Figure it Out — Mean, Median and Data Puzzles — Tales by Dots and Lines

1. 13 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 113, Q1

   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.

   Hint. Each of these lists is evenly spaced, so look at what the balance point of an evenly spaced list must be.

   (i) First 50 natural numbers: 1, 2, 3, …, 50. Sum = 50 × 51/2 = 1275, so the mean is 1275/50 = 25.5.

   (ii) First 50 odd numbers: 1, 3, 5, …, 99. The sum of the first n odd numbers is n², so the sum is 50² = 2500 and the mean is 2500/50 = 50.

   (iii) First 50 multiples of 4: 4, 8, 12, …, 200. This is 4 × (1 + 2 + … + 50) = 4 × 1275 = 5100, so the mean is 5100/50 = 102.

   The observation. Each list is evenly spaced, and for evenly spaced data the dots are symmetric about the middle, so the balance point is simply the midpoint of the first and last value:

   · (1 + 50)/2 = 25.5 ✓ · (1 + 99)/2 = 50 ✓ · (4 + 200)/2 = 102 ✓

   No addition is needed at all. There is a second pattern too: the mean of the multiples of 4 is exactly 4 times the mean of the natural numbers (4 × 25.5 = 102), which is the doubling rule from earlier in the chapter — multiplying every value by 4 multiplies the mean by 4.

   And the mean of the first 50 odd numbers being exactly 50 is a small surprise worth keeping: the mean of the first n odd numbers is always n.

   ✦ (i) 25.5 (ii) 50 (iii) 102. For any evenly spaced list the mean is the midpoint of the smallest and largest value, so it can be written down without adding anything up.

2. 23 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 113, Q2

   The dot plot shows values 4, 7, 8, 8 and five dots at 9, and 11, with one dot missing. Mark the missing value so that the mean is 9.

   Hint. Turn the required mean into a required total, or add up how far the visible dots sit from 9.

   Read the plot. Measuring the ten printed dots against the scale gives

   4, 7, 8, 8, 9, 9, 9, 9, 9, 11

   Method 1 — through the total. With the missing dot there will be 11 values, and a mean of 9 needs a total of 11 × 9 = 99.

   Sum of the ten visible values = 4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 = 4 + 7 = 11; 11 + 16 = 27; 27 + 45 = 72; 72 + 11 = 83

   So the missing value is 99 − 83 = 16.

   Method 2 — through the balance. Measure each visible dot from 9: · below 9: 4 is 5 away, 7 is 2 away, 8 and 8 are 1 each → shortfall 9 · above 9: 11 is 2 away → surplus 2 · the five dots at 9 contribute nothing

   The left side is heavier by 9 − 2 = 7, so the missing dot must sit 7 above 9 to restore the balance, i.e. at 9 + 7 = 16. Both methods agree.

   Sanity check. With 16 included the values are 4, 7, 8, 8, 9, 9, 9, 9, 9, 11, 16 with sum 99 and 99/11 = 9 ✓.

   ✦ The missing dot goes at 16.

3. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 114, Q3

   Shreyas measured the heights of all 24 students and reported the average as 150.2 cm, but everyone was wearing shoes that add 1 cm. (i) Must the class be re-measured, or is there a simpler way? (ii) What is the correct average height? (a) 174.2 (b) 126.2 (c) 150.2 (d) 149.2 (e) 151.2 (f) None (g) Insufficient information.

   Hint. Every single measurement is wrong by the same amount — recall what that does to the mean.

   (i) No re-measuring is needed. Every one of the 24 heights is too big by exactly the same amount, 1 cm. From the rule established earlier in the chapter, subtracting a fixed number from every value subtracts that same number from the mean:

   ((x₁ − 1) + (x₂ − 1) + … + (x₂₄ − 1))/24 = (x₁ + … + x₂₄ − 24)/24 = (x₁ + … + x₂₄)/24 − 1 = a − 1

   So the corrected average is obtained by a single subtraction. Re-measuring 24 students would be a great deal of work for an answer already available.

   Why the shoes are the same for everyone matters. If different students had worn different shoes the correction would differ from child to child and the shortcut would fail — the whole class really would have to be measured again. The problem says uniform shoes, all adding 1 cm, which is exactly what makes one subtraction enough.

   (ii) The corrected average. 150.2 − 1 = 149.2 cm, which is option (d).

   Checking the other options. 174.2 (a) would mean adding 24 cm, as if the 24 cm of total error were applied to one child; 126.2 (b) subtracts 24 cm from the average, making the same mistake downwards — but the 24 cm of total error is spread over 24 students, so it costs only 1 cm each. 150.2 (c) ignores the shoes altogether, and 151.2 (e) corrects in the wrong direction.

   ✦ (i) No — subtract 1 cm from the reported average; the whole class does not need re-measuring, because every reading is inflated by the same 1 cm. (ii) (d) 149.2 cm.

4. 44 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 114, Q4

   Three dot plots A, B and C show the lengths in minutes of the songs on three albums. Which album has a mean of 5.57 minutes? Explain how you arrived at the answer.

   Hint. You can rule out two of the plots by eye before doing any arithmetic — a mean must lie inside the data.

   Reading the three plots against the scale (measured off the printed figure):

   · A: 5, 5, 5.25, 5.5, 5.75, 6, 6.5 — 7 songs · B: 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5 — 8 songs · C: 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5 — 8 songs

   Eliminate by eye first. The mean always lies between the smallest and the largest value, and it sits near where the dots are bunched. · C has every dot between 3.5 and 4.5, so its mean cannot possibly reach 5.57 — the mean can never exceed the largest value. · B is bunched down at the low end with only a couple of dots past 4, so its mean is clearly well under 5. · A is the only plot whose dots cluster in the 5-to-6 region, so A is the candidate.

   Confirm by calculating.

   A: 5 + 5 + 5.25 + 5.5 + 5.75 + 6 + 6.5 = 39, and 39/7 = 5.5714…, which rounds to 5.57 ✓

   B: 0.5 + 0.75 + 1.5 + 1.5 + 2 + 3.75 + 4.25 + 5 = 19.25, and 19.25/8 = 2.41

   C: 3.5 + 3.5 + 3.5 + 4 + 4 + 4 + 4.25 + 4.5 = 31.25, and 31.25/8 = 3.91

   Only A matches.

   A useful check on the answer. 5.57 is a slightly awkward decimal, and 39/7 explains why: seven songs rarely divide a total evenly, so the mean of a 7-song album usually comes out as a recurring decimal (5.571428…). Albums B and C have 8 songs and their means, 2.40625 and 3.90625, are exact.

   ✦ Album A has mean 39/7 ≈ 5.57 minutes. B averages 2.41 minutes and C averages 3.91 minutes, and C can be ruled out immediately since all its songs are under 4.5 minutes and a mean can never lie outside the data.

5. 55 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 114, Q5

   Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. Then find (i) a value that can be included without changing the median, (ii) two values that can be included without changing it, (iii) a value that can be removed without changing it.

   Hint. Look carefully at the two middle values — the answer to all three parts depends on what you notice about them.

   The median. The list is already sorted and has 16 values, so the median is the average of the 8th and 9th:

   8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92

   Median = (41 + 41)/2 = 41

   The key observation: the two middle values are the same number. That makes this data unusually forgiving, as the next three parts show.

   (i) Including one value. With 17 values the median becomes the single 9th value. Where does the newcomer land? · If it is ≤ 41, it slots in at or before position 9; the old 8th and 9th values (both 41) shift right, and the new 9th value is 41. · If it is ≥ 41, it slots in at or after position 9 and the 9th value stays 41. Either way the 9th value is 41. Check: adding 5 gives 5, 8, 10, 19, 23, 26, 34, 40, 41, 41, … ✓; adding 100 gives …, 40, 41, 41, 48, … ✓

   So any value at all may be included.

   (ii) Including two values. With 18 values the median is the average of the 9th and 10th. Now the choice matters: · two values both below 41 push the pair of 41s to positions 10 and 11, so the 9th value becomes 40 and the median drops — adding 1 and 2 gives (40 + 41)/2 = 40.5 ✗ · two values both above 41 pull the 41s down to positions 7 and 8, so the median becomes (48 + 51)/2 = 44.5 ✗ · one at most 41 and one at least 41 leaves the two 41s at positions 9 and 10 — adding 5 and 100 gives (41 + 41)/2 = 41 ✓

   So the two values must straddle the median: one ≤ 41 and one ≥ 41 (for instance 5 and 100, or 41 and 41).

   (iii) Removing one value. With 15 values left the median is the single 8th value. Removing anything from positions 1–7 slides both 41s left to positions 7 and 8, so the 8th value is 41; removing anything from positions 10–16 leaves the 41s at 8 and 9, so the 8th is still 41; and removing one of the 41s themselves leaves the other one sitting at position 8. Check: remove 8 → 41 ✓, remove 41 → 41 ✓, remove 92 → 41 ✓.

   So any value at all may be removed.

   ✦ Median = 41. (i) Any value may be included. (ii) The two values must be one ≤ 41 and one ≥ 41 (e.g. 5 and 100). (iii) Any value may be removed. Parts (i) and (iii) are so free precisely because the two middle values are both 41 — there is a spare 41 to take over the middle position.

6. 65 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 115, Q6

   Decide whether each statement is always true, sometimes true, or never true: (i) Removing a value less than the median will decrease the median. (ii) Including a value less than the mean will decrease the mean. (iii) Including any 4 values will not affect the median. (iv) Including 4 values less than the median will increase the median.

   Hint. Removing a value from the low side takes weight off the low side — think about which way the middle then has to move.

   (i) Removing a value less than the median will decrease the median — NEVER TRUE. Taking away a low value removes one of the values below the middle, so there are now fewer values on the low side and the middle position slides upwards, never downwards. The median therefore either increases or stays the same, and can never fall. · Increase: 1, 2, 3, 4, 5 has median 3; remove 1 → 2, 3, 4, 5 has median 3.5. · Stays the same: 1, 3, 3, 3, 5 has median 3; remove 1 → 3, 3, 3, 5 has median 3. It never decreases, so the statement is never true.

   (ii) Including a value less than the mean will decrease the mean — ALWAYS TRUE. The new value sits to the left of the balance point and adds weight there, so the balance point must move left. In algebra: with n values of mean a and a newcomer y < a, the new mean is (na + y)/(n + 1), and

   (na + y)/(n + 1) − a = (na + y − na − a)/(n + 1) = (y − a)/(n + 1) < 0 since y < a.

   So the mean strictly decreases, every time.

   (iii) Including any 4 values will not affect the median — SOMETIMES TRUE. It depends entirely on where the four values land. · Unaffected: to 10, 20, 30, 40, 50 (median 30) add 1, 2, 90, 100 → 1, 2, 10, 20, 30, 40, 50, 90, 100, median still 30 ✓ (two on each side). · Affected: add 1, 2, 3, 4 → 1, 2, 3, 4, 10, 20, 30, 40, 50, median 20 ✗. Since it happens for some choices and not others, the statement is sometimes true.

   (iv) Including 4 values less than the median will increase the median — NEVER TRUE. Four extra values on the low side make the low side heavier, so the middle position slides down towards them. The median can decrease or stay put, but it cannot rise. · Decrease: 10, 20, 30, 40, 50 (median 30) plus 1, 2, 3, 4 → median 20. · Stays the same: 10, 30, 30, 30, 50 (median 30) plus 1, 2, 3, 4 → 1, 2, 3, 4, 10, 30, 30, 30, 50, median 30. It never increases, so never true.

   ✦ (i) Never true (ii) Always true (iii) Sometimes true (iv) Never true. The pattern behind (i) and (iv): adding values on one side pulls the median towards that side, and removing values from one side pushes it away — the median can also refuse to move, but it never moves the wrong way.

7. 72 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 115, Q7

   The mean of 8, 13, 10, 4, 5, 20, y and 10 is 10.375. Find y.

   Hint. Count how many values there are, then convert the mean into a total.

   Count the values. The list 8, 13, 10, 4, 5, 20, y, 10 has 8 entries, one of which is unknown.

   Turn the mean into a total. Sum of all values = mean × count = 10.375 × 8 = 83.

   Add the seven known values. 8 + 13 = 21 21 + 10 = 31 31 + 4 = 35 35 + 5 = 40 40 + 20 = 60 60 + 10 = 70

   Solve. 70 + y = 83, so y = 13.

   Check by the balance idea. Deviations from 10.375 for the seven known values: −2.375, +2.625, −0.375, −6.375, −5.375, +9.625, −0.375. These total −2.625, so y must be +2.625 above the mean: 10.375 + 2.625 = 13 ✓.

   ✦ y = 13

8. 81 markGanita Prakash Cl-8 Part 2, Figure it Out, page 115, Q8

   The mean of a set of data with 15 values is 134. Find the sum of the data.

   Hint. The definition of the mean can be read in either direction.

   Read the definition backwards. Mean = (sum of the values)/(number of values), so

   sum of the values = mean × number of values

   Substitute. Sum = 134 × 15.

   134 × 15 = 134 × 10 + 134 × 5 = 1340 + 670 = 2010

   A useful remark. This is worth noticing on its own: an average never tells you the individual values, but it always tells you the total, because the total is just the average shared out in reverse. That single fact is what makes Q3, Q7 and the Venkayya coconut problem solvable without knowing any individual reading.

   ✦ The sum of the data is 2010.

9. 94 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 115, Q9

   For the data 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p, which of these could p be if the median is 29? (i) 10 (ii) 25 (iii) 40 (iv) 100 (v) 29 (vi) 47 (vii) 30

   Hint. Sort the ten known values first and see where the 6th position falls as p moves about.

   Sort the ten known values.

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

   Where is the median? Adding p gives 11 values, so the median is the single 6th value in sorted order.

   Without p, the 6th value is already 29 — but p may push things about, so check what happens as p moves.

   Case 1: p < 29. Then p slots in somewhere among the first five, pushing 29 out to position 7 and bringing 25 down to position 6. The median becomes 25, not 29. · p = 10 → 8, 8, 10, 12, 18, 25, 29, … median 25 ✗ · p = 25 → 8, 8, 12, 18, 25, 25, 29, … median 25 ✗

   Case 2: p ≥ 29. Then p sits at position 6 or later, leaving 29 exactly where it was. · p = 29 → 8, 8, 12, 18, 25, 29, 29, 35, … median 29 ✓ · p = 30 → 8, 8, 12, 18, 25, 29, 30, 35, … median 29 ✓ · p = 40 → 8, 8, 12, 18, 25, 29, 35, 39, 40, 47, 73 median 29 ✓ · p = 47 → median 29 ✓ · p = 100 → median 29 ✓

   So the condition is simply p ≥ 29, and every listed value from 29 upwards works while everything below 29 fails.

   Testing the seven options.

   p	median	works?
   (i) 10	25	✗
   (ii) 25	25	✗
   (iii) 40	29	✓
   (iv) 100	29	✓
   (v) 29	29	✓
   (vi) 47	29	✓
   (vii) 30	29	✓

   ✦ p can be (iii) 40, (iv) 100, (v) 29, (vi) 47 or (vii) 30 — that is, any value with p ≥ 29. The options 10 and 25 fail because a value below 29 pushes 29 out of the middle and drops the median to 25.

10. 106 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 115-116, Q10

    A dot plot shows how many times students rode their cycles in a week: 0 rides (3 students), 1 (1), 2 (4), 3 (7), 4 (7), 5 (5), 6 (4), 7 (6), 8 (3), 9 (0), 10 (2). (i) Find the mean. (ii) Find the median. (iii) Which statements are valid: (a) Everyone used their cycle at least once. (b) Almost everyone used their cycle a few times. (c) Some students cycled more than once on some days. (d) Exactly 5 students used their cycles more than once on some days. (e) If everyone cycled one more time next week, what would the mean and median be?

    Hint. Use the frequencies to get the total; for (c) and (d), remember how many days there are in a week.

    Set up the frequency table. (The book's own hint — four students rode twice — matches the plot, which is a good check that the dots have been read correctly.)

    Rides	0	1	2	3	4	5	6	7	8	9	10
    Students	3	1	4	7	7	5	4	6	3	0	2

    Number of students = 3 + 1 + 4 + 7 + 7 + 5 + 4 + 6 + 3 + 0 + 2 = 42

    (i) The mean. Total rides = (0×3) + (1×1) + (2×4) + (3×7) + (4×7) + (5×5) + (6×4) + (7×6) + (8×3) + (9×0) + (10×2) = 0 + 1 + 8 + 21 + 28 + 25 + 24 + 42 + 24 + 0 + 20 = 193

    Mean = 193/42 = 4.60 rides per student (to 2 d.p.).

    (ii) The median. With 42 values the median is the average of the 21st and 22nd. Running totals from the bottom: 0 → 3, 1 → 4, 2 → 8, 3 → 15, 4 → 22. Positions 16 to 22 are all 4, so both the 21st and 22nd values are 4 and the median is 4.

    (iii) The statements.

    (a) Everyone used their cycle at least once — NOT VALID. Three students are sitting on 0, so three students did not ride at all.

    (b) Almost everyone used their cycle a few times — VALID. 39 of the 42 students (about 93%) rode at least once, and the bulk of the class is spread over 2 to 7 rides, which is fairly described as 'a few times'.

    (c) Some students cycled more than once on some days — VALID. A week has only 7 days, so a student recording more than 7 rides must have ridden twice or more on at least one day. Three students rode 8 times and two rode 10 times, so at least 5 students did so.

    (d) Exactly 5 students used their cycles more than once on some days — NOT VALID. The count 5 is only a lower bound. A student who rode exactly 7 times might have ridden twice on Monday and skipped Tuesday, which would also count; the dot plot records only the weekly total and cannot tell us how the rides were spread across the days. So the data supports 'at least 5' but not 'exactly 5'.

    (e) One extra ride each next week. Every value increases by 1, which by the rule from earlier in the chapter shifts both measures up by exactly 1: · new mean = 4.60 + 1 = 5.60 (check: total 193 + 42 = 235, and 235/42 = 5.595 ✓) · new median = 4 + 1 = 5

    ✦ (i) Mean = 193/42 ≈ 4.60 (ii) Median = 4 (iii) (b) and (c) are valid; (a) fails because 3 students rode 0 times, and (d) fails because the plot gives only the weekly totals, so 5 is a minimum and not an exact count. (e) Next week the mean would be ≈ 5.60 and the median 5.

11. 114 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 116, Q11

    In a dart competition, the number of throws taken to hit the bull's eye was: 1 throw (1 student), 2 (0), 3 (0), 4 (1), 5 (4), 6 (9), 7 (12), 8 (15), 9 (10), 10 (10). Describe the data using its minimum, maximum, mean and median.

    Hint. Build the total from the frequencies, then use running totals to locate the middle positions.

    Number of students. 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62

    Minimum and maximum. The smallest number of trials recorded is 1 (one lucky student hit the bull's eye first throw) and the largest is 10 (ten students needed all ten throws). So the data spans a range of 9.

    Mean. Total throws = (1×1) + (4×1) + (5×4) + (6×9) + (7×12) + (8×15) + (9×10) + (10×10) = 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473

    Mean = 473/62 = 7.63 throws (to 2 d.p.).

    Median. With 62 values the median is the average of the 31st and 32nd. Running totals from the smallest value up:

    Trials	1	4	5	6	7	8	9	10
    Students	1	1	4	9	12	15	10	10
    Running total	1	2	6	15	27	42	52	62

    Positions 28 to 42 all hold the value 8, so the 31st and 32nd values are both 8 and the median is 8.

    Describing the data in words. Nobody found it very easy — only 2 of the 62 students managed it in 4 throws or fewer, and the marks pile up heavily at the difficult end, peaking at 8 trials. The mean 7.63 sits a little below the median 8 because the handful of quick successes at 1 and 4 pull the balance point leftwards, while the middle position is anchored in the crowded block at 8. Twenty students — nearly a third of the group — used 9 or 10 throws, and since 10 was the maximum allowed, some of those may not have hit the target at all within their throws.

    ✦ Minimum 1, maximum 10, mean 473/62 ≈ 7.63, median 8. The data is bunched towards the high end, with most students needing 7 to 10 throws, and the mean falls slightly below the median because of the two early successes at 1 and 4 trials.

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.