NCERT Solutions: Tales by Dots and Lines
Page No. 103
Q: Calculate and mark the mean of each collection of data below.
Ans:
(a) Collection: 5, 7, 9
Mean = (5 + 7 + 9) ÷ 3 = 21 ÷ 3 = 7
The mean is 7.
(b) Collection: 2, 4, 6, 8
Mean = (2 + 4 + 6 + 8) ÷ 4 = 20 ÷ 4 = 5
The mean is 5.
(c) Collection: 10, 10, 11, 17
Mean = (10 + 10 + 11 + 17) ÷ 4 = 48 ÷ 4 = 12
The mean is 12.
Page No. 104
Math Talk
Q: Can you explain how the mean is the centre of each collection?
Ans: The mean is the centre because the sum of distances of all points to the left of the mean equals the sum of distances of all points to the right of the mean.
For example, in collection (c): 10, 10, 11, 17 with mean = 12:
- Distance on left: (12 – 10) + (12 – 10) + (12 – 11) = 2 + 2 + 1 = 5
- Distance on right: (17 – 12) = 5
Both distances are equal (5 units each), showing the mean balances the data.
Q: Mark the mean for the collections below.
Ans: (a) Collection: 3, 5, 7, 9, 11
Mean = (3 + 5 + 7 + 9 + 11) ÷ 5 = 35 ÷ 5 = 7
(b) Collection: 1, 3, 5, 7
Mean = (1 + 3 + 5 + 7) ÷ 4 = 16 ÷ 4 = 4
Math Talk
Q: Can you explain how the mean is the centre of each collection?
Ans: For collection (a): 3, 5, 7, 9, 11 with mean = 7:
- Distances on left: (7 – 3) + (7 – 5) = 4 + 2 = 6
- Distances on right: (9 – 7) + (11 – 7) = 2 + 4 = 6
For collection (b): 1, 3, 5, 7 with mean = 4:
- Distances on left: (4 – 1) + (4 – 3) = 3 + 1 = 4
- Distances on right: (5 – 4) + (7 – 4) = 1 + 3 = 4
In both cases, the total distances on both sides are equal, confirming the mean is the centre.
Q: Is the mean the midpoint of the two endpoints/extremes of the data?
Ans: No, the mean is not always the midpoint of the two extremes.
For example, in the collection 10, 10, 11, 17:
- Midpoint of extremes = (10 + 17) ÷ 2 = 27 ÷ 2 = 13.5
- Mean = 12
The mean (12) is different from the midpoint of extremes (13.5).
Instead, the mean is the point where the sum of distances on the left equals the sum of distances on the right.
Q: Can there be more than one such ‘centre’? In other words, is there any other value such that the sum of the distances to the values lower than it and the values higher than it will still be equal?
Ans: No, there cannot be more than one such centre.
Proof: Consider the collection 10, 10, 11, 17 with mean = 12.
If we take any value greater than 12:
- All distances on the left side will increase
- All distances on the right side will decrease
- The balance is broken
If we take any value less than 12:
- All distances on the left side will decrease
- All distances on the right side will increase
- Again, the balance is broken
Therefore, there is only one unique centre (the mean) where the sum of distances on both sides is equal.
Page No. 105
Q: Will including a new value in the data increase or decrease the mean?
Ans: It depends on the value being included:
1. If the new value is greater than the current mean: The mean will increase.
- Example: Data: 4, 6, 8 (mean = 6)
- Adding 10: New data: 4, 6, 8, 10
- New mean = (4 + 6 + 8 + 10) ÷ 4 = 28 ÷ 4 = 7
- The mean increased from 6 to 7.
2. If the new value is less than the current mean: The mean will decrease.
- Example: Data: 4, 6, 8 (mean = 6)
- Adding 2: New data: 2, 4, 6, 8
- New mean = (2 + 4 + 6 + 8) ÷ 4 = 20 ÷ 4 = 5
- The mean decreased from 6 to 5.
3. If the new value equals the mean: The mean will remain the same.
Math Talk
Q: What happens to the mean when an existing value is removed? When will the mean increase, decrease, or stay the same?
Ans: It depends on which value is removed:
- If a value greater than the mean is removed: The mean will decrease.
- Example: Data: 2, 4, 6, 10 (mean = 5.5)
- Removing 10: New data: 2, 4, 6
- New mean = (2 + 4 + 6) ÷ 3 = 12 ÷ 3 = 4
- The mean decreased.
- If a value less than the mean is removed: The mean will increase.
- Example: Data: 2, 4, 6, 10 (mean = 5.5)
- Removing 2: New data: 4, 6, 10
- New mean = (4 + 6 + 10) ÷ 3 = 20 ÷ 3 = 6.67
- The mean increased.
- If a value equal to the mean is removed: The mean will stay the same.
Q: What happens to the mean if a value equal to the mean is included or removed?
Ans: Including a value equal to the mean: The mean remains unchanged.
- Example: Data: 3, 5, 7 (mean = 5)
- Adding 5: New data: 3, 5, 5, 7
- New mean = (3 + 5 + 5 + 7) ÷ 4 = 20 ÷ 4 = 5
- Mean stays 5.
Removing a value equal to the mean: The mean remains unchanged.
- Example: Data: 3, 5, 5, 7 (mean = 5)
- Removing 5: New data: 3, 5, 7
- New mean = (3 + 5 + 7) ÷ 3 = 15 ÷ 3 = 5
- Mean stays 5.
Fair-share interpretation: When we include or remove a value equal to the mean, we’re not changing the “fair share” amount. Each person already has their fair share, so adding or removing someone with exactly the fair share doesn’t change what everyone else has.
Page No. 106
Q: Explore if it is possible to include or remove 2 values such that the mean is unchanged.
Ans: Yes, it is possible. We need to include or remove two values whose average equals the current mean.
Example: Data: 4, 5, 6, 9 (mean = 6)
To keep the mean at 6, we can:
- Include 5 and 7 (their average = 6)
- New data: 4, 5, 5, 6, 7, 9
- New mean = (4 + 5 + 5 + 6 + 7 + 9) ÷ 6 = 36 ÷ 6 = 6
Or we can remove values:
- Remove 4 and 8 if 8 was in the data (their average = 6)
- This keeps the mean unchanged.
Q: How about including or removing 3 values without changing the mean? Is it possible?
Ans: Yes, it is possible. We need to include or remove three values whose average equals the current mean.
Example: Data: 3, 6, 9 (mean = 6)
To keep the mean at 6:
- Include 4, 6, and 8 (their average = 6)
- New data: 3, 4, 6, 6, 8, 9
- New mean = (3 + 4 + 6 + 6 + 8 + 9) ÷ 6 = 36 ÷ 6 = 6
Q: Can we include 2 values less than the mean and 1 value greater than the mean, so that the mean remains the same?
Ans: Yes, this is possible if the total of the three values equals 3 times the mean.
Example: Data: 5, 7, 9 (mean = 7)
To keep mean at 7:
- We need 3 values that total to $3 \times 7$ = 21
- Include 4, 6 (both less than 7) and 11 (greater than 7)
- Check: 4 + 6 + 11 = 21
- New data: 4, 5, 6, 7, 9, 11
- New mean = (4 + 5 + 6 + 7 + 9 + 11) ÷ 6 = 42 ÷ 6 = 7
Q: Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same.
Ans:
Example: Data: 5, 7, 9 (mean = 7)
To keep mean at 7:
- We need 3 values that total to $3 \times 7$ = 21
- Include 8, 10 (both greater than 7) and 3 (less than 7)
- Check: 8 + 10 + 3 = 21
- New data: 3, 5, 7, 8, 9, 10
- New mean = (3 + 5 + 7 + 8 + 9 + 10) ÷ 6 = 42 ÷ 6 = 7
Relatively Unchanged!
Q: Consider the data: 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Calculate its mean.
Ans: Sum = 8 + 3 + 10 + 13 + 4 + 6 + 7 + 7 + 8 + 8 + 5 = 79
Number of values = 11
Mean = 79 ÷ 11 = 7.18 (approximately)
Q: Now, consider this data with every value increased by 10: 18, 13, 20, 23, 14, 16, 17, 17, 18, 18, 15. What is its mean? Is there a quicker way to find out? [Hint: Observe the following dot plots corresponding to the two data collections.]
Ans: Method 1 (Direct calculation):Sum = 18 + 13 + 20 + 23 + 14 + 16 + 17 + 17 + 18 + 18 + 15 = 189 Mean = 189 ÷ 11 = 17.18 (approximately)
Method 2 (Quicker way):Since every value increased by 10, the mean also increases by 10. New mean = Original mean + 10 = 7.18 + 10 = 17.18
This is much quicker!
Observation: The relative position of the mean stays the same. The entire data set shifts up by 10, so the mean also shifts up by 10.
Page No. 107Try This
Q: Try to explain, using algebra, what the average is when a fixed number, e.g., 2 is subtracted from every value in the collection.
Ans: Let there be n values: x₁, x₂, x₃, … xₙ
Original average = (x₁ + x₂ + x₃ + … + xₙ) ÷ n = a
When 2 is subtracted from every value:
New average = [(x₁ – 2) + (x₂ – 2) + (x₃ – 2) + … + (xₙ – 2)] ÷ n
= [x₁ + x₂ + x₃ + … + xₙ – 2n] ÷ n
= [(x₁ + x₂ + x₃ + … + xₙ) ÷ n] – (2n ÷ n)
= a – 2
Therefore, the new average is 2 less than the original average.
Q: Try to explain this using the fair-share interpretation of average that you learnt last year.
Ans: Do it Yourself.
Q: What happens to the average if every value in the collection is doubled?
Ans: The average also doubles.
Example: Consider data: 3, 5, 7
Original mean = (3 + 5 + 7) ÷ 3 = 15 ÷ 3 = 5
Doubled data: 6, 10, 14
New mean = (6 + 10 + 14) ÷ 3 = 30 ÷ 3 = 10
New mean = 2 × Original mean
Algebraic Proof:
Let the n values be x₁, x₂, x₃, … xₙ with average a.
Original average: (x₁ + x₂ + x₃ + … + xₙ) ÷ n = a
When every value is multiplied by 5:
New average = [(5x₁) + (5x₂) + (5x₃) + … + (5xₙ)] ÷ n
= [5(x₁ + x₂ + x₃ + … + xₙ)] ÷ n (using distributive property)
= 5 × [(x₁ + x₂ + x₃ + … + xₙ) ÷ n]
= 5 × a
= 5a
Therefore, when every value is multiplied by 5, the new average is 5 times the original average.
Similarly, when every value is doubled, the average also doubles.
Page No. 113
Figure it Out
Math Talk
Q. Find the mean of the following data and share your observations:
(i) The first 50 natural numbers.
Ans: The first 50 natural numbers are: 1, 2, 3, 4, …, 50
Sum of first n natural numbers = n(n + 1) ÷ 2
Sum = $50 \times 51$ ÷ 2 = 2550 ÷ 2 = 1275
Mean = 1275 ÷ 50 = 25.5
Observation: The mean of the first n natural numbers is always (n + 1) ÷ 2. For 50 numbers, it’s 51 ÷ 2 = 25.5.
(ii) The first 50 odd numbers.
Ans: The first 50 odd numbers are: 1, 3, 5, 7, …, 99
Sum of first n odd numbers = n²
Sum = 50² = 2500
Mean = 2500 ÷ 50 = 50
Observation: The mean of the first n odd numbers is always n. For 50 odd numbers, the mean is 50.
(iii) The first 50 multiples of 4.
Ans: The first 50 multiples of 4 are: 4, 8, 12, 16, …, 200
This is 4 times the first 50 natural numbers.
Sum = 4 × (sum of first 50 natural numbers) = $4 \times 1275$ = 5100
Mean = 5100 ÷ 50 = 102
Observation: The mean of the first n multiples of 4 is 4 times the mean of the first n natural numbers. Mean = $4 \times 25$.5 = 102.
Q. 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).
Ans: From the dot plot, the data values
4, 7, 8, 8, 9, 9, 9, 9, 9, 11, x
Number of observations = 11
Mean = 9
Sum of data values = 4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 = 83
Total sum = 9 × 11 = 99
Missing number = 99 – 83 = 16
Page No. 114
Q. 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?
Ans: (i) The teacher does not need to remeasure all the heights. Since the shoes add 1 cm to every student’s height, the average height with shoes is 1 cm more than the actual average height. So the correct average height is 1 cm less than the measured average height.
(ii) What is the correct average height of the class?
(ii) Correct average height = 150.2 cm – 1 cm = 149.2 cm
∴ Option (d) is correct.
Q. 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.
Ans: We can determine it by examining the dot plots. In dot plot A, most song lengths are between 5 and 6.5 minutes, so the mean is likely around 5.57 minutes. In dot plots B and C, all song lengths are below 5.57 minutes, so their means cannot exceed 5.57. Therefore, dot plot A is the one with a mean of 5.57 minutes.
Check:
For Album A
Data values = 5, 5, 5.25, 5.5, 5.75, 6, 6.5
= 5.57 minutes
For Album B
Data values = 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5
= 2.41 minutes
For Album C
Data values = 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5
= 3.9 minutes
Album (A) has a mean of 5.57 minutes.
Math Talk
Q. Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92.
Ans: The data is already in ascending order.
Number of values = 16 (even)
Median = Average of 8th and 9th values
8th value = 41, 9th value = 41
Median = (41 + 41) ÷ 2 = 82 ÷ 2 = 41
(i) If we include one value to the data (in the given list) without affecting the median, what could that value be?
Ans: To keep the median at 41, the new value should be 41 itself.
When we add 41, the data becomes: 8, 10, 19, 23, 26, 34, 40, 41, 41, 41, 48, 51, 55, 70, 84, 91, 92 (17 values)
The 9th value (middle value) is still 41.
Median remains 41
(ii) If we include two values to the data without affecting the median, what could the two values be?
Ans: With n = 18 (even) the median is the average of the 9th and 10th values.
To keep the median 41, we need to add two numbers whose sum is 82.
For this, one value should be less than 41 and the other greater than 41.
For example, the two values could be 40 and 42.
(iii) If we remove one value from the data without affecting the median, what could the value be?
Ans: With n = 15 (odd) the median is the 8th value that is 41.
So we can remove another 41 from the ordered list.
Page No. 115
Q. 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.
Ans: (i) Sometimes true: Removing a value less than the median can decrease the median. But if the data set has an even number of elements and the removed value is below the lower of the two middle values, the median may stay the same.
(ii) Always true: The mean is the sum of all values divided by the number of values. Adding a value less than the mean decreases the total sum less than proportionally to the increase in the number of values, thus decreasing the mean.
(iii) Sometimes true: Including any 4 values could shift the median depending on whether those values are above or below the median and on the original number of observations.
(iv) Never true: Including values less than the median will either keep the median the same or decrease it, but it will never increase it.
Q. The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.
Ans: Mean = Sum of all values ÷ Number of values
10.375 = (8 + 13 + 10 + 4 + 5 + 20 + y + 10) ÷ 8
10.375 = (70 + y) ÷ 8
Multiplying both sides by 8:
83 = 70 + y
y = 83 – 70 = 13
The value of y is 13.
Q8. The mean of a set of data with 15 values is 134. Find the sum of the data.
Ans: Mean = Sum of data ÷ Number of values
134 = Sum ÷ 15
Sum = $134 \times 15$ = 2010
The sum of the data is 2010.
Q. 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
Ans: Arranging the data (excluding p)
8, 8, 12, 18, 25, 29, 35, 39, 47, 73
Total observations = 11 (odd)
Median = (n+1/2)th term
= 6th term
= 29
So, p must be ≥ 29 to keep the value 29 in the middle.
So possible values of p are (iii) 40, (iv) 100, (v) 29, (vi) 47, and (vii) 30.
Q. 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.
Ans:
Sum = (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
Total number of students = 42
Average = 193/42 = 4.59 times
(ii) Find the median number of times students rode their cycles.
Ans: (ii) Total number of students = 42 (even)
So Median = average of 21st and 22nd terms
The 21st and 22nd terms lie in the cumulative frequency corresponding to 4.
So median = 4+4/2 = 4
Median number of times students rode their cycles = 4
(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?
(iii) (a) Invalid
(b) Valid
(c) Valid
(d) Invalid
(e) Next week’s mean and median will be Mean = previous mean + 1
= 4.59 + 1
= 5.59
Median: Now the 21st and 22nd terms lie in the cumulative frequency corresponding to 5.
So Median = 5+5/2 = 5
Page No. 116
Q. 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.
Given table:
Ans: The minimum number of trials is 1.
The maximum number of trials is 10.
Median:
Median = average of (62/2)th term and (62/2+1)th term
= average of 31th and 32th terms
The 31st and 32nd terms lie in the cumulative frequency corresponding to 8.
Page No. 122-123Figure it Out
Q: 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.
Ans: 
Q2: 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?
Ans: Line graph
(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.
Ans: 
(iii) Based on the line for New Delhi in the graph, fill the data in the table.
Ans: 
(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?
Ans: Mangaluru receives the most number of days of rainfall per year, and Rameswaram receives the least.
(v) Looking at the table, when is the rainy season in New Delhi and Rameswaram?
Ans: In New Delhi rainy season is from June to August, and in Rameswaram, it is from September to December.
Q: 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.
Ans: (i) There is both increasing and decreasing trends in the number of births in every month.
(ii) Approx 1.8 M.
(iii) The graph captures the time period from July 2017 to Jan 2020.
(iv) The number of births in Jan 2018 is approx 1.6 M, in Jan 2019 approx 1.8 M, and in Jan 2020 approx 1.9 M. That means it is increasing continuously.
(v) The number of births in the year 2019 is approx 1.9 M.Page No. 127Figure it Out
Q: Mean Grids:
(i) Fill the grid with 9 distinct numbers such that the average along each row, column, and diagonal is 10.
Ans:
(ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions?
Ans: Yes, we can fill by changing any number that requires adjusting other numbers to keep row, column, and diagonal sums equal to 3 × average = 3 × 10 = 30.
Q: Give two examples of data that satisfy each of the following conditions:
(i) 3 numbers whose mean is 8.
Ans: 3 numbers whose mean is 8 are
(a) 6, 8, and 10
(b) 7, 8, and 9.
(ii) 4 numbers whose median is 15.5.
Ans: 4 numbers whose median is 15.5 are
(a) 10, 15, 16, and 20
(b) 12, 15, 16, and 18.
(iii) 5 numbers whose mean is 13.6.
Ans: 5 numbers whose mean is 13.6 are
(а) 10, 12, 13, 15, and 18.
(b) 11, 12, 13, 14, and 18.
(iv) 6 numbers whose mean = median.
Ans: 6 numbers whose Mean = Median
(a) 2, 4, 6, 8, 10, 12 (Mean = Median = 7)
(b) 1, 3, 5, 7, 9, 11 (Mean = Median = 6)
(v) 6 numbers whose mean > median.
Ans: 6 numbers whose Mean > Median
(a) 1, 2, 3, 4, 5, 30 (Mean = 7.5, Median = 3.5)
(b) 2, 3, 4, 5, 6, 20 (Mean = 6.67, Median = 4.5)
Q: 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?
Ans: The collection is (5, 21, 14, _____, ______, ______) with median 13.
For a set of 6 numbers,
the median (even) = average of (6/2)th and (6/2+1)th term
= average of 3rd and 4th terms
Let other terms be a, b, and c.
Arranging in order with a median of 13.
So, 14 must be the fourth term.
5, a, b, 14, 21, c
OR
5, a, b, 14 c, 21
6 + 14
Median = b+14/2 = 13
b = 26 – 14 = 12
∴ a < b and c > 14
One possible answer is 5, 8, 12, 14, 21, 28
If all three blanks are counting numbers and the median is 13, then infinite possibilities can exist as c > 14.
Q: 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?
Ans: The collection is (3, 11, x, y, 15, 6)
Mean = 6.5, let the other numbers be x and y.
For 6 numbers = 3+11+x+y+15+6/66 = 6.5
⇒ 35 + x + y = 6.5 × 6
⇒ 35 + x + y = 39
⇒ x + y = 39 – 35 = 4
With counting numbers, the pairs (x, y) satisfying x + y = 4 are (1, 3), (3, 1), and (2, 2).
So, the number of possibilities is 2.
Q: 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.
Ans: True
Let the two even numbers be 2a and 2b.
Average = (2a+2b)/2 = a + b
Since (a + b) is an integer and the sum of two integers is an integer, the average is even.
(ii) The average of any two multiples of 5 will be a multiple of 5.
Ans: True
Let the two multiples of 5 be 5m and 5n.
Average = 5m+5n/2=5[(m+n)/2]
This is a multiple of 5 if [(m+n)/2] is an integer. That is m + n is even.
(iii) The average of any 5 multiples of 5 will also be a multiple of 5.
Ans: True
Let the five multiples of 5 be 5a, 5b, 5c, 5d, 5e.
Average = 5a+5b+5c+5d+5e/5
= 5[(a+b+c+d+e)/5]
= a + b + c + d + e
Since a + b + c + d + e is a whole number, the average of any 5 multiples of 5 will also be a multiple of 5.
Page No. 128
Q: 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.
Ans: (c) The heights of the new students have to be measured to find out the new average height.
The average may increase, decrease, or stay the same depending on the heights of the other 2 new students.
(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.
Ans: (b) The average will increase.
Old mean = 150.2 cm
Let the number of students be x
150.5 cm > 150.2 cm
As the average of 2 new added heights is greater than old average, the class average increases.
Option (b) is correct.
(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.
Ans: (d) The information is not sufficient to make a claim about the median.
The median depends on the arrangement of all heights in the ordered list.
Adding two values changes the position of the middle value.
Option (d) is correct.
Math Talk
Q: Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question.
Ans: Method: Count and Calculate
Sum = (14 × 2) + (15 × 2) + (16 × 3) + (17 × 5) + (18 × 4) + (19 × 4) + (20 × 3) + 21 + 23
= 28 + 30 + 48 + 85 + 72 + 76 + 60 + 21 + 23
= 443
Total numbers = 2 + 2 + 3 + 5 + 4 + 4 + 3 + 1 + 1 = 25
Mean = 443/25 = 17.72
The average of the data is 17.72, or on a dot plot, 17.
Math Talk
Q: 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?
Ans: Original average weight = 65.3 kg
Let there be n people in the group.
Total weight = n × 65.3 kg
After one person loses 2 kg and two people each gain 1 kg, the net change in total weight is
Total weight = 65.3n + 1 + 1 – 2 = 65.3n
New mean = 65.3n/n = 65.3 kg
The new mean weight will remain unchanged.
The effect on the median depends on the original position of these three people, whose weight gains or lose.
So new median weight cannot be determined exactly without knowing more data.
Page No. 129
Q: 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.
Ans: (i) 
(ii) What do you find interesting in this data? Share your observations.
Ans: Price generally increases over the years for most states.
Mizoram shows the highest price jump and the highest overall prices in later years.
Gujarat has relatively stable prices compared to other states.
(iii) Compare the price variation in Gujarat and Uttar Pradesh.
Ans: Uttar Pradesh has a large price increase compared to Gujarat.
(iv) In which state has the price increased the most from 2016 to 2025?
Ans: Calculating the price increase for each state.
Andaman and Nicobar Islands = 20.99 – 16 = 4.99
Assam = 12.35 – 6 = 6.35
Gujarat = 19.2 – 16.5 = 2.7
Mizoram = 29.8 – 20 = 9.8
Uttar Pradesh = 24.81 – 16.15 = 8.66
West Bengal = 23.99 – 9.47 = 14.52
The price in West Bengal increases the most (₹ 14.42).
(v) What are you curious to explore further?
Ans: We want to explore the reason behind the sharp price rise in West Bengal.
Page No. 130 – 131
Q: Referring to the graph below, which of the following statements are valid? Why?
[Graph showing percentage of households using electricity and kerosene as primary lighting source in rural and urban areas from 1981 to 2023]
(i) In 1983, the majority in rural areas used kerosene as a primary lighting source while the majority in urban areas used electricity.
Ans: In 1983, approx 85% majority in rural areas used kerosene, whereas 65% majority in urban areas used electricity. So the statement is valid.
(ii) The use of kerosene as a primary lighting source has decreased over time in both rural and urban areas.
Ans: Valid statement as the graph shows a decreasing trend of kerosene in both areas.
(iii) In the year 2000, 10% of the urban households used electricity as a primary lighting source.
Ans: Invalid statement.
The graph shows that more than 80% urban majority use electricity as a primary lighting source.
(iv) In 2023, there were no power cuts.
Ans: The graphs do not give any information about the power cut. So the statement is invalid.
Q: 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?
Ans: (i) Children aged 10 years in urban areas spend approximately 2 hours 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?
Options:(a) 8 years (b) 10 years (c) 12 years (d) 14 years (e) 18 years
Ans: (d) 14 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.
Ans: (a) and (b) both statements are correct.
Q: Individual project: Make your own activity strip for different days of the week.
(i) Do you eat and sleep at regular times every day? Typically how long do you spend outdoors?
(ii) Calculate the average time spent per activity. Represent this average day using a strip.
(iii) Similarly, track the activities of any adult at home. Compare your data with theirs.
Ans: Collect and analyse your own data to answer the question.
Q: Small group project: Make a group of 3 – 4 members. Do at least one of the following:
(i) Track daily sleep time of all your family members for a week.
(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? Collect information on the daily timings of different schools for Grade 8, including class time and break time. Analyse and present the data collected.
Ans: Collect and analyse your own data to answer the question.
Page No. 131 – 132
Q14: The following graphs show the sunrise and sunset times across the year at 4 locations in India. Observe how the graphs are organised.
Are you able to identify which lines indicate the sunrise and which indicate the sunset?
Q: 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?
Ans: Yes, we can identify the lines.
(i) In January, the sunrise was earliest in Kibithu. (approx 6:00)
The day length is (16:00-6:00) = 10 hours
(ii) The place with the longest day length over the year is Ghuar Moti.
(iii) Observations are
Longest day length – Ghuar Moti
Shortest day length – Kanyakumari
Earliest sunrise in Jan – Kibithu
Latest sunset in Jan – Ghuar Moti
I am curious to find out the seasonal variation of sunrise and sunset, or day length, for any particular location.
Q: We all know the typical sunrise and sunset timings. Do you know when the moon rises and sets? Does it follow a regular pattern like the sun? Let’s find out.
The following graph shows the moonrise and moonset times 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?
Ans: (i) Amavasya – Day 21
Purnima – Day 7-8
(ii) Observation: The difference between moonrise and moonset changes throughout the month.
Around the middle of the month, the gap between moonrise and moonset is the maximum.
I wonder
Why do the moonrise and moonset times change every day?
Why is the time gap between moonrise and moonset maximum on some days and minimum on others?