Rameswaram: October to December.

3. 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) Observations:
1. Births peak in the mid-to-late months of each year and dip around the beginning of the year.
2. The peak is close to 1.9 million per month, while the lowest is around 1.4 million per month.
(ii) Approximately 1.8 million.
(iii) April 2017 to March 2020.
(iv)

Births in January showed a slight steady increase from 2018 to 2020.
(v) Average monthly births in 2019 ≈ 1.7 million. Estimated annual births = 1.7 × 12 ≈ 20.4 million (about 20 million).

Figure it Out (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) A 3×3 grid requires a row/column/diagonal sum of 3 × 10 = 30.

(ii) Yes, by rotating or reflecting the numbers:

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 = 3 × 8 = 24. Examples: (6, 8, 10) and (5, 7, 12).
(ii) Middle two numbers must sum to 31. Examples: (12, 15, 16, 25) and (10, 15, 16, 20).
(iii) Sum = 5 × 13.6 = 68. Examples: (10, 12, 14, 15, 17) and (8, 12, 14, 16, 18).
(iv) Examples: (2, 4, 6, 8, 10, 12) [Mean = Median = 7] and (5, 7, 9, 11, 13, 15) [Mean = Median = 10].
(v) Examples: (1, 2, 3, 4, 5, 15) [Mean = 5, Median = 3.5] and (2, 4, 6, 8, 10, 30) [Mean = 10, Median = 7].

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:
Total numbers = 6. Median is the average of the 3^rd and 4^th terms in sorted order.
(3^rd term + 4^th term) / 2 = 13 → 3^rd term + 4^th term = 26.
Any choice of counting numbers where the two middle values add up to 26 will work (e.g., 5, 10, 13, 13, 14, 21).
Therefore, infinitely many possibilities exist.

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:
Let the missing numbers be a and b.
(3 + 11 + a + b + 15 + 6) / 6 = 6.5
35 + a + b = 39 → a + b = 4.
Possible pairs of counting numbers (natural numbers): (1, 3), (2, 2), and (3, 1).
Therefore, there are 3 possibilities.

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: (2x + 2y) / 2 = x + y, which can be odd (e.g., (2 + 4) / 2 = 3).
(ii) False: (5m + 5n) / 2 = 5(m + n)/2, which is an integer multiple of 5 only when m + n is even (e.g., (5 + 10) / 2 = 7.5).
(iii) True: (5a + 5b + 5c + 5d + 5e) / 5 = a + b + c + d + e, which is always an integer, so the result is 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 statement is correct?
(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.
Solution: (c)

(ii) The heights of the two new joinees are 149 cm and 152 cm. Which statement is correct?
(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.
Solution: Average of the new joinees = (149 + 152) / 2 = 150.5 cm. Since 150.5 > 150.2, the overall average increases. (b)

(iii) Which statement about the new class median height is correct?
(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.
Solution: (d)

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

Solution:

Total observations = 25
Sum = (14 × 2) + (15 × 2) + (16 × 3) + (17 × 5) + (18 × 4) + (19 × 4) + (20 × 3) + (21 × 1) + (23 × 1)
= 28 + 30 + 48 + 85 + 72 + 76 + 60 + 21 + 23 = 443
Mean = 443 / 25 = 17.72 ≠ 17.
Therefore, 17 is not the average.

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:
Net weight change = (−2) + 1 + 1 = 0 kg. Since the total sum of weights is unchanged, the mean remains 65.3 kg.
The change in the median cannot be determined without knowing the exact positions of the affected individuals in the ordered list.

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.
(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)

(ii) Observations: Overall salt prices show an upward trend over the decade; Mizoram has the highest prices; Gujarat shows the least variation.
(iii) The price variation is significantly higher in Uttar Pradesh than in Gujarat.
(iv)

West Bengal had the largest price increase.
(v) Exploring transportation, state taxation, and local production differences affecting commodity pricing.

10. Referring to the graph below, which of the following statements are valid? Why?

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

Solution:
(i) Valid: ~83% rural households used kerosene, while ~65% urban households used electricity.
(ii) Valid: Both areas show a consistent downward trajectory for kerosene use.
(iii) Not valid: In 2000, ~90% of urban households used electricity.
(iv) Not valid: The chart only covers the primary energy source for lighting, not power outage frequency.

11. Answer the following questions based on the line graph.

(i) How long do children aged 10 in urban areas spend each day on hobbies and games?
(ii) At what age is the average time spent daily on hobbies and games by rural kids 1.5 hours?
(a) 8 years (b) 10 years (c) 12 years (d) 14 years (e) 18 years
(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 every day.

Solution:
(i) A little more than 2 hours.
(ii) (d) 14 years.
(iii) (a) Incorrect. (b) Correct.

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

Solution:
Do it yourself.

13. Small group project: Sleep time and school timing tracking.

Solution:
Do it yourself.

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 (~6:00 a.m.). Approximate day length = 4:30 p.m. − 6:00 a.m. = 10 hours 30 minutes.
(ii) Srinagar in June (Sunrise ~5:00 a.m., Sunset ~7:45 p.m., Day length ≈ 14 hours 45 minutes).
(iii) Srinagar shows the greatest day-length variation, while Kanyakumari shows the least.

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) Purnima: 7^th day; Amavasya: 21^st day.
(ii) The Moon rises and sets approximately 50 minutes later each day, completing a cycle of about 29.5 days between corresponding lunar phases.