Chapter 5 exercise answers: Connecting the Dots...

Open NCERT p. 101Checked by computer

1. Add up all 8 bounce counts: 6+2+9+5+4+6+3+5 = 40.
2. Divide the total by the number of attempts: 40 ÷ 8 = 5.

Answer5 bounces on average.

Watch this explained “It need not be one of your numbers”, 4:37 into The arithmetic mean as fair-share · हिंदी में देखें

Open NCERT p. 101One way to think about it

1. Bounce the ball on the bat and count the bounces before it falls. Do this for at least 7 attempts and write down each count.
2. Add all the counts to get the total number of bounces.
3. Divide the total by the number of attempts. That is the average (mean).
4. Your numbers will be your own, so every student's answer is different. Here is one example: 7 attempts gave 4, 7, 3, 8, 5, 6, 9 bounces.
5. Total = 4 + 7 + 3 + 8 + 5 + 6 + 9 = 42. Average = 42 ÷ 7 = 6 bounces.

In shortAverage = total bounces ÷ number of attempts. The answer depends on your own data; for example, 4, 7, 3, 8, 5, 6, 9 gives 42 ÷ 7 = 6 bounces.

Watch this explained “It need not be one of your numbers”, 4:37 into The arithmetic mean as fair-share · हिंदी में देखें

Open NCERT p. 101One way to think about it

1. Choose a flowering plant near you. For 7 days in its flowering season, count the flowers that bloom each day and write down each count.
2. Add the 7 daily counts to get the total for the week.
3. Divide the total by 7 to get the average number of flowers per day.
4. Your counts will be your own, so answers will differ. Here is one example: 3, 5, 4, 6, 2, 5, 3 flowers.
5. Total = 3 + 5 + 4 + 6 + 2 + 5 + 3 = 28. Average = 28 ÷ 7 = 4 flowers per day.

In shortAverage = total flowers in the week ÷ 7. The answer depends on your own counts; for example, 3, 5, 4, 6, 2, 5, 3 gives 28 ÷ 7 = 4 flowers a day.

Watch this explained “It need not be one of your numbers”, 4:37 into The arithmetic mean as fair-share · हिंदी में देखें

Open NCERT p. 101Checked by computer

1. Add Nikhil's 7 times: 17+18+17+16+19+17+18 = 122. Average = 122 ÷ 7 = 17 3/7 ≈ 17.43 s.
2. Add Sunil's 7 times: 20+18+18+17+16+16+17 = 122. Average = 122 ÷ 7 = 17 3/7 ≈ 17.43 s.
3. Both totals (122) and both counts (7 runs) are the same, so both averages are exactly equal.
4. Since a smaller time means a quicker run, and the averages are identical, neither runner is quicker than the other on average.

AnswerIt's a tie — both averaged 17 3/7 ≈ 17.43 seconds.

Watch this explained “Two runners over seven days”, 6:12 into The arithmetic mean as fair-share · हिंदी में देखें

Open NCERT p. 101Checked by computer

1. Add all six enrolment figures: 1555+1670+1750+2013+2040+2126 = 11154.
2. Divide by the number of years, 6: 11154 ÷ 6 = 1859.

AnswerMean enrolment = 1859 students.

Watch this explained “Always between the ends”, 5:24 into The arithmetic mean as fair-share · हिंदी में देखें

Open NCERT p. 112Checked by computer

1. Sort Yahapur's 12 monthly prices: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59.
2. There are 12 values (an even count), so the median is the average of the 6th and 7th values: (35 + 39) ÷ 2 = 37.
3. Sort Wahapur's 12 monthly prices: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60.
4. Median = average of the 6th and 7th values: (38 + 39) ÷ 2 = 38.5.

AnswerMedian price: ₹37 per kg in Yahapur and ₹38.5 per kg in Wahapur.

Watch this explained “Two rules on the line”, 5:58 into Dot plots: seeing spread and clustering at a glance · हिंदी में देखें

Open NCERT p. 112Checked by computer

1. The dashes (—) stand for absent students, not a count of zero — leave them out, leaving 20 actual values.
2. Add the 20 values: 0+1+0+4+8+0+0+2+1+1+5+3+4+0+0+10+25+2+2+4 = 72.
3. Mean = 72 ÷ 20 = 18/5 = 3.6.
4. Sort the 20 values: 0,0,0,0,0,0,1,1,1,2,2,2,3,4,4,4,5,8,10,25.
5. With 20 values, median = average of the 10th and 11th = (2+2)/2 = 2.

AnswerMean = 3.6, median = 2. Most students have 0-2 pets, but two unusually large values (10 and 25) pull the mean well above the median, so the median describes a 'typical' student better here.

Watch this explained “A nought is not a blank”, 5:58 into Why one number is never enough to describe a data set · हिंदी में देखें

Open NCERT p. 112Checked by computer

1. Dot plot: on the 0 to 80 line, put one dot above each tree's height, stacking dots when a height repeats (60 ft gets 4 dots; 52, 54, 55, 56 and 61 ft get 2 each). The dots run from 43 ft to 67 ft.
2. Mean: the 29 heights add up to 1621 ft, so mean = 1621 ÷ 29 ≈ 55.9 ft. Mark it with a line at about 55.9.
3. Median: in sorted order the middle (15th of 29) height is 56 ft. Mark it with a dashed line at 56.
4. Description: the heights spread from 43 ft to 67 ft (a range of 24 ft), and 20 of the 29 trees are between 50 ft and 62 ft. The mean and median are almost equal, so no unusually short or tall tree is pulling the mean either way.
5. A quicker way to find the mean: subtract 50 from every height (−7, −5, …, 17), add these small numbers to get 171, and divide: 171 ÷ 29 ≈ 5.9. So mean ≈ 50 + 5.9 = 55.9 ft. Grouping equal heights (60 × 4 instead of 60 + 60 + 60 + 60) also saves time.
6. Trees shorter than 55.9 ft: 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55. That is 13 trees.

AnswerMean = 1621/29 ≈ 55.9 ft, median = 56 ft, and 13 trees are shorter than the average height.

Watch this explained “Two rules on the line”, 5:58 into Dot plots: seeing spread and clustering at a glance · हिंदी में देखें

Open NCERT p. 112Checked by computer

(a) Can the mean or median daily usage lie between 25 and 30?

1. The mean is a total shared out equally, so it can never exceed the largest reading; the median is one of the sorted values (or midway between two), which also always sits inside the data's range.
2. The largest reading here is 20.5 L, so no average of these numbers can reach as high as 25.

AnswerNo — since the highest reading is only 20.5 L, neither the mean nor the median can lie between 25 and 30.

(b) Can the mean or median be lesser than the minimum value

1. By definition, the mean always lies between the smallest and largest values of a data set, and so does the median.
2. So neither measure can ever fall below the minimum or above the maximum of any data set — this is true for any data, not just this tap's readings.

AnswerNo — the mean and median always lie between the minimum and maximum of a data set.

Watch this explained “What the ends carry”, 3:21 into Why one number is never enough to describe a data set · हिंदी में देखें

Open NCERT p. 112One way to think about it

1. On the 0 to 4.5 kg line, put one dot for each baby, using one colour or shape for boys and another for girls. Boys: 2.6, 3.2, 3.4, 3.5, 3.8, 4.1. Girls: 2.5, 3.1, 3.4, 3.4 (two dots stacked), 3.7, 4.0.
2. Ends: boys go from 2.6 kg to 4.1 kg, girls from 2.5 kg to 4.0 kg. Both have a range of 1.5 kg.
3. Mean: boys 20.6 ÷ 6 ≈ 3.43 kg; girls 20.1 ÷ 6 = 3.35 kg.
4. Median: boys (3.4 + 3.5) ÷ 2 = 3.45 kg; girls (3.4 + 3.4) ÷ 2 = 3.4 kg.
5. In both groups the mean is a little below the median, because the lightest baby (2.6 kg for boys, 2.5 kg for girls) sits apart from the rest.

In shortOther observations are possible. One: the two groups are very alike. Both have a range of 1.5 kg, and the boys are only slightly heavier (mean ≈ 3.43 kg against 3.35 kg; median 3.45 kg against 3.4 kg). With only 6 babies in each group, this data cannot tell us that boy babies are heavier in general.

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Open NCERT p. 113One way to think about it

1. This section's summaries: whole class mean 141.21 cm and median 142.5 cm; boys 142.05 cm and 143 cm; girls 140.14 cm and 140 cm.
2. Here the boys are taller on average: both their mean and their median are higher than the girls'.
3. Spread: the boys are packed closely, mostly between about 142 cm and 148 cm, with a few shorter boys down to about 130 cm. The girls are spread widely, from about 126 cm to about 158 cm.
4. The boys' mean (142.05) is a little below their median (143) because the few short boys pull it down. The girls' mean and median are almost equal.
5. The first Grade 5 section (page 109) had the opposite result: girls' mean 146.9 cm and median 148 cm, against boys' 142.94 cm and 144 cm. There the boys were the more spread out group (128 cm to 158 cm).
6. Whole class: the first section's mean is 144.4 cm and median 145 cm, against 141.21 cm and 142.5 cm here, so the first section is about 3 cm taller on average.

In shortObservations can differ; here is one set. In this section the boys are taller on average, but in the other section the girls were, so one section cannot tell us whether boys or girls of this age are taller in general. Overall, the first section is about 3 cm taller on average.

Watch this explained “The arrows will not agree”, 1:27 into Telling tall tales: how a truthful graph still misleads · हिंदी में देखें

Open NCERT p. 113Checked by computer

1. There are 5 wrestlers but 6 dancers, so dividing one group's total by the other's would be unfair. Compare a typical wrestler with a typical dancer by using averages.
2. Mean wrestler weight = (295.2 + 250.7 + 234.1 + 221.0 + 200.9) ÷ 5 = 1201.9 ÷ 5 = 240.38 kg.
3. Mean dancer weight = (40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2) ÷ 6 = 254.5 ÷ 6 ≈ 42.42 kg.
4. 240.38 ÷ 42.42 ≈ 5.67.
5. Using medians instead gives nearly the same: 234.1 ÷ 42.2 ≈ 5.55.

AnswerComparing average weights, a sumo wrestler is 72114/12725 ≈ 5.67 times as heavy as a ballet dancer, that is, roughly 6 times.

Watch this explained “How many times heavier”, 8:17 into Why one number is never enough to describe a data set · हिंदी में देखें

Open NCERT p. 122Checked by computerAnswers can differ: one example

Bar graph?

1. Each animal has one bar, all bars are the same width, and each bar's length is measured against one scale that starts at 0 and has equally spaced marks.
2. The colours only group the animals (air, land, water), and the numbers printed on the bars are extra labels.

AnswerYes. It is a bar graph drawn as an infographic.

(a) What is the scale used in this graph?

1. The scale marks go 0, 16, 32, 48, … up to 322 kph in 20 equal steps.
2. 322 ÷ 20 = 16.1, and the printed labels are this rounded to whole numbers.

Answer1 division = 16.1 kph (about 16 kph).

(b) What did you find interesting in this infographic?

1. This part has no single answer. For example, the peregrine falcon (322 kph) is almost twice as fast as the next fastest animal, the spine-tailed swift (170 kph), and a human (37 kph) is faster than the Australian tiger beetle (8 kph), even though the beetle covers about 120 of its body lengths each second.
2. Things to explore further: how these top speeds were measured, and how long each animal can keep them up.

AnswerAnswers will vary; one example is given above.

(c) Identify a pair of creatures where one's speed is about twice

1. Several pairs work. The closest to 2 times is sailfish (109 kph) and flying fish (56 kph): 109 ÷ 56 ≈ 1.95.
2. Other pairs: spine-tailed swift (170) and pronghorn antelope (88), 170 ÷ 88 ≈ 1.93; peregrine falcon (322) and spine-tailed swift (170), 322 ÷ 170 ≈ 1.89.

AnswerSailfish (109 kph) and flying fish (56 kph). Other pairs are also possible.

(d) Can we say that a sailfish is about 4 times faster than

1. Sailfish 109 kph, humpback whale 26 kph: 109 ÷ 26 ≈ 4.2, so about 4 times.
2. The infographic shows only five water animals (sailfish, dolphin, California sea lion, Gentoo penguin, humpback whale). The sailfish is the fastest of these five, but the data says nothing about the many other aquatic animals in the world.

AnswerYes, a sailfish is about 4 times as fast as a humpback whale. No, we cannot say from this data that it is the fastest aquatic animal in the world.

Watch this explained “Reading a value off the line”, 2:38 into Clustered bar graphs: comparing across categories and across time · हिंदी में देखें

Open NCERT p. 124One way to think about it

1. Tally Grade 5 (25 answers): aerial (a) 13, aquatic (w) 6, none (n) 4, spaceborne (s) 2.
2. Tally Grade 9 (25 answers): spaceborne (s) 9, aerial (a) 8, aquatic (w) 6, none (n) 2.
3. Choose a scale: the largest count is 13, so 1 unit = 1 student with the axis going from 0 to 14 works well (1 unit = 2 students is also fine).
4. For each option (aquatic, aerial, spaceborne, none), draw two bars side by side, one for Grade 5 and one for Grade 9, using different colours and a key.

In shortBar heights (Grade 5, Grade 9): aquatic 6 and 6, aerial 13 and 8, spaceborne 2 and 9, none 4 and 2. Grade 5 students mostly chose aerial, Grade 9 students mostly chose spaceborne, and both grades chose aquatic equally.

Watch this explained “Slide them together”, 1:14 into Clustered bar graphs: comparing across categories and across time · हिंदी में देखें

Open NCERT p. 124One way to think about it

1. Draw the times (12 am, 3 am, … 9 pm) along the bottom and temperature up the side, with 1 unit = 4°C; the axis must reach at least 44°C (11 units).
2. Day 1 bar heights in units: 5, 4.5, 4, 5, 6.5, 8.5, 7.5, 6 (for 20, 18, 16, 20, 26, 34, 30, 24°C).
3. Day 2 bar heights in units: 9.25, 8.5, 7.5, 8.25, 9.25, 10.75, 10.5, 9.75 (for 37, 34, 30, 33, 37, 43, 42, 39°C).
4. At each time, draw the Day 1 and Day 2 bars side by side in two colours, with a key.
5. Day 1 goes from 16°C to 34°C: warm in the afternoon but cool at night. Day 2 goes from 30°C to 43°C and stays very hot even at night.

In shortThis is a guess, so other answers are possible. Day 1 could be from a month like March or October–November, and Day 2 from May or June, Jodhpur's hottest months.

Watch this explained “Slide them together”, 1:14 into Clustered bar graphs: comparing across categories and across time · हिंदी में देखें

Open NCERT p. 124Checked by computer

(a) Mark the corresponding bars on the bar graph.

1. The graph has a guideline every 5000. Gujarat: 69000 between the 65000 and 70000 lines, 89000 between 85000 and 90000, 78000 between 75000 and 80000.
2. Delhi: 62000 between 60000 and 65000, 74000 between 70000 and 75000, 81000 between 80000 and 85000.

AnswerDraw each bar with its top between the two guidelines named above.

(b) Notice how the graph is organised, what scale is used

1. There is one cluster for each state, with three bars: blue for 2022, red for 2023, yellow for 2024.
2. The vertical scale goes from 0 to 100000, labelled every 25000, with a guideline every 5000.
3. Pattern: in every state, 2024 has more registrations than 2022. Gujarat is the only one that fell from 2023 to 2024 (89000 to 78000).

AnswerOne cluster per state, scale 0 to 100000 with guidelines every 5000; registrations grew in every state from 2022 to 2024.

(c) How would you describe the change for various states between

1. Reading the bars (approximately): West Bengal about 11000 → 43000; Odisha about 28000 → 63000; Andhra Pradesh about 29000 → 55000; Assam about 40000 → 65000; Uttarakhand about 15000 → 19000; Delhi 62000 → 81000; Gujarat 69000 → 78000.

AnswerAll states grew. West Bengal grew the most (about 4 times), Odisha more than doubled, Andhra Pradesh nearly doubled, and Uttarakhand and Gujarat grew the least.

(d) Approximately how many more registrations did Assam get in 2023 compared to

1. Assam: about 40000 in 2022 and about 60000 in 2023.
2. 60000 − 40000 = 20000.

AnswerAbout 20000 more registrations.

(e) How many times more did the registrations in West Bengal increase from

1. West Bengal: about 11000 in 2022 (just above the 10000 line) and about 43000 in 2024.
2. 43000 ÷ 11000 ≈ 3.9, which is about 4.

AnswerAbout 4 times as many registrations in 2024 as in 2022.

(f) There were very few new registrations in Uttarakhand in 2023 and 2024

1. Each bar shows the number of new registrations in that year: about 16000 in 2023 and about 19000 in 2024 in Uttarakhand.
2. Those are thousands of new vehicles each year. The bars change only a little because the yearly number grew slowly (about 15000 → 16000 → 19000), not because there were few registrations.

AnswerNo, the statement is not correct.

Watch this explained “Identify before you infer”, 4:45 into Clustered bar graphs: comparing across categories and across time · हिंदी में देखें

Open NCERT p. 129Checked by computer

(a) The data varies more for the boys than for the girls.

1. Read the boys' dot plot: pockets 3 (1 boy), 4 (4 boys), 5 (5 boys), 6 (2 boys) — range = 6−3 = 3.
2. Read the girls' dot plot: pockets 0 (1 girl), 2 (1 girl), 3 (4 girls), 4 (5 girls), 5 (1 girl), 6 (1 girl) — range = 6−0 = 6.
3. The girls' data is spread over a wider range (6) than the boys' (3), so the data varies MORE for girls, not boys.

AnswerNo — the girls' pockets vary more (range 6) than the boys' (range 3).

(b) The median number of pockets for the boys is more than

1. Boys sorted (12 values): 3,4,4,4,4,5,5,5,5,5,6,6 — median = average of 6th and 7th = (5+5)/2 = 5.
2. Girls sorted (13 values): 0,2,3,3,3,3,4,4,4,4,4,5,6 — median = 7th value = 4.
3. 5 is more than 4.

AnswerYes — the boys' median (5) is more than the girls' median (4).

(c) The mean number of pockets for the girls is more than

1. Boys' mean = (3+4×4+5×5+6×2)/12 = 56/12 ≈ 4.67.
2. Girls' mean = (0+2+3×4+4×5+5+6)/13 = 45/13 ≈ 3.46.
3. 3.46 is not more than 4.67.

AnswerNo — the girls' mean (≈3.46) is actually less than the boys' mean (≈4.67).

(d) The maximum number of pockets for boys is greater than that

1. Boys' maximum = 6 pockets. Girls' maximum = 6 pockets.
2. They are equal, so the boys' maximum is not greater.

AnswerNo — both boys and girls have the same maximum, 6 pockets.

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Open NCERT p. 129Checked by computer

(a) Find the average number of points scored per game by A.

1. Add A's scores: 14+16+10+10 = 50.
2. Divide by 4 games: 50 ÷ 4 = 12.5.

Answer12.5 points per game.

(b) would you divide the total points by 3 or by 4?

1. C has 'Did not play' in Game 3 — that game was never played, so it does not belong on the bottom of the average (a blank is not a zero); divide C's total by 3, the number of games C actually played.
2. B has a 0 in one game, but 0 is a real score for a game that WAS played, so it counts; divide B's total by 4, the number of games played.

AnswerDivide C's total by 3 (games actually played); divide B's total by 4 (B played all 4 games, and a 0 still counts as a played game).

(c) Who is the best performer?

1. A's average = 50 ÷ 4 = 12.5.
2. B's average = (0+8+6+4) ÷ 4 = 18 ÷ 4 = 4.5.
3. C's average = (8+11+13) ÷ 3 = 32 ÷ 3 ≈ 10.67.
4. A's average is the highest of the three.

AnswerPlayer A is the best performer, with the highest average of 12.5 points per game.

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Open NCERT p. 130Checked by computer

1. First group (10 marks): sum = 85+76+90+85+39+48+56+95+81+75 = 730; mean = 730 ÷ 10 = 73.
2. Sort the first group: 39,48,56,75,76,81,85,85,90,95 — with 10 values, median = average of 5th and 6th = (76+81)/2 = 78.5.
3. Second group (9 marks): sum = 68+59+73+86+47+79+90+93+86 = 681; mean = 681 ÷ 9 ≈ 75.67.
4. Sort the second group: 47,59,68,73,79,86,86,90,93 — with 9 values, median = 5th value = 79.

AnswerFirst group: mean 73, median 78.5. Second group: mean 227/3 ≈ 75.67, median 79. The second group scored a little higher on both measures, and in both groups the mean sits below the median, showing a few low scores are pulling each group's mean down.

Watch this explained “The lean in both groups”, 2:40 into Why one number is never enough to describe a data set · हिंदी में देखें

Open NCERT p. 130One way to think about it

1. The largest value is 1240, so a scale of 1 unit = 100 people works well (the cricket 'Watching' bar is 12.4 units tall). 1 unit = 200 people also works.
2. For each sport, draw a 'Watching' bar and a 'Participating' bar side by side, in two colours, with a key.
3. Compare the two bars for each sport, then compare across sports.

In shortOther observations are possible. In every sport more people watch than take part. Cricket is the most popular both to watch (1240) and to play (620, exactly half of those who watch it). Athletics is the least popular for both (250 and 105). Basketball and swimming have the same number of participants (320), but swimming has more watchers (510 against 470). Watchers are about 1.5 times participants for basketball and about 2.4 times for athletics.

Watch this explained “Slide them together”, 1:14 into Clustered bar graphs: comparing across categories and across time · हिंदी में देखें

Open NCERT p. 130Checked by computer

1. Sort the 17 heights: 101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125.
2. The middle (9th) height is the median, 115 cm: 8 students come before it and 8 after it. Use 115 cm as the dividing height.
3. The 7 students shorter than 115 cm form one group and the 5 taller than 115 cm form the other. Five students are exactly 115 cm, so the teacher places them to balance the groups: 1 joins the shorter group and 3 join the taller group, giving 8 and 8. Because 17 is odd, one student (also 115 cm) is left in the middle and can join either group.
4. Mean height = 1930 ÷ 17 ≈ 113.5 cm, and the median is 115 cm.
5. In the 2019 columns of the 'Telling Tall Tales' table, 5-year-olds average about 107 cm, 6-year-olds about 113 cm (boys 113.1, girls 112.9) and 7-year-olds about 118 cm. This class's heights centre on 113–115 cm.

AnswerDivide at the median height, 115 cm (8 students on each side of the middle student). The students are probably about 6 years old.

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Open NCERT p. 130One way to think about it

1. Measure every student in your class in centimetres and list the heights.
2. Mean = sum of all heights ÷ number of students.
3. Median = the middle height after sorting (the average of the two middle heights if the class size is even).
4. Make a dot plot of the heights and mark the mean (solid line) and median (dashed line). Look at the ends, where the dots cluster, and any height far from the rest.
5. Every class has different data, so answers will differ. Here is one example with 9 students: 132, 135, 138, 140, 141, 141, 144, 147, 150 cm. Mean = 1268 ÷ 9 ≈ 140.9 cm, median = 141 cm (5th value). The heights range from 132 to 150 cm, and the mean and median are almost equal, so no height is pulling the mean away.

In shortMean = total height ÷ number of students; median = middle sorted height. The answer depends on your class; in the example, mean ≈ 140.9 cm and median = 141 cm.

Watch this explained “A question with no answer”, 0:00 into Why one number is never enough to describe a data set · हिंदी में देखें

Open NCERT p. 130Checked by computer

1. The mean height of ONE section is a fact about the 30 students in THAT section only — it says nothing about a different, independent group of 30 students.
2. Nothing in the question links the two sections' heights (no shared total, no rule connecting them), so the other section could have a higher, lower, or identical mean.
3. Options (a), (b) and (c) each claim something is 'must be true' about the other section, but none of them can be guaranteed from the given information.

Answer(d) The mean height of students in the other section cannot be determined from this information alone.

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Open NCERT p. 131Checked by computer

(a) Write estimated values for the number of skyscrapers in New York, Tokyo

1. Estimates will vary a little. Compare each unlabelled bar with the labelled bars near it.
2. New York's bar is between Shenzhen (367) and Dubai (251), a little nearer Shenzhen: about 315.
3. Tokyo's bar is between Shanghai (183) and Kuala Lumpur (154), nearer Kuala Lumpur: about 165.
4. London's bar is the shortest, a little more than half the length of Moscow's (46): about 28.

AnswerNew York about 315, Tokyo about 165, London about 28.

(b)(i) Only 12 cities have more skyscrapers than Mumbai.

1. Mumbai is 13th on the chart, so 12 cities are above it.
2. The chart lists the 20 cities with the most skyscrapers, so a city left off the chart has fewer than London, which is below Mumbai. No other city can have more than Mumbai.

AnswerYes, valid.

(b)(ii) Only 7 cities have fewer skyscrapers than Mumbai.

1. 7 cities are below Mumbai on the chart.
2. But every city that is not on the chart also has fewer skyscrapers than Mumbai, so many more than 7 cities have fewer.

AnswerNo, not valid.

(b)(iii) The tallest building in the world is in Hong Kong.

1. The chart counts buildings taller than 150 m. It does not show how tall any building is.
2. Having the most tall buildings is not the same as having the single tallest one. (The world's tallest building, the Burj Khalifa, is in Dubai.)

AnswerNo, not valid.

Watch this explained “The chart that counts”, 9:19 into Telling tall tales: how a truthful graph still misleads · हिंदी में देखें

Open NCERT p. 131One way to think about it

1. For each object (pen, eraser, palm, geometry box, maths notebook), first write your estimate in cm, then measure it with a ruler.
2. Positive difference = the larger of the two numbers minus the smaller (it doesn't matter whether you guessed too high or too low).
3. Draw a double bar graph: one pair of bars for each object, 'Estimate' and 'Measure' side by side.
4. Average difference = sum of the 5 positive differences ÷ 5. The smaller it is, the more accurate your estimates were.
5. Everyone's numbers are different. Here is one example. Estimates 15, 5, 16, 18, 25 cm against measurements 14, 4, 18, 17, 28 cm give differences 1, 1, 2, 1, 3. Average difference = 8 ÷ 5 = 1.6 cm.

In shortAverage difference = total of the positive differences ÷ 5. The answer depends on your own data; in the example it is 8 ÷ 5 = 1.6 cm.

Watch this explained “Slide them together”, 1:14 into Clustered bar graphs: comparing across categories and across time · हिंदी में देखें

Open NCERT p. 132Checked by computer

(a) Construct a dot plot below showing the data for both weeks.

1. Draw one number line for seconds (e.g. 200 to 420).
2. Plot Week 1's 9 times (410,400,370,340,360,400,320,330,310) as one kind of marker, and Week 2's 8 times (320,290,380,280,270,230,220,240) as another, on the same line.

AnswerA shared dot plot with Week 1's 9 points and Week 2's 8 points marked with different symbols on one number line from 200 to 420 seconds.

(b) Describe the mean, median, and any observations you may have about

1. Week 1: sum = 410+400+370+340+360+400+320+330+310 = 3240; mean = 3240 ÷ 9 = 360. Sorted: 310,320,330,340,360,370,400,400,410; median (5th of 9) = 360.
2. Week 2: sum = 320+290+380+280+270+230+220+240 = 2230; mean = 2230 ÷ 8 = 278.75. Sorted: 220,230,240,270,280,290,320,380; median = average of 4th and 5th = (270+280)/2 = 275.
3. Both mean and median dropped from Week 1 to Week 2 (360→278.75 and 360→275), so Aditi is solving puzzles noticeably faster in her second week — she's improving with practice.

AnswerWeek 1: mean = median = 360 s. Week 2: mean = 278.75 s, median = 275 s. Aditi got about 80 seconds quicker on average in Week 2 — a clear sign of improvement.

Watch this explained “Both towns, one scale”, 5:05 into Dot plots: seeing spread and clustering at a glance · हिंदी में देखें

Open NCERT p. 132One way to think about it

1. Pick at least one of the two projects. Your data is your own, so every answer will be different; one example is shown for (a).
2. (a)(i) On each page, count the words in every sentence and make a dot plot of the counts for each page on one shared scale.
3. (a)(ii) Find the mean (total words ÷ number of sentences) and median (middle sorted count) for each page and compare. Example: page 1 sentences 12, 8, 15, 20, 9, 14, 11 words: mean = 89 ÷ 7 ≈ 12.7, median = 12. Page 2 sentences 18, 25, 16, 22, 30, 19 words: mean = 130 ÷ 6 ≈ 21.7, median = (19 + 22) ÷ 2 = 20.5. The second book uses much longer sentences.
4. (b)(i)–(ii) Count the letters in each classmate's name, find the mean and median length, and make a dot plot to see the spread and centre.
5. (b)(iii) Tally the first letters to see which are common and which are rare.
6. (b)(iv) Sort the names alphabetically and take the middle one's first letter. If that median letter comes before N, more than half the names start with A–M; if it comes after M, more than half start with N–Z.
7. (b)(v) Put each name into one of the four vowel/consonant start-and-end groups, count boys' and girls' names in each, and draw a double bar graph.

In shortAn open project: the numbers depend on your books or your class. In the example for (a), page 1 has mean ≈ 12.7 and median 12 words per sentence, and page 2 has mean ≈ 21.7 and median 20.5.

Watch this explained “Run the ring yourself”, 8:14 into Why every answer from data raises the next question · हिंदी में देखें

Open NCERT p. 133One way to think about it

1. Every day for a month, write down how many times you stepped out of the house.
2. Make a dot plot of the daily counts, one dot per day.
3. Find the mean (total ÷ number of days), the median (middle sorted count) and the smallest and largest counts.
4. Your data will be your own, so answers will differ. Here is one example for 30 days: 0 outings on 2 days, 1 on 6 days, 2 on 10 days, 3 on 7 days, 4 on 3 days, 6 on 2 days. Total = 71, so mean = 71 ÷ 30 ≈ 2.4; the 15th and 16th sorted values are both 2, so median = 2; counts range from 0 to 6.
5. Look for patterns: for example, whether weekends differ from school days, and what caused any unusual day. You can compare with a family member's or friend's month.

In shortDescribe your month with its mean, median, smallest and largest count and dot plot. In the example: mean ≈ 2.4, median 2, range 0 to 6. The two busy days (6 outings) pull the mean a little above the median.

Watch this explained “Run the ring yourself”, 8:14 into Why every answer from data raises the next question · हिंदी में देखें

Open NCERT p. 133One way to think about it

1. Pick at least one option. Your data is your own, so every answer will be different; one example is shown for (b).
2. (a) Each student measures their family's heights, makes a dot plot of them and finds their mean and median. Then the group draws a double bar graph of each student's height beside their family's mean height and discusses the patterns.
3. (b) Each student, and each family member, closes their eyes and opens them when they think 1 minute has passed; note the seconds. Repeat for 3 minutes. Make a dot plot for the 1-minute guesses and one for the 3-minute guesses, mark the mean and median, then draw a double bar graph of each family's mean 1-minute and 3-minute guesses.
4. Example for (b), a family of six. 1-minute guesses: 48, 52, 55, 58, 62, 70 s, so mean = 345 ÷ 6 = 57.5 s, median = (55 + 58) ÷ 2 = 56.5 s, spread 22 s. 3-minute guesses: 150, 165, 170, 185, 190, 210 s, so mean = 1070 ÷ 6 ≈ 178.3 s, median = (170 + 185) ÷ 2 = 177.5 s, spread 60 s.

In shortAn open group project: results depend on your families. In the example, both averages are a little under the true time (57.5 s against 60 s, and about 178.3 s against 180 s), and the 3-minute guesses are spread much more widely (60 s against 22 s).

Watch this explained “Run the ring yourself”, 8:14 into Why every answer from data raises the next question · हिंदी में देखें