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

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Class 8 Maths Ganita Prakash Part 2 Chapter 5 Solutions

Class 8 Maths Tales by Dots and Lines Solutions

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

NCERT (Page 103-105)

Question 1.
Calculate and mark the mean of each collection of data below.

Solution:
From the figure, the observations are 6,7 and 8.
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
Mean = \(\frac{6+7+8}{3}=\frac{21}{3}\) = 7

Solution:
From the figure, the observations are 3, 6 and 9.
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
Mean = \(\frac{3+6+9}{3}=\frac{18}{3}\) = 6

Solution:
From the figure, the observations are 2, 4 and 9.
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
Mean = \(\frac{2+4+9}{3}=\frac{15}{3}\) = 5

Solution:
From the figure, the observations are 4, 11 and 15.
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
Mean = \(\frac{4+11+15}{3}=\frac{30}{3}\) = 10

Question 2.
Can you explain, how the mean is the centre of each collection?
Solution:
The mean is the centre of a collection because it is the balance point of the data.
When all data values are placed on a number line, the total distance of values to the left of the mean equals the total distance to the right.
Thus, the mean balances the data and represents its centre.

Question 3.
Mark the mean for the collections below.

Solution:
We Know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
(i) From the figure, the observations are 11, 13, 17 and 19.
So,
Mean = \(\frac{11+13+17+19}{4}\)
= \(\frac{60}{4}\)
= 15

(ii) From the figure, the observations are 5, 6, 15 and 16
So,
Mean = \(\frac{5+6+15+16}{4}\)
= \(\frac{42}{4}\)
= 10.5

(iii) From the figure, the observations are 10, 10, 11 and 17.
So,
Mean = \(\frac{3+5+10+12}{4}\)
= \(\frac{30}{4}\)
= 7.5

Question 4.
Is the mean the midpoint of the two endpoints/extremes of the data? It is not always so. Instead, the total distances are equal on both the sides of the mean. This is illustrated through the following dot plots.

Solution:
Sol. The mean is the centre of a collection because the total distances of data points on both sides of the mean are equal. It is not necessarily the midpoint of the extreme values of the data.
In all the given collections, the sum of distances of data values from the mean on the left side equals the sum on the right side.
Hence, the mean is the centre of the collection for all the data sets shown earlier.

Question 5.
What happens to the mean, when an existing value is removed? When will the mean increase, decrease or stay the same?
Solution:
The mean will increase, decrease or stay the same depending on the value removed.

• If the removed value is less than the original mean, the sum decreases by a smaller proportion than the number of items, so the new mean will increase.
• If the removed value is greater than the original mean, the sum decreases by a larger proportion than the number of items, so the new mean will decrease.
• If the removed value is equal to the original mean, the new mean will stay the same.

Question 6.
What happens to the mean if a value equal to the mean is included or removed? Try to explain this using the fair-share interpretation of mean that we studied last year.
Solution:
If a value exactly equal to the existing mean is included or removed from a data set, the mean remains unchanged. Using the fair-share interpretation.

• The mean represents the value each person would get if the total “shared amount” was distributed equally among all people.
• Adding a person, who brings a value exactly equal to this fair-share (the mean) does not alter the fair-share for everyone else. The total shared amount and the number of people both increase proportionally.
• Similarly, removing a person, who only had their fair-share also keeps the average constant, as nothing “extra” or “less” than the average amount was removed from the total pool.

NCERT (Page 106)

Question 1.
Explore if it is possible to include or remove 2 values such that the mean is unchanged.
You may use the following data to experiment with.

Solution:
Yes, it is possible to include or remove 2 values without changing the mean.
We know that the mean remains unchanged if the two values are balanced about the mean.
Here, the mean = 9

Case 1: Including two values
If we include one value less than 9 and one value greater than 9 such that their average is 9 then the mean does not change.
e.g. Include 7 and 11.
\(\frac{7+11}{2}\) = 9
Since, their average equals the mean, the overall mean stays 9.

Case 2: Removing two values
If we remove one value below 9 and one value above 9, with equal distances from 9, the mean remains unchanged, e.g Remove 8 and 10.
They balance each other around 9, so the mean stays 9.

Question 2.
Try to include 2 values greater than the mean and 1 value less than the mean so that the mean stays the same.

Solution:
Yes, it is possible.
If two values greater than the mean and one value less than the mean are added such that their total equals three times the mean, the mean remains unchanged.
Here, mean = 9

Let two values greater than 9 be 10 and 11, and one value less than 9 be 6.
10 + 11 + 6 = 27
Here, the sum of new values is equal to 3 times of the original mean.
Since, the total of the added values equals the required fair-share total, the overall mean remains 9.

NCERT (Page 107)

Question 1.
(i) Try to explain, using algebra, what the average is when a fixed number, e.g. 2 subtracted from every value in the collection,
(ii) Try to, explain this using the fair-share interpretation of average that you learnt last year.
Solution:
(i) Let the number of values be n and the original average be x.
Then, sum of all values = nx̄
If 2 is subtracted from every value, the new sum becomes nx̄ – 2n
New average = \(\frac{n \bar{x}-2 n}{n}\) = x̄ – 2
Hence, using algebra, we see that when 2 is subtracted from every value, the average also decreases by 2.

(ii) Fair-share (equal sharing) interpretation
Suppose, 10 students share 5 chocolates equally.
So, each student gets the average number of chocolates. Now, imagine that 2 chocolates are taken away from each student.
Since, everyone loses the same amount i.e. 2 chocolates.
So, the fair-share for each student also reduces by 2.
Hence, subtracting a fixed number from every value simply shifts the fair-share down by that amount.

NCERT (Page 112-113)

Question 1.

(i) Can you tell, what data is in Column B?
(ii) In which subjects has Ashwin scored more than 30 marks?
Solution:
(i) From the table, Column B shows the marks obtained in Odia (Rl).
(ii) From the table, Ashwin scored more than 30 marks in Telugu, English and Maths.

Question 2.
What formula would you type to find out the class average marks in Science?
Solution:
The average (or mean) of a set of numbers in a spreadsheet program is calculated using the AVERAGE function.
After inserting the data in a spreadsheet, the data for the Science marks are located in Column G, starting from Cell G2 and ending at Cell G8.
Therefore, the formula to find the class average for Science
= AVERAGE(G2:G8)

Question 3.
Show the average marks in other subjects after the last row by typing the appropriate formulae.
Try these out on a computer. You can download the tabular data from this QR code.
Solution:
Do Yourself

Question 4.
Get the total scores of each student by typing the appropriate formulae.
Solution:
Do Yourself

Figure it Out (NCERT Page 113-116)

Question 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) We know that the sum of first n natural numbers
= \(\frac{n(n+1)}{2}\)
For n = 50
Sum of first 50 natural numbers
= \(\frac{50(50+1)}{2}=\frac{50 \times 51}{2}\) = 1275

Now, the mean (x̄) is the sum divided by the number of terms.
x̄ = \(\frac{\text { Sum of first } 50 \text { natural numbers }}{\text { Total numbers }}\)
= \(\frac{1275}{50}\)
= 25.5
Hence, the mean of the first 50 natural numbers is 25.5.

(ii) We know that the sum of first n odd numbers = n2
For n = 50, Sum of first 50 odd numbers = 502 = 2500
The mean (x) is the sum divided by the number of terms.
x̄ = \(\frac{\text { Sum of first } 50 \text { odd numbers }}{\text { Total numbers }}\)
= \(\frac{2500}{50}\)
= 50
Hence, the mean of the first 50 odd numbers is 50.

(iii) We know that if each observation in a data set is
multiplied by a constant, the new mean is the original mean multiplied by that same constant.
So, the first 50 multiples of 4 can be represented as 4 × (first 50 natural numbers).
The mean of the first 50 natural numbers is 25.5 (from part i).
Using the property, the mean of the first 50 multiples of 4 is 25.5 × 4 i.e. 102.

Question 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:
Given, mean = 9
From the figure, the data points currently shown on the dot plot are 4 (one dot), 7 (one dot), 8 (two dots), 9 (five dots) and 11 (one dot).
The sum of these values
= (1 × 4) + (1 × 7) + (2 × 8) + (5 × 9) + (1 × 11)
= 4 + 7 + 16 + 45 + 11
= 83
There are 1 + 1 + 2 + 5 + 1 = 10 dots currently.

With one dot missing, the total number of data points will be 10 + 1 = 11.
We know that
Mean = \(\frac{\text { Total sum }}{\text { Total count }}\)
To find the required total sum for a mean of 9 with 11 data points.
Total sum = Mean × Total count = 9 × 11 = 99
Let the missing value be x then
x = Required total sum – Sum of known values x = 99 – 83 = 16
Hence, the missing value is 16. A dot should be marked above the value 16 on the dot plot.

Question 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) Since, the shoes add a uniform 1 cm to each student’s height measurement.
We know that when a constant value is added to every observation in a data set, the mean (average) of the data set also increases by that same constant value.
No, the teacher does not need to measure all the heights again. A simpler way is to subtract the shoe height of 1 cm from the reported average height of 150.2 cm.

(ii) (d) Correct average height = Measured average height – Height of shoe
= 150.2 cm – 1cm
= 149.2 cm
So, option (d) is correct.

Question 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 min? Explain, how you arrived at the answer?

Solution:
From the Album A
Given data 5, 5, 5.25, 5.5, 5.75, 6, 6.5.
So, Mean = \(\frac{5+5+5.25+5.5+5.75+6+6.5}{7}=\frac{39}{7}\) = 557

From the Album B
Given, data 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5.
So, Mean = \(\frac{0.5+0.75+1.5+1.5+2+3.75+4.25+5}{8}\)
= \(\frac{19.25}{8}\)
= 2.41

From the Album C
Given data 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5.
So, Mean = \(\frac{0.5+0.75+1.5+1.5+2+3.75+4.25+5}{8}\)
= \(\frac{31.25}{8}\)
= 3.91
Hence, Album A has a mean of 5.57 min.

Question 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 in 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:
(i) If adding one value in the data set changes the total number of values from 16 (even) to 17 (odd).
So, the new median, if n is odd
= \(\frac{(n+1) \text { th term }}{2}\)
= \(\frac{(17+1) \text { th term }}{2}\)
= \(\frac{18 \mathrm{th} \text { term }}{2}\)
= 9th term
To keep the median at 41, the 9th term in the new sorted list must be 41.
Hence, the value could be any number less than, greater than or equal to 41.

(ii) If adding two values keeps the total number of values even i.e. 16 + 2 = 18.
So, the new median = \(\frac{1}{2}\)
= \(\frac{\left(\frac{18}{2}\right) \text { th term }+\left(\frac{18}{2}+1\right) \text { th term }}{2}\)
To keep the median at 41 then the average of the new 9th and 10th terms must be 41.
Hence, the two values could be any number less than or equal to 41.

(iii) If removing one value, changes the total number of values from 16 to 15.
So, the new median = \(\frac{(15+1) \text { th term }}{2}\) = 8th term
To keep the median at 41 then the new 8th term must be 41.
Hence, the value could be any number greater than or equal to 41.

Question 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
Justification Removing a value less than the median does not shift the middle position to the left. The median either remains the same or increases but it cannot decrease.
(ii) Always true
Justification The mean is the fair-share. Adding a value less than the current mean reduces the overall average.
(iii) Sometimes true
Justification The median remains unchanged only if the added values are placed symmetrically around the median.
(iv) Never true
Justification Adding values below the median shifts the data towards the lower side, so the median cannot increase.

Question 7.
The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.
Solution:
Given numbers are 8, 13, 10, 4, 5, 20, y, 10.
Now, number of observations = 8 and mean = 10.375
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
⇒ 10.375 = \(\frac{8+13+10+4+5+20+y+10}{8}\)
⇒ 10375 = \(\frac{70+y}{8}\)
⇒ 70 + y = 10.375 × 8
⇒ 70 + y = 83
⇒ y = 83 – 70 = 13
Hence, the value of y is 13.

Question 8.
The mean of a set of data with 15 values is 134. Find the sum of the data.
Solution:
Given, number of values =15
and mean = 134
∴ Mean = \(\frac{\text { Sum of all values }}{\text { Number of values }}\)
⇒ 134 = \(\frac{\text { Sum }}{15}\)
⇒ Sum = 134 × 15
⇒ Sum = 2010
Hence, the sum of the data is 2010.

Question 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:
The given data set has 11 values including p.
12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p

First arrange the 10 known values in ascending order.
8, 8, 12, 18, 25, 29, 35, 39, 47, 73

The total number of data points isN = 11(10 known values plus p).

Since, N is odd, the median is the value at the \(\left(\frac{N+1}{2}\right)\)th position.
Median position = \(\left(\frac{11+1}{2}\right)\)th = (\(\frac{12}{2}\))th = 6th position

The median is given as 29, so the 6th value in the ordered list be 29.
If p ≥ 29 then p will be one of the last 5 positions and 29 will still be the 6th value.
Therefore, for the median to be 29, p can be any number that allows 29 to remain the 6th value, when all 11 numbers are ordered. This means p can be 40,100,29,47 and 30. Hence, the options (iii), (iv), (v), (vi) and (vii) are possible values for p.

Question 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:
Given data are 0,0,0,1,2, 2,2,2,3, 3,3, 3, 3, 3, 3,4,4,4,
4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 10, 10.

(i) Average (Mean)

Hence, the average number of times students rode their cycles is 4.6 times.

(ii) Table for finding the median.

Number of Weeks (xi)	Number of Students (fi)	Cumulative Frequency (cf)
0	3	3
1	1	4
2	4	8
3	7	15
4	7	22
5	5	27
6	4	31
7	6	37
8	3	40
10	2	42

Here, n = 42 (even)
Median = \(\frac{1}{2}\)[Value of{(\(\frac{n}{2}\))th + (\(\frac{n}{2}\) + 1)th} observation]
Median = \(\frac{1}{2}\)[Value of{(\(\frac{42}{2}\))th + (\(\frac{42}{2}\) + 1)th} observation]
= \(\frac{1}{2}\) [Value of 21th observation + Value of 22th observation]
= \(\frac{1}{2}\) [4 + 4]
= 4

(iii) (a) This is invalid because 3 students rode their cycles 0 times.
(b) This is valid as the majority of students (28 out of 31) rode their cycles between 2 and 8 times, which can be considered “a few times”.
(c) This is invalid. The dot plot shows the total number of times students rode their cycles in a week, not per day. The data does not provide enough information to determine if a student rode more than once on a single day.
(d) Exactly 5 students have used their cycles more than once on some days. This is invalid for the same reason as (c), the data is aggregated weekly, not daily.
(e) If every student cycles exactly 1 more time, the entire data increases by 1.
∴ New average = Old average + 1= 4.6 + 1= 5.6
and new median = Old median + 1 = 4 + 1 = 5

Question 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:
The data represents the number of throws (trials) taken by students.
The minimum number of trials is the smallest value in the “Number of trials” row, which is 1.
The maximum number of trials is the largest value in the “Number of trials” row which is 10.

Table for Finding the Mean
Number of Trials (xi)	Number of Students (fi)	fixi
1	1	1
2	0	0
3	0	0
4	1	4
5	4	20
6	9	54
7              .	12	84
8	15	120
9	10	90
10	10	100
Total	Σfi = 62	Σfixi = 473

Here, Σf_i = 62 and Σf_ix_i = 473
∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{473}{62}\) = 7.63 ≈ 7.6
Hence, the mean is 7.6.

Table for Finding the Median

Number of Trials (xi)	Number of Students (fi)	Cumulative Frequency (cf)
1	1	1
2	0	1
3	0	1
4	1	2
5	4	6
6	9	15
7	12	27
8	15	42
9	10	52
10	10	62

Here, N = 62
⇒ \(\frac{N}{2}=\frac{62}{2}\) = 31
The 31st observation lies in the cumulative frequency 42, which corresponds to 8.
∴ Median of the data is 8.

NCERT (Page 116-120)

Question 1.
The following figure is a clustered-column graph that shows the monthly maximum temperature in Kerala and Punjab in 2023.


Now, observe the following graph. Do both these graphs represent the same information?
Solution:
Yes, both the graphs represent the same information.
In both graphs
(i) The X-axis in both graphs represents the months from January to December.
(ii) The Y-axis in both graphs represents the temperature in degrees Celsius.
(iii) The blue color/line consistently represents Kerala, while the red color/line represents Punjab in both visualisations.

Question 2.
Take a look at the line graph below showing the annual number of objects launched into space. Annual number of objects launched into space including satellites, landers crewed spacecrafts and space station fight elements launched into Earth orbit or beyond.

Which of the following statements are valid inferences?
(i) From 2012 till 2024,the worldwide count of space object launches increased every year.
Solution:
Not valid
From the graph, the worldwide count of space objects launches does not increase every single year.
There is a slight dip/stagnation around 2015-2016 and again a small decrease from 2023 to 2024.
Hence, the increase is overall, but not year-by-year.

(ii) USA is a major contributor in the years 2022-24, launching about \(\frac{3}{4}\)th of the worldwide count.
Solution:
Not valid
Although the USA is a major contributor, it does not
account for about \(\frac{3}{4}\) (75%) of the worldwide launches.
In 2022-24, the world total is much higher than the USA count, mainly due to contributions from China and others as well.
So, the fraction is less than \(\frac{3}{4}\).

(iii) Nepal did not launch any object in the period 2012-24.
Solution:
Not valid
The graph only shows data for World, USA, China and Russia.
There is no information about Nepal, so we cannot conclude anything about Nepal’s launches from this graph.

(iv) The combined count of object launches by China and Russia in 2024 is about 400.
Solution:
Valid
From the graph for 2024,

• China ≈ 300
• Russia ≈ 100

Combined ≈ 400, which matches the statement.

Figure it Out (Page 122 123)

Question 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:

Question 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?
Solution:
The data is likely compiled by observing and recording the number of rainy days in these cities each month over many years. This long term data is then averaged to provide a reliable estimate of the typical number of rainy days per month.

(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.

Solution:

(iii) Based on the line for New Delhi in the graph fill the data in the table.
Solution:

(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?
Solution:
From the table,
For Mangaluru 0.1+ 0 + 0.1 + 1.8 + 6.2 + 24.1 + 27.7 + 24.5 + 14 + 8.8 + 3.9 + 0.9 = 112.1
For New Delhi L5 + 1.6 + 1.5 + 1.3 + 1.5 + 4.2 + 9.8 + 9.8 + 4.4 + L2 + 02 + 0.9 = 37.9
For Port Blair 2.4 + L3 + 0.9 + 33 + 153 + 18.7 + 173 + 18.8 + 16.8 + 14.1 + 113 + 5.4 = 125.8
For Rameswaram 2.6 + 13 + L9 + 3.4 + 25 + 0.4 + 1 + 1 + L9 + 8.1 + 10.4 + 7.8 = 423
Hence, Port Blair receives the most number of days of rainfall per year and New Delhi 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:
Rainy season in New Delhi and Rameswaram.
New Delhi The rainy season is from July to September, with maximum rainfall in July and August.
Rameswaram The rainy season occurs mainly from October to December, due to the North-East monsoon.

Question 3.
The following line graph shows the number of births in every month in India over a time period.

(i) What are your observations?
Solution:
From the graph, the graph shows a clear cyclical pattern, with peaks occurring around the middle of each year (July-August) and throughs around January. The number of births is consistently above 1.4 M per month throughout the period shown.

(ii) What was the approximate number of births in July 2017?
Solution:
From the graph, the data point for July 2017 is just below the 2 M line, approximately at the 1.8 M mark.
Hence, the approximate number of births in July 2017 is 1.8 M.

(iii) What time period does the graph capture?
Solution:
From the graph, the graph starts at July 2017 and the last visible data point appears to be around March 2020.
Hence, time period the graph captures July 2017 to March 2020.

(iv) Compare the number of births in the month of January in the years 2018, 2019 and 2020.
Solution:
Births in January 2018 were approximately 1.7 M, January 2019 were approximately 1.8 M and January 2020 were approximately 1.8 M.
The number of births in January 2018,2019, and 2020 all appear to be roughly the same, at approximately 1.8 M.

(v) Estimate the number of births in the year 2019.
Solution:
From the graph, the number of births in January 2019 = 1.8 M,
the number of births in July 2019 = 1.85 M and
the number of birth in January 2020 = 1.8 M.
So, the number of births in the year 2019 = 1.8 M + 1.85 M + 1.8 M
= 5.45 M

NCERT (Page 124)

Question 1.

Share your observations. Based on this infographic, answer the following.
(i) The value of Karnataka is hidden. Can you guess what it could be?
Solution:
From the infographic, the value near Karnataka is shown close to other southern states like Tamil Nadu (+ 85), Kerala (+ 79) and Undivided Andhra (+ 92).
A reasonable estimate for Karnataka is around + 80 to + 90.
So, Karnataka = +85 (strong preference for rice).

(ii) Which are the top 5 states, where rice is the most popular?
Solution:
From the infographic, Top 5 states, where rice is most popular.
States with the highest positive values (deep purple shades)
1. Manipur (+ 100)
2. Nagaland (+ 99)
3. Mizoram (+ 97)
4. Tripura (+ 96)
5. Meghalaya (+ 95)
These states show a very strong preference for rice over wheat.

(iii) Which are the top 5 states, where wheat is the most popular?
Solution:
From the infographic, states with the most negative values (dark yellow/orange shades).
1. Rajasthan (- 93)
2. Haryana(- 81)
3. Punjab (-78)
4. Madhaya Pradesh (- 60)
5. Delhi (- 45)
These states show a strong preference for wheat over rice.

(iv) List a few states, where the preference between rice and wheat is more or less balanced.
Solution:
From the infographic, states where the difference between per capita consumption of rice and wheat is close to zero indicate a balanced preference for both cereals.
From the infographic, such states are
1. Bihar (+3)
2. Maharashtra (-15)
3. Uttarakhand (-18)
4. Himachal Pradesh (-19)
5. Uttar Pradesh (-30)
In these states, neither rice nor wheat is strongly preferred and both are consumed in relatively similar amounts.

NCERT (Page 125-26)

Question 1.
Manoj has an interesting hobby. He makes note of what he does throughout the day. He records his activities by colouring a strip of paper with 48 boxes, marking time in 30 min intervals from midnight to mid-night.

He has recorded five types of activities for three different days of the week on three coloured strips, shown to the right.

• Sleeping
• Eating
• Meeting friends, hobbies media, time with family.
• Attending classes, studying and homework.
• Showering and getting dressed, yoga or exercise.
• Travelling

Look at the three coloured strips carefully.
(i) What activity does each colour stand for?
Solution:
From the figure,
By observing time of day, duration and repetition of colours.

• Light blue → Sleeping
• Green → Eating
• Purple → Showering and getting dressed, yoga or exercise.
• Yellow → Meeting friends, hobbies, media, time with family.
• Orange → Attending Classes, studying and homework.
• Gray→ Travelling

(ii) The three strips correspond to the days Friday-Sunday in some order. Which day do you think each strip represents?
Solution:
First strip → Friday

• Clear school/study (orange) blocks
• Regular routine
• Less leisure time Second strip → Saturday
• Some study time
• More leisure and social activities then Friday
• Slightly relaxed routine Third strip → Sunday
• Some study time
• Maximum leisure and family time
• Irregular routine

(iii) On one of these days, he went out with friends to watch a long movie. When do you think this happened?
Solution:
A long continuous yellow blocks at evening is visible on the Saturday strip. So, he went to watch a long movie on Saturday at the evening.

(iv) At what time does his school break for lunch?
Solution:
A short green (eating) block appears in the middle of the school day.
This occurs around 1:00 PM.
So, lunch break is at about 1: 00 PM.

(v) What more can the strips tell us?
Solution:
From the strips, we learn that

• Manoj follows a regular routine on school days.
• His sleep and lesiure time increase on weekends.
• Meals are taken at fixed times.
• He balances study, rest and recreation.
• His daily routine changes clearly from weekday to weekend.

Question 2.
What would your strip for a weekday look like?
How similar or different is it to Manoj’s?
Solution:
Do yourself

Question 3.
What would a strip of your typical day during your vacation look like? How similar/different would it look?
Solution:
Do yourself

Question 4.
What would a strip for any of the adults in your family look like? Make a strip of a day for any adult at home. Compare your strip with theirs. What do you find interesting?
Solution:
Do yourself

Figure it Out (NCERT Page 127-132)

Question 1.
Mean Grids
(i) Fill the grid with 9 distinct numbers such that the average along each row, column and diagonal is 10.
Solution:
Here, the average along each row, column and diagonal is 10.
Since, there are 3 numbers in each line, the sum (magic constant) must be 3 × 10 = 30.
In a 3 × 3 magic square, the center number is always one-third of the magic constant.
Therefore, the centre number must be \(\frac{30}{3}\) = 10
There are many possible solutions. One example uses numbers centred around 10.
A standard magic square with numbers 1 – 9 has a centre of 5 and a sum of 15.
Adding 5 to each number yields a valid solution with a sum of 30.

(ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions?

Solution:
Since, there are multiple possible magic squares that sum to 30, it is possible to change some numbers and still maintain the average of 10 in all directions. As long as the centre number remains 10 and all sums continue to equal 30, various combinations of distinct numbers can be used. Here, is the another example

Question 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) Given, mean = 8
Since, the mean of 3 numbers is 8, so their sum is 3 x 8 = 24.
Example 1 Choose any two numbers, for example,
7 and 9. The third number is 24 – 7 – 9 = 8.
e.g. 1 (7, 8, 9)
Example 2 Choose another two numbers, for example, 5 and 10.
The third number is 24 – 5 – 10 = 9. e.g. 2 (5, 9, 10)

(ii) For 4 numbers, the median is the average of the two middle numbers. The average is 15.5, so the sum of the two middle numbers is 15.5 x 2 = 31.
Example 1 Choose two numbers that sum to 31 ,for example, 15 and 16.Choose a smaller number (e.g. 10) and a larger number (e.g. 20).
eg. 1 (10, 15, 16, 20)
Example 2 Choose another two numbers that sum to 31, for example, 14 and 17.Choose a smaller number (e.g. 12) and a larger number (e.g. 18). e.g. 2 (12, 14, 17, 18)

(iii) Since, the mean of 5 numbers is 13.6, so their sum is 5 x 13.6 = 68
Example 1 Choose four numbers, for example, 10, 12, 14, and 16.
Their sum is 52. The fifth number is 68 – 52 = 16.’ e.g. 1(10, 12, 14, 16, 16)
Example 2 Choose four different numbers, for example, 8,11,15, and 19.Their sum is 53. The fifth number is 68 – 53 = 15.
e.g. 2 (8, 11, 15, 15, 19)

(iv) The sum of the 6 numbers must be 6 × mean. The median is the average of the 3rd and 4th sorted numbers. A simple set of consecutive integers works well.
Example 1 The numbers 1, 2, 3, 4, 5, 6.
Mean = \(\frac{21}{6}\) = 3.5 6
Median = \(\frac{3+4}{2}\) = 3.5

Example 2 The numbers 5,10,15,20,25,30.
Mean = \(\frac{105}{6}\) = 17.5
Median = \(\frac{15+20}{2}\) = 175

(v) Example 1 Choose a set of 6 numbers, for example, {1, 2, 3, 4, 5, 20}.
The numbers in order are 1, 2, 3, 4, 5, 20.
The median is the average of the 3rd and 4th numbers.
Median = \(\frac{3+4}{2}\) = 3.5

The mean is the sum divided by 6.
Mean = \(\frac{1+2+3+4+5+20}{6}=\frac{35}{6}\) = 5.833
Since, 5.833 > 3.5, the condition mean > median is satisfied.

Example 2. Choose another set of 6 numbers, for example, {10, 11, 12, 13, 14, 100}.
The numbers in order are 10, 11, 12, 13, 14, 100.
The median is the average of the 3rd and 4th numbers.
Median = \(\frac{12+13}{2}\) = 125

The mean is the sum divided by 6
Mean = \(\frac{10+11+12+13+14+100}{6}\)
= \(\frac{160}{6}\)
= 26.667
Since, 26.667 > 12.5, the condition mean > median is satisfied.

Question 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:
Given collection 5, 21, 14,…,…….
There are 6 numbers in total.
When the number of observations is even (6), the median is the average of the 3rd and 4th numbers after arranging in ascending order.
Median = \(\frac{1}{2}\)
Given, \(\frac{1}{2}\) = 13
So, 3rd number + 4th number = 26
Given number in ascending order 5, 14, 21.
To place the 3rd and 4th numbers, we must insert three counting numbers in such a way that the middle two add upto 26.
The middle pair must be counting numbers, whose sum is 26.

Possible pairs

• (13, 13)
• (12, 14)
• (11, 15)
• (10, 16)
• (9, 17)
• (8, 18)
• (7,19)
• (6,20)

(All are counting numbers.)
Choose the middle pair 13 and 13 and one small number to keep order correct.
Filled collection 5, 12,13, 13, 14, 21.

Question 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:
Given collection 3, 11, ……. 15, 6.
There are 6 numbers.
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
Here, mean = 6.5
∴ Required sum = 65 × 6 = 39
Now, the sum of known numbers,
3 + 11 + 15 + 6 = 35
Let the blanks be x and y.
35 + x + y = 39
⇒ x + y = 39 – 35
⇒ x + y = 4
Now, find all pairs of counting numbers, whose sum is 4.
1 + 3 = 4,
2 + 2 = 4
and 3 + 1 = 4.
Hence, the number of possibilities are 3.

Question 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
Let the two even numbers be 2n and 2m, where n and m are integers.
The average of these two numbers
= \(=\frac{2 n+2 m}{2}=\frac{2(n+m)}{2}\) = n + m
The sum of two integers (M + m) is always an integer, but it is not necessarily an even integer,
e.g. If the even numbers are 2 and 4.
The average is \(\frac{2+4}{2}=\frac{6}{2}\) = 3, which is odd.
Thus, the statement is false.

(ii) False
Let the two multiples of 5 be 5m and 5M, where m and n are integers.
The average is given by \(\frac{5 m+5 n}{2}=\frac{5(m+n)}{2}\)
This result is a multiple of 5 only if \(\frac{m+n}{2}\) is an integer (i.e. m + n is even).
e.g. The average of 5 and 10 is 7.5, which is not a multiple of 5.

(iii) False
Let the five multiples of 5 be 5m, 5n,5p, 5q and 5r, where m, n, pq and r are integers.
The average is given by
= \(\frac{5 m+5 n+5 p+5 q+5 r}{5}=\frac{5(m+n+p+q+r)}{5}\)
= m + n + p + q + r
The result is an integer but is not necessarily a multiple of 5.
e.g. The average of 5,10,10,10 and 10 is 9, which is not a multiple of 5.

Question 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.
Solution:
(c) Without knowing the heights of the two new students, the new average cannot be calculated.

(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.
Solution:
(b) Given, the heights of the two new students are 149 cm and 152 cm.
Average of new students = \(\frac{149+152}{2}\)
= 150.5 cm
Since, 150.5 > 150.2
The class average will increase slightly.

(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.
Solution:
(d) The median is the middle value in an ordered data set. Knowing the original mean and the two new data points is not enough to determine the new median. The median depends on the specific distribution and ordering of all individual heights in the original class, which are not provided. Hence, the information is not sufficient to make a claim about median.

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

Solution:
From the figure, observations are 14, 14, 15, 15, 16, 16, 16, . 17, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 21, 23
We know that

So, 17 is not the average (mean) of the data shown in the dot plot.

Question 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:
Given,
Previous month’s mean weight = 65.3 kg
Previous month’s median weight = 67 kg This month
Since, one person lost 2 kg and two people gained 1 kg each.
Total change in weight = (-2) + 1 + 1 = 0 kg
Since, the total weight of the group remains the same, the mean weight does not change.
Hence, mean weight this month = 65.3 kg (unchanged)
The median depends on the order of weights, not on the total.
We are not told whether the person, who lost weight or the people, who gained weight were below, above or equal to the median.
Therefore, the median may remain the same, increase or decrease.
No definite conclusion can be made about the median.

Question 9.
The following table shows the retail price (in ₹) of iodised salt in the month of January in a few states over 10 yr. For your calculations and plotting you may round off values to the nearest counting number.

(i) Choose data from any 3 states you find interesting and present it through a line graph using an appropriate scale.
Solution:

(ii) What do you find interesting in this data? Share your observations.
Solution:
Interesting observations from the data.

• Prices in all states show an overall increasing trend over the years.
• Mizoram shows a sharp and rapid increase, especially after 2021.
• Uttar Pradesh shows a steady rise, becoming much steeper after 2019.
• Gujarat shows comparatively smaller fluctuations and slower growth.
• West Bengal shows a significant rise after 2021, indicating faster price increase in recent years.

(iii) Compare the price variation in Gujarat and Uttar Pradesh.
Solution:
Comparison of price variation in Gujarat and Uttar Pradesh.
Gujarat

• Price in 2016= 16.5
• Price in 2025 = 19.2
• Increase = 19.2 – 16.5 = 2.7

Uttar Pradesh

• Price in 2016= 16.15
• Price in 2025 = 24.81
• Increase =24.81 – 16.15 = 8.66

Comparison

• Uttar Pradesh shows a much larger increase than Gujarat.
• Price variation in Uttar Pradesh is more rapid and pronounced.

(iv) In which state has the price increased the most from 2016 to 2025?
Solution:

State	Increase
Andaman and Nicobar Islands	4.99
Assam	6.35
Gujarat	2.7
Mizoram	9.8
Uttar Pradesh	8.66
West Bengal	14.52

Hence, West Bengal has the maximum price increase from 2016 to 2025.

(v) What are you curious to explore further?
Solution:
Do yourself

Question 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.
Solution:
Valid
In 1983, the rural graph shows kerosene at approx 85% and electricity at approx 15%. The urban graph shows electricity at approx 65% and kerosene at approx 35%.

(ii) The use of kerosene as a primary lighting source has decreased overtime in both rural and urban areas.
Solution:
Valid
The orange line representing kerosene in both rural and urban graphs shows a consistent downward trend from 1983 to 2023.

(iii) In the year 2000, 10% of the urban households used electricity as a primary lighting source.
Solution:
Invalid
In 2000, the urban graph shows electricity at approx 85% (blue line), while kerosene is at approx 10% (orange line).

(iv) In 2023, there were no power cuts.
Solution:
Invalid
The provided graph does not contain any data regarding power cuts. It only illustrates the share of primary energy sources used for household lighting over time, with the latest data point being 2013.

Question 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?
Solution:
From the urban (blue) line at age 10, the average time spent is approximately 2 h.

(ii) At what age is the average time spent daily on hobbies and games by rural kids 1.5 h?

(a) 8 yr
(b) 10 yr
(c) 12 yr
(d) 14 yr
(e) 18 yr
Solution:
From the graph, the age at which the average time spent daily on hobbies and games by rural kids 1.5 h is 12 yr.

(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 h on hobbies and games everyday.
Solution:
(a) Correct
From the graph,
The average time spent by kids aged 10 = 2 h and the average time spent by kids aged 15 = 1h.
So, the average time spent daily on hobbies and games by kids aged 15 is twice that of kids 10.

(b) Correct
From the graph, all rural kids aged 15 spend atleast 1 h on hobbies and games everyday.

Question 12.
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.
Solution:
Do yourself.

Question 13.
Small group project: Make a group of 3 – 4 members. Do atleast one of the following.
(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 h, 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.
Solution:
Do yourself

Question 14
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?

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?
Solution:
From the graph, the sun rises the earliest in January in kibithu.
The timing of sunrise in Kibithu = 5 : 45 AM
and the timing of sunset in Kibithu = 4:45 PM
So, the length of this day = 5: 45 AM – 4:45 PM = 12h

(ii) Which place has the longest day length over the year?
Solution:
Fromt he graph,
In Kibithu,
The earliest time of sunrise = 4:00 AM
and the latest time of sunset =6:00 PM
So, length of longest day over the year in Kibithu = 4:00 AM -6:00 PM
= 14h

In Ghuar Moti,
The earliest time of sunrise = 6 AM
and the latest time of sunset = 7:45 PM
So, length of the longest day over the year in Ghuar Moti = 6.00 AM – 7.45 PM
= 13 h and 45 min.

In Srinagar,
The earliest time of sunrise = 5:45 AM
and the latest time of sunset = 7:45 PM
So, the length of the longest day over the year in Srinagar = 5 : 45 AM -7:45 PM = 12h
In Kanyakumari,
The earliest time of sunrise = 6: 00 AM and the latest time of sunset =6:30 PM
So, the length of the longest day over the year = 6:00 AM -6:30 PM
= 12 h and 30 min
Hence, the longest day length over the year at kibithu.

(iii) Share your observations, what do you find interesting? What are you curious to find out?
Solution:
Do yourself

Question 15.
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 time over a month.
(i) Find out on what dates amavasya (new moon) and purnima (full moon) were in this month.
Solution:
According to the graph,
Amavsya was on the 8th and Purnima was on the 22th.

(ii) What do you notice? What do you wonder?

Solution:
The moonrise time gets later each day by about 45-50 min.
The moonset time also shifts later every day. Moonrise and moonset follow a regular pattern but it is different from sun’s pattern.

Wonders

• Why does the moonrise get delayed every day?
• Why does the moon follow a cycle of about one month?
• How is this pattern connected to the movement of the moon around the Earth?