14 marksGanita Prakash Cl-8 Part 2, Section 5.1, pages 103-105
The mean of 10, 10, 11 and 17 is 12, which is not the midpoint of the smallest and largest values. In what sense, then, is the mean the 'centre' of the data? Show that there can be only one such centre.
Hint. Add up the gaps on the left of the mean and the gaps on the right, and compare the two totals.
The mean is a balance point, not a midpoint.
The midpoint of the two extremes 10 and 17 is (10 + 17)/2 = 13.5, and the mean is 12, so the two ideas are certainly not the same thing. What is true of the mean is something better.
Check the distances for 10, 10, 11, 17. Mean = (10 + 10 + 11 + 17)/4 = 48/4 = 12.
· Values below 12: 10, 10, 11 → distances 2, 2, 1 → total 5
· Values above 12: 17 → distance 5 → total 5
The two totals match. Imagine the number line as a see-saw with one unit of weight sitting on each dot; the plank balances exactly at 12. That is the sense in which the mean is the centre.
Why this always happens. For values x₁, x₂, …, xₙ with mean a, the sum of the signed distances is
(x₁ − a) + (x₂ − a) + … + (xₙ − a) = (x₁ + x₂ + … + xₙ) − na = na − na = 0.
A signed total of zero means the negative part (the left distances) and the positive part (the right distances) are equal in size, which is exactly the balancing property.
There is only one such centre. Suppose someone proposes a balance point c larger than 12 — say 12.5. Moving right from 12:
· every distance on the left grows (10 → 2.5, 10 → 2.5, 11 → 1.5, total 6.5)
· every distance on the right shrinks (17 → 4.5, total 4.5)
so the left total becomes strictly bigger and the balance is lost. Move left of 12 instead and the reverse happens — the right total wins. Since any shift away from 12 breaks the balance in a definite direction, 12 is the only balance point.
✦ The mean is the point where the total distance to the values on its left equals the total distance to the values on its right — for 10, 10, 11, 17 both totals are 5 at the mean 12. It is not generally the midpoint of the extremes (that would be 13.5). Because the sum of signed distances from the mean is always 0, and because moving away from it increases one side while decreasing the other, this balance point is unique.
23 marksGanita Prakash Cl-8 Part 2, Section 5.1, page 105
What happens to the mean when a new value is included in the data, and what happens when an existing value is removed? When does the mean stay the same? Explain using both the balance picture and the fair-share picture.
Hint. Compare the new value with the current mean, and think of the mean as everyone's equal share.
Including a value. Let the current mean be a.
· If the new value is greater than a, extra weight lands to the right of the balance point, so the plank tips right and the balance point must move right — the mean increases.
· If the new value is less than a, the mean decreases for the mirror-image reason.
· If the new value is exactly a, it sits on the balance point itself and adds no turning effect at all, so the mean is unchanged.
Removing a value works exactly the other way round: removing a value greater than the mean lowers the mean, removing a value less than the mean raises it, and removing a value equal to the mean leaves it alone.
The fair-share picture. The mean is what each member would get if the total were shared out equally. Suppose 4 friends have 12 sweets each — a fair share of 12.
· A fifth friend arrives carrying 20 sweets. Pooling and resharing gives (48 + 20)/5 = 13.6. The newcomer brought more than a fair share, so there is surplus to spread around and everybody's share rises.
· If instead the newcomer brought only 7, the pool is (48 + 7)/5 = 11, and the shortfall has to be made up by the others, so every share falls.
· If the newcomer brought exactly 12, they bring precisely their own share and nobody else is affected, which is why the mean does not move.
A quick numerical check on 10, 10, 11, 17 (mean 12): include 20 → 68/5 = 13.6 (up ✓); include 4 → 52/5 = 10.4 (down ✓); include 12 → 60/5 = 12 (unchanged ✓).
✦ Including a value above the mean raises it, below the mean lowers it, and equal to the mean leaves it unchanged; removing a value does the opposite in each case. In fair-share language, a newcomer who brings more than a fair share raises everyone's share, one who brings less lowers it, and one who brings exactly a fair share changes nothing.
34 marksGanita Prakash Cl-8 Part 2, Section 5.1, page 106 — 'Unchanging Mean!'
Is it possible to include two values, or three values, in a collection without changing its mean? In particular, can two values less than the mean and one value greater than the mean be included so that the mean stays the same? Can it be done the other way round?
Hint. The mean stays put exactly when the new values are themselves balanced about it — their surpluses must cancel their shortfalls.
The condition. Adding k new values keeps the mean at a exactly when the new values themselves have mean a — equivalently when their deviations from a add up to zero. If the new values are y₁, …, y_k, the new mean is (na + y₁ + … + y_k)/(n + k), and this equals a precisely when y₁ + … + y_k = ka.
Two values. Take 4, 6, 8, 10, 12 with mean 8. Any pair whose deviations cancel will do:
· 5 and 11 → deviations −3 and +3 → new mean 56/7 = 8 ✓
· 1 and 15 → deviations −7 and +7 → 56/7 = 8 ✓
· 8 and 8 → deviations 0 and 0 → 56/7 = 8 ✓
So there are infinitely many possibilities, one for every pair symmetric about 8.
Three values — two below the mean and one above. The single surplus must pay for both shortfalls, so the value above has to be far enough above.
· 6 and 7 are 2 and 1 below 8; the third value must be 3 above, i.e. 11. Check: (40 + 6 + 7 + 11)/8 = 64/8 = 8 ✓
· 3 and 5 are 5 and 3 below; the third must be 8 above, i.e. 16. Check: 64/8 = 8 ✓
The other way round — two above and one below. Now the single shortfall must cover both surpluses.
· 10 and 12 are 2 and 4 above 8; the third must be 6 below, i.e. 2. Check: (40 + 10 + 12 + 2)/8 = 64/8 = 8 ✓
· 9 and 13 are 1 and 5 above; the third must be 6 below, i.e. 2 again. Check: 64/8 = 8 ✓
Note the price of the arrangement: with two values on one side, the lone value on the other side has to be pushed further from the mean, because it is carrying the whole imbalance by itself.
✦ Yes to all three. Values can be included without changing the mean exactly when their deviations from the mean cancel out. Two below and one above works — e.g. add 6, 7 and 11 to 4, 6, 8, 10, 12 (mean stays 8); and so does two above and one below — e.g. add 10, 12 and 2 to the same data.
44 marksGanita Prakash Cl-8 Part 2, Section 5.1, pages 106-108 — 'Relatively Unchanged!'
The data 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5 has a certain mean. What happens to the mean if every value is increased by 10? Reduced by 1? Doubled? Prove each rule using algebra, and explain it with the fair-share idea.
Hint. Work out the original mean first, then think about what happens to the total when every single value changes in the same way.
The original mean. Sum = 8 + 3 + 10 + 13 + 4 + 6 + 7 + 7 + 8 + 8 + 5 = 79, and there are 11 values, so the mean is 79/11 = 7.18 (to 2 d.p.).
Add 10 to every value. The new data is 18, 13, 20, 23, 14, 16, 17, 17, 18, 18, 15, with sum 79 + 11 × 10 = 189, so the new mean is 189/11 = 17.18 — exactly 10 more. The quick way is simply 7.18 + 10, no re-adding needed.
The algebra. Let the values be x₁, x₂, …, xₙ with (x₁ + x₂ + … + xₙ)/n = a. Adding a fixed number, say 3, to each:
((x₁ + 3) + (x₂ + 3) + … + (xₙ + 3))/n = (x₁ + x₂ + … + xₙ + 3n)/n = (x₁ + … + xₙ)/n + 3n/n = a + 3
Subtract 2 from every value. The same argument with −2 in place of +3 gives (x₁ + … + xₙ − 2n)/n = a − 2, so the mean drops by 2. Reducing every value by 1 therefore gives 7.18 − 1 = 6.18.
Double every value. Sum becomes 2 × 79 = 158 and the mean 158/11 = 14.36 — exactly twice 7.18. In general, multiplying every value by 5:
(5x₁ + 5x₂ + … + 5xₙ)/n = (x₁ + x₂ + … + xₙ) × 5/n (distributive property) = 5 × (x₁ + … + xₙ)/n = 5a
The fair-share reading. If every person in a group is handed 10 extra sweets, the pool grows by 10 per person, so the equal share must grow by exactly 10. If everybody's pile is doubled, the pool doubles and so must each fair share. In the dot-plot picture, adding 10 slides the whole plot 10 steps to the right without changing any of the gaps, which is why the balance point slides 10 steps too — the mean keeps the same relative position inside the data.
✦ Original mean 79/11 ≈ 7.18. Adding 10 to every value gives 17.18, reducing every value by 1 gives 6.18, and doubling every value gives 14.36. In general, adding c to every value adds c to the mean and multiplying every value by c multiplies the mean by c, because the total changes in exactly the same way while the count n stays fixed.
Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp205.pdf), Chapter 5 'Tales by Dots and Lines', pages 103-133. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every numeric answer was derived from first principles and independently recomputed in Python. MEASURED OFF THE PRINTED FIGURES at 300-1200 dpi: the page-113 dot plot reads 4, 7, 8, 8, 9, 9, 9, 9, 9, 11 (ten dots, so the missing eleventh is 16); the three page-114 album dot plots read A = 5, 5, 5.25, 5.5, 5.75, 6, 6.5 (mean 39/7 = 5.5714), B = 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5 (mean 2.406) and C = 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5 (mean 3.906), so A is the 5.57 album; the page-115 cycle dot plot reads 0:3, 1:1, 2:4, 3:7, 4:7, 5:5, 6:4, 7:6, 8:3, 9:0, 10:2 (N = 42, sum 193, mean 4.595, median 4 — and the book's own hint that four students rode twice confirms the reading); the page-128 dot plot reads 14:2, 15:2, 16:3, 17:5, 18:4, 19:4, 20:3, 21:1, 22:0, 23:1 (N = 25, sum 443, mean 17.72); the page-122 New Delhi rainfall line reads 1.3, 1.5, 1.5, 1.1, 1.5, 3.8, 9.7, 9.7, 4.0, 1.0, 0.4, 1.0 days, total about 37, which makes New Delhi the least-rainy of the four cities and Port Blair the most at 125.8; the page-123 births line graph was found to span April 2017 to March 2020 (36 monthly points) with July 2017 about 1.77 M, January 1.67/1.75/1.77 M in 2018/19/20 and a 2019 total of about 21.4 M; the page-124 Wheat-vs-Rice infographic was read state by state and Karnataka's hidden shade was matched against the colour bar (calibrated exactly on Kerala +79 and Chhattisgarh +80) to about +68; the page-125 activity strips were decoded box by box across all 48 boxes of all three strips, giving Friday/Saturday/Sunday and an identical 10.5 hours of sleep and 1.5 hours of eating on each day; the page-130 hobbies line graph gives urban age 10 about 2 h 06 min and rural 1.5 h at age about 14.3, so option (d); the page-132 sunrise/sunset charts give Kibithu the earliest January sunrise (05:57, day length 10 h 31 min) and Srinagar the longest day of the year (14 h 25 min in June); and the page-132 moon chart gives purnima about the 14th and amavasya about the 28th-29th with a measured daily lag of 49 minutes. TWO SLIPS IN THE PRINTED BOOK ARE FLAGGED: page 125 says Manoj recorded 'five types of activities' and then lists six (the strips do use six colours), and the page-122 Figure it Out numbers two different questions as '2'.. Questions are referenced from the NCERT textbook for identification.
