Chapter 5 exercise answers: Tales by Dots and Lines

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(i) The first 50 natural numbers.

1. The first 50 natural numbers are 1, 2, 3, ..., 50.
2. Their sum is 50 × 51 ÷ 2 = 1275.
3. Mean = 1275 ÷ 50 = 25.5.

Answer25.5

(ii) The first 50 odd numbers.

1. The first 50 odd numbers are 1, 3, 5, ..., 99.
2. Their sum is 50² = 2500.
3. Mean = 2500 ÷ 50 = 50.

Answer50

(iii) The first 50 multiples of 4.

1. The first 50 multiples of 4 are 4, 8, 12, ..., 200.
2. Each one is exactly 4 times the matching natural number, so their sum is 4 × 1275 = 5100.
3. Mean = 5100 ÷ 50 = 102.

Answer102

Watch this explained “Multiply every value, and two sums nobody did”, 6:43 into Which added values move the mean, and in which direction

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1. Read the dot plot carefully: there is one dot at 4, one at 7, two at 8, five stacked on the mean line at 9, and one at 11 — 10 dots in all.
2. Their total is 4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 = 83.
3. With the missing dot there will be 11 values, and their mean must be 9, so all 11 values must add up to 11 × 9 = 99.
4. The missing value is 99 − 83 = 16.
5. Check: (83 + 16) ÷ 11 = 99 ÷ 11 = 9.

Answer16 — mark the missing dot at 16.

Watch this explained “Nine weights, and a letter for the tenth”, 1:23 into Working backwards from an average to a missing value

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(i) Should the teacher get all the heights measured again without the …

1. Every one of the 24 heights was measured with the same 1 cm of shoe added, so the recorded total was inflated by exactly 24 × 1 = 24 cm.
2. Subtracting that fixed amount from the recorded average removes the error for the whole class in one step — there is no need to remeasure anyone.

AnswerNo — simply subtract 1 cm from the recorded average; a simpler way exists.

(ii) What is the correct average height of the class?

1. Correct average = recorded average − 1 cm = 150.2 − 1 = 149.2 cm.

Answer149.2 cm, option (d).

Watch this explained “Shoes, and two students joining”, 7:45 into Which added values move the mean, and in which direction

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1. Album A's seven points read 5, 5, 5.25, 5.5, 5.75, 6 and 6.5 minutes.
2. Their total is 5+5+5.25+5.5+5.75+6+6.5 = 39 minutes.
3. Mean = 39 ÷ 7 = 5.571..., which rounds to 5.57 minutes.
4. Every point on Album B is at or below 5 minutes, and every point on Album C is at or below 4.5 minutes, so neither of their means can possibly reach 5.57 — a mean can never sit above every single value in its own collection.

AnswerAlbum A.

Watch this explained “Rejecting an average without dividing”, 8:32 into The mean as the point where the distances balance

Open NCERT p. 114Checked by computerAnswers can differ: one example

(i) include one value to the data … without affecting the median

1. Put the 16 values in order: 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92.
2. With 16 values (an even count) the median is the average of the 8th and 9th values. Both are 41, so the median is (41 + 41) ÷ 2 = 41.
3. Adding one value gives 17 values, and the median becomes the 9th value. If the new value is 41 or less, the first 41 moves up into 9th place; if it is more than 41, the 9th value is still the second 41. Either way the median stays 41.
4. So any single value can be added. For example, add 0: the list 0, 8, 10, …, 92 has 41 as its 9th value.

AnswerThe median is 41. Any one value can be added without changing it — for example 0.

(ii) include two values to the data without affecting the median

1. Adding two values gives 18 values, so the median is the average of the 9th and 10th values.
2. If one new value is 41 or less and the other is 41 or more, one lands on each side of the two 41s, so the 9th and 10th values are both 41 and the median stays 41.
3. For example, add 30 and 60: 8, 10, 19, 23, 26, 30, 34, 40, 41, 41, 48, 51, 55, 60, 70, 84, 91, 92 — the 9th and 10th values are 41 and 41.
4. If both new values were below 41, the middle pair would become 40 and 41; if both were above 41, it would become 41 and 48. Either way the median would change.

Answer30 and 60 (any pair with one value at most 41 and the other at least 41 works).

(iii) remove one value from the data without affecting the median

1. Removing one value leaves 15 values, so the median is the 8th value.
2. If the removed value is 41 or below, the second 41 moves down into 8th place; if it is above 41, the 8th value is still the first 41. Either way the median stays 41.
3. For example, remove one of the 41s: 8, 10, 19, 23, 26, 34, 40, 41, 48, 51, 55, 70, 84, 91, 92 has 41 as its 8th value.

AnswerOne of the 41s — in fact, removing any one value leaves the median at 41.

Watch this explained “The middle that will not budge”, 5:07 into Whether adding a value raises or lowers the median

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(i) Removing a value less than the median will decrease the median.

1. Removing a value below the median takes a value away from the lower side only, so the middle position shifts towards the larger values. The median can go up or stay the same, but it can never go down.
2. Example: 1, 2, 3, 4, 5 has median 3. Remove 1: 2, 3, 4, 5 has median 3.5 — it went up.
3. Example: 1, 3, 3, 3, 5 has median 3. Remove 1: 3, 3, 3, 5 has median 3 — it stayed the same.

AnswerNever true.

(ii) Including a value less than the mean will decrease the mean.

1. Say n values have mean m, so their total is n × m. Add a value m − t, where t is more than 0 (so the new value is below the mean).
2. New mean = (n × m + m − t) ÷ (n + 1) = m − t ÷ (n + 1), which is always less than m.
3. Example: 4, 6, 8 have mean 6. Add 2: the mean becomes (4 + 6 + 8 + 2) ÷ 4 = 20 ÷ 4 = 5.

AnswerAlways true.

(iii) Including any 4 values will not affect the median.

1. Some sets of 4 values leave the median unchanged: 10, 20, 30, 40, 50, 60 has median 35, and adding 5, 15, 45 and 55 gives 5, 10, 15, 20, 30, 40, 45, 50, 55, 60, whose median is still (30 + 40) ÷ 2 = 35.
2. Other sets change it: adding 1, 2, 3 and 4 instead gives 1, 2, 3, 4, 10, 20, 30, 40, 50, 60, whose median is (10 + 20) ÷ 2 = 15.

AnswerSometimes true.

(iv) Including 4 values less than the median will increase the median.

1. Adding values below the median puts more values on the lower side, so the middle position moves towards the smaller values. The median can go down or stay the same, but never up.
2. Example: 1, 2, 3, 4, 5 has median 3. Add 0, 0, 0, 0: 0, 0, 0, 0, 1, 2, 3, 4, 5 has median 1 — it went down.
3. Example: 3, 3, 3, 3, 3 has median 3. Add 1, 1, 1, 1: 1, 1, 1, 1, 3, 3, 3, 3, 3 has median 3 — it stayed the same.

AnswerNever true.

Watch this explained “When it is not settled”, 7:42 into Whether adding a value raises or lowers the median

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1. There are 8 numbers in all, so their total must be 8 × 10.375 = 83.
2. The 7 known numbers add up to 8+13+10+4+5+20+10 = 70.
3. So y = 83 − 70 = 13.

Answer13

Watch this explained “Two quick unknowns”, 5:23 into Working backwards from an average to a missing value

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1. Sum = mean × count = 134 × 15 = 2010.

Answer2010

Watch this explained “Two quick unknowns”, 5:23 into Working backwards from an average to a missing value

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1. Sort the 10 known values: 8, 8, 12, 18, 25, 29, 35, 39, 47, 73.
2. Adding p makes 11 values, so the median is fixed as the 6th value once everything is sorted.
3. Insert each candidate and read off the new 6th value: p=10 → median 25; p=25 → median 25; p=40 → median 29; p=100 → median 29; p=29 → median 29; p=47 → median 29; p=30 → median 29.
4. So every option except 10 and 25 keeps the median at 29.

Answer(iii) 40, (iv) 100, (v) 29, (vi) 47 and (vii) 30 all work; (i) 10 and (ii) 25 do not.

Watch this explained “The missing value”, 6:44 into Whether adding a value raises or lowers the median

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(i) Find the average number of times students rode their cycles.

1. Multiply each value by how many dots sit above it and add: 0×3+1×1+2×4+3×7+4×7+5×5+6×4+7×6+8×3+9×0+10×2 = 193 rides, from the 42 dots on the plot.
2. Average = 193 ÷ 42 ≈ 4.60 times.

Answer193/42 ≈ 4.60 times.

(ii) Find the median number of times students rode their cycles.

1. There are 42 students, so the median sits between the 21st and 22nd values once sorted.
2. Running totals from the smallest value up (3, 4, 8, 15, 22, ...) show the 21st and 22nd values both fall on 4.
3. So the median is 4.

Answer4 times.

(iii) Which of the following statements are valid? Why?

1. (a) is false: 3 students are at 0, so not everyone used their cycle even once.
2. (b) is true: only 3 of 42 students (about 7%) never rode at all, so 'almost everyone' really did use their cycle, several times over.
3. (c) is true: a week has only 7 days, so any student above 7 rides (5 students, at 8 or 10 rides) must have ridden more than once on at least one day.
4. (d) cannot be confirmed: the plot only guarantees the 5 students above 7 rides rode twice on some day — a student with, say, 4 rides could also have ridden twice on one day and none on others, so the true count could be more than 5. 'At least 5' is safe; 'exactly 5' is not.

Answer(b) and (c) are valid; (a) and (d) are not.

(iii)(e) if all of them cycled 1 more time than …

1. Adding exactly 1 ride to every single student's count shifts the whole distribution up by 1, without changing how many students there are.
2. So both the average and the median simply move up by 1: new average = 193/42 + 1 = 235/42 ≈ 5.60; new median = 4 + 1 = 5.

AnswerAverage ≈ 5.60 (235/42); median = 5.

Watch this explained “Forty-two students and a week of rides”, 6:59 into Mean and median from a frequency table, by hand and in a spreadsheet

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1. Expand the frequency table into a sorted list of 62 trial-counts (1 student needed 1 throw, none needed 2 or 3, 1 needed 4, 4 needed 5, 9 needed 6, 12 needed 7, 15 needed 8, 10 needed 9, 10 needed 10).
2. Minimum = 1, maximum = 10 (the smallest and largest values that actually occur).
3. Total throws = 1×1+4×1+5×4+6×9+7×12+8×15+9×10+10×10 = 473; mean = 473 ÷ 62 ≈ 7.63.
4. With 62 values the median is the average of the 31st and 32nd; the running totals show both fall on 8, so the median is 8.

AnswerMinimum 1, maximum 10, mean 473/62 ≈ 7.63, median 8.

Watch this explained “Sixty-two students and two empty rows”, 8:02 into Mean and median from a frequency table, by hand and in a spreadsheet

Open NCERT p. 122One way to think about it

1. Draw one horizontal axis for the seven days (Mon to Sun) and one vertical axis for number of customers, scaled to fit the largest value (35).
2. Plot the seven 'Visiting' values (16, 19, 10, 14, 20, 22, 35) as one set of points and join them with straight segments.
3. Plot the seven 'Purchasing' values (10, 8, 7, 11, 12, 16, 26) as a second set of points, in a different colour, and join them too.
4. Add a legend so the two lines can be told apart, and look at the GAP between them on each day — that gap is how many visitors did not buy anything.

In shortA two-line graph over Mon-Sun: one line for 'Visiting' (16,19,10,14,20,22,35) and one for 'Purchasing' (10,8,7,11,12,16,26), with a legend to tell them apart.

Watch this explained “Building one from a table”, 6:57 into Line graphs, and what change over time looks like

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(iii) Based on the line for New Delhi in the graph …

1. The New Delhi row is blank in the table, but its line is drawn on the graph. Read each dot against the 0 to 30 days axis (gridlines every 5 days).
2. The dots sit at about: Jan 1.3, Feb 1.5, Mar 1.5, Apr 1.1, May 1.5, Jun 3.8, Jul 9.7, Aug 9.7, Sep 4, Oct 1, Nov 0.3, Dec 1 days.
3. Readings off a graph are estimates. Rounded to the nearest whole day they are 1, 2, 2, 1, 2, 4, 10, 10, 4, 1, 0, 1.

AnswerAbout 1.3, 1.5, 1.5, 1.1, 1.5, 3.8, 9.7, 9.7, 4, 1, 0.3 and 1 days (January to December).

(iv) Which city among these receives the most number of days …

1. Add each city's 12 monthly figures: Mangaluru 112.1, Port Blair 125.8, Rameswaram 42.3 and New Delhi about 36.4 days.
2. Port Blair has the most rainy days in a year.
3. New Delhi has the fewest. This does not depend on how exactly the graph is read: even rounding every New Delhi reading up to the next whole day gives only 41, still less than Rameswaram's 42.3.

AnswerMost: Port Blair. Least: New Delhi.

(v) when is the rainy season in New Delhi and Rameswaram?

1. New Delhi: June to September stand out (about 4, 10, 10 and 4 days, against 1.5 days or fewer in every other month), so its rainy season is June to September, heaviest in July and August.
2. Rameswaram: October, November and December stand out (8.1, 10.4 and 7.8 days, against at most 3.4 in any other month), so its rainy season is October to December.

AnswerNew Delhi: June to September. Rameswaram: October to December.

Watch this explained “A blank row, read off a line”, 8:07 into Interrogating a real dataset: traffic, and rainfall

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(iii) What time period does the graph capture?

1. The labelled ticks run from Jul 2017 to Jan 2020, six labels each 6 months apart.
2. Counting the plotted points from the first one on the left to the last one on the right gives 36 monthly points in all.
3. Since the 4th point lines up with the Jul 2017 tick and the 34th lines up with the Jan 2020 tick, the first point must be April 2017 and the last must be March 2020.

AnswerApril 2017 to March 2020 (36 months, exactly 3 years).

Watch this explained “Reading one you did not draw, in two passes”, 2:19 into Line graphs, and what change over time looks like

Open NCERT p. 127Checked by computerAnswers can differ: one example

(i) Fill the grid with 9 distinct numbers such that the average

1. Each row, column and diagonal holds 3 numbers with average 10, so each of these 8 lines must add up to 3 × 10 = 30.
2. One filling that works, row by row: 7, 12, 11 / 14, 10, 6 / 9, 8, 13.
3. Rows: 7 + 12 + 11 = 30, 14 + 10 + 6 = 30, 9 + 8 + 13 = 30. Columns: 7 + 14 + 9 = 30, 12 + 10 + 8 = 30, 11 + 6 + 13 = 30. Diagonals: 7 + 10 + 13 = 30, 11 + 10 + 9 = 30. All nine numbers are different.

Answer7, 12, 11 / 14, 10, 6 / 9, 8, 13 (one of many possible grids).

(ii) Can we fill the grid by changing a few numbers …

1. Yes. For example, swap the left and right columns: 11, 12, 7 / 6, 10, 14 / 13, 8, 9. Six numbers have moved, and every row, column and diagonal still adds up to 30, so every average is still 10.
2. Many other grids work too, but one number can never change: the centre. The middle row, the middle column and the two diagonals all pass through the centre. Together they cover every cell once and the centre 3 extra times, and they add up to 4 × 30 = 120.
3. All nine cells add up to 3 × 30 = 90 (the three rows). So 3 × centre = 120 − 90 = 30, and the centre is always 10.

AnswerYes — for example 11, 12, 7 / 6, 10, 14 / 13, 8, 9 also works — but the centre must always be 10.

Watch this explained “The grid whose centre is forced”, 7:32 into Working backwards from an average to a missing value

Open NCERT p. 127Checked by computerAnswers can differ: one example

(i) 3 numbers whose mean is 8.

1. Any 3 numbers totalling 3×8=24 will do.
2. 7, 8, 9 add to 24; 4, 8, 12 also add to 24 but is otherwise a different set.

Answer7, 8, 9 and 4, 8, 12.

(ii) 4 numbers whose median is 15.5.

1. With 4 numbers the median is the average of the middle two, so those two must add to 31.
2. 10, 15, 16, 40 and 12, 15, 16, 19 both have middle pair 15,16 (sum 31), but are otherwise different sets.

Answer10, 15, 16, 40 and 12, 15, 16, 19.

(iii) 5 numbers whose mean is 13.6.

1. Any 5 numbers totalling 5×13.6=68 will do.
2. 10, 12, 13, 15, 18 and 5, 10, 15, 18, 20 both add to 68.

Answer10, 12, 13, 15, 18 and 5, 10, 15, 18, 20.

(iv) 6 numbers whose mean = median.

1. A symmetric set automatically has mean = median.
2. 4, 6, 9, 11, 14, 16 is symmetric about 10 (mean 60÷6=10, median average of 9,11 = 10).
3. 1, 2, 3, 4, 5, 6 is symmetric about 3.5 (mean 21÷6=3.5, median average of 3,4 = 3.5).

Answer4, 6, 9, 11, 14, 16 and 1, 2, 3, 4, 5, 6.

(v) 6 numbers whose mean > median.

1. Start from a symmetric set (mean = median) and drag its LARGEST value far out — that raises the mean but leaves the median untouched.
2. 4, 6, 9, 11, 14, 76: median stays at 10, mean rises to 120÷6=20.
3. 1, 2, 3, 4, 5, 100: median stays at 3.5, mean rises to 115÷6≈19.17.

Answer4, 6, 9, 11, 14, 76 and 1, 2, 3, 4, 5, 100.

Watch this explained “Building data to order”, 6:07 into Working backwards from an average to a missing value

Open NCERT p. 127Checked by computerAnswers can differ: one example

1. With the three blanks filled there are 6 numbers, so the median is the average of the 3rd and 4th numbers once they are in order. For the median to be 13, those two middle numbers must add up to 26.
2. One filling: 3, 12, 40. In order the six numbers are 3, 5, 12, 14, 21, 40; the 3rd and 4th are 12 and 14, and (12 + 14) ÷ 2 = 13.
3. Now keep 3 and 12 and change 40 to any counting number from 14 upwards — 14, 15, 100, 5000, and so on. The two middle numbers stay 12 and 14, so the median stays 13.
4. Counting numbers never end, so the fillings never end either. (If the numbers were allowed only up to 20, there would be 97 different sets of three; up to 40, there would be 337 — the count keeps growing with the limit.)

AnswerOne filling is 3, 12, 40. There are infinitely many possibilities, because the last blank can be any counting number from 14 upwards.

Watch this explained “How many is a question too”, 8:32 into Working backwards from an average to a missing value

Open NCERT p. 127Checked by computerAnswers can differ: one example

1. Six numbers averaging 6.5 must total 6×6.5=39.
2. The 4 given numbers (3, 11, 15, 6) already total 35, so the two blanks together must add to 39−35=4.
3. With counting numbers, the only ways to split 4 between the two (ordered) blanks are 1&3, 2&2, and 3&1 — three fillings in all.

AnswerOne example: 1 and 3 (in either blank). There are 3 possible fillings.

Watch this explained “How many is a question too”, 8:32 into Working backwards from an average to a missing value

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(i) The average of two even numbers is even.

1. Two even numbers can be written as 2a and 2b; their average is (2a+2b)/2 = a+b, which is just some whole number — nothing forces it to be even.
2. Counterexample: 2 and 4 are both even, but their average is 3, which is odd.

AnswerNo — e.g. 2 and 4 average to 3.

(ii) The average of any two multiples of 5 will be …

1. Two multiples of 5 can be written as 5a and 5b; their average is 5(a+b)/2, which is only a whole multiple of 5 when a+b happens to be even.
2. Counterexample: 5 and 10 are both multiples of 5, but their average is 7.5 — not even a whole number.

AnswerNo — e.g. 5 and 10 average to 7.5.

(iii) The average of any 5 multiples of 5 will also …

1. Five multiples of 5 can be written as 5a₁,...,5a₅; their average is (5a₁+...+5a₅)/5 = a₁+...+a₅, simply the sum of 5 whole numbers — nothing forces that to be a multiple of 5.
2. Counterexample: 5, 5, 5, 5 and 10 are all multiples of 5, but their average is (5+5+5+5+10)/5 = 6, which is not a multiple of 5.

AnswerNo — e.g. 5, 5, 5, 5, 10 average to 6.

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(i) Which of the following statements are correct? Why?

1. Nothing is known yet about the two new students' heights, so we cannot say whether the average will go up, go down or stay the same: two new values can push it either way, or leave it unchanged. So (a) and (b) are not correct.
2. The new average is (old total + the two new heights) ÷ 26. The old total is already known from the old average: Sudhakar's class has 24 students (question 3), so it is 24 × 150.2 = 3604.8 cm.
3. So only the two new students need to be measured: (c) is correct, and (d) is not.

AnswerOnly (c) is correct.

(ii) The heights of the two new joinees are 149 cm and …

1. New total = 3604.8 + 149 + 152 = 3905.8 cm, for 26 students.
2. New average = 3905.8 ÷ 26 ≈ 150.22 cm, which is more than 150.2 cm.
3. Reason without dividing: 149 is 1.2 below the old average and 152 is 1.8 above it. Together they bring 0.6 cm more than their share, so the average rises, by 0.6 ÷ 26 ≈ 0.02 cm.

AnswerOnly (b) is correct: the average increases slightly.

(iii) Which of the following statements about the new class average height …

1. The median depends on where the individual heights lie in order, and we know only the average of the 24 heights, not the heights themselves.
2. Three classes of 24, all with average 150.2 cm, show that anything can happen when 149 cm and 152 cm join. 12 students at 150 cm and 12 at 150.4 cm: the median stays 150.2 cm. 12 at 148 cm and 12 at 152.4 cm: the median rises from 150.2 cm to 150.5 cm. 11 at 145 cm and 13 at 154.6 cm: the median falls from 154.6 cm to 153.3 cm.
3. So the information is not enough to say what happens to the median.

AnswerOnly (d) is correct.

Watch this explained “Shoes, and two students joining”, 7:45 into Which added values move the mean, and in which direction

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1. Weigh the data against 17 directly instead of adding everything up: for each value below 17 note how far below, and for each above, how far above.
2. Values below 17 (14×2, 15×2, 16×3) are 3,3,2,2,2,1,1,1 away — a total of 13.
3. Values above 17 (18×4, 19×4, 20×3, 21×1, 23×1) are 1,1,1,1,2,2,2,2,3,3,3,4,6 away — a total of 31.
4. The high side outweighs the low side by 31−13=18, so the true average must sit above 17.
5. Sharing that surplus of 18 over all 25 values gives average = 17 + 18/25 = 17.72.

AnswerNo — the actual average is 17.72 (443÷25), not 17.

Watch this explained “Weighing twenty-five throws against 17”, 7:43 into The mean as the point where the distances balance

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1. The three changes add up to −2+1+1=0, so the TOTAL weight of the group has not changed at all — meaning the mean, total÷count, cannot have changed either.
2. The median, though, depends on exactly WHICH three people changed and how their new weights sit relative to everyone else's — two groups can share the same previous mean (65.3 kg) and median (67 kg), yet the same three changes (−2, +1, +1) can leave one group's median unchanged while shifting the other's, so the median's fate cannot be pinned down from what's given.

AnswerMean: no change (the total is unchanged). Median: cannot be determined from the given information.

Watch this explained “When the information runs out”, 8:52 into Which added values move the mean, and in which direction

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(iii) Compare the price variation in Gujarat and Uttar Pradesh.

1. Gujarat's price ranges from a low of ₹13 (2020) to a high of ₹19.2 (2025) — a spread of ₹6.20.
2. Uttar Pradesh's price ranges from a low of ₹16.15 (2016) to a high of ₹26.9 (2024) — a spread of ₹10.75, noticeably wider.

AnswerUttar Pradesh shows the greater price variation.

(iv) In which state has the price increased the most …

1. Work out each state's rise from 2016 to 2025 (2025 price minus 2016 price): Andaman & Nicobar ₹4.99, Assam ₹6.35, Gujarat ₹2.70, Mizoram ₹9.80, Uttar Pradesh ₹8.66, West Bengal ₹14.52.
2. West Bengal's rise of ₹14.52 (from ₹9.47 to ₹23.99) is the largest of all six states.

AnswerWest Bengal had the greatest price increase.

Watch this explained “What one number hides”, 8:02 into Line graphs, and what change over time looks like

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1. (i) In 1983 rural kerosene use starts at about 85% (a clear majority) while urban electricity use starts at about 65% (also a majority) — both halves of the statement hold.
2. (ii) The kerosene line falls steadily toward 0% in BOTH panels from 1983 to 2023 — the decline is genuine in both areas.
3. (iii) By the year 2000 urban electricity use is already close to 90%, nowhere near 10% — the 10% figure belongs to the KEROSENE line at that time, not electricity, so this is false.
4. (iv) The graph only records which SOURCE lit each household, never whether the power supply was reliable — it cannot support or contradict any claim about power cuts.

Answer(i) and (ii) are valid; (iii) is false, and (iv) is not something this graph can address.

Watch this explained “The right number off the wrong line”, 8:57 into Interrogating a real dataset: traffic, and rainfall

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(i) How long do children aged 10 in urban areas spend …

1. Find age 10 on the Age axis and go up to the blue (Urban) curve. It meets the curve at about the 2-hour gridline, very slightly above it.

AnswerAbout 2 hours.

(ii) At what age is the average time spent daily on hobbies …

1. Go across from 1.5 hours (halfway between the 1h and 2h gridlines) to the red (Rural) curve. It meets the curve a little after age 14.
2. Of the options, 14 years is the one that fits: at 12 the rural curve is at about 2 hours, and at 18 it is below 1 hour.

Answer(d) 14 years.

(iii) Are the following statements correct?

1. (a) At age 10 children spend about 2.1 hours (urban) and 2.4 hours (rural); at age 15, about 1.1 hours and 1.3 hours. The time at 15 is less than at 10 (roughly half), not twice, so (a) is not correct.
2. (b) The graph shows only the average for rural 15-year-olds, about 1.3 hours. An average can hide children who play much less: four children averaging 1.3 hours could include one who plays 0 hours. So the graph cannot show that all rural 15-year-olds spend at least 1 hour, and (b) is not correct.

AnswerNeither (a) nor (b) is correct.

Watch this explained “The right number off the wrong line”, 8:57 into Interrogating a real dataset: traffic, and rainfall

Open NCERT p. 131One way to think about it

1. For a few days, colour a strip of 48 boxes for each day — each box is half an hour, from midnight to midnight — using one colour per activity (sleeping, eating, school or study, travel, playing outdoors, and so on).
2. (i) Compare the strips: if the sleeping and eating colours sit in the same places each day, you eat and sleep at regular times. Count the outdoor boxes each day and halve the count to get the hours spent outdoors.
3. (ii) For each activity, add its hours over all the days and divide by the number of days. Turn each average into boxes (hours × 2), round to whole boxes so that they still total 48, and colour one strip with them.
4. Example: if you slept 9, 8.5 and 9.5 hours on three days, the average is 27 ÷ 3 = 9 hours, which is 18 boxes.
5. (iii) Make the same strips and averages for an adult at home, put the two average strips side by side, and compare which activities take more or less of each day.

In shortAnswers depend on your own data. Record each day as a strip of 48 half-hour boxes, average each activity's time over the days, draw the average day as one 48-box strip, and do the same for an adult to compare.

Watch this explained “Three days averaged into a day nobody lived”, 9:29 into Infographics and activity strips: what a visual shows and what it hides

Open NCERT p. 131One way to think about it

1. Choose one of the two projects.
2. Sleep project: each member records the daily sleep of everyone in their family for 7 days, counting night sleep, naps and any daytime sleep, as the question says, and draws each person's days as strips.
3. Pool the group's data into three groups — children, adults and elderly. For each group, find the average (total hours of sleep ÷ number of daily figures) and the median (the middle value when all the daily figures are put in order).
4. Compare the groups, and compare each average with its median: if they differ a lot, a few unusually long or short sleepers are pulling the average.
5. School-timings project: record the start time, end time, class time and break time of several Grade 8 schools. For each school, the school-day length is end time − start time (Manoj's school: 9:30 am to 4:30 pm = 7 hours). Find the average and the median length and present them in a table or bar graph.
6. Example: five schools with days of 7, 7, 7, 6.75 and 6.5 hours have median 7 hours and average 34.25 ÷ 5 = 6.85 hours.

In shortAnswers depend on the data your group collects. Report both the average and the median for each group (children, adults, elderly) or for the school-day lengths, and say what the difference between them shows.

Watch this explained “Three bands, two summaries, and the next figure”, 8:18 into Telling a story with data, and letting it raise the next question

Open NCERT p. 131Checked by computer

(i) At which place does the sun rise the earliest in January?

1. Reading each location's sunrise curve in January: Kibithu ≈ 5:58 am, Kanyakumari ≈ 6:41 am, Srinagar ≈ 7:40 am, Ghuar Moti ≈ 7:44 am.
2. Kibithu's sunrise is the earliest of the four.
3. Kibithu's January sunset reads about 4:29 pm, so its day runs from about 6:00 am to about 4:30 pm — about 10.5 hours.

AnswerKibithu; day length about 10.5 hours in January.

(ii) Which place has the longest day length over the year?

1. Finding each location's longest day of the year (around June): Kibithu ≈ 13.9 hours, Ghuar Moti ≈ 13.5 hours, Srinagar ≈ 14.4 hours, Kanyakumari ≈ 12.6 hours.
2. Srinagar's June day is the longest of the four — and, being the most northerly of the four places, it also has the shortest day in January (about 10 hours), the biggest swing across the year of all four locations.

AnswerSrinagar.

Watch this explained “Bounds, day lengths and a moon that drifts”, 8:55 into Line graphs, and what change over time looks like

Open NCERT p. 132Checked by computerReads two ways: both answers shown

(i) Find out on what dates amavasya (new moon) and purnima …

1. At amavasya the moon is on the same side as the sun: it rises and sets with the sun. At purnima it is opposite the sun: it rises around sunset and sets around sunrise, so it is up all night.
2. The graph does not give the sun's times, so we take sunrise at about 7 am and sunset at about 6 pm (typical winter times in India).
3. Amavasya: on the 29th the moon rises at about 7:01 am and sets at about 6:03 pm, both almost exactly with the sun. (On the 28th it rises at 6:12 am but sets at 4:59 pm, an hour before sunset, so the 29th fits better.)
4. Purnima: on the 13th the moon rises at about 5:24 pm and sets the next morning at about 7:31 am. On the 14th it rises at about 6:26 pm. So the full moon falls in the night between the 13th and the 14th, and the date can be read two ways.
5. Reading 1 (the date on whose evening the full-moon night begins): on the 13th the moon is already up before sunset and still up after sunrise, the whole night. We lead with this because it is the only night the moon is up from sunset to sunrise; the book gives no answer key. Purnima = 13th.
6. Reading 2 (the day the moon rises closest to sunset): 6:26 pm on the 14th is 26 minutes after 6 pm, while 5:24 pm on the 13th is 36 minutes before. Purnima = 14th.

AnswerAmavasya: date 29. Purnima: read as the night the full moon is up from sunset to sunrise, date 13; read as the day the moon rises closest to sunset, date 14.

(ii)

1. The moon rises later every day: from about 8:25 am on the 1st to about 7:46 am on the 30th, after going once all the way round the clock. That is 23 hours 21 minutes in 29 days, about 48 minutes later each day.
2. The moonset times move later in the same way.
3. Because of this delay, one day has no moonrise at all (the 21st: the moon rises at 11:43 pm on the 20th and next at 12:40 am on the 22nd), and one day has no moonset (the 6th).
4. Things to wonder about: why is the moon about 50 minutes late each day? Why does the delay change a little through the month? After how many days does the pattern repeat?

AnswerThe moon rises about 48 minutes later each day, so there is no moonrise on date 21 and no moonset on date 6.

Watch this explained “Bounds, day lengths and a moon that drifts”, 8:55 into Line graphs, and what change over time looks like