Exercise 5.3

Tales by Dots and Lines Exercise 5.3 Class 8 Maths – Chapter 5 NCERT Solutions

NCERT Class 8 Maths Tales by Dots and Lines Term 2 Exercise 5.3 detailed solutions, prepared by the ALLEN educational team according to the newly updated syllabus, are provided on this webpage. These solutions provide clear explanations of the exercise questions for revision and exam preparation.Exercise 5.3 Infographics Mean Grids Generation of Data Sets on the Basis of Given Mean and Median Conditions Mean and Median Equal or Related in a Particular Manner

1.0Download Class 8 Maths Chapter 5 Ex 5.3 NCERT Solutions PDF

Get the Exercise 5.3 NCERT Solutions PDF for Class 8 Maths Chapter 5 to check answers and work through the questions with clarity.

Chapter 5 : Tales by Dots and Lines Exercise 5.3

2.0Detailed NCERT Solutions for Tales by Dots and Lines Exercise 5.3 Class 8 Maths (Ganita Prakash Term 2)

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 direction? Sol. (i) Here, the average of 3 number is 10, So, their sum must be 30

13	6	11
8	10	12
9	14	7

Here is one way to fill the grid

Row 1: 13, 6, 11
Row 2: 8, 10, 12
Row 3: 9, 14, 7 (ii) Yes, but you would have to change all the numbers to a completely new sequence that is still perfectly balanced around 10.

For example: We could use even numbers: $2, 4, 6, 8, 10, 12, 14, 16, 18$ . As long as the numbers jump by a consistent amount and center on 10, you can build a new magic square to We use only even numbers Our new 9 numbers will be: $2, 4, 6, 8$ , 10, 12, 14, 16, 18

And, 10 our target average is at the center So, our grid will be

16	2	12
6	10	14
8	18	4

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.

Sol. (i) 3 numbers whose mean is 8 are (1) 6,8 , and 10 (2) 7,8, and 9 . (ii) 4 numbers whose median is 15.5 are (1) $10, 15, 16$ , and 20 (2) $12, 15, 16$ , and 18 . (iii) 5 numbers whose mean is 13.6 are (1) $10, 12, 13, 15$ , and 18 . (2) $11, 12, 13, 14$ , and 18. (iv) 6 numbers whose Mean = Median (1) $2, 4, 6, 8, 10, 12$ (Mean $=$ Median $= 7$ ) (2) $1, 3, 5, 7, 9, 11$ (Mean $=$ Median $= 6$ ) (v) 6 numbers whose Mean > Median (1) $1, 2, 3, 4, 5, 30$ (Mean=7.5, Median=3.5) (2) $2, 3, 4, 5, 6, 20$ (Mean = 6.67, Median = 4.5)

3. Fill in the blanks such that the median of the collection is $13 : 5, 21, 14$ , $____$ . How many possibilities exist if only counting numbers are allowed? Sol. Our data is Number of observations $= 6$ Since number of observations is even Median is the average of middle two terms Since we are given 14 as a term, but median is lesser than 14

So, 14 can be our $4^{th}$ term Thus, Median $= \frac{3 ^{rd} term + 4 ^{th} term}{2}$ Putting $4^{th}$ term $= 14$ , Median $= 13$ $13 = \frac{3 ^{rd} term + 14}{2}$ $13 \times 2 = 3^{rd}$ term +14 $26 = 3^{rd}$ term +14 $26 - 14 = 3^{rd}$ term $12 = 3^{rd}$ term $3^{rd}$ term $= 12$ Given terms were 5, 21, 14 and $3^{rd}$ term = 12

Let $5^{th}$ term $= 21$ Arranging data in ascending order, our data becomes

5, $____$ , 12, 14, 21, $____$ Thus, $2^{nd}$ term can be any number from 5 to 12 i.e. $2^{nd}$ term can be $5, 6, 7, 8, 9, 10, 11, 12$ and $6^{th}$ term can be any number greater than 21 . But there are infinitely many terms greater than 21 , like $22, 100, 100000, 99$ , 99000000.

So, we can say that 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?

Sol. The collection is $(3, 11, x, y, 15, 6)$ Mean $= 6.5$ , let the other numbers be x and y . For 6 numbers $= 6.5$ $\Rightarrow 35 + x + y = 6.5 \times 6$ $\Rightarrow 35 + x + y = 39$ $\Rightarrow x + y = 39 - 35 = 4$ With counting numbers, the pairs ( $x, y$ ) satisfying $x + y = 4$ are ( 1,3 ) and ( 2,2 ).So, the number of possibilities is 2 .

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 .

Sol. (i) False Let the two even numbers be 2 a and 2b. Average $= (\frac{2 a + 2 b}{2}) = a + b$

Since 2 a and 2 b is even but we cannot say about parity(odd/even) of ' a ' and ' b '

So, $a + b$ can be odd or even. We take an example Let the two even numbers be 2 & 4 Average $= \frac{2 + 4}{2} = \frac{6}{2} = 3$ Since the average is odd, the given statement is false

(ii) False

Let the two multiples of 5 be 5 m and 5n.

Average $= \frac{5 m + 5 n}{2} = 5 (\frac{m + n}{2})$ $'^{'} (\frac{m + n}{2})^{'}$ can be any number other than integer We take an example Let the two multiples of 5 be 5 & 10 . Average $= \frac{5 + 10}{2} = \frac{15}{2} = 7.5$ Since 7.5 is not a multiple of 5, The given statement is false

(iii)True

Let the five multiples of 5 be 5a, 5b, 5c, 5d, 5e.

Average $= \frac{5 a + 5 b + 5 c + 5 d + 5 e}{5}$ $= 5 (\frac{a + b + c + d + e}{5})$ $= a + b + c + d + e$ Since $a + b + c + d + e$ can be or cannot be multiple of 5 .

We take an example Let five multiples of 5 be $5, 10, 15$ , 20, & 25 Average $= \frac{5 + 10 + 15 + 20 + 25}{5} = \frac{75}{5} = 15$ Since 15 is not a multiple of 5, the given statement is true.

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. (2) The average height of the class will remain the same. (3) The heights of the new students have to be measured to find out the new average height. (4) The heights of everyone in the class has to be measured again to calculate the new average height. (ii) The heights of the two new joinees are 149 cm and 152 cm . Which of the following statements about the class' average height are correct? Why? (a) The average will remain the same. (b) The average will increase. (c) The average will decrease. (d) The information is not sufficient to make a claim about the average. (iii) Which of the following statements about the new class average height are correct? Why? (a) The median will remain the same. (b) The median will increase. (c) The median will decrease. (d) The information is not sufficient to make a claim about median.

Sol. (i) To find the average, we need Sum of all heights.

Thus, we cannot find average without knowing the heights of the new students.

So, option (c) is correct (ii) Old mean $= 150.2 cm$

Let the number of students be x New mean $=$ $\frac{Sum of heights of x + 149 + 152 students}{x + 2}$ Average of new two heights $= \frac{149 + 152}{2} = \frac{301}{2} = 150.5$ $150.5 cm > 150.2 cm$ As the average of 2 new added heights is greater than old average, the class average increases.

Option (b) is correct. (iii) We do not know the individual heights of the original 24 students. Without seeing the list of numbers, we can't determine how adding two new numbers will shift the middle position.

So, option (d) is correct

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

Sol. Method use: Count and Calculate

Value $(x)$	Frequency $(f)$
$14$	2
$15$	2
$16$	3
$17$	5
$18$	4
$19$	4
$20$	3
$21$	1
$23$	1

Sum $= (14 \times 2) + (15 \times 2) + (16 \times 3) + (17 \times 5) + (18 \times 4) + (19 \times 4) + (20 \times 3) + (21 \times 1) + (23 \times 1) = 25 + 30 + 48 + 8572 + 76 + 60 + 21 + 23 = 443$

Total numbers $= 2 + 2 + 3 + 5 + 4 + 4 + 3 + 1 + 1 = 25$ Mean $= \frac{443}{25} = 17.72$ The average of the data is 17.72 or on a dot plot 17.

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? Sol. Original average weight $= 65.3 kg$ Let there be n people in the group. Total weight $= n \times 65.3 kg$ After one person loses 2 kg and two people each gain 1 kg , the net change in total weight is Total weight $= 65.3 n + 1 + 1 - 2 = 65.3 n$ New mean $= \frac{65.3}{n} = 65.3 kg$

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.

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

	Andaman and Nicobar Islands	Assam	Gujarat	Mizoram	Uttar Pradesh	West Bangal
2016	16	6	16.5	20	16.15	9.47
2017	12	12	14.75	20	16.97	11.65
2018	12	12	14.75	22	16.18	11.63
2019	12	12	14.75	22	18.24	11.43
2020	13.88	12	13	20	18.96	11.11
2021	18.22	15	14.45	22	20.63	12.79
2022	18.73	14	14.28	25	21.3	16.14
2023	20.63	12.02	14.54	27.65	25.39	18.43
2024	19.73	13.72	14.8	29.03	26.9	21.06
2025	20.99	12.35	19.2	29.8	24.81	23.99

(i) Choose data from any 3 states you find interesting and present it through a line graph using an appropriate scale. (ii) What do you find interesting in this data? Share your observations. (iii) Compare the price variation in Gujarat and Uttar Pradesh. (iv) In which state has the price increased the most from 2016 to 2025? (v) What are you curious to explore further?

Sol. (i) Let's choose Andaman & Nicobar, Assam and Gujarat While drawing a line graph, we follow these steps.

First, we decide what to put in x-axis and $y$ -axis
We always use time in x -axis so Year is in x -axis
So, Price of Salt is in $y$ -axis
Then, we find out our Scale
Since numbers go from 6 to 20.99
We choose Scale 1 -unit length $= 1$ unit
Lastly, we decide our legend - which colour to denote
We choose Blue for Andaman, Red for Assam and Green for Gujrat
(ii) 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) Gujarat's price is incredibly stable, staying right around 14 to 16 for an entire decade.

Uttar Pradesh shows a steep, relentless upward climb from 16 all the way to nearly 27 .

Uttar Pradesh has a large price increase compared to Gujarat. (iv) 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 $= 24.81 - 16.15 =$ 8.66West Bengal $= 23.99 - 9.47 =$ 14.52The price in West Bengal increases the most (₹ 14.42). (v) Why is Mizoram's salt so expensive? Is it because it is a remote, landlocked state where transportation costs heavily impact retail prices?

What happened in West Bengal around 2021? The price was stable around 11-12 for years, and then suddenly doubled over the next four years. Was there a local supply chain issue or a change in state taxes?

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.

Sol. (i) The graph clearly shows Kerosene > Electricity for rural and Electricity > Kerosene for urban in 1983 Thus, given statement is valid (ii) This is true as Both yellow lines trend downward toward zero.

Thus, given statement is valid (iii) In the year 2000, the Urban electricity line is way up at roughly $92%$ , not $10%$ . Thus, given statement is invalid (iv) This graph tracks which energy source is used, not whether that source is reliable. It says nothing about power cuts. Thus, given statement is invalid

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

Average Daily Time spent on Hobbies and Games

(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.

Sol. (i)

Looking where the blue line hits the 10 years mark

Urban children aged 10 spend little more than 2 hours. each day on hobbies and games. (ii)

The rural (red) line hits the 1.5 - hour mark at (d) 14 years. (iii) Statement (a)

At age 15, the average is roughly 1 hour. At age 10, it is 2 hours. 1 is half of 2, not twice!

Thus, given statement is incorrect. Statement (b) The graph shows the average is 1 hour. Averages are made of highs and lowssome kids probably spend 0 hours, while others spend 3.

Thus, given statement is incorrect.

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.

Sol. Do it yourself

13. 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. Daily sleep time includes night sleep, naps, and any sleep during the day. (1) Represent this on strips. (b) Put together the data of all your group members. Calculate the average and median sleep time of children, adults, elderly. (c) Share your findings and observations. (ii) When do schools start and end? On a weekday, Manoj's school starts at 9:30 am and ends at 4:30 pm, i.e., 7 hours which include class time and breaks. Collect information on the daily timings of different schools for Grade 8, including class time and break time (the schools can be anywhere in the country. You can ask your neighbours, relatives, parents and friends to find out). Analyse and present the data collected.

Sol. Do it yourself

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? (ii) Which place has the longest day length over the year? (iii) Share your observation- what do you find interesting? What are you curious to find out?

Sol. The sun rises in the early morning and sets in the evening.

Therefore, the bottom lines (which hover between 04:00 and 08:00) represent sunrise, and the top lines (which hover between 16:00 and 20:00) represent sunset. (i)

Earliest Sunrise: We need to look at the bottom line for the month of January across all four graphs. Kibithu has the lowest starting point in January, sitting exactly on the 06 : $00 (6$ : 00 AM $)$ mark.
Day Length: To find the day length, we subtract the sunrise time from the sunset time. In January at Kibithu, the sun rises at roughly 06:00 and the top line (sunset) is around 16 : $30 (4 : 30 PM)$ .
Calculation: From 06:00 to 16:30 is a gap of 10.5 hours (or 10 hours and 30 minutes). We are looking for the graph with the widest vertical gap between its top line and bottom line. That happens in Srinagar during the summer month of June. The sun rises very early (around 05:00) and sets very late (around 19:45), creating a massive daylight window of nearly 14.75 hours.

Observations: Kanyakumari: Its lines are almost completely flat! The length of the day barely changes from winter to summer. Kibithu vs. Ghuar Moti: Kibithu's entire day is shifted "early" (early to rise, early to set), while Ghuar Moti's day is shifted "late" (late to rise, late to set). Curiosities (The "Why"): As your teacher, I would hope this makes you wonder why Kanyakumari is so stable while Srinagar varies so much! It is entirely based on geography. Kanyakumari is very close to the equator, where day length stays a consistent 12 hours.

Srinagar is much further north, meaning it experiences extreme summer daylight and shorter winter daylight due to the tilt of the Earth.

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

Moonrise - Moonset

Sol. (i) We need to understand the relationship between the Sun and Moon during these two phases:

Amavasya (New Moon): The Moon rises and sets at approximately the same time as the Sun (around 06:00 and 18:00).
Purnima (Full Moon): The Moon rises when the Sun sets (around 18:00) and sets when the Sun rises (around 06:00). Finding Purnima: Look at the yellow line (moonrise). When does it hit the 18:00 mark? It crosses 18:00 exactly on Day 15.

Finding Amavasya: Look at the yellow line again. When does it hit the 06:00 morning mark? It crosses 06:00 exactly on Day 30. (You can also verify this by checking the blue moonset line for Day 30, which hits exactly at 18:00!) (ii) Notice: The lines form perfect diagonal staircases that repeat themselves. The moon does not rise at the same time every day. Instead, the yellow line creeps higher and higher up the graph. Wonder: By looking at the grid, you can wonder (and mathematically prove!) exactly how much the moon is delayed each day. Because the line travels a full 24 hours (from 00:00 to 24:00) over roughly 29.5 days, it gets delayed by about 50 minutes every single day. That is why the lines slope upward.

3.0Key Concepts of Chapter 5 Tales by Dots and Lines Exercise 5.3

Completing mean grids using row, column, and diagonal averages
Constructing data sets with a specified mean
Constructing data sets with a specified median
Creating data sets where the mean and median are equal
Creating data sets where the mean is greater than the median
Finding missing values using a given mean

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.3

Completes the given mean grids using the required values.
Determines missing values in rows, columns, and diagonals of the mean grid.
Forms data sets with a specified mean.
Forms data sets with a specified median.
Identifies data sets in which the mean and median have a given relationship.
Works with different possible data sets that satisfy the conditions given in the questions.

Tales by Dots and Lines Exercise 5.3 Class 8 Maths – Chapter 5 NCERT Solutions

1.0Download Class 8 Maths Chapter 5 Ex 5.3 NCERT Solutions PDF

Get the Exercise 5.3 NCERT Solutions PDF for Class 8 Maths Chapter 5 to check answers and work through the questions with clarity.

Chapter 5 : Tales by Dots and Lines Exercise 5.3

2.0Detailed NCERT Solutions for Tales by Dots and Lines Exercise 5.3 Class 8 Maths (Ganita Prakash Term 2)

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 direction? Sol. (i) Here, the average of 3 number is 10, So, their sum must be 30

13	6	11
8	10	12
9	14	7

Here is one way to fill the grid

Row 1: 13, 6, 11
Row 2: 8, 10, 12
Row 3: 9, 14, 7 (ii) Yes, but you would have to change all the numbers to a completely new sequence that is still perfectly balanced around 10.

And, 10 our target average is at the center So, our grid will be

16	2	12
6	10	14
8	18	4

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.

Let $5^{th}$ term $= 21$ Arranging data in ascending order, our data becomes

So, we can say that 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?

Sol. (i) False Let the two even numbers be 2 a and 2b. Average $= (\frac{2 a + 2 b}{2}) = a + b$

Since 2 a and 2 b is even but we cannot say about parity(odd/even) of ' a ' and ' b '

So, $a + b$ can be odd or even. We take an example Let the two even numbers be 2 & 4 Average $= \frac{2 + 4}{2} = \frac{6}{2} = 3$ Since the average is odd, the given statement is false

(ii) False

Let the two multiples of 5 be 5 m and 5n.

(iii)True

Let the five multiples of 5 be 5a, 5b, 5c, 5d, 5e.

Average $= \frac{5 a + 5 b + 5 c + 5 d + 5 e}{5}$ $= 5 (\frac{a + b + c + d + e}{5})$ $= a + b + c + d + e$ Since $a + b + c + d + e$ can be or cannot be multiple of 5 .

We take an example Let five multiples of 5 be $5, 10, 15$ , 20, & 25 Average $= \frac{5 + 10 + 15 + 20 + 25}{5} = \frac{75}{5} = 15$ Since 15 is not a multiple of 5, the given statement is true.

Sol. (i) To find the average, we need Sum of all heights.

Thus, we cannot find average without knowing the heights of the new students.

So, option (c) is correct (ii) Old mean $= 150.2 cm$

So, option (d) is correct

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

Sol. Method use: Count and Calculate

Value $(x)$	Frequency $(f)$
$14$	2
$15$	2
$16$	3
$17$	5
$18$	4
$19$	4
$20$	3
$21$	1
$23$	1

Sum $= (14 \times 2) + (15 \times 2) + (16 \times 3) + (17 \times 5) + (18 \times 4) + (19 \times 4) + (20 \times 3) + (21 \times 1) + (23 \times 1) = 25 + 30 + 48 + 8572 + 76 + 60 + 21 + 23 = 443$

Total numbers $= 2 + 2 + 3 + 5 + 4 + 4 + 3 + 1 + 1 = 25$ Mean $= \frac{443}{25} = 17.72$ The average of the data is 17.72 or on a dot plot 17.

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.

	Andaman and Nicobar Islands	Assam	Gujarat	Mizoram	Uttar Pradesh	West Bangal
2016	16	6	16.5	20	16.15	9.47
2017	12	12	14.75	20	16.97	11.65
2018	12	12	14.75	22	16.18	11.63
2019	12	12	14.75	22	18.24	11.43
2020	13.88	12	13	20	18.96	11.11
2021	18.22	15	14.45	22	20.63	12.79
2022	18.73	14	14.28	25	21.3	16.14
2023	20.63	12.02	14.54	27.65	25.39	18.43
2024	19.73	13.72	14.8	29.03	26.9	21.06
2025	20.99	12.35	19.2	29.8	24.81	23.99

Sol. (i) Let's choose Andaman & Nicobar, Assam and Gujarat While drawing a line graph, we follow these steps.

First, we decide what to put in x-axis and $y$ -axis
We always use time in x -axis so Year is in x -axis
So, Price of Salt is in $y$ -axis
Then, we find out our Scale
Since numbers go from 6 to 20.99
We choose Scale 1 -unit length $= 1$ unit
Lastly, we decide our legend - which colour to denote
We choose Blue for Andaman, Red for Assam and Green for Gujrat
(ii) 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) Gujarat's price is incredibly stable, staying right around 14 to 16 for an entire decade.

Uttar Pradesh shows a steep, relentless upward climb from 16 all the way to nearly 27 .

Uttar Pradesh has a large price increase compared to Gujarat. (iv) Calculating the price increase for each state.

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

Average Daily Time spent on Hobbies and Games

Sol. (i)

Looking where the blue line hits the 10 years mark

Urban children aged 10 spend little more than 2 hours. each day on hobbies and games. (ii)

The rural (red) line hits the 1.5 - hour mark at (d) 14 years. (iii) Statement (a)

At age 15, the average is roughly 1 hour. At age 10, it is 2 hours. 1 is half of 2, not twice!

Thus, given statement is incorrect. Statement (b) The graph shows the average is 1 hour. Averages are made of highs and lowssome kids probably spend 0 hours, while others spend 3.

Thus, given statement is incorrect.

Sol. Do it yourself

Answer the following questions based on the graphs:

Sol. The sun rises in the early morning and sets in the evening.

Therefore, the bottom lines (which hover between 04:00 and 08:00) represent sunrise, and the top lines (which hover between 16:00 and 20:00) represent sunset. (i)

Earliest Sunrise: We need to look at the bottom line for the month of January across all four graphs. Kibithu has the lowest starting point in January, sitting exactly on the 06 : $00 (6$ : 00 AM $)$ mark.
Day Length: To find the day length, we subtract the sunrise time from the sunset time. In January at Kibithu, the sun rises at roughly 06:00 and the top line (sunset) is around 16 : $30 (4 : 30 PM)$ .
Calculation: From 06:00 to 16:30 is a gap of 10.5 hours (or 10 hours and 30 minutes). We are looking for the graph with the widest vertical gap between its top line and bottom line. That happens in Srinagar during the summer month of June. The sun rises very early (around 05:00) and sets very late (around 19:45), creating a massive daylight window of nearly 14.75 hours.

Srinagar is much further north, meaning it experiences extreme summer daylight and shorter winter daylight due to the tilt of the Earth.

Moonrise - Moonset

Sol. (i) We need to understand the relationship between the Sun and Moon during these two phases:

Amavasya (New Moon): The Moon rises and sets at approximately the same time as the Sun (around 06:00 and 18:00).
Purnima (Full Moon): The Moon rises when the Sun sets (around 18:00) and sets when the Sun rises (around 06:00). Finding Purnima: Look at the yellow line (moonrise). When does it hit the 18:00 mark? It crosses 18:00 exactly on Day 15.

3.0Key Concepts of Chapter 5 Tales by Dots and Lines Exercise 5.3

Completing mean grids using row, column, and diagonal averages
Constructing data sets with a specified mean
Constructing data sets with a specified median
Creating data sets where the mean and median are equal
Creating data sets where the mean is greater than the median
Finding missing values using a given mean

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.3

Completes the given mean grids using the required values.
Determines missing values in rows, columns, and diagonals of the mean grid.
Forms data sets with a specified mean.
Forms data sets with a specified median.
Identifies data sets in which the mean and median have a given relationship.
Works with different possible data sets that satisfy the conditions given in the questions.

Frequently Asked Questions

What is covered in Exercise 5.3 of Class 8 Maths Chapter 5?

What is a mean grid?

How are missing values found in a mean grid?

How are data sets created with a given mean?

How are data sets created with a given median?

Can different data sets have the same mean or median?

Where can students find NCERT Solutions for Exercise 5.3?

Join ALLEN!

Tales by Dots and Lines Exercise 5.3 Class 8 Maths – Chapter 5 NCERT Solutions

1.0Download Class 8 Maths Chapter 5 Ex 5.3 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Tales by Dots and Lines Exercise 5.3 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 5 Tales by Dots and Lines Exercise 5.3

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.3

Frequently Asked Questions

What is covered in Exercise 5.3 of Class 8 Maths Chapter 5?

What is a mean grid?

How are missing values found in a mean grid?

How are data sets created with a given mean?

How are data sets created with a given median?

Can different data sets have the same mean or median?

Where can students find NCERT Solutions for Exercise 5.3?

Join ALLEN!

Tales by Dots and Lines Exercise 5.3 Class 8 Maths – Chapter 5 NCERT Solutions

1.0Download Class 8 Maths Chapter 5 Ex 5.3 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Tales by Dots and Lines Exercise 5.3 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 5 Tales by Dots and Lines Exercise 5.3

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 5 Exercise 5.3