Tales by Dots and Lines
Complete worked solutions for Chapter 5 — every in-text question and Figure-it-Out exercise on mean, median, and data visualisation, with step-by-step working and diagrams.
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The Balancing Act — Mean as the Centre
For any two numbers, their average always lands exactly halfway between them on the number line. For 3 and 7: mean \(=\frac{3+7}{2}=5\), which sits 2 units from each. For 8 and 9: mean \(=\frac{8+9}{2}=8.5\), again exactly midway. This will be true no matter which pair you pick — the mean of two numbers is always their midpoint.
| Collection | Values | Mean |
|---|---|---|
| Plot 1 | 6, 7, 8 | \(\frac{6+7+8}{3}=7\) |
| Plot 2 | 3, 6, 9 | \(\frac{3+6+9}{3}=6\) |
| Plot 3 | 2, 4, 8 | \(\frac{2+4+8}{3}=4.67\) |
| Plot 4 | 3, 11, 15 | \(\frac{3+11+15}{3}=9.67\) |
The mean balances the data — it is the point where the total distance of all values below it equals the total distance of all values above it. It is not simply the midpoint of the smallest and largest value; it’s a true “balance point,” like the fulcrum of a see-saw, that accounts for every single value’s pull on either side.
No — the mean is not generally the midpoint of the extremes. For example, in the collection 9, 11, 11, 17 the mean is 12, but the midpoint of the extremes (9 and 17) is also 13, which is close but not exactly equal in general, and for skewed data they can differ a lot. Instead, what’s always true is:
| Collection | Mean | LHS distances | RHS distances |
|---|---|---|---|
| 6, 7, 8 | 7 | 1 | 1 |
| 2, 4, 4, 8, 8 | 5.2 (example) | balanced | balanced |
| 9, 11, 11, 17 | 12 | 2+1+1=4 | 5 |
| 1, 3, 5, 9, 11 | 5.8 | 4.8+2.8+0.8=8.4 (approx) | balanced |
- If the new value is greater than the current mean, the mean increases — to restore balance, the centre must shift right toward this added weight.
- If the new value is less than the current mean, the mean decreases — the centre shifts left.
- Removing a value greater than the mean — the mean decreases (we’ve taken away weight from the heavier side).
- Removing a value less than the mean — the mean increases.
- Removing a value exactly equal to the mean — the mean stays the same, since that value contributed zero distance on either side; removing it doesn’t disturb the balance.
The mean stays exactly the same. Using the fair-share idea: imagine the mean is the amount everyone gets if the total is shared equally. If a new person joins with exactly that fair-share amount, the total share-per-person doesn’t change at all — they simply bring along precisely their “fair” contribution, neither adding nor subtracting any imbalance. The same logic applies in reverse when removing such a value.
Unchanging Mean!
Relatively Unchanged! (Adding, Subtracting, Scaling)
Think of the mean as what everyone would get if the total were shared equally. If every single value increases by 2 (say everyone receives ₹2 extra), then naturally each person’s fair share also goes up by exactly ₹2 — nothing about how the total is redistributed has changed relatively, only an equal bonus has been layered on top of everyone. So the fair share (mean) simply shifts by the same fixed amount.
The average also doubles. Using the same data (mean 7.18), doubling every value gives a new mean of \(2\times 7.18 = 14.36\) — confirmed directly in the chapter’s example.
Tinkering with Median
Finding the Unknown
Mean and Median with Frequencies
Spreadsheets
Cell B7 contains 27 — Gowri’s marks in Odia (column B = Odia, row 7 = Gowri).
Looking at Ashwin’s row (row 4: 29, 31, 33, 34, 30, 28): Ashwin scored more than 30 in Telugu (31), English (33), and Maths (34). (Social Science is exactly 30, not “more than” 30.)
(Science is column G; the student data spans rows 2 to 23 for all 22 students listed.) Similarly, the Odia average is =AVERAGE(B2:B23) and the Telugu average is =AVERAGE(C2:C23). Comparing the two computed values directly tells us which subject has the higher class average — this requires actually running the formulas on the full dataset to confirm numerically.
| Task | Formula (example row/col) |
|---|---|
| Total marks for a student (e.g. Ratna, row 2) | =SUM(B2:G2) |
| Average marks in a subject (e.g. Maths, column E) | =AVERAGE(E2:E23) |
Once written for one row/column, the same formula can be copied down or across to every other student/subject — spreadsheets automatically adjust the cell references, saving enormous time compared to manual calculation.
Mean & Median — Exercise
| Data | Sum | Mean |
|---|---|---|
| (i) 1, 2, 3, …, 50 | 1275 | 25.5 |
| (ii) 1, 3, 5, …, 99 | 2500 | 50 |
| (iii) 4, 8, 12, …, 200 | 5100 | 102 |
Plot A has all its dots clustered tightly between 5 and 6.5 minutes — a fairly narrow, high cluster. Since the mean of such tightly-bunched-high values will itself land in that high range (around 5.5–6), Plot A is the one with mean 5.57 minutes. Plots B and C have dots spread more toward lower values (B has many dots below 2, and C is clustered around 3.5–4.5), so their means would be noticeably lower than 5.57.
| Part | Strategy | Example answer |
|---|---|---|
| (i) Add 1 value | Add any value equal to 41 — this keeps both middle values at 41, so the median (their average) stays 41. | Add another 41 |
| (ii) Add 2 values | Add one value below 41 and one value above 41 symmetrically placed around the current two middle 41’s, so the new middle pair (now positions 9, 10 of 18) is still 41 & 41. | Add 40 and 41 (or any pair keeping the two new middle values at 41) |
| (iii) Remove 1 value | Remove one of the two values that are already 41 — then with 15 values left, the new middle (8th) value is still 41. | Remove one of the 41s |
| # | Statement | Verdict | Reasoning |
|---|---|---|---|
| (i) | Removing a value less than median decreases median | Sometimes true | It can increase, decrease, or stay the same depending on which values surround the median — removing a low value shifts more “weight” to one side, but the exact effect on the middle position depends on the full data layout. |
| (ii) | Including a value less than mean decreases mean | Always true | Adding a below-mean value always pulls the balance point down — this follows directly from the balancing principle we proved earlier. |
| (iii) | Including any 4 values won’t affect median | Sometimes true | It’s only true for specially chosen values (e.g. 2 above + 2 below median symmetrically); most arbitrary sets of 4 values will shift the median. |
| (iv) | Including 4 values less than median increases median | Never true | Adding 4 values, all below the median, pushes more data to the lower side — the median can only stay the same or decrease, never increase. |
| p | 10 | 25 | 40 | 100 | 29 | 47 | 30 |
|---|---|---|---|---|---|---|---|
| Median | 25 | 25 | 29 | 29 | 29 | 29 | 29 |
Using the same method throughout: read off each value and its frequency (number of dots stacked above it) directly from the plot, then:
For the median, build a running total of frequencies from the smallest value upward, and find which value(s) fall at the middle position(s) (depending on whether the total count is odd or even).
| Statement | Valid? | Why |
|---|---|---|
| (a) Everyone used their cycle at least once | Depends on plot — Invalid if any dots appear at 0 | If there’s a dot at the value 0, some students never rode their cycle that week. |
| (b) Almost everyone used their cycle a few times | Valid (generally) | Most of the dots cluster around the middle values (e.g. 3–6 times), suggesting most students cycled “a few” times, not zero and not extremely often. |
| (c) Some students cycled more than once on some days | Valid | Since “more than once a day” would require well over 7 rides/week, and several students have totals like 8 or 10 — more than the 7 days in a week — meaning they must have ridden more than once on at least one day. |
| (d) Exactly 5 students used their cycles more than once on some days | Invalid (too specific/unverifiable) | The dot plot only tells us the total weekly count per student, not how those rides were distributed across days — we cannot determine an exact count of “more than once in a day” from this data alone. |
Line Graphs — Temperature, Space & Rainfall
Multiple weather stations spread across the state continuously record local temperatures. The “monthly maximum” for the whole state is obtained by taking the single highest temperature reading recorded by any station, anywhere in the state, during that month.
- The horizontal axis shows months (Jan–Dec); the vertical axis shows temperature in °C, scaled in steps of 10.
- Kerala’s points are connected with a blue line, Punjab’s with a red line — using distinct colours (and ideally distinct marker shapes) lets us track each state’s trend separately even when the lines cross.
- Punjab shows a sharp rise-and-fall pattern (a clear summer peak); Kerala’s line stays much flatter throughout the year.
This is open-ended — possible questions: Why does Punjab show such extreme seasonal swings while Kerala stays stable? Is this related to distance from the coast, or latitude? A plot of monthly minimum temperatures would likely show a similar overall shape (Punjab swinging widely, Kerala staying flatter) but with every value shifted down — since minimum temperatures are always lower than maximum temperatures for the same month and place.
Space agencies and international bodies (like the UN Office for Outer Space Affairs) maintain official registries where every satellite, probe, lander, or crewed mission is logged at launch. The yearly count is simply the total number of such registered launches recorded for each country (and the world) within that calendar year.
- “Worldwide count increased every single year” — Not valid. The graph shows the count actually decreased from 2023 to 2024 (from ~2900 down to ~2800), so it didn’t increase every year.
- “USA launched about 3/4 of the worldwide count in 2022–24” — Valid (approximately) — the USA’s line sits very close to the World line in those years, visually confirming it’s the dominant contributor at roughly that proportion.
- “Nepal did not launch any object 2012–24” — Not a valid inference. The graph only shows USA, China, and Russia individually plus the World total — it gives us no information whatsoever about Nepal specifically.
- “Combined China + Russia count in 2024 is about 400” — Valid (approximately) — reading both countries’ lines at 2024 and adding their values gives a figure in this range.
Looking at the steepest jump in the World line: between 2019 and 2020 (or thereabouts), the count roughly doubles — jumping from several hundred to over a thousand. This is the sharpest year-on-year rise visible on the graph, corresponding to the surge in commercial satellite launches (e.g. large satellite constellations) during this period.
Method: Rainfall is measured daily at weather stations in each city over many years; for each calendar month, the total rainfall recorded in that month is averaged across all the years of data — giving the “monthly average rainfall” used in the graph.
Grouping: Kovalam, Udupi, and Mumbai lie along India’s west coast; Rameswaram, Chennai, and Puri lie along the east coast. The west-coast cities generally receive much heavier rainfall (up to ~800–900 mm peaks) than most east-coast cities — except Puri and Chennai, which still show clear but smaller peaks.
| City | Peak rainfall months | Low rainfall months |
|---|---|---|
| Kovalam, Udupi, Mumbai | June – August | January – March |
| Rameswaram | October – December | January – September (very low all through) |
| Chennai | Rises from June, peaks in November, continues to December | January – March |
| Puri | July – September | January – March |
Infographics & Activity Strips
| (ii) Top 5 — Rice | Value |
|---|---|
| Manipur | +100 |
| Nagaland | +99 |
| Mizoram | +97 |
| Tripura | +96 |
| Assam | +95 |
| (iii) Top 5 — Wheat | Value |
|---|---|
| Rajasthan | −93 |
| Haryana | −81 |
| Punjab | −78 |
| Madhya Pradesh | −60 |
| Delhi | −45 |
This is a visual-reasoning exercise based on Manoj’s coloured 48-box strips (30-min intervals). General approach:
- (i) Identify colours: The largest continuous block, usually overnight, is sleeping. Short, repeated small blocks around typical meal times are eating. A long uninterrupted weekday block (mid-morning to afternoon) is attending classes. Scattered colourful blocks in the evening/weekend are meeting friends/hobbies. Brief blocks near waking time are showering/exercise, and short blocks at the start/end of the school block represent travelling.
- (ii) Identify the days: The strip with the longest unbroken “class” block on a weekday pattern is likely a school day (Friday); strips with much longer “friends/hobbies” blocks and later wake-times are the weekend (Saturday, Sunday).
- (iii) The long movie: Look for an unusually long, single uninterrupted block of “meeting friends/media” time (longer than a typical TV episode) — this is most likely to appear on a weekend afternoon/evening strip.
- (iv) Lunch break: A short “eating” block sandwiched in the middle of the long weekday “classes” block marks the lunch break time.
Visualising & Interpreting Data — Exercise
Plot both rows as two separate lines sharing the same days-of-week axis: “Visiting” (16, 19, 10, 14, 20, 22, 35) and “Purchasing” (10, 8, 7, 11, 12, 16, 26), each with their own colour/marker. Notice that “Purchasing” is always lower than “Visiting” — not everyone who visits ends up buying — and both numbers rise sharply on Saturday and especially Sunday.
| Condition | Example 1 | Example 2 |
|---|---|---|
| (i) 3 numbers, mean 8 | 6, 8, 10 | 4, 8, 12 |
| (ii) 4 numbers, median 15.5 | 10, 15, 16, 20 (middle two: 15,16 → avg 15.5) | 5, 14, 17, 30 (middle two: 14,17 → avg 15.5) |
| (iii) 5 numbers, mean 13.6 | 10, 12, 13, 14, 19 (sum=68) | 5, 10, 15, 20, 18 (sum=68) |
| (iv) 6 numbers, mean = median | 1, 2, 3, 4, 5, 6 (mean=3.5, median=3.5) | 2, 4, 6, 8, 10, 12 (mean=7, median=7) |
| (v) 6 numbers, mean > median | 1, 2, 3, 4, 5, 50 (mean≈10.8, median=3.5) | 2, 3, 4, 5, 6, 100 (mean≈20, median=4.5) |
| Statement | Verdict | Justification |
|---|---|---|
| (i) Average of two even numbers is even | Sometimes true (not always) | Counter-example: \(\frac{4+6}{2}=5\), which is odd! It’s only even when the two numbers, divided by 2, have the same parity (e.g. \(\frac{2+8}{2}=5\) — wait, let’s verify: \(\frac{4+10}{2}=7\), odd. \(\frac{2+10}{2}=6\), even). So whether the average is even depends on the specific pair chosen. |
| (ii) Average of any two multiples of 5 is a multiple of 5 | Sometimes true (not always) | Let the two multiples be \(5a\) and \(5b\). Average \(=\frac{5a+5b}{2}=\frac{5(a+b)}{2}\). This is a multiple of 5 only when \(a+b\) is even. Counter-example: \(\frac{5+10}{2}=7.5\) — not even a multiple of 5 at all (not even an integer)! |
| (iii) Average of any 5 multiples of 5 is also a multiple of 5 | Sometimes true (not always) | Let the 5 multiples be \(5k_1,\dots,5k_5\). Average \(=\frac{5(k_1+\dots+k_5)}{5}=k_1+\dots+k_5\), which is always an integer — but it equals a multiple of 5 only if \(k_1+\dots+k_5\) is itself a multiple of 5. Counter-example: \(5,10,10,10,10\) → average \(=\frac{45}{5}=9\), not a multiple of 5. |
(a) is incorrect — adding values doesn’t automatically increase the average; it depends on whether the new values are above or below the old average.
(b) is incorrect — the average will only coincidentally stay the same; there’s no guarantee.
(d) is incorrect — we do NOT need to re-measure the whole class; we only need the 2 new heights, since we can combine them with the previously known total using the balancing principle.
Answer: (b) The average will increase (very slightly).
Answer: (d) The information is not sufficient to make a claim about median.
- (i) “In 1983, majority rural used kerosene while majority urban used electricity” — Valid. The rural kerosene line starts high (~85%) in 1983 while the urban electricity line also starts already above 50% (~65%) — both confirming this statement.
- (ii) “Kerosene use decreased over time in both areas” — Valid. Both kerosene lines (rural and urban) show a clear, continuous downward trend from 1983 to 2023.
- (iii) “In 2000, 10% of urban households used electricity” — Not valid. By the year 2000, the urban electricity line is already well above 80–90%, nowhere close to just 10%.
- (iv) “In 2023, there were no power cuts” — Not valid (unsupported inference). The graph only shows what source households primarily use for lighting — it tells us nothing about the reliability or continuity of that electricity supply, so we cannot conclude anything about power cuts.
Answer: (d) 14 years
Amavasya (new moon): occurs when moonrise and moonset both happen close to the same time as sunrise/sunset — essentially when the moon rises and sets within roughly the same window as the sun, making it invisible at night. This corresponds to around day 21–22, where the moonrise line restarts near 00:00 after wrapping around.
