NCERT Solutions Ganita Prakash (Part 2) Chapter 5 –132Section 5.2 Visualising and Interpreting Data — Figure it Out

Book page 127 Updated on2026-09-05

Q1.
Mean Grids: (i) Fill the grid with 9 distinct numbers such that the average along each row, column, and diagonal is 10. (ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions?
Answer

(i) Each line holds 3 numbers, so an average of 10 means every row, column and diagonal must add up to 30. Take the ordinary 3 × 3 magic square (which sums to 15 in every direction) and add 5 to each entry:

13611
81012
9147
Rows: 13 + 6 + 11 = 30,   8 + 10 + 12 = 30,   9 + 14 + 7 = 30
Columns: 13 + 8 + 9 = 30,   6 + 10 + 14 = 30,   11 + 12 + 7 = 30
Diagonals: 13 + 10 + 7 = 30,   11 + 10 + 9 = 30
All nine numbers 6, 7, 8, 9, 10, 11, 12, 13, 14 are distinct ✓
Why adding 5 works: every line of the magic square has 3 entries, so adding 5 to each entry adds 15 to every line total — 15 + 15 = 30. This is the ‘add a constant’ rule of page 107, used nine times at once.

(ii) Yes, and there are endless ways. The centre is forced to be 10, but everything else is free. Fill the grid like this, choosing any two numbers a and b:

10 + a10 − a − b10 + b
10 − a + b1010 + a − b
10 − b10 + a + b10 − a

Every row, column and diagonal here adds to 30 whatever a and b are — the a’s and b’s cancel out. Taking a = 3, b = 1 gives the grid above; taking a = 4, b = 1 gives a fresh one:

14511
71013
9156
Why the centre must be 10: add the middle row, the middle column and both diagonals. That is 4 lines, total 4 × 30 = 120. Every cell of the grid is used once in that sum except the centre, which is used four times. So (sum of all 9 cells) + 3 × (centre) = 120. But the three rows show the sum of all 9 cells is 90, so 3 × (centre) = 30 and the centre is 10 — it can never be anything else.
Q2.
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.
Answer
ConditionExample 1Example 2How it was built
(i) 3 numbers, mean 86, 8, 101, 8, 15Total must be 3 × 8 = 24
(ii) 4 numbers, median 15.510, 15, 16, 201, 15, 16, 100The two middle values must average 15.5, so 15 + 16 = 31
(iii) 5 numbers, mean 13.610, 12, 14, 16, 1613, 13, 14, 14, 14Total must be 5 × 13.6 = 68
(iv) 6 numbers, mean = median1, 2, 3, 4, 5, 62, 4, 6, 8, 10, 12Evenly spaced data: both come out at 3.5 and 7
(v) 6 numbers, mean > median1, 2, 3, 4, 5, 1001, 1, 2, 2, 3, 20Put one very large value far out to the right
Checks
(i) 6 + 8 + 10 = 24, 24 ÷ 3 = 8 ✓   1 + 8 + 15 = 24 ✓
(ii) (15 + 16) ÷ 2 = 15.5 ✓ in both
(iii) 10 + 12 + 14 + 16 + 16 = 68, 68 ÷ 5 = 13.6 ✓   13 + 13 + 14 + 14 + 14 = 68 ✓
(iv) 1+2+3+4+5+6 = 21, mean 3.5; median = (3 + 4) ÷ 2 = 3.5 ✓
(v) 1+2+3+4+5+100 = 115, mean ≈ 19.2; median = (3 + 4) ÷ 2 = 3.5, and 19.2 > 3.5 ✓
What (iv) and (v) are really about: when the data is spread symmetrically the mean and median land in the same place. Pull one value far out to the right and the mean chases it — because the mean measures distance — while the median stays put, because it only counts positions. That gap between mean and median is a signal that the data has an extreme value.
Q3.
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?
Answer

There will be six numbers, so the median is the average of the 3rd and 4th when sorted.

Median = 13  →  (3rd value) + (4th value) = 2 × 13 = 26

The fixed values sorted are 5, 14, 21, and 13 lies between 5 and 14. Only two pairs of whole numbers can occupy the two middle places:

  • 12 and 14. Put 12 in one blank; then one blank must be 12 or less and the other 14 or more, so 12 is 3rd and 14 is 4th.
    Example: 5, 21, 14, 12, 3, 40 → sorted 3, 5, 12, 14, 21, 40 → median (12 + 14) ÷ 2 = 13
  • 13 and 13. Put 13 in two blanks; the third must be 13 or less.
    Example: 5, 21, 14, 13, 13, 2 → sorted 2, 5, 13, 13, 14, 21 → median 13

How many possibilities? Infinitely many. In the first family the largest blank may be any counting number from 14 upwards — 14, 15, 16, …, 1000, … — and each choice gives a fresh answer with the median still 13.

Why any pair other than 12&14 or 13&13 fails: suppose you tried 11 and 15 (they also add to 26). The value 14 is already in the collection and lies between 11 and 15, so 14 would have to take one of the two middle places itself. The middle pair must therefore fit into the gap between 5 and 14, or use 14 as its upper member — which leaves exactly the two families above.
Q4.
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?
Answer
There will be 6 numbers, so the total must be
6 × 6.5 = 39
Sum of the four known numbers = 3 + 11 + 15 + 6 = 35
So the two blanks must add up to 39 − 35 = 4

With counting numbers (1, 2, 3, …) there are only 3 possibilities:

First blankSecond blankCollectionCheck
133, 11, 1, 3, 15, 639 ÷ 6 = 6.5 ✓
223, 11, 2, 2, 15, 639 ÷ 6 = 6.5 ✓
313, 11, 3, 1, 15, 639 ÷ 6 = 6.5 ✓

If you count 1 and 3 as the same answer whichever blank they go in, there are just 2 possibilities.

Why so few, when Q3 had endlessly many? The mean fixes the total, so the two blanks are tied to each other: choose one and the other is decided, and neither can exceed 3. The median fixes only a position, which leaves the far values completely free. This is the sharpest illustration in the chapter of how differently the two measures behave.
Q5.
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.
Answer

(i) False.

Counter-example: 2 and 4 → (2 + 4) ÷ 2 = 3, which is odd.

In algebra, let the numbers be 2m and 2n.
Average = (2m + 2n) ÷ 2 = m + n
This is always a whole number, but it is even only when m + n is even.
2 and 4 give m = 1, n = 2, so m + n = 3 — odd.

(ii) False.

Counter-example: 5 and 10 → (5 + 10) ÷ 2 = 7.5, not even a whole number.

Let the numbers be 5a and 5b.
Average = (5a + 5b) ÷ 2 = 5(a + b) ÷ 2
This is a multiple of 5 only when a + b is even. With 5 and 10, a + b = 1 + 2 = 3, which is odd.
It does work for 10 and 20: (10 + 20) ÷ 2 = 15 ✓

(iii) False.

Counter-example: 5, 5, 5, 5, 10 → sum 30, average = 30 ÷ 5 = 6, not a multiple of 5.

Let the numbers be 5a1, 5a2, 5a3, 5a4, 5a5.
Average = 5(a1 + a2 + a3 + a4 + a5) ÷ 5 = a1 + a2 + a3 + a4 + a5
So the average is always a whole number — but the 5 has cancelled away, and there is nothing left to force the answer to be a multiple of 5.
What the algebra reveals: dividing by 5 removes exactly one factor of 5 from the total. What survives is the sum of the a’s, which can be anything at all. It happens to be a multiple of 5 for 5, 10, 15, 20, 25 (average 15) — but one lucky example is never a proof. To disprove a claim you need only one counter-example; to prove one, you need the algebra.
Tip: a good habit with an ‘always true?’ question — first hunt for a counter-example with small numbers. If you cannot find one after a few honest tries, then look for the algebraic reason why none exists.
Q6.
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. (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.
Answer

(i) Only (c) is correct.

  • (a) Wrong. Adding values does not automatically raise a mean. Two very short students would pull it down.
  • (b) Wrong. It stays the same only in the special case where the two new heights add to 2 × 150.2 = 300.4 cm. There is no reason to expect that.
  • (c) Correct. The two new heights are the only missing information — until they are measured, the new average cannot be found.
  • (d) Wrong. Nobody needs re-measuring. The old average already carries the old total: 150.2 × 24 = 3604.8 cm.

(ii) Statement (b) is correct — the average will increase, very slightly.

Old total = 150.2 × 24 = 3604.8 cm
New total = 3604.8 + 149 + 152 = 3905.8 cm
New average = 3905.8 ÷ 26 = 150.22 cm
The quick way: 149 is 1.2 cm below the old mean and 152 is 1.8 cm above it. The two pulls do not cancel — there is 0.6 cm of surplus, shared among 26 students, which lifts the mean by 0.6 ÷ 26 ≈ 0.023 cm. Tiny, but upward. Had the two heights been 149 and 151.4 the mean would not have moved at all.

(iii) Statement (d) is correct — there is not enough information.

Why: we know the class average, but nothing about the 24 individual heights. The median might be 150.2 cm, or 147 cm, or 153 cm — the average tells us nothing about the middle student. Without the sorted list of heights we cannot say where the median was, let alone where it moves when two students join. Remember that 149 and 152 straddle the mean; there is no guarantee they straddle the median.
Q7.
Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question.
Answer

No. The average is 17.72, not 17.

Reading the crosses column by column:

Value14151617181920212223Total
Frequency223544310125

Method 1 — the balancing method (no long addition needed). Measure every cross from 17 and see whether the two sides balance.

Below 17: 3 × 2 (the 14s) + 2 × 2 (the 15s) + 1 × 3 (the 16s) = 6 + 4 + 3 = 13
Above 17: 1 × 4 + 2 × 4 + 3 × 3 + 4 × 1 + 6 × 1 = 4 + 8 + 9 + 4 + 6 = 31
The right side is 31 − 13 = 18 heavier, so 17 cannot be the balance point.
Correct mean = 17 + 18 ÷ 25 = 17 + 0.72 = 17.72

Method 2 — the direct calculation, as a check.

Total = (14×2) + (15×2) + (16×3) + (17×5) + (18×4) + (19×4) + (20×3) + (21×1) + (23×1)
= 28 + 30 + 48 + 85 + 72 + 76 + 60 + 21 + 23 = 443
Mean = 443 ÷ 25 = 17.72
Why 17 looks tempting: 17 is the tallest column, and it is roughly in the middle of the range 14 to 23. But the crosses are not spread evenly around it — twelve of them lie above 17 and only seven below, and one straggler out at 23 adds 6 to the right-hand pull all by itself. The mean has to slide up to restore the balance.
Tip: the balancing method is often quicker than adding everything, and it also tells you which way the mean is wrong and by how much — here, 0.72 too low.
Q8.
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?
Answer

The mean does not change at all. About the median we cannot be sure.

Mean. Change in the total weight = −2 + 1 + 1 = 0 kg
The number of people is the same, so
New mean = old total ÷ n = 65.3 kg, exactly as before

Median. The median depends on which people changed and where they sat in the sorted order.

  • If the three people who changed are all well away from the middle, the middle person is untouched and the median stays 67 kg.
  • If one of them is near the middle, the median can shift — by at most a kilogram or two, and it can move either way.
A small illustration with three people: 66, 67, 68 → median 67
If the middle person loses 2 kg → 65, 66, 68 → median 66 (changed)
If the lightest person loses 2 kg → 64, 67, 68 → median 67 (unchanged)
The lesson of this question: the mean is completely determined by the total, so knowing the changes add to zero settles it at once. The median depends on who changed, not on how much changed in total — so the same piece of information that pins the mean down leaves the median open.
Q9.
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. (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?
Answer

(i) Mizoram, Uttar Pradesh and West Bengal make an interesting trio — the most expensive state, a steady riser and the fastest riser. Years along the horizontal axis, price in ₹ up the vertical axis, scale 0 to 30 in steps of 5.

30 25 20 15 10 5 0 2016 2018 2020 2022 2024 Mizoram Uttar Pradesh West Bengal
Retail price of iodised salt in January, 2016 to 2025, for three states. West Bengal starts lowest and climbs fastest; Mizoram is dearest throughout.

(ii) Observations. Salt costs very different amounts in different states in the same month — in 2016 it was ₹6 in Assam and ₹20 in Mizoram, more than three times as much. Mizoram is the dearest every single year. Prices are not always rising: Andaman and Nicobar fell from ₹16 to ₹12 between 2016 and 2017 and stayed there for three years, and Gujarat dipped to ₹13 in 2020. But from about 2021 onwards almost every state rises together, which suggests a common cause such as higher transport and packaging costs.

(iii) Gujarat compared with Uttar Pradesh.

GujaratUttar Pradesh
Lowest price₹13.00 (2020)₹16.15 (2016)
Highest price₹19.20 (2025)₹26.90 (2024)
Range (highest − lowest)₹6.20₹10.75
Change 2016 → 2025+₹2.70+₹8.66

Gujarat’s price hardly moved for nine years — it sat between ₹13 and ₹16.5 right up to 2024, and only jumped to ₹19.2 in 2025. Uttar Pradesh climbed steadily year after year, and its total rise is more than three times Gujarat’s. So Gujarat is the steadier of the two and Uttar Pradesh the more variable.

(iv) West Bengal, by a clear margin.

State20162025IncreaseAs a percentage
West Bengal9.4723.99+14.52about 153%
Mizoram2029.80+9.80about 49%
Uttar Pradesh16.1524.81+8.66about 54%
Assam612.35+6.35about 106%
Andaman and Nicobar1620.99+4.99about 31%
Gujarat16.519.20+2.70about 16%
Worth noticing: West Bengal tops both columns, so the answer is not in doubt here. But the two columns can disagree — Assam’s rise of ₹6.35 is smaller than Mizoram’s ₹9.80, yet in percentage terms Assam’s price more than doubled (106%) while Mizoram’s rose by only about half (49%). Always say which kind of ‘most’ you mean.

(v) Things to explore. Why is salt so much dearer in Mizoram than in Assam next door — is it the cost of carrying goods into the hills? Why did West Bengal’s price more than double while Gujarat’s barely moved? Do coastal states where salt is made pay less than landlocked ones? And how do these rises compare with the rise in the price of other everyday things over the same ten years?

Q10.
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.
Answer

The two graphs show, for rural and for urban households, the percentage using each source as their primary lighting source.

(i) Valid. Reading the two graphs at 1983:

1983KeroseneElectricity
Ruralabout 84%about 15%
Urbanabout 34%about 64%

In rural areas kerosene was well above half, and in urban areas electricity was well above half. Both halves of the statement hold.

(ii) Valid. The kerosene line slopes downward throughout in both graphs — from about 84% to nearly 0% in rural areas, and from about 34% to nearly 0% in urban areas.

(iii) Not valid. In 2000 about 88–90% of urban households used electricity; it was kerosene that was down to roughly 10%. The statement has the two lines swapped.

(iv) Not valid. The graph is about which source a household uses as its main one for lighting. It says nothing about how reliable the supply is. A household counted under ‘electricity’ may still face power cuts every day.

The habit being trained here: statement (iii) fails on careless reading — the right year, the wrong line. Statement (iv) fails on a different and deeper count: the graph does not measure that quantity at all. Before judging any claim, ask first “does this graph even measure this?” and only then “does the reading support it?”
Q11.
Answer the following questions based on the line graph. (i) How long do children aged 10 in urban areas spend each day on hobbies and games? (ii) At what age is the average time spent daily on hobbies and games by rural kids 1.5 hours? (a) 8 years (b) 10 years (c) 12 years (d) 14 years (e) 18 years (iii) Are the following statements correct? (a) The average time spent daily on hobbies and games by kids aged 15 is twice that of kids aged 10. (b) All rural kids aged 15 spend at least 1 hour on hobbies and games everyday.
Answer

The blue line is Urban and the orange line is Rural. Both fall steadily with age.

(i) Reading the blue line above age 10: about 2 hours a day (a shade over 2 hours).

(ii) Follow the orange line down to a height of 1.5 hours:

Age1012131415
Rural (hours)2.42.11.851.61.3

The orange line crosses 1.5 hours a little after age 14, so the answer is (d) 14 years.

(iii)(a) Not correct — it is the other way round, and the other way about.

Urban at age 10 ≈ 2.1 h, urban at age 15 ≈ 1.1 h
Rural at age 10 ≈ 2.4 h, rural at age 15 ≈ 1.3 h
In both cases the 15-year-olds spend about half as long, not twice as long.

(iii)(b) Not correct. The graph gives an average of about 1.3 hours for rural 15-year-olds. An average of 1.3 hours is perfectly consistent with some children spending 3 hours and others spending none at all. A statement about all children can never be read off a graph of averages.

The idea underneath (iii)(b): this is the same warning as in Q10 of the previous set. A single summary number tells you where the balance point of the data lies; it tells you nothing about the smallest value. To claim that every child clears an hour you would need the whole data set, not its mean.
Q12.
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.
Answer

How to do it. Draw a strip of 48 boxes for each day, one box per half hour from midnight to midnight, and fix a colour for each activity before you start. Fill the strip in the same evening while you still remember the day; filling in a whole week at the weekend gives poor data.

(i) Look down the strips at the same hour on different days. If your green meal boxes and your blue sleep boxes fall in the same places all week, your routine is regular. Count the outdoor boxes and halve the count to get hours.

(ii) For each activity, add the boxes across all seven strips and divide by 7.

Example: sleeping = 20, 19, 20, 19, 18, 22, 23 boxes on the seven days
Total = 141 boxes, average = 141 ÷ 7 = 20.1 boxes ≈ 20 boxes = 10 hours a day

Do the same for every activity. Because a day has exactly 48 boxes, the seven averages must themselves add up to 48 — a useful check on your arithmetic. Round each average to the nearest box and colour a single ‘average day’ strip. Note that this average strip is not any real day of yours: it puts the same amount of sleep and study into one picture, but the order of the blocks is your choice.

(iii) Make an average strip for an adult at home the same way and set the two side by side. Compare the sleep block, the largest daytime block, and how many blocks each day is broken into. Adults usually sleep fewer boxes and have far more, shorter blocks.

Tip: record before you calculate, and do not tidy up a day to make it look better. Data collected honestly is the only kind worth averaging.
Q13.
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. (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 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.
Answer

Option (i): sleep across three age groups.

  1. Collect. Each member records every family member’s sleep for seven days, in half hours, counting naps. Agree beforehand on the age boundaries — for example children up to 18, adults 19 to 59, elderly 60 and above — so that everyone’s data can be pooled.
  2. (a) Strips. One 48-box strip per person per day, with sleep in one colour. Stacking a person’s seven strips shows at a glance whether their bedtime is regular.
  3. (b) Averages and medians. Find each person’s average over the week, then pool the group. A worked example for a group of 12 children:
    Daily sleep (hours): 9, 9.5, 8, 10, 9, 8.5, 9, 10.5, 8, 9.5, 9, 7.5
    Total = 107.5, mean = 107.5 ÷ 12 = 8.96 h
    Sorted: 7.5, 8, 8, 8.5, 9, 9, 9, 9.5, 9.5, 10, 10.5
    For 12 values the median is the average of the 6th and 7th → (9 + 9) ÷ 2 = 9 h
    Repeat for adults and for the elderly.
  4. (c) Observations. Expect children to sleep the most and adults the least, with the elderly somewhere between — the same shape as the graph on page 127. Compare the mean and the median in each group: if one person sleeps unusually little, the mean will sit below the median, and the median is then the better summary.

Option (ii): school timings. Collect start time, end time and total break time for at least ten Grade 8 schools. Set them out like this:

SchoolStartsEndsTotal hoursBreak timeClass time
Manoj’s9:30 am4:30 pm7 h1 h6 h
Sample A8:00 am2:00 pm6 h45 min5 h 15 min
Sample B7:30 am1:30 pm6 h40 min5 h 20 min

Then find the mean and median class time, note the earliest and latest start, and draw a dot plot of the total hours. Look for a pattern: schools in hotter regions often start earlier, and schools that run two shifts have shorter days.

Tip: in a group project, agree on the exact definitions first — does ‘break time’ include the short gaps between periods? If different members count differently, the pooled average means nothing.
Q14.
The following graphs show the sunrise and sunset times across the year at 4 locations in India. Observe how the graphs are organised. Are you able to identify which lines indicate the sunrise and which indicate the sunset? Answer the following questions based on the graphs: (i) At which place does the sun rise the earliest in January? What is the approximate day length at this place in January? (ii) Which place has the longest day length over the year? (iii) Share your observations — what do you find interesting? What are you curious to find out?
Answer

Reading the graphs. Each place has two lines. The lower pair, between about 5:00 and 8:00, are the sunrise times; the upper pair, between about 17:00 and 20:00, are the sunset times. The gap between a place’s two lines is the length of its day.

(i) Kibithu, in the first graph, where the sun rises at about 5:50 am in January — nearly two hours before Ghuar Moti (about 7:35 am).

Kibithu in January: sunrise ≈ 5:50, sunset ≈ 16:30
Day length ≈ 16:30 − 5:50 = about 10 hours 40 minutes
Why so early: Kibithu in Arunachal Pradesh is the easternmost inhabited place in India and Ghuar Moti in Gujarat the westernmost, yet the whole country runs on one clock, Indian Standard Time. The sun reaches the east first, so Kibithu’s clock reads about 5:50 at sunrise while Ghuar Moti’s reads 7:35. Note that Kibithu’s sun also sets early — by 4:30 in the afternoon.

(ii) Srinagar. Its two lines spread furthest apart in June:

Srinagar in June: sunrise ≈ 5:15, sunset ≈ 19:40 → day ≈ 14 hours 25 minutes
Kanyakumari in June: sunrise ≈ 6:00, sunset ≈ 18:40 → day ≈ 12 hours 40 minutes

(iii) Observations.

  • Kanyakumari’s two lines stay almost flat all year — its day length hardly changes, staying near 12 hours. Srinagar’s lines bow strongly apart in summer and close in winter, from about 14½ hours in June down to about 10 hours in January.
  • So the further north a place is, the more its day length swings; near the equator it barely moves.
  • Kibithu and Ghuar Moti have curves of almost the same shape, just shifted up or down the clock — because they are at similar latitudes but far apart in longitude. Srinagar and Kanyakumari have curves of very different shapes, because they differ in latitude.
  • Longitude shifts the clock times; latitude changes the length of the day. The four graphs separate these two effects rather neatly.
  • Curious to find out: should India have more than one time zone, given that sunrise differs by nearly two hours from end to end? How do people in Kibithu arrange a school day when it is dark by 4:30 pm? And what would these graphs look like for a place in the southern hemisphere?
Q15.
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?
Answer

(i) Use what each phase means in terms of rising and setting:

  • Purnima (full moon) — the moon is opposite the sun, so it rises around sunset and sets around sunrise, staying up the whole night.
  • Amavasya (new moon) — the moon is in the same direction as the sun, so it rises around sunrise and sets around sunset, keeping the sun’s company all day and never showing at night.

Now read the graph. Counting the tick marks from the left edge:

Day of the monthMoonriseMoonsetWhat it means
13about 5:25 pmabout 6:35 amUp nearly all night
14about 6:30 pmabout 7:30 amRises at sunset — purnima
28about 6:10 amabout 5:05 pmRises at sunrise — amavasya
29about 7:00 amabout 6:05 pmUp only during the day

So purnima falls around the 13th–14th of this month and amavasya around the 28th–29th — about a fortnight apart, as it should be.

(ii) What to notice

  • The moon rises about 50 minutes later each day. Over the first twenty days the moonrise slides from about 8:25 am to about 11:45 pm — 15 hours and 20 minutes later in 19 days, an average of 48 minutes a day.
  • Because of that slide, both lines climb steadily and then drop right off the top of the graph and restart at the bottom. That is not an error: the moonrise has simply passed midnight and moved into the next day.
  • There is one day with no moonrise at all (the 21st here) and one day with no moonset (the 6th). If moonrise falls just before midnight one day, the next rise is after midnight of the following day, so one calendar date gets skipped.
  • The moon is often up in broad daylight — around amavasya it is in the sky only during the day. We simply do not notice it against a bright sky.
  • The pattern is regular, but its period is about 29½ days, not 24 hours — so unlike the sun it does not repeat with the calendar day.
Why the 50-minute delay: while the earth spins once, the moon has moved a little further along its own orbit, so the earth must turn a bit extra to bring the moon back into view — roughly 50 minutes extra each day. Twelve or thirteen such delays add up to half a day, which is exactly why the moon that rose at sunset at purnima rises at sunrise a fortnight later at amavasya. The Indian calendar’s tithis are built on this same cycle.
Check it yourself: look up tonight’s moonrise time for your town, then look again tomorrow. The difference should be around three quarters of an hour.
Was this helpful? Report an error