13 marksGanita Prakash Cl-8 Part 2, Section 5.1, pages 108-109 — 'Tinkering with Median'
A collection has median 8. The value 11 is now included. Explain why the median must change, and find its new value for the data 2, 5, 7, 8, 9, 10 with 11 included.
Hint. The median has as many values below it as above it — check whether that is still true after the new value arrives.
Why the median has to move. The data 2, 5, 7, 8, 9, 10 has 6 values, so the median is the average of the 3rd and 4th, that is (7 + 8)/2 = 7.5. To match the chapter's example take the collection whose median is exactly 8, with an equal count of values below and above it.
Now include 11. Since 11 > 8, the count of values greater than the median has gone up by one while the count below is unchanged. The two sides no longer match, so 8 can no longer be the middle — the middle must shift towards the new value, i.e. the median increases.
The new value for 2, 5, 7, 8, 9, 10, 11. Sorted, there are now 7 values, so the median is the single middle one — the 4th:
2, 5, 7, 8, 9, 10, 11 → median 8
Adding 11 has pushed the median up from 7.5 to 8, which agrees with the rule.
The book's own case. Its data has 8 as the median with an even count on either side; after 11 joins, the two middle values become 8 and 11, so the median becomes (8 + 11)/2 = 9.5 — again an increase.
The mirror rule. Include a value less than the median and the same argument runs backwards: there are now more values below than above, so the median slides down. A value equal to the median leaves it where it is.
✦ Including a value greater than the median makes the median increase, because the balance between the counts on either side is broken in favour of the upper side. For 2, 5, 7, 8, 9, 10 the median rises from 7.5 to 8; for the book's collection with median 8 it rises to (8 + 11)/2 = 9.5.
23 marksGanita Prakash Cl-8 Part 2, Section 5.1, page 109 — 'Finding the Unknown'
Coach Balwan recorded the weights of 10 kushti players as 42, 40, 39, 33, 48, 38, 42, 35, 32 and one smudged value, with mean 39.2 kg. Find the missing weight.
Hint. Turn the mean back into a total: the mean times the count is the sum of all the values.
Use the definition backwards. Average weight = (sum of all weights)/(number of players), so the sum must be the average multiplied by the count.
Let the unknown weight be w kg. Then
(42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 + w)/10 = 39.2
Add the nine known weights.
42 + 40 = 82
82 + 39 = 121
121 + 33 = 154
154 + 48 = 202
202 + 38 = 240
240 + 42 = 282
282 + 35 = 317
317 + 32 = 349
Solve. (349 + w)/10 = 39.2, so 349 + w = 392, giving w = 392 − 349 = 43.
Check by the balance idea. Deviations of the nine known weights from 39.2 are +2.8, +0.8, −0.2, −6.2, +8.8, −1.2, +2.8, −4.2, −7.2, which total −3.8. The tenth deviation must cancel this, so it is +3.8, giving 39.2 + 3.8 = 43 ✓.
✦ The missing weight is 43 kg.
33 marksGanita Prakash Cl-8 Part 2, Section 5.1, page 109
Venkayya calculated the average coconut harvest per tree as 25.6 across 15 trees. His son then found that one tree's count had been written down as 3 more than the actual number. Find the correct average, without knowing the individual counts.
Hint. The individual counts are not needed — only the total is, and you can recover the total from the average.
Recover the total from the average. Average = (total coconuts)/(number of trees), so with z as the recorded total,
25.6 = z/15, giving z = 25.6 × 15 = 384
So 384 coconuts were recorded in all. Notice we never needed the individual tree counts — the average already carries the total inside it.
Correct the total. One tree's figure was 3 too high, so the true total is 384 − 3 = 381. Only one number changed, and it changed by 3, so the total changes by exactly 3.
Recompute the average. 381/15 = 25.4.
Shortcut worth remembering. An error of 3 in a single value is shared out over all 15 trees, so the average shifts by only 3/15 = 0.2. That is why 25.6 becomes 25.6 − 0.2 = 25.4 — one large-looking mistake barely moves an average taken over many values.
✦ The correct average harvest is 25.4 coconuts per tree (total corrected from 384 to 381).
44 marksGanita Prakash Cl-8 Part 2, Section 5.1, page 110 — 'Mean and Median with Frequencies'
The family sizes of a class are: 3 (3 students), 4 (11), 5 (9), 6 (7), 7 (3), 8 (1), 9 (1), 10 (1). Find the mean and the median family size. Explain why (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) ÷ 8 = 6.5 is wrong, and find the median without writing out all 36 values.
Hint. Each family size occurs several times — those repeats have to appear in the sum. For the median, add up the frequencies as you go.
Why 6.5 is wrong. The calculation (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)/8 = 52/8 = 6.5 averages the eight different sizes, not the 36 students. Family size 4 was reported by 11 students and family size 10 by just 1, yet this sum counts each of them once. An average must add up all the values in the data, so every repeat must be counted.
Mean, done properly. Multiply each value by its frequency:
(3 × 3) + (4 × 11) + (5 × 9) + (6 × 7) + (7 × 3) + (8 × 1) + (9 × 1) + (10 × 1)
= 9 + 44 + 45 + 42 + 21 + 8 + 9 + 10 = 188
Number of students = 3 + 11 + 9 + 7 + 3 + 1 + 1 + 1 = 36
Mean = 188/36 = 5.22 (to 2 d.p.)
Notice how far this is from 6.5 — the wrong method gave a badly inflated answer because the rare large families were given the same weight as the very common size 4.
Median without listing 36 numbers. With 36 values the median is the average of the 18th and 19th values in sorted order. Build a running total of the frequencies from the smallest value up:
| Family size | Frequency | Positions occupied |
|---|
| 3 | 3 | 1 – 3 |
| 4 | 11 | 4 – 14 |
| 5 | 9 | 15 – 23 |
| 6 | 7 | 24 – 30 |
| 7 | 3 | 31 – 33 |
| 8 | 1 | 34 |
| 9 | 1 | 35 |
| 10 | 1 | 36 |
Positions 18 and 19 both fall inside the block 15–23, so both are 5, which means the median is (5 + 5)/2 = 5.
Reading the answer. The mean 5.22 sits a little above the median 5 because the few large families (8, 9, 10) pull the balance point rightwards while leaving the middle position untouched.
✦ Mean family size = 188/36 ≈ 5.22; median family size = 5. The value 6.5 is wrong because it averages the eight distinct sizes instead of the 36 students' answers.
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.
