Chapter 5: Tales by Dots and Lines Class 8th Mathematics (Ganita Prakash-II) NCERT Solution

Tales by Dots and Lines — Ch.5 Solutions | EduGrown
Ganita Prakash · Grade 8 · Part II

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.

Chapter 5NCERT · 2 Topics
5.1The Balancing Act
5.2Visualising & Interpreting Data
40+Questions solved
5.1 · IN-TEXT

The Balancing Act — Mean as the Centre

Q
Consider any 2 numbers. Find their average. Repeat with other pairs. What do you observe?
Answer

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.

Q
Calculate and mark the mean of each collection of data shown (4 dot plots of 3 numbers each).
Answer
CollectionValuesMean
Plot 16, 7, 8\(\frac{6+7+8}{3}=7\)
Plot 23, 6, 9\(\frac{3+6+9}{3}=6\)
Plot 32, 4, 8\(\frac{2+4+8}{3}=4.67\)
Plot 43, 11, 15\(\frac{3+11+15}{3}=9.67\)
Exact values read off a dot plot can vary slightly depending on the printed scale — the method (sum ÷ count) is what matters.
Q
Can you explain how the mean is the centre of each collection? Mark the mean for further collections, then explain again.
Answer

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.

Q
Is the mean the midpoint of the two endpoints/extremes of the data? Verify that the “equal total distances” property holds for all the collections shown earlier.
Answer

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:

$$\text{Sum of distances on the LHS of mean} = \text{Sum of distances on the RHS of mean}$$
CollectionMeanLHS distancesRHS distances
6, 7, 8711
2, 4, 4, 8, 85.2 (example)balancedbalanced
9, 11, 11, 17122+1+1=45
1, 3, 5, 9, 115.84.8+2.8+0.8=8.4 (approx)balanced
This balance property is exactly why the mean is called a “fair-share” or “centre of mass” of the data — pulling any single value further away from the mean must be compensated by the rest of the data, keeping the two sides in balance.
Q
Can there be more than one such ‘centre’? For the collection 10, 10, 11, and 17 (mean 12), suppose there is a different centre larger than 12 — what happens?
Answer
CASE: centre > 12
All four values (10, 10, 11, 17) are now further from this new, larger centre on the left side (since 10, 10, 11 are all still less than it, and even further away than before), while 17 is now closer on the right. So LHS total distance increases and RHS total distance decreases — the two sides are no longer balanced.
CASE: centre < 12
By a mirror argument, all the values on the left get closer (LHS distance decreases) while 17 gets further away (RHS distance increases) — again unbalanced.
Only at exactly 12 are the two sides balanced. There is only one such centre — the mean is unique.
Q
Will including a new value in the data increase or decrease the mean?
Answer
  • 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.
Q
What happens to the mean when an existing value is removed? When will the mean increase, decrease, or stay the same?
Answer
  • 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.
Q
What happens to the average if a value equal to the mean is included or removed? Explain using the fair-share interpretation.
Answer

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.

5.1 · IN-TEXT

Unchanging Mean!

Q
Explore if it is possible to include or remove 2 values such that the mean is unchanged.
Answer
Yes, it’s possible. The key rule: the values added (or removed) must themselves average out to exactly the original mean. For example, if the mean is 9, including the pair 7 and 11 (whose own average is \(\frac{7+11}{2}=9\)) keeps the overall mean unchanged — because together they contribute zero net imbalance.
Q
How about including or removing 3 values without changing the mean? Is it possible?
Answer
Yes. The same rule generalises: any group of new values can be added without changing the mean, as long as their own average equals the original mean. For mean = 9, the triple 6, 9, 12 has average \(\frac{6+9+12}{3}=9\) — so including all three keeps the mean at 9.
Q
Can we include 2 values less than the mean and 1 value greater than the mean, so that the mean remains the same?
Answer
SETUP
Let the mean be 9. Pick 2 values below 9, say 7 and 8 (distances 2 and 1 below), totalling 3 units of “deficit.”
BALANCE
To cancel this deficit, the 1 value above the mean must contribute exactly 3 units of “surplus” — so pick 9+3 = 12.
CHECK
Average of the three new values: \(\frac{7+8+12}{3}=9\) ✓ — exactly the original mean, so including them leaves the overall mean unchanged.
Yes — e.g. include 7, 8 (below mean) and 12 (above mean).
Q
Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same.
Answer
SETUP
Mean = 9. Pick 2 values above 9, say 10 and 13 (surplus of 1 and 4, totalling 5 units).
BALANCE
The 1 value below the mean must contribute a deficit of exactly 5 units — so pick \(9-5=\)4.
CHECK
\(\frac{10+13+4}{3}=9\) ✓
Yes — e.g. include 10, 13 (above mean) and 4 (below mean).
5.1 · IN-TEXT

Relatively Unchanged! (Adding, Subtracting, Scaling)

Q
Consider the data: 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Calculate its mean. Then find the mean after adding 10 to every value. Is there a quicker way?
Answer
ORIGINAL MEAN
$$\text{Sum} = 8+3+10+13+4+6+7+7+8+8+5 = 79,\quad n=11$$ $$\text{Mean} = \frac{79}{11} = 7.18$$
AFTER +10 TO EACH
New sum \(= 79 + 10\times 11 = 79+110=189\). New mean \(=\frac{189}{11}=17.18\).
QUICKER WAY
Yes — just add 10 to the original mean directly: \(7.18+10=17.18\). No need to recompute the whole sum!
Original mean = 7.18; new mean = 17.18 (exactly 10 more).
Q
Try to explain, using algebra, what the average is when a fixed number (e.g. 2) is subtracted from every value in the collection.
Answer
SETUP
Let the original average be \(a = \dfrac{x_1+x_2+\dots+x_n}{n}\).
SUBTRACT 2 FROM EACH
$$\text{New average} = \frac{(x_1-2)+(x_2-2)+\dots+(x_n-2)}{n} = \frac{(x_1+x_2+\dots+x_n) – 2n}{n}$$ $$= \frac{x_1+x_2+\dots+x_n}{n} – \frac{2n}{n} = a – 2$$
The new average is exactly 2 less than the original average — subtracting a fixed number from every value shifts the mean down by that same amount.
Q
Try to explain this (adding/subtracting a constant) using the fair-share interpretation of average.
Answer

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.

Q
What happens to the average if every value in the collection is doubled?
Answer

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.

5.1 · IN-TEXT

Tinkering with Median

Q
Will including a new value in the data increase or decrease the median?
Answer
EXAMPLE
Data: 3, 5, 7, 7, 11, 13 has median 8 (chapter’s example before inserting). Median of original list 3,5,7,7,11,13 → average of two middle values 7 and 7? Let’s use the chapter’s actual case: data with median 8, then inserting 11.
RULE
Inserting a value greater than the current median pushes the count of “high” values up, so the median increases (or stays the same momentarily before settling on the new middle value/average of two middle values). Inserting a value less than the median makes the median decrease.
Including a value above the median → median rises (or stays the same); including a value below the median → median falls (or stays the same). Including a value equal to the median leaves it unchanged.
5.1 · IN-TEXT

Finding the Unknown

Q
Coach Balwan noted weights of 10 kushti players; one value got smudged. The mean is 39.2 kg. Weights given: 42, 40, 39, 33, 48, 38, 42, 35, 32, and an unknown w. Find w.
Answer
SET UP EQUATION
$$\frac{42+40+39+33+48+38+42+35+32+w}{10} = 39.2$$
SIMPLIFY
$$349 + w = 392 \implies w = 392-349 = 43$$
The missing weight is 43 kg.
Q
Venkayya’s average coconut harvest per tree is 25.6 over 15 trees, but one tree’s count was incorrectly noted as 3 more than actual. Find the correct average.
Answer
FIND RECORDED TOTAL
$$25.6 = \frac{z}{15} \implies z = 25.6\times 15 = 384$$
CORRECT THE ERROR
Since one tree’s count is 3 more than actual, the true total is \(384-3=381\).
RECOMPUTE AVERAGE
$$\text{Correct average} = \frac{381}{15} = 25.4$$
The correct average harvest per tree is 25.4.
5.1 · IN-TEXT

Mean and Median with Frequencies

Q
What is the average family size of the class, given the frequency table (3→3, 4→11, 5→9, 6→7, 7→3, 8→1, 9→1, 10→1)?
Answer
COMMON MISTAKE
Simply averaging the 8 distinct numbers \(\frac{3+4+5+6+7+8+9+10}{8}=6.5\) is wrong — it ignores how many students actually have each family size.
CORRECT METHOD
Multiply each value by its frequency before summing: $$\text{Mean} = \frac{(3{\times}3)+(4{\times}11)+(5{\times}9)+(6{\times}7)+(7{\times}3)+(8{\times}1)+(9{\times}1)+(10{\times}1)}{3+11+9+7+3+1+1+1}$$
CALCULATE
$$=\frac{9+44+45+42+21+8+9+10}{36}=\frac{188}{36}=5.22$$
The correct average family size is 5.22 (not 6.5).
Q
What is the median family size of this class? Do we need to write all 36 numbers in order?
Answer
QUICKER METHOD
No — use a running (cumulative) total of frequencies instead of listing every value:
CUMULATE
Up to value 3: 3 students (positions 1–3). Up to value 4: \(3+11=14\) students (positions 1–14). Up to value 5: \(14+9=23\) students (positions 15–23).
LOCATE MIDDLE
With \(n=36\), the median is the average of the 18th and 19th values. Both the 18th and 19th positions fall within the range 15–23 — i.e. both equal 5.
The median family size is 5.
5.1 · IN-TEXT

Spreadsheets

Q
Can you tell which cell has the marks obtained by Farooq in Mathematics?
Answer
Cell E5 — Farooq is in row 5, and Maths is column E.
Q
Can you tell what data is in column B7? In which subjects has Ashwin scored more than 30 marks?
Answer

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.)

Q
What formula would you type to find out the class average marks in Science? Find out if the class average in Odia is greater than in Telugu.
Answer
=AVERAGE(G2:G23)

(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.

Q
Show the average marks in other subjects after the last row, and get the total scores of each student, by typing the appropriate formulae.
Answer
TaskFormula (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.

5.1 · FIGURE IT OUT

Mean & Median — Exercise

1
Find the mean of: (i) the first 50 natural numbers (ii) the first 50 odd numbers (iii) the first 50 multiples of 4. Share your observations.
Answer
DataSumMean
(i) 1, 2, 3, …, 50127525.5
(ii) 1, 3, 5, …, 99250050
(iii) 4, 8, 12, …, 2005100102
Observation: In each case, since the data forms an evenly-spaced (arithmetic) sequence, the mean simply equals the average of the first and last terms — e.g. for the odd numbers, \(\frac{1+99}{2}=50\); for the multiples of 4, \(\frac{4+200}{2}=102\). This works because evenly spaced data is perfectly symmetric around its centre.
2
The dot plot shows a collection of data and its average, but one dot is missing. Mark the missing value so that the mean is 9.
Answer
READ VISIBLE DOTS
From the plot: visible values are 4, 7, 7, 8, 9, 9, 9, 11 (8 dots shown, one more dot missing, so \(n=9\) total).
SET UP EQUATION
$$\frac{4+7+7+8+9+9+9+11+m}{9} = 9 \implies 64+m = 81 \implies m=17$$
The missing value should be marked at 17 (the exact value depends on reading the visible dots precisely off the printed plot — use this same balancing method with your own reading).
3
Shreyas measured 24 students’ heights with shoes on (average 150.2 cm); shoes add 1 cm. (i) Does the teacher need to re-measure? (ii) What is the correct average?
Answer
(i) NO NEED TO RE-MEASURE
Since every single student wore the same uniform shoes adding exactly 1 cm, we can simply subtract 1 cm from the average itself — no need to re-measure anyone.
(ii) CORRECT AVERAGE
$$150.2 – 1 = 149.2 \text{ cm}$$
Correct average height = (d) 149.2 cm
4
Three dot plots (A, B, C) show song lengths in minutes for different albums. Which has a mean of 5.57 minutes?
Answer

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.

Method: estimate the mean visually by finding the “balance point” of each cluster of dots — the plot whose dots balance closest to 5.57 is the answer. Precise calculation requires summing all dot positions and dividing by the count for each plot.
5
Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. (i) Include 1 value without affecting median. (ii) Include 2 values without affecting median. (iii) Remove 1 value without affecting median.
Answer
FIND MEDIAN
There are 16 values (already sorted). Median = average of 8th and 9th values = \(\frac{41+41}{2}=41\).
Median = 41
PartStrategyExample answer
(i) Add 1 valueAdd 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 valuesAdd 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 valueRemove 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
6
Examine each statement: always true, sometimes true, or never true? (i) Removing a value less than median decreases median. (ii) Including a value less than mean decreases mean. (iii) Including any 4 values won’t affect median. (iv) Including 4 values less than median increases median.
Answer
#StatementVerdictReasoning
(i)Removing a value less than median decreases medianSometimes trueIt 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 meanAlways trueAdding 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 medianSometimes trueIt’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 medianNever trueAdding 4 values, all below the median, pushes more data to the lower side — the median can only stay the same or decrease, never increase.
7
The mean of 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.
Answer
SET UP
There are 8 values total. $$\frac{8+13+10+4+5+20+y+10}{8}=10.375$$
SOLVE
$$70+y = 10.375\times 8 = 83 \implies y = 83-70 = 13$$
\(y = 13\)
8
The mean of a set of data with 15 values is 134. Find the sum of the data.
Answer
$$\text{Sum} = \text{Mean} \times n = 134 \times 15 = 2010$$
Sum of the data = 2010
9
Consider the data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of the following could be p if the median is 29? (i) 10 (ii) 25 (iii) 40 (iv) 100 (v) 29 (vi) 47 (vii) 30
Answer
SORT THE 10 KNOWN VALUES
8, 8, 12, 18, 25, 29, 35, 39, 47, 73 — and now we insert \(p\) to make 11 values total, so the median is the 6th value after sorting.
TEST EACH OPTION
Insert each candidate \(p\) into the sorted list and check what lands in the 6th position.
p102540100294730
Median25252929292929
Valid values of \(p\): 40, 100, 29, 47, 30 — i.e. options (iii), (iv), (v), (vi), (vii). Options (i) 10 and (ii) 25 do NOT give median 29.
10
The dot plot shows the number of times students rode their cycles in a week. Four students rode their cycles twice. (i) Find the average. (ii) Find the median. (iii) Which of the listed statements are valid?
Answer

Using the same method throughout: read off each value and its frequency (number of dots stacked above it) directly from the plot, then:

$$\text{Average} = \frac{\sum (\text{value} \times \text{its frequency})}{\text{total number of students}}$$

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).

StatementValid?Why
(a) Everyone used their cycle at least onceDepends on plot — Invalid if any dots appear at 0If there’s a dot at the value 0, some students never rode their cycle that week.
(b) Almost everyone used their cycle a few timesValid (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 daysValidSince “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 daysInvalid (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.
(iv) If everyone cycles 1 more time the following week, every value shifts up by 1 — so both the average and median increase by exactly 1 (using the “adding a constant” rule learned earlier in the chapter).
11
A dart competition: number of trials to hit the bullseye. Trials: 1–10, Students: 1, 0, 0, 1, 4, 9, 12, 15, 10, 10. Describe using minimum, maximum, mean and median.
Answer
MIN & MAX
Minimum number of trials taken = 1 (one student got it on the very first try). Maximum = 10 (some students needed all the way up to 10 trials).
TOTAL STUDENTS
\(n = 1+0+0+1+4+9+12+15+10+10 = 62\)
MEAN
$$\text{Mean}=\frac{(1{\times}1)+(2{\times}0)+(3{\times}0)+(4{\times}1)+(5{\times}4)+(6{\times}9)+(7{\times}12)+(8{\times}15)+(9{\times}10)+(10{\times}10)}{62}$$ $$=\frac{1+0+0+4+20+54+84+120+90+100}{62}=\frac{473}{62}\approx 7.63$$
MEDIAN
With \(n=62\) (even), median = average of 31st and 32nd values. Cumulative counts: up to trial 6 → \(1+0+0+1+4+9=15\); up to trial 7 → \(15+12=27\); up to trial 8 → \(27+15=42\). Both the 31st and 32nd positions fall within the “trial 8” range (positions 28–42), so both middle values are 8.
Minimum = 1, Maximum = 10, Mean ≈ 7.63, Median = 8
5.2 · IN-TEXT

Line Graphs — Temperature, Space & Rainfall

Q
Observe the clustered-column graph and the line graph of Kerala/Punjab temperatures. Do both graphs represent the same information?
Answer
Yes — both graphs encode exactly the same underlying data (monthly maximum temperatures for Kerala and Punjab). They differ only in presentation style: the clustered-column graph uses paired bars for each month, while the line graph connects each city’s monthly points with lines — making it easier to see the overall trend and direction of change across the year.
Q
How do we get the maximum temperature over a month in a state?
Answer

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.

Q
Notice how the graph is organised — what scale is used, and what patterns does the data show?
Answer
  • 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.
Q
What thoughts or questions occur to you, looking at this temperature data? What would a plot of monthly minimum temperatures look like?
Answer

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.

Q
Space Jam: What could be the possible method used to derive the “objects launched into space” data?
Answer

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.

Q
Which of the following statements are valid inferences from the space-launch graph? (a) Worldwide count increased every year 2012–24 (b) USA launched ~3/4 of worldwide count in 2022–24 (c) Nepal launched nothing 2012–24 (d) China+Russia combined ≈400 in 2024
Answer
  • “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.
Q
Identify two consecutive years where the worldwide count increased by 2 times or more.
Answer

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.

Q
Rainfall graphs: What could be the possible method to compile this data? Mark cities on a map — what’s common to how they’re grouped?
Answer

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.

Q
Identify the peak months and low months of rainfall for each city.
Answer
CityPeak rainfall monthsLow rainfall months
Kovalam, Udupi, MumbaiJune – AugustJanuary – March
RameswaramOctober – DecemberJanuary – September (very low all through)
ChennaiRises from June, peaks in November, continues to DecemberJanuary – March
PuriJuly – SeptemberJanuary – March
5.2 · IN-TEXT

Infographics & Activity Strips

Q
Rice vs. Wheat infographic: (i) Guess Karnataka’s hidden value. (ii) Top 5 rice-preferring states. (iii) Top 5 wheat-preferring states. (iv) States with balanced preference.
Answer
(i) KARNATAKA
Karnataka’s southern neighbours (Kerala +79, Tamil Nadu +85) and its own coastal/southern geography strongly favour rice. A reasonable estimate is somewhere in the high +70s to +80s range — consistent with its southern, rice-favouring neighbours.
(ii) Top 5 — RiceValue
Manipur+100
Nagaland+99
Mizoram+97
Tripura+96
Assam+95
(iii) Top 5 — WheatValue
Rajasthan−93
Haryana−81
Punjab−78
Madhya Pradesh−60
Delhi−45
(iv) Balanced states (values closest to 0): Bihar (+3), Uttarakhand (−18), Himachal Pradesh (−19) — these show little overall preference between rice and wheat.
Q
What can a Strip Say? (i) What activity does each colour represent? (ii) Which day does each strip represent? (iii) When did he watch a movie? (iv) When is his lunch break?
Answer

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.
Since the exact colours in your printed copy may render differently, match each colour to the activity using the block lengths and positions described above as your guide.
5.2 · FIGURE IT OUT

Visualising & Interpreting Data — Exercise

1
Visualise the shop’s “Visiting” and “Purchasing” customer data (Mon–Sun) on a line graph.
Answer

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.

2
Monthly average rainy days table for Mangaluru, New Delhi (blank), Port Blair, Rameswaram. (i) Method to compile? (ii)–(v) various reading/plotting tasks.
Answer
(i) METHOD
For each city, weather records over many years are checked — for every month, count how many days had measurable rainfall each year, then average that count across all the years to get the typical number of rainy days for that month.
(ii) PLOTTING
Mark each city’s 12 monthly values on the shared graph and connect with lines — Mangaluru will show a sharp single peak around June–August (over 24 days/month!), Port Blair will show a broader, more spread-out wet season, and Rameswaram will show its (smaller) peak around October–December instead.
(iii) NEW DELHI
Reading the New Delhi line from the provided graph axis and filling in the blank row requires the actual rendered chart — but New Delhi typically shows a short, sharp monsoon peak around July–August (similar timing to Mangaluru, but smaller magnitude) and very low rainy days otherwise.
(iv) MOST & LEAST RAINY DAYS
Adding up each city’s 12 monthly values: Port Blair has rain spread generously across nearly every month, giving it the highest annual total of rainy days. Rameswaram, with very low values for 8+ months of the year, has the lowest annual total.
(v) RAINY SEASONS
New Delhi’s rainy season is concentrated in July–August (the south-west monsoon). Rameswaram’s rainy season falls in October–December (the north-east monsoon, since it lies on the south-east coast).
2(births)
Line graph of monthly births in India (Jul 2017 – ~2020). (i) Observations (ii) Births in July 2017 (iii) Time period (iv) Compare January across 2018/19/20 (v) Estimate births in 2019.
Answer
(i) OBSERVATIONS
The data shows a clear repeating seasonal wave pattern — births rise from around January, peak in the middle of the year, and dip again toward the year’s end, repeating this cycle every year. This matches the title clue (“seasonal variations in births”).
(ii) JULY 2017
Reading the first plotted point: births in July 2017 were approximately 1.4 million.
(iii) TIME PERIOD
The graph spans from July 2017 to roughly early 2020 — about 2.5 years.
(iv) COMPARE JANUARYS
January values across the three years (2018, 2019, 2020) all sit in a similar range, roughly 1.6–1.7 million — reflecting the recurring seasonal pattern rather than a long-term trend; each January falls partway up the rising slope of that year’s cycle.
(v) ESTIMATE FOR 2019
Summing or averaging the approximate monthly values across 2019 (reading each of the 12 points for that year and totaling), the estimated total births for 2019 comes out to roughly 20 million, consistent with India’s well-known annual birth count in this period.
1 (Mean Grids)
(i) Fill a 3×3 grid with 9 distinct numbers so the average along each row, column, and diagonal is 10. (ii) Can we change a few numbers and still keep average 10 everywhere?
Answer
KEY IDEA
For the average of 3 numbers to be 10, their sum must be 30. So we need a grid where every row, column, and both diagonals sum to exactly 30 — this is a classic magic square with magic constant 30.
CONSTRUCTION
Start from the standard 3×3 magic square using 1–9 (magic sum 15, centre 5), then add 5 to every cell — each row/column/diagonal sum becomes \(15+5{\times}3=30\), giving average \(30/3=10\) everywhere.
(ii) Yes — there are infinitely many valid grids, since you can rotate/reflect the magic square, or add/subtract any value \(d\) from a cell while compensating elsewhere (e.g. swapping two diagonally-symmetric cells), as long as every row, column, and diagonal sum stays exactly 30.
2 (Examples)
Give two examples of data satisfying: (i) 3 numbers, mean 8 (ii) 4 numbers, median 15.5 (iii) 5 numbers, mean 13.6 (iv) 6 numbers, mean = median (v) 6 numbers, mean > median.
Answer
ConditionExample 1Example 2
(i) 3 numbers, mean 86, 8, 104, 8, 12
(ii) 4 numbers, median 15.510, 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 13.610, 12, 13, 14, 19 (sum=68)5, 10, 15, 20, 18 (sum=68)
(iv) 6 numbers, mean = median1, 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 > median1, 2, 3, 4, 5, 50 (mean≈10.8, median=3.5)2, 3, 4, 5, 6, 100 (mean≈20, median=4.5)
For (v), a single very large “outlier” value pulls the mean far above the median — this is a classic example of why median is often a more “robust” measure of centre than mean when extreme values are present.
3
Fill in the blanks so the median of 5, 21, 14, ___, ___, ___ is 13. How many possibilities exist with only counting numbers?
Answer
SETUP
With 6 values total, median = average of the 3rd and 4th sorted values. We need this average to equal 13, i.e. the two middle sorted values must sum to 26.
EXAMPLE
Fill blanks with 12, 13, 90: sorted list becomes 5, 12, 13, 14, 21, 90 → middle two are 13, 14 → median \(=\frac{13+14}{2}=13.5\)… Let’s instead try blanks 1, 12, 90: sorted → 1, 5, 12, 14, 21, 90 → middle two 12,14 → median 13 ✓.
One valid filling: 1, 12, 90 (median = 13). Since counting numbers are unbounded, there are infinitely many possibilities — any three numbers that, combined with 5, 21, 14, produce two middle sorted values summing to 26 will work.
4
Fill in the blanks so the mean of 3, 11, ___, ___, 15, 6 is 6.5. How many possibilities exist with only counting numbers?
Answer
SETUP
There are 6 values total, so total sum must be \(6.5\times 6 = 39\).
KNOWN SUM
\(3+11+15+6=35\). The two blanks must together sum to \(39-35=4\).
Possible counting-number pairs summing to 4: (1,3), (2,2), (3,1) — exactly 3 possibilities (since each blank must be a positive counting number, i.e. ≥1).
5
Check if each statement is true; justify using algebra if necessary. (i) Average of two even numbers is even. (ii) Average of any two multiples of 5 is a multiple of 5. (iii) Average of any 5 multiples of 5 is also a multiple of 5.
Answer
StatementVerdictJustification
(i) Average of two even numbers is evenSometimes 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 5Sometimes 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 5Sometimes 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.
6
2 new admissions joined Sudhakar’s class right after the average height was found to be 150.2 cm. (i) Which statements about whether average will change are correct? (ii) New joinees are 149 cm and 152 cm — what happens to average? (iii) What happens to the median?
Answer
(i)
(c) is correct — we genuinely need the new students’ heights to compute the new average; we cannot assume it stays the same or automatically increases just because new values were added.
(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.
(ii)
The two new heights, 149 cm and 152 cm, themselves average to \(\frac{149+152}{2}=150.5\) cm — extremely close to (and slightly above) the existing class average of 150.2 cm. Since their own average is just barely above 150.2, including them will increase the overall class average by a very small amount.
Answer: (b) The average will increase (very slightly).
(iii)
We don’t know the exact position of these two new heights relative to the median (which depends on the full sorted list of all individual heights, not just the average) — so we cannot determine whether the median increases, decreases, or stays the same just from this information.
Answer: (d) The information is not sufficient to make a claim about median.
7
Is 17 the average of the data shown in the dot plot (×’s from 14 to 23)? Share your method.
Answer
METHOD: BALANCE CHECK
Rather than computing the full sum and dividing, use the balancing principle: check whether the total distance of all values below 17 equals the total distance of all values above 17. If they balance exactly, 17 is the mean; if not, it isn’t.
APPLY
Count the ×’s below 17 and their distances from 17, sum them up; do the same for ×’s above 17. Compare the two totals.
Using the visible distribution (more ×’s clustered slightly above 17 than directly matched below it), the two sides do not balance exactly at 17 — so 17 is not exactly the average of this data; the true mean is slightly different from 17 depending on the precise distances. (Apply the balance-check directly to your printed plot for the exact verdict.)
8
Previous month: mean weight 65.3 kg, median 67 kg. This month: one person lost 2 kg, two people gained 1 kg each. What happens to mean and median weight?
Answer
MEAN
Net change in total weight = \(-2 + 1 + 1 = 0\) kg. Since the total sum is unchanged and the number of people is unchanged, the mean stays exactly the same at 65.3 kg.
MEDIAN
The median depends on the middle position in the sorted order, not the total sum. Small individual changes like ±1–2 kg could shift people’s relative positions in the sorted list slightly — but without knowing exactly which specific individuals changed weight (and where they sit in the sorted order), we cannot be certain whether the median shifts at all.
Mean weight: unchanged (stays 65.3 kg, since the total sum is unaffected). Median weight: cannot be determined with certainty from the given information alone — it depends on each person’s exact position in the sorted list.
9
Retail price of iodised salt across 6 states, 2016–2025. (i) Plot any 3 states (ii) Observations (iii) Compare Gujarat vs UP (iv) Which state increased most? (v) Curiosities.
Answer
(i) PLOTTING
Choosing Mizoram, Uttar Pradesh, and West Bengal as 3 interesting states: plot each state’s 10 yearly prices (2016–2025) as a separate line on a shared chart with year on the x-axis and price (₹) on the y-axis.
(ii) OBSERVATIONS
Most states show a generally rising trend in salt prices over the decade, though Assam and Gujarat show relatively flat/stable prices in the earlier years before rising later. Mizoram shows the steepest overall climb.
(iii) GUJARAT vs UP
Gujarat’s price stays relatively stable, hovering mostly between ₹13–19 across the decade (low variation). Uttar Pradesh shows a more steady, continuous climb from ₹16.15 in 2016 to ₹26.9 in 2024 (higher overall variation/range).
(iv) MOST INCREASED
Comparing the 2016 and 2025 values: Mizoram rose from ₹20 to ₹29.8 (an increase of ₹9.8) and West Bengal rose from ₹9.47 to ₹23.99 (an increase of nearly ₹14.5) — West Bengal shows the largest absolute increase, and also one of the largest percentage increases (more than doubling) over this period.
(v) Open-ended — students might be curious whether these price changes track general inflation, transportation costs, or specific state taxes/subsidies on essential goods.
10
Referring to the household lighting-source graphs (Rural vs Urban, Electricity vs Kerosene over 1983–2023), which statements are valid?
Answer
  • (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.
11
Based on the “hobbies and games” line graph (urban vs rural, by age): (i) Time spent by 10-year-olds in urban areas? (ii) At what age is rural average 1.5 hours? (iii) Check two more statements.
Answer
(i)
Reading the urban line at age 10: roughly 2 hours per day spent on hobbies and games.
(ii)
Tracing the rural line down to the 1.5-hour mark on the y-axis and reading across to the x-axis: this occurs at approximately age 14 years.
Answer: (d) 14 years
(iii)(a)
“Time spent by 15-year-olds is twice that of 10-year-olds” — Likely false. Both lines show a gradual, smooth decline with age (not a sudden halving), so the 15-year mark is typically only modestly lower than the 10-year mark, not exactly half.
(iii)(b)
“All rural kids aged 15 spend at least 1 hour on hobbies daily” — Cannot be confirmed. The graph shows only the average across all rural 15-year-olds — averages can hide wide individual variation, so we cannot conclude that every single child meets this minimum.
14
Sunrise/sunset graphs at 4 Indian locations (Kibithu, Ghuar Moti, Srinagar, Kanyakumari). Identify which lines show sunrise vs sunset, and answer location-based questions.
Answer
IDENTIFYING THE LINES
In each pair of plots, the lower-positioned line (closer to the morning hours, roughly 04:00–08:00) represents sunrise time, while the upper-positioned line (roughly 16:00–20:00) represents sunset time.
(i) EARLIEST SUNRISE IN JANUARY
Comparing the January sunrise values across all 4 locations, Kibithu (in India’s far north-east) sees the sun rise earliest, since it’s the easternmost location and sunrise times move earlier as you go east within the same time zone. Its approximate day length in January (sunset minus sunrise) works out to roughly 10–10.5 hours.
(ii) LONGEST DAY LENGTH OVER THE YEAR
Srinagar, being the northernmost location among the four, shows the widest swing between its longest summer days and shortest winter days — its June day length (sunset−sunrise gap) is the largest of all four locations, since locations farther from the equator experience more extreme seasonal day-length variation.
(iii) OBSERVATIONS
Kanyakumari (closest to the equator) shows the flattest, most stable sunrise/sunset pattern all year — consistent with locations near the equator experiencing little seasonal variation in day length. This naturally raises curiosity about how latitude affects day-length variation, and how locations even further north or south (e.g. near the poles) would look on a similar graph.
15
Moonrise and moonset graph over a month. (i) Find amavasya and purnima dates. (ii) What do you notice/wonder?
Answer
(i) FINDING THE LUNAR DATES
Purnima (full moon): occurs when the moon rises right around sunset and sets around sunrise — i.e., when moonrise and moonset times are roughly 12 hours apart, with moonrise happening in the evening. From the graph, this occurs around day 6–7, where moonset crosses past midnight (00:00) into the next day.
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.
(ii) NOTICE & WONDER
Both moonrise and moonset times shift steadily later each day — by roughly 45–50 minutes daily — creating these smooth, steadily climbing diagonal lines, unlike the sun’s sunrise/sunset times which barely shift day-to-day. This naturally makes one wonder: why does the moon’s rise time shift so much faster than the sun’s? (The answer relates to the Moon’s own ~27.3-day orbit around Earth, on top of Earth’s daily rotation.)
Ganita Prakash · Grade 8 · Part II · Chapter 5 Solutions — compiled for study reference · @edugrown

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