Chapter 5 Tales by Dots and Lines (Class 8 - Latest Maths NCERT (Ganita Prakash II) Solutions)

(i) Mean of the first $50$ natural numbers

Given: The data consists of natural numbers $1, 2, 3, \dots, 50$.

To Find: Mean of these numbers.

Solution:

The sum of the first $n$ natural numbers is given by the formula $\frac{n(n+1)}{2}$.

(ii) Mean of the first $50$ odd numbers

Given: The first $50$ odd numbers are $1, 3, 5, \dots, 99$.

To Find: Mean of these numbers.

Solution:

The sum of the first $n$ odd numbers is given by the formula $n^2$.

(iii) Mean of the first $50$ multiples of $4$

Given: The multiples of $4$ are $4, 8, 12, \dots, 200$.

To Find: Mean of these numbers.

Solution:

The sum can be found by taking $4$ common: $4(1 + 2 + 3 + \dots + 50)$.

Observations:

1. For a sequence of numbers in Arithmetic Progression (where the difference between terms is constant), the mean is simply the average of the first and the last term.

2. For the first $n$ odd numbers, the mean is always equal to $n$.

3. In all cases, the mean represents the central value or the "balance point" of the data set.

Given:

The dot plot shows the following existing data points:

Value $4$: $1$ dot

Value $7$: $1$ dot

Value $8$: $2$ dots

Value $9$: $5$ dots

Value $11$: $1$ dot

Target Mean = $9$.

To Find: The position of the missing dot ($x$).

Solution:

First, let's calculate the number of existing dots: $1 + 1 + 2 + 5 + 1 = 10$ dots.

With the missing dot, the total number of observations will be $11$.

Let the value of the missing dot be $x$.

Final Answer: The missing dot should be marked at the value $16$ on the number line.

Given:

Total students = $24$

Initial average height (with shoes) = $150.2$ cm

Height added by shoes = $1$ cm

Solution (i):

No, the teacher does not need to measure all the heights again. There is a much simpler way based on the properties of the mean. If a constant value is subtracted from every observation in a data set, the mean of the data set also decreases by that same constant value.

Solution (ii):

Since every student's height was recorded as $1$ cm more than their actual height, we can find the correct average by subtracting $1$ cm from the calculated average.

Thus, the correct option is (d) $149.2$ cm.

Given: Three dot plots A, B, and C representing song lengths. We need to identify which one has a mean of $5.57$ minutes.

Solution:

We can find the answer by observing the range and distribution of the data in each plot without performing complex calculations.

1. Observation of Plot B: The data points in Plot B range from approximately $0.5$ to $5$ minutes. Since all the values are less than or equal to $5$, the mean cannot be $5.57$.

2. Observation of Plot C: The data points in Plot C are clustered between $3.5$ and $4.5$ minutes. Since the highest value is $4.5$, the mean must be somewhere between $3.5$ and $4.5$. Therefore, it cannot be $5.57$.

3. Observation of Plot A: The data points in Plot A range from $5$ to $6.5$ minutes. The target mean of $5.57$ falls within this range ($5 < 5.57 < 6.5$).

Verification for Plot A:

Let's list the values in Plot A: $5, 5, 5.25, 5.5, 5.75, 6, 6.5$. (Note: precise decimals are estimated from the grid).

Final Answer: Plot A has a mean of $5.57$ minutes. This is because it is the only plot where the data values are distributed around the $5.57$ mark, while the other plots contain values that are strictly lower than $5.57$.

Given:

The data set is: $8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92$

To Find:

The median and the effect of adding/removing values on it.

Solution:

First, we count the number of observations ($n$).

Since $n$ is even, the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.

(i) Including one value without affecting the median:

If we add one value, $n$ becomes $17$ (odd). The median will be the $(\frac{17+1}{2})^{th} = 9^{th}$ observation. In the original sorted list, the $9^{th}$ observation is $41$. If we include the value $41$, the new $9^{th}$ observation remains $41$.

(ii) Including two values without affecting the median:

If we add two values, $n$ becomes $18$ (even). To keep the median as $41$, the average of the $9^{th}$ and $10^{th}$ terms must be $41$. This happens if we add one value $\leq 41$ and one value $\geq 41$ (for example, $41$ and $41$, or $40$ and $42$).

(iii) Removing one value without affecting the median:

If we remove one value, $n$ becomes $15$ (odd). The median will be the $8^{th}$ observation. In the original list, the $8^{th}$ observation is $41$. So, if we remove one $41$, the new $8^{th}$ observation will still be $41$.

(i) Removing a value less than the median:

Statement: Never True.

Justification: Removing a smaller value from the left side of the data set shifts the center towards the right (larger values). This will either keep the median same or increase it, but never decrease it.

(ii) Including a value less than the mean:

Statement: Always True.

Justification: The mean is the balance point. Adding a value smaller than the current average "pulls" the balance point down towards the smaller value.

(iii) Including any $4$ values will not affect the median:

Statement: Sometimes True.

Justification: If we include $2$ values less than the median and $2$ values greater than the median, the middle position stays the same. However, if all $4$ values are very large or very small, the median will shift.

(iv) Including $4$ values less than the median:

Statement: Never True.

Justification: Adding values to the left side of the data set shifts the center towards the smaller values. This will either decrease the median or keep it the same, but it can never increase it.

Given:

Observations: $8, 13, 10, 4, 5, 20, y, 10$

Total number of observations ($n$) = $8$

Mean = $10.375$

To Find:

The value of $y$.

Solution:

Final Answer: The value of $y$ is $13$.

Given:

Number of values ($n$) = $15$

Mean = $134$

To Find:

Sum of the data.

Solution:

Calculation:

Final Answer: The sum of the data is $2010$.

Given:

Data without $p$: $12, 47, 8, 73, 18, 35, 39, 8, 29, 25$

Total observations including $p$ ($n$) = $11$

Target Median = $29$

To Find:

Possible values of $p$ from the given options.

Solution:

First, let us arrange the existing $10$ numbers in ascending order:

$8, 8, 12, 18, 25, 29, 35, 39, 47, 73$

Since $n = 11$, the median is the $(\frac{11+1}{2})^{th} = 6^{th}$ observation.

Currently, the $6^{th}$ value in the sorted list is $29$.

1. If $p < 29$, then $p$ will be placed before $29$. The $6^{th}$ observation would then become $25$ (if $p \leq 25$) or $p$ itself. The median would not be $29$.

2. If $p \geq 29$, then $p$ will be placed at or after the $6^{th}$ position. In this case, $29$ will remain as the $6^{th}$ observation.

Therefore, for the median to be $29$, we must have $p \geq 29$.

Checking Options:

(i) $10$ (No), (ii) $25$ (No), (iii) $40$ (Yes), (iv) $100$ (Yes), (v) $29$ (Yes), (vi) $47$ (Yes), (vii) $30$ (Yes).

Final Answer: The possible values for $p$ are $29, 30, 40, 47, 100$.

Given:

From the provided dot plot, we can extract the following frequency table:

(i) Average number of times students rode their cycles:

(ii) Median number of times:

Since the total number of students $N = 44$ (even), the median is the average of the $22^{nd}$ and $23^{rd}$ observations.

By calculating cumulative frequency:

Up to 3 rides, there are $3+1+4+8 = 16$ students. The next 8 students (up to $24^{th}$ student) rode 4 times.

So, both the $22^{nd}$ and $23^{rd}$ values are 4.

(iii) Validity of Statements:

(a) Invalid: There are 3 students who rode 0 times, so not everyone used their cycle.

(b) Valid: The majority of the dots are clustered between 2 and 7, showing most students cycled a few times.

(c) Valid: There are students who cycled 8 or 10 times in a week. Since a week has 7 days, they must have cycled more than once on some days.

(d) Invalid: More than 5 students cycled more than 7 times (3 students at 8 times and 2 at 10 times, total 5), but even those cycling 4 or 5 times might have cycled twice on a single day. The data doesn't provide enough detail to say "exactly 5".

(e) Average and Median for next week:

If every student cycles 1 more time, every observation $x$ increases by 1.

To Find: Minimum, Maximum, Mean, and Median of the trials taken by students.

Solution:

First, let's calculate the total number of students ($N$) and the sum of trials ($\sum fx$).

1. Minimum: The fewest trials taken by a student to hit the bull's eye was 1.

2. Maximum: The most trials taken was 10.

3. Mean:

4. Median:

Since $N = 62$, the median is the average of the $31^{st}$ and $32^{nd}$ observations. Looking at the cumulative frequencies:

Up to 7 trials: $1+0+0+1+4+9+12 = 27$ students.

The $28^{th}$ to $42^{nd}$ students all took 8 trials.

Since both the $31^{st}$ and $32^{nd}$ values are 8:

Conclusion: On average, students hit the bull's eye in about 7.63 trials, with 50% of the students taking 8 or fewer trials.

Given:

Customer footfall (Visiting) and conversion (Purchasing) data for seven days of a week.

To Find:

A double line graph to visualize the relationship between visiting and purchasing patterns.

Solution:

To construct the line graph, we follow these steps:

1. Axes Setup: Represent the Days of the week on the horizontal x-axis and the Number of Customers on the vertical y-axis (Scale: 1 unit = 5 customers).

2. Plotting Points: For each day, mark two points—one for 'Visiting' and one for 'Purchasing'.

3. Joining Lines: Use a solid line to connect 'Visiting' points and a dashed line (or different color) for 'Purchasing' points.

Observations from the Graph:

1. The gap between the two lines represents the number of customers who visited but did not buy anything.

2. In the Indian Perspective, we see a massive spike on Sunday, as it is a public holiday and families often go out for shopping. The visiting count peaks at $35$ and purchasing at $26$.

3. The "conversion rate" is highest on Sunday and Saturday, likely due to more serious shoppers during the weekend.

(i) Possible method to compile this data:

The data is likely compiled by recording the actual number of rainy days every month over a period of many years (usually 30 years or more) and then calculating the arithmetic mean for each month. Meteorological departments like the IMD (India Meteorological Department) maintain these historical records.

(ii) Graphical Analysis:

By rounding the values for the line graph (as per the requested image template):

1. Mangaluru would show a sharp peak in June, July, and August (reaching $\approx 28$ days in July) representing the heavy South-West Monsoon.

2. Port Blair shows high rainfall from May to November, as it is an island territory receiving rain from both monsoon branches.

3. Rameswaram shows a different pattern with a peak in November, which is typical for the retreating monsoon on the Coromandel Coast.

(iii) Data for New Delhi (extracted from Fig 4.7):

Based on the provided line graph (blue line), the values for New Delhi are approximately:

(iv) Most and Least Rainfall Days:

1. Most Days: By summing the values, Mangaluru or Port Blair has the most rainy days. Calculation for Mangaluru: $0.1+0+0.1+1.8+6.2+24.1+27.7+24.5+14+8.8+3.9+0.9 \approx 112$ days.

2. Least Days: New Delhi has the least number of rainy days per year compared to these coastal cities.

(v) Rainy Seasons:

1. New Delhi: The rainy season is primarily during July and August (Monsoon season in North India).

2. Rameswaram: The rainy season is during October, November, and December. This is due to the North-East Monsoon (Retreating Monsoon) which brings rain to the southern eastern coast of India.

(i) General Observations:

The graph shows a distinct seasonal trend in birth rates in India. There are peaks in late summer/early autumn (around August to October) and troughs in early spring (around March to April). The values fluctuate between approximately $1.4$ Million and $1.9$ Million births per month.

(ii) Births in July $2017$:

Looking at the point for July $2017$ on the x-axis, the corresponding value on the y-axis is approximately $1.6$ Million ($1.6$ M).

(iii) Time Period:

The graph captures data from approximately early $2017$ to early $2020$.

(iv) Comparison of January Births:

1. Jan $2018$: Approximately $1.8$ M.

2. Jan $2019$: Approximately $1.7$ M (A slight decrease from previous year).

3. Jan $2020$: Approximately $1.8$ M.

The birth rate in January remains relatively high but showed a small dip in $2019$.

(v) Estimated Births in $2019$:

To estimate the total births in $2019$, we add the approximate values for each month of that year:

Based on the peaks and valleys, the total annual births in India for $2019$ were roughly $20$ to $22$ Million.

Given: A $3 \times 3$ grid and a target average of $10$ for each row, column, and diagonal.

To Find: Fill the grid with 9 distinct numbers.

Solution:

If the average of 3 numbers is $10$, their sum must be:

This is equivalent to creating a Magic Square with a magic constant of $30$. In a $3 \times 3$ magic square, the middle number is always $\frac{1}{3}$ of the magic constant. So, the center number must be $10$.

One such combination of 9 distinct numbers is:

Check:

Row 1: $(8+15+7) \div 3 = 10$

Row 2: $(9+10+11) \div 3 = 10$

Diagonal 1: $(8+10+12) \div 3 = 10$

(ii) Answer:

Yes, we can fill the grid in different ways. By keeping $10$ at the center and choosing different pairs $(a, b)$ that sum to $20$ for the opposite cells, we can create multiple variations. For example, replacing $15$ and $5$ with $16$ and $4$ (and adjusting others) would still work.

(i) 3 numbers whose mean is 8 (Sum must be 24):

Example 1: $\{7, 8, 9\}$

Example 2: $\{4, 10, 10\}$

(ii) 4 numbers whose median is 15.5 (Average of middle two must be 15.5):

Example 1: $\{10, 15, 16, 20\}$ (Median = $\frac{15+16}{2} = 15.5$)

Example 2: $\{5, 14, 17, 30\}$ (Median = $\frac{14+17}{2} = 15.5$)

(iii) 5 numbers whose mean is 13.6 (Sum must be 68):

Example 1: $\{10, 12, 13, 13, 20\}$

Example 2: $\{1, 1, 1, 1, 64\}$

(iv) 6 numbers whose mean $=$ median:

Example 1: $\{1, 2, 3, 4, 5, 6\}$ (Mean = $3.5$, Median = $3.5$)

Example 2: $\{10, 10, 10, 10, 10, 10\}$ (Mean = $10$, Median = $10$)

(v) 6 numbers whose mean $>$ median:

Example 1: $\{1, 2, 3, 4, 5, 100\}$ (Median = $3.5$, Mean $\approx 19.16$)

Example 2: $\{10, 10, 10, 10, 10, 70\}$ (Median = $10$, Mean = $20$)

Given: Data is $\{5, 14, 21, \text{blank}, \text{blank}, \text{blank}\}$. Median = $13$.

Solution:

There are $6$ numbers in total. For an even number of observations, the median is the average of the $3^{rd}$ and $4^{th}$ terms when arranged in ascending order.

We already have $\{5, 14, 21\}$. Since $14$ and $21$ are greater than $13$, they must occupy positions $x_4, x_5,$ or $x_6$. The smallest possible position for $14$ is $x_4$.

If $x_4 = 14$, then $x_3 = 26 - 14 = 12$.

One possible set is: $5, 12, 12, 14, 15, 21$. (Blanks are $12, 12, 15$).

Possibilities:

Since we only have the constraint that the $3^{rd}$ and $4^{th}$ terms sum to $26$ (and $x_4$ must be $\le 14$ because $14$ is in the list), there are infinitely many possibilities if we consider the choice of the other two blanks (one $\le 12$ and one $\ge 14$). Even with restricted counting numbers, there are hundreds of combinations for the other blanks.

Given: Data is $\{3, 11, x, y, 15, 6\}$. Mean = $6.5$.

Solution:

Sum of existing numbers $= 3 + 11 + 15 + 6 = 35$.

Since $x$ and $y$ must be counting numbers ($1, 2, 3, \dots$), the possible pairs $(x, y)$ are:

1. $(1, 3)$

2. $(2, 2)$

3. $(3, 1)$

Final Answer: There are 2 unique sets of numbers possible: $\{1, 3\}$ and $\{2, 2\}$. (If the order of blanks matters, there are 3 possibilities).

(i) The average of two even numbers is even.

Result: False.

Justification: Let the numbers be $2$ and $4$ (both even). Average $= \frac{2+4}{2} = 3$, which is odd. Algebraically, $\frac{2n + 2m}{2} = n + m$. If $n$ is even and $m$ is odd (or vice-versa), the result is odd.

(ii) The average of any two multiples of 5 will be a multiple of 5.

Result: False.

Justification: Let the numbers be $5$ and $10$. Average $= \frac{5+10}{2} = 7.5$, which is not even an integer, let alone a multiple of $5$.

(iii) The average of any 5 multiples of 5 will also be a multiple of 5.

Result: False.

Justification: Let the five multiples be $5, 5, 5, 5,$ and $10$.

Since $6$ is not a multiple of $5$, the statement is false. Algebraically, the average is $(n_1 + n_2 + n_3 + n_4 + n_5)$, which only becomes a multiple of $5$ if the sum of the coefficients is divisible by $5$.

(i) Analysis of the first set of statements:

Correct Statement: (c) The heights of the new students have to be measured to find out the new average height.

Justification: The new average depends on whether the new heights are greater than, less than, or equal to the existing average. Simply adding two values does not guarantee an increase or decrease. Furthermore, statement (d) is incorrect because if we know the previous sum (from the old average and old count), we only need the new heights to calculate the new total sum and new average.

(ii) Impact of $149$ cm and $152$ cm on Average:

Correct Statement: (b) The average will increase.

Justification: To see how the mean changes, we compare the average of the new values with the old mean.

Since the new students' average ($150.5$ cm) is greater than the old class average ($150.2$ cm), the overall mean of the class will increase.

(iii) Impact on Median:

Correct Statement: (d) The information is not sufficient to make a claim about median.

Justification: The Median is a positional value. It depends on the specific heights of every student in the class when arranged in order. While we know the mean is $150.2$ cm, we do not know the individual heights of the other $24$ students. Without knowing their exact values and distribution, we cannot determine how the middle value(s) will shift when $149$ and $152$ are inserted into the list.

Given: A dot plot representing a data set.

To Find: Check if the mean (average) of this data is $17$.

Solution (Tabulation Method):

Let's convert the dot plot into a frequency table to calculate the mean.

Final Answer: No, $17$ is not the average of the data. The actual average is $17.72$. From an Indian perspective of visual estimation, we can see that more dots are spread out on the right side of $17$ (up to $23$) compared to the left (only down to $14$), which "pulls" the average higher than $17$.

Given:

Previous Mean = $65.3$ kg

Previous Median = $67$ kg

Changes this month: One person $-2$ kg, Two people $+1$ kg each.

Analysis of Mean:

The mean depends on the sum of all values. Let's calculate the net change in the total sum:

Since the total sum of the weights remains the same and the number of people hasn't changed, the Mean weight will remain exactly the same ($65.3$ kg).

Analysis of Median:

The median is the middle value of the sorted data. A small change of $1$ or $2$ kg for three people may or may not change the median. For example, if the people who changed weights were already far from the middle position, the median stays $67$. If one of them was the middle person, it might shift slightly. However, with the limited information, we can only say that the Median is likely to remain $67$ or very close to it, but it cannot be determined with $100\%$ certainty.

Final Conclusion: The Mean weight remains unchanged, whereas the Median weight is generally stable but cannot be confirmed without the full list of weights.

Given:

A table showing retail prices of iodised salt in six Indian states/UTs from $2016$ to $2025$.

(i) Line Graph Presentation:

We choose Mizoram, Uttar Pradesh, and West Bengal for the line graph. For plotting, we round the values to the nearest integer.

On a graph, the x-axis would represent the Year and the y-axis would represent Price in $\textsf{₹}$. Mizoram would show the highest starting and ending points, while West Bengal would show the steepest climb.

(ii) Observations:

1. Salt prices, which are generally considered stable, have shown a significant upward trend in almost all states over the last decade.

2. Mizoram consistently has the highest price, likely due to its hilly terrain and higher transportation costs from coastal salt-producing regions like Gujarat.

3. Prices in Assam and Gujarat have remained relatively lower and more stable compared to other states.

(iii) Comparison of Gujarat and Uttar Pradesh:

In Gujarat, the price started at $\textsf{₹} 16.5$ in $2016$ and remained extremely stable around $\textsf{₹} 14$ for nearly 8 years, only rising to $\textsf{₹} 19.2$ in $2025$. This is because Gujarat is the largest producer of salt in India.

In Uttar Pradesh, the price has seen a more consistent and higher increase, moving from $\textsf{₹} 16.15$ to $\textsf{₹} 24.81$, reflecting the added costs of logistics to a landlocked state.

(iv) State with Maximum Increase:

Let's calculate the increase for West Bengal and Mizoram:

West Bengal has seen the most significant increase in price (nearly $154\%$) from $2016$ to $2025$.

(v) Further Exploration:

I am curious to explore whether the implementation of GST (Goods and Services Tax) or changes in fuel prices for transportation directly correlate with the sudden price jumps observed around $2022-2023$ in most states.

(i) Statement: In $1983$, the majority in rural areas used kerosene while the majority in urban areas used electricity.

Validity: Valid. By observing the graphs for $1983$: In Rural areas, Kerosene was at $\approx 84\%$, which is a clear majority. In Urban areas, Electricity was at $\approx 64\%$, which is also a majority (greater than $50\%$).

(ii) Statement: The use of kerosene as a primary lighting source has decreased over time in both rural and urban areas.

Validity: Valid. In both graphs, the orange lines (Kerosene) show a continuous downward slope from $1983$ to $2023$, indicating a consistent decline in usage.

(iii) Statement: In the year $2000$, $10\%$ of the urban households used electricity as a primary lighting source.

Validity: Invalid. In the Urban graph for the year $2000$, the Electricity line (blue) is near the $90\%$ mark. It was Kerosene usage that had dropped to nearly $10\%$.

(iv) Statement: In $2023$, there were no power cuts.

Validity: Invalid. The graph only shows the "Primary source of energy used for lighting". Even if a household uses electricity as its primary source, the graph does not provide data on the reliability of the power supply or the frequency of power cuts.

(i) Time spent by Urban children aged 10:

By observing the blue line (Urban) at the point corresponding to Age 10 on the x-axis, the y-axis value is exactly $2$ hours ($2$h).

(ii) Age for Rural kids spending 1.5 hours:

We look for the $1.5$h mark on the y-axis (midway between $1$h and $2$h) and see where the orange line (Rural) crosses it. This happens at approximately Age $14$.

The correct option is (d) $14$ years.

(iii) Analysis of Statements:

(a) The average time spent by kids aged 15 is twice that of kids aged 10:

Incorrect. At age 10, the time is $\approx 2$h. At age 15, the time is $\approx 1$h. The time spent by 15-year-olds is actually half that of 10-year-olds, not twice.

(b) All rural kids aged 15 spend at least 1 hour on hobbies and games everyday:

Incorrect. The graph shows the average daily time. An average of $1$ hour means some children might spend $3$ hours while others spend $0$ hours. We cannot conclude that "all" children spend at least $1$ hour based only on the mean.

Given:

This is an individual project requiring the tracking of personal daily activities, specifically eating, sleeping, and outdoor time.

(i) Eating, Sleeping, and Outdoor Habits:

In the Indian student context, a typical routine involves fixed times for meals and sleep to maintain discipline for school. For example, sleeping at $10:00$ pm and waking up at $6: 00$ am. Time spent outdoors for sports or play is usually $1$ to $2$ hours in the evening.

(ii) Average Time per Activity:

To calculate the average time, we sum the hours spent on a specific activity over $7$ days and divide by $7$.

If a student spends $14$ hours outdoors in a week, the average is:

(iii) Comparison with an Adult:

Tracking an adult (like a parent) usually reveals that they spend more time on household chores or office work and significantly less time on outdoor games compared to a Grade $8$ student. Adults also tend to have a shorter average sleep duration.

(i) Tracking Sleep Time:

By collecting data from family members, we can categorize them. For example, an elderly grandparent might take more naps during the day, whereas an adult may only sleep at night.

Average and Median Sleep Time:

Assuming a group collected data for children ($8$ hours), adults ($7$ hours), and elderly ($6$ hours), the findings usually show that children require the most sleep for growth and development.

(ii) School Timings Information:

To Find: Comparison of school durations.

Solution: In India, school timings for Grade $8$ vary by state and type of school (Government vs. Private).

Observations show that while total duration is similar ($\approx 6-7$ hours), the start times vary significantly based on local climate and transportation.

Given:

Graphs for Kibithu (Arunachal Pradesh), Ghuar Moti (Gujarat), Srinagar (J&K), and Kanyakumari (Tamil Nadu).

Identification: The bottom lines in each set represent Sunrise (earlier times, $04:00$ to $08:00$), and the top lines represent Sunset (later times, $16:00$ to $20:00$).

(i) Earliest Sunrise in January:

By observing the January data on the first graph, the sun rises earliest at Kibithu (the easternmost point of India). The sunrise is approximately at $06:00$ am.

(ii) Longest Day Length:

The Srinagar graph shows the widest gap between the top and bottom lines in June. In northern latitudes, summer days are much longer than in southern parts of India.

(iii) Observations:

1. Kanyakumari has the most stable lines, meaning day length doesn't change much because it is closer to the Equator.

2. Ghuar Moti has the latest sunrise and sunset because it is in the far West of India.

I am curious to find out why India uses a single Standard Time (IST) despite such a large difference (nearly $2$ hours) in sunrise times between the East and West.

(i) Finding Amavasya and Purnima:

Purnima (Full Moon): On this day, the moon rises roughly when the sun sets (around $18:00$). Observing the graph, the yellow diamond (Moonrise) is at $18: 00$ around Date 14.

Amavasya (New Moon): On this day, the moon rises and sets at almost the same time as the sun (around $06:00$ am). In the graph, the yellow diamond is near $06:00$ am on Date 28 or 29 and Date 1.

(ii) Observations and Wonders:

Observations:

1. The moonrise time is not constant; it shifts by approximately $50$ minutes later every day.

2. The lines are diagonal, showing a steady progression through the $24$-hour cycle over the month.

I wonder: Since the moon's orbit is tilted, does the path look the same in the Southern Hemisphere? And how does the moon sometimes appear in the sky during broad daylight?