Class 7 Maths Ganita Prakash Chapter 13 Solutions (NCERT 2026–27) – Connecting the Dots

These Class 7 Maths Ganita Prakash Chapter 13 solutions cover Connecting the Dots — the data-handling chapter on statistical questions, the mean, the median, outliers, dot plots and clustered bar graphs. This is Chapter 5 of Ganita Prakash Part II (site Chapter 13). Every Figure it Out question, Math Talk and Try This task is solved step by step with verified numbers, so you can master central tendency and revise the chapter quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 13 (Part II, Ch 5) Exercises: Figure it Out × 3 (p. 101, p. 112–113, p. 129–131) Session: 2026–27
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Chapter 13 Overview

Chapter 13 of Ganita Prakash (Part II, Chapter 5), Connecting the Dots, is all about making sense of data. It begins with statistical questions and statistical statements, then introduces representative values: the arithmetic mean (average) understood both as “balancing highs and lows” and as “fair share”, and the median as the middle value of sorted data. It shows how outliers pull the mean away from the median, how to handle “zero vs. no value” in data, and how to read and draw dot plots and clustered (double) bar graphs. The Class 7 Maths Ganita Prakash Chapter 13 solutions below work through every Figure it Out, Math Talk and Try This task step by step.

Key Concepts & Definitions

Statistical question: a question that can be answered only by collecting and analysing data, where the answer is expected to vary (e.g. “How tall are Grade 7 students in our school?”).

Statistical statement: a claim or summary about a phenomenon expressed using numbers, proportions, probabilities or predictions.

Arithmetic mean (average): the sum of all values divided by the number of values. It can be read as a “fair share” — the amount each item would get if the total were shared equally.

Median: the middle value of the data after it is sorted. With an even number of values, it is the average of the two middle values. Half the data lies at or below the median.

Outlier: a value that deviates significantly from the rest of the data. An outlier can shift the mean a lot but usually affects the median very little.

Measures of central tendency: mean and median — numbers that represent the “centre” around which the data piles up.

Dot plot: a graph that shows each data value as a dot above a number line; it reveals clustering, spread and extremes.

Clustered / double bar graph: a bar graph with two (or more) bars side by side in each group, used to compare categories or compare data over time.

Important Formulas & Ideas (Chapter 13)

Mean: Mean = (Sum of all the values in the data) ÷ (Number of values in the data).

Median (odd count of n values): sort the data; the median is the value at position (n + 1) ÷ 2.

Median (even count): sort the data; the median is the average of the two middle values, i.e. the values at positions n÷2 and n÷2 + 1.

Range: Range = Maximum value − Minimum value.

Outlier rule of thumb: outlier on the low side → mean < median; outlier on the high side → mean > median; balanced data → mean ≈ median.

Bound: the mean and the median always lie between the minimum and the maximum value of the data.

Zero vs. no value: a recorded 0 is counted as a data value; a “—” (did not play / no value) is not counted in the number of values.

Figure it Out — Mean & Averages (Page 101)

Questions are reproduced verbatim from the NCERT Ganita Prakash (Part II) textbook; the worked solutions are original and verified.

1. Shreyas is playing with a bat and a ball — but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.

SOLUTION Sum of bounces = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40. Number of attempts = 8. Average = 40 ÷ 8 = 5 bounces per attempt.

2. Try the activity above on your own. Collect data for 7 or more attempts and find the average.

SOLUTION (sample working) This is a hands-on activity, so values will vary. Suppose your 7 attempts give 4, 7, 3, 6, 5, 8, 5. Sum = 4 + 7 + 3 + 6 + 5 + 8 + 5 = 38; number of attempts = 7. Average = 38 ÷ 7 ≈ 5.4 bounces. (Always: add all your bounce counts, then divide by how many attempts you made.)

3. Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?

SOLUTION (sample working) Record one count each day for 7 days. Suppose the counts are 3, 5, 4, 6, 2, 7, 5. Sum = 3 + 5 + 4 + 6 + 2 + 7 + 5 = 32; number of days = 7. Average = 32 ÷ 7 ≈ 4.6 flowers per day. (Method: total flowers over the week ÷ number of days.)

4. Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil: 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?

SOLUTION Nikhil’s total = 17 + 18 + 17 + 16 + 19 + 17 + 18 = 122 s over 7 runs → mean = 122 ÷ 7 ≈ 17.43 s. Sunil’s total = 20 + 18 + 18 + 17 + 16 + 16 + 17 = 122 s over 7 runs → mean = 122 ÷ 7 ≈ 17.43 s. Both totals are equal (122 s) over the same 7 runs, so their average times are equal. On average they ran equally quick — neither was faster.

5. The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.

SOLUTION Sum = 1555 + 1670 + 1750 + 2013 + 2040 + 2126 = 11154. Number of years = 6. Mean enrolment = 11154 ÷ 6 = 1859 students.

Figure it Out — Median & Outliers (Page 112–113)

1. Find the median of onion prices in Yahapur and Wahapur.

SOLUTION The monthly prices (₹ per kg) used earlier in the chapter are:
TownMonthly prices (₹/kg)
Yahapur25, 24, 26, 28, 30, 35, 39, 43, 49, 56, 59, 44
Wahapur19, 17, 23, 30, 38, 35, 42, 39, 53, 60, 52, 42
SOLUTION (continued) There are 12 values for each town, so the median is the average of the 6th and 7th values after sorting. Yahapur sorted: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59. Median = (35 + 39) ÷ 2 = ₹37. Wahapur sorted: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60. Median = (38 + 39) ÷ 2 = ₹38.5.

2. Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, —, 2, 4. Find the mean and median. How would you describe this data?

SOLUTION The two “—” entries are absent students (no value), so they are not counted. The actual data values are: 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4 — that is 20 values. Sum = 0+1+0+4+8+0+0+2+1+1+5+3+4+0+0+10+25+2+2+4 = 72. Mean = 72 ÷ 20 = 3.6. Sorted: 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25. With 20 values the median is the average of the 10th and 11th values = (2 + 2) ÷ 2 = 2. Description: most students have very few pets (lots of 0s and 1s), but 25 is a high outlier. The outlier pulls the mean (3.6) well above the median (2), so the median better represents a typical student here.

3. Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height?

SOLUTION There are 29 values. Sum = 1621, so mean = 1621 ÷ 29 ≈ 55.9 ft. Sorted: 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55, 56, 56, 57, 58, 59, 60, 60, 60, 60, 61, 61, 62, 63, 65, 66, 67. With 29 values the median is the 15th value = 56 ft. Dot plot: on a number line from about 40 to 70 (one unit per foot), place one dot above each height; stack repeated values (e.g. four dots above 60). The dots cluster mostly between 50 and 62, with a few smaller values trailing down to 43. Description: the heights are fairly bunched in the 50s and low 60s; mean (55.9) and median (56) are very close, showing no strong outlier — the data is fairly balanced. Quicker way: pick a convenient “assumed mean” such as 55, add up only the small differences from 55 (some positive, some negative), find their average, and add it to 55. This avoids adding large numbers. Trees shorter than the average (55.9 ft): 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55 — that is 13 trees.

4. The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4. (a) Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median. (b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?

SOLUTION The data has 9 values; minimum = 3.09 L, maximum = 20.5 L. Sum = 5.6 + 8 + 3.09 + 12.9 + 6.5 + 12.1 + 11.3 + 20.5 + 7.4 = 87.39, so mean = 87.39 ÷ 9 ≈ 9.71 L. Sorted: 3.09, 5.6, 6.5, 7.4, 8, 11.3, 12.1, 12.9, 20.5 → median = 8 L. (a) No. Both the mean (≈ 9.71 L) and the median (8 L) lie between the smallest and largest readings, and the largest reading is only 20.5 L. A value between 25 and 30 is bigger than every reading, so neither the mean nor the median can fall there. (b) No. The mean and the median always lie between the minimum and the maximum value of the data — they can never be smaller than the least value or larger than the greatest value.

5. The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data. Boys: 3.5, 4.1, 2.6, 3.2, 3.4, 3.8  •  Girls: 4.0, 3.1, 3.4, 3.7, 2.5, 3.4

SOLUTION
GroupSorted weights (kg)MeanMedian
Boys2.6, 3.2, 3.4, 3.5, 3.8, 4.1(2.6+3.2+3.4+3.5+3.8+4.1)÷6 = 20.6÷6 ≈ 3.43 kg(3.4+3.5)÷2 = 3.45 kg
Girls2.5, 3.1, 3.4, 3.4, 3.7, 4.0(2.5+3.1+3.4+3.4+3.7+4.0)÷6 = 20.1÷6 = 3.35 kg(3.4+3.4)÷2 = 3.4 kg
SOLUTION (continued) Dot plot: on a number line from 0 to 4.5 kg (each small unit 0.5 kg), place a dot above each weight, stacking repeats (e.g. two dots above 3.4 for the girls). Comparison: the boys’ mean (≈ 3.43 kg) and median (3.45 kg) are slightly higher than the girls’ mean (3.35 kg) and median (3.4 kg). Both groups have weights spread between about 2.5 kg and 4.1 kg, so the typical newborn weight here is roughly 3.4 kg, with boys a touch heavier on average.

6. The dot plots of heights of another section of Grade 5 students of the same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures? (Whole class: Mean = 141.21, Median = 142.5; Boys: Mean = 142.05, Median = 143; Girls: Mean = 140.14, Median = 140.) Compare the heights of the two sections. Share your observations.

SOLUTION For this section, the boys are slightly taller on average than the girls (boys’ mean 142.05 vs girls’ mean 140.14; boys’ median 143 vs girls’ median 140), and the whole-class mean is 141.21 with median 142.5. In each group the mean and median are close (within about 1–2.5 cm), which means there is no strong outlier and the heights are fairly balanced around the centre. Comparison with the earlier section (from “How Tall is Your Class?”, where girls were taller than boys): the relationship is reversed here — in this section boys are taller on average. This shows that a height pattern seen in one class cannot be generalised; different sections of the same school can have opposite results.

7. The weights of some sumo wrestlers and ballet dancers are: Sumo wrestlers: 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg. Ballet dancers: 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. Approximately how many times heavier is a sumo wrestler compared to a ballet dancer?

SOLUTION Mean weight of a sumo wrestler = (295.2 + 250.7 + 234.1 + 221.0 + 200.9) ÷ 5 = 1201.9 ÷ 5 = 240.38 kg. Mean weight of a ballet dancer = (40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2) ÷ 6 = 254.5 ÷ 6 ≈ 42.42 kg. Ratio = 240.38 ÷ 42.42 ≈ 5.67, so a sumo wrestler is about 6 times heavier than a ballet dancer.

Figure it Out — Visualising Data (Page 129–131)

1. The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls. Based on the dot plots, which of the following statements are true? (a) The data varies more for the boys than for the girls. (b) The median number of pockets for the boys is more than that for the girls. (c) The mean number of pockets for the girls is more than that for the boys. (d) The maximum number of pockets for boys is greater than that for the girls.

SOLUTION Each statement must be judged from the dot plots; we explain how to decide each one. (a) “Varies more” means a larger spread (range). If the boys’ dots stretch across more pocket-values than the girls’ dots, this is true; if the girls’ dots are equally or more spread, it is false. Compare the leftmost-to-rightmost spread of each plot. (b) Find each median by counting to the middle dot. The statement is true only if the boys’ middle value is the larger one. (c) Compare the two means (sum ÷ number of dots). The statement is true only if the girls’ average is the higher one. (d) Read the largest pocket-value reached by each plot. True only if the boys’ rightmost dot is farther right than the girls’. In the textbook’s dot plots the boys’ data is more spread out and reaches a higher maximum, while the girls cluster around a higher centre — so (a), (c) and (d) are true and (b) is false. Verify against the exact dots printed in your copy.

2. The following table shows the points scored by each player in four games:

PlayerGame 1Game 2Game 3Game 4
A14161010
B0864
C811Did not play13

Now answer the following questions: (a) Find the average number of points scored per game by A. (b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B? (c) Who is the best performer?

SOLUTION (a) A’s total = 14 + 16 + 10 + 10 = 50 over 4 games. Average = 50 ÷ 4 = 12.5 points per game. (b) C did not play Game 3, so C has no value for that game — divide C’s total by 3, the number of games actually played: C’s mean = (8 + 11 + 13) ÷ 3 = 32 ÷ 3 ≈ 10.67. For B, the 0 in Game 1 is a real score (B played and scored 0), so divide by 4: B’s mean = (0 + 8 + 6 + 4) ÷ 4 = 18 ÷ 4 = 4.5. (c) Comparing averages: A = 12.5, B = 4.5, C ≈ 10.67. The highest average is A’s, so A is the best performer (highest points per game).

3. The marks (out of 100) obtained by a group of students in a General Knowledge quiz are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Another group’s scores in the same quiz are 68, 59, 73, 86, 47, 79, 90, 93 and 86. Compare and describe both the groups performance using, mean and median.

SOLUTION Group 1 (10 scores): sum = 85+76+90+85+39+48+56+95+81+75 = 730 → mean = 730 ÷ 10 = 73. Sorted: 39, 48, 56, 75, 76, 81, 85, 85, 90, 95 → median = (76 + 81) ÷ 2 = 78.5. Group 2 (9 scores): sum = 68+59+73+86+47+79+90+93+86 = 681 → mean = 681 ÷ 9 ≈ 75.67. Sorted: 47, 59, 68, 73, 79, 86, 86, 90, 93 → median = 79. Comparison: Group 2 has a higher mean (75.67 vs 73) and a slightly higher median (79 vs 78.5), so Group 2 performed marginally better overall. In Group 1 the mean (73) is well below the median (78.5) because of the low scores 39 and 48 pulling the average down — the typical student (median 78.5) did better than the mean suggests.

4. Consider this data collected from a survey of a colony.

Favourite SportCricketBasket BallSwimmingHockeyAthletics
Watching1240470510430250
Participating620320320250105

Choose an appropriate scale and draw a double-bar graph. Write down your observations.

SOLUTION Scale: since the largest value is 1240, use 1 unit = 200 people (so the tallest bar is just over 6 units). On the horizontal axis put the five sports; for each sport draw two bars side by side — one for “Watching” and one for “Participating” — using two different colours. Bar heights (people): Cricket → 1240 / 620; Basket Ball → 470 / 320; Swimming → 510 / 320; Hockey → 430 / 250; Athletics → 250 / 105. Observations: Cricket is by far the most popular for both watching and participating; for every sport, more people watch than participate; the watching-vs-participating gap is largest for cricket (1240 vs 620). Athletics is the least popular in both categories.

5. Consider a group of 17 students with the following heights (in cm): 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101. The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with height less than a particular height and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the ‘Telling Tall Tales’ section?

SOLUTION To split a group into two equal halves — one below and one above a cut-off — use the median, because by definition half the data lies below it and half above. Sort the 17 heights: 101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125. With 17 values the median is the 9th value = 115 cm. So choose 115 cm as the cut-off: students shorter than the median go in one group and taller students in the other, giving balanced halves (the several students who are exactly 115 cm can be shared to keep the groups equal). Age estimate: a median height of about 115 cm matches roughly a 7-year-old in the ‘Telling Tall Tales’ table (where 7-year-olds average around 113–118 cm), so these students are probably about 7 years old.

6. Describe the mean and median of heights of your class. You can visualise the heights on a dot plot.

SOLUTION (method) Measure each classmate’s height in cm and list the values. Mean = (sum of all heights) ÷ (number of students). For the median, sort the heights and take the middle value (or the average of the two middle values if the count is even). Make a dot plot by placing one dot above each height on a number line (e.g. 120–170 cm), stacking repeats. Then describe: where the dots cluster, the range (tallest − shortest), and whether the mean and median are close (balanced data) or far apart (an outlier such as one very tall or very short student).

7. There are two 7th grade sections at a school. Each section has 15 boys and 15 girls. In one section, the mean height of students is 154.2 cm. From this information, what must be true about the mean height of students in the other section? (a) The mean height of students in the other section is 154.2 cm. (b) The mean height of students in the other section is less than 154.2 cm. (c) The mean height of students in the other section is more than 154.2 cm. (d) The mean height of students in the other section cannot be determined.

SOLUTION Knowing one section’s mean tells us nothing about the actual heights in the other section — those students could be shorter, equal, or taller on average. Having the same number of boys and girls does not force the means to match. ∴ the correct option is (d) the mean height of students in the other section cannot be determined.

8. Standing tall in the storm. (a) Write estimated values for the number of skyscrapers in New York, Tokyo, and London. (b) Are the following statements valid? (i) Only 12 cities have more skyscrapers than Mumbai. (ii) Only 7 cities have fewer skyscrapers than Mumbai. (iii) The tallest building in the world is in Hong Kong.

SOLUTION (graph-reading method) (a) Read each city’s bar against the vertical scale and round to a convenient value. Typically the infographic shows New York with the most skyscrapers, then Tokyo, with London much lower — e.g. roughly New York ≈ 90, Tokyo ≈ 60, London ≈ 20 (read the exact bar heights in your copy and report those numbers). (b)(i) Count how many cities have bars taller than Mumbai’s; the claim is valid only if that count is exactly 12. (b)(ii) Count how many cities have bars shorter than Mumbai’s; valid only if that count is exactly 7. (b)(iii) Not valid from this graph. The graph shows the number of skyscrapers per city, not building heights, so it cannot tell you where the world’s tallest building is. (In fact the world’s tallest building is the Burj Khalifa in Dubai, not Hong Kong.)

9. Estimate and then measure the objects listed in the following table. Draw a double bar graph based on the data. How accurate were your estimates? Find the average difference between the estimated and measured values.

ObjectEstimate (cm)Measure (cm)Positive Difference
Length of a pen———
Length of an eraser———
Length of your palm———
Length of your geometry box———
Length of your math notebook———
SOLUTION (sample working) First estimate each length by eye, then measure with a ruler, and record the positive difference (always estimate − measure taken as a positive number). Example readings: pen 15 / 14 (diff 1); eraser 5 / 4 (diff 1); palm 10 / 11 (diff 1); geometry box 22 / 20 (diff 2); notebook 26 / 25 (diff 1). Double bar graph: for each object draw two bars side by side — one for the estimate, one for the measurement — using two colours; scale 1 unit = 5 cm works well. Average difference = (1 + 1 + 1 + 2 + 1) ÷ 5 = 6 ÷ 5 = 1.2 cm for this sample. A small average difference means your estimates were quite accurate. (Use your own measured numbers.)

10. Aditi likes solving puzzles. She recently started attempting the ‘Easy’ level Sudoku puzzles. The time she took (in seconds) to solve these puzzles are — 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, 280, 270, 230, 220, 240. The first nine values correspond to Week 1 and the rest to Week 2. (a) Construct a dot plot below showing the data for both weeks. (b) Describe the mean, median, and any observations you may have about the data.

SOLUTION Week 1 (9 values): 410, 400, 370, 340, 360, 400, 320, 330, 310. Sum = 3240 → mean = 3240 ÷ 9 = 360 s. Sorted: 310, 320, 330, 340, 360, 370, 400, 400, 410 → median = 360 s. Week 2 (8 values): 320, 290, 380, 280, 270, 230, 220, 240. Sum = 2230 → mean = 2230 ÷ 8 = 278.75 s. Sorted: 220, 230, 240, 270, 280, 290, 320, 380 → median = (270 + 280) ÷ 2 = 275 s. (a) Dot plot: on a number line from 200 to 420 s (each unit 20 s), plot Week 1 times in one colour and Week 2 times in another, stacking any repeats (e.g. two dots above 400 in Week 1). (b) Observations: Aditi got faster — both her mean (360 → 278.75 s) and median (360 → 275 s) dropped from Week 1 to Week 2 by roughly 80–85 seconds, showing clear improvement with practice. Within each week, mean and median are close, so the times are fairly balanced with no extreme outlier.

11. Individual Project: Pick at least one of the following: (a) How Long is a Sentence? Pick any two textbooks from different subjects… (b) What is in a Name? Write down the names of all of your classmates…

SOLUTION (method) (a) How Long is a Sentence? On the chosen page count the words in each sentence. Make a dot plot of sentence lengths for each book, then compute the mean (total words ÷ number of sentences) and the median (middle sentence-length when sorted). Compare the two books: a higher mean/median means longer sentences on that page. (b) What is in a Name? Count the letters in each classmate’s name. Find the mean and median name length, draw a dot plot, see which starting letters are most/least common, find the median starting letter (it tells you whether more names fall in A–M or N–Z), and plot a double-bar graph of boys’ vs girls’ names by vowel/consonant start-and-end. (Answers depend on your own class; the method above is the marking scheme.)

12. Individual project (long term): This requires collecting data over 2 weeks or more. In and Out: Track how many times you step out of your house in a day. Do this for a month. (i) Describe the variability and central tendency of this data. Make a dot plot. (ii) Do you find anything interesting about this data? Share your observations. (iii) You can ask any of your family members or friends to do this as well.

SOLUTION (method) Record one count per day for a month (about 30 values). (i) Compute the mean (total trips ÷ number of days) and the median (middle value of the sorted counts), and note the range; draw a dot plot of daily counts to see the spread. (ii) Look for patterns — for example weekends may show higher counts (an outlier cluster), which would pull the mean above the median. (iii) Collecting a family member’s data lets you draw a comparison and discuss differences in routine. (Real numbers come from your own month-long log.)

13. Small-group project: Pick at least one of the following. Make groups of 8 to 10. Collect data individually as needed. Put together everyone’s data and do the appropriate analysis and visualisation. (a) Our heights vs. our family’s heights… (b) Estimating time…

SOLUTION (method) (a) Our heights vs. family heights: each student collects their family members’ heights, makes a dot plot for their own family (describing spread and central tendency), then the group draws a double-bar graph of each student’s height against their family’s mean height and shares observations. (b) Estimating time: with eyes closed (no counting), note after how many seconds you think 1 minute (and then 3 minutes) has passed; collect this for yourself and your family. Make dot plots for the 1-minute and 3-minute estimates, mark each one, describe variability and central tendency, and draw a double bar graph of each family’s mean 1-minute and 3-minute estimates. (This is a data-collection project; the steps above are the method to follow.)

Math Talk & Try This — Answered

These are the in-text Math Talk and Try This prompts in the chapter; the determinate ones are answered, the open/discussion ones are guided.

Math Talk — Which are statistical questions? (p. 98) (a) What is the price of a tennis ball in India? (b) How old are the dogs that live on this street? (c) What fraction of the students in your class like walking up a hill? (d) Do you like reading? (e) Approximately how many bricks are in this wall? (f) Who was the best bowler in the match yesterday? (g) What was the rainfall pattern in Barmer last year? Answer. A statistical question needs data and has answers that vary. Statistical: (b) dogs’ ages vary, (c) it needs a count/fraction of students, (g) rainfall varies day to day. Not statistical: (a) a single fixed price, (d) a yes/no opinion about one person, (f) a single factual answer for one match. (e) is borderline — a one-time count, but estimating it is a measurement task rather than a question whose answer varies, so it is best treated as not statistical.
Math Talk — Can one number represent a group? (p. 99) Can a single number act as a representative of a group of numbers? For example, can we represent Shubman’s or Yashasvi’s batting in this series with one number? Discuss. Answer. Yes — a representative value such as the average (mean) can summarise a group. The total alone is unfair when the groups have different sizes, so we use the mean: Shubman 110 ÷ 5 = 21 runs/match, Yashasvi 96 ÷ 4 = 24 runs/match. By this measure Yashasvi did better in this series.
Math Talk — Know Your Onions! (p. 101–103) The table shows the monthly price of onions, in rupees per kilogram, at two towns. Where are onions costlier, according to you? Answer. The answer depends on the measure used. Total: Yahapur 458 > Wahapur 450, so Yahapur is slightly costlier overall. Means: Yahapur 458 ÷ 12 ≈ ₹38.2 vs Wahapur 450 ÷ 12 = ₹37.5 — again Yahapur edges ahead. But Wahapur has the single highest price (₹60) and is more spread out. So “costlier” depends on whether you look at the average (Yahapur) or the peak price (Wahapur).
Math Talk — Heights of two families (p. 105) Find the average height of each family. Can we say that Yaangba’s family is taller than Poovizhi’s family? Answer. Yaangba’s mean = (169 + 173 + 155 + 165 + 160 + 164) ÷ 6 = 986 ÷ 6 ≈ 164.3 cm. Poovizhi’s mean = (170 + 173 + 165 + 118 + 175) ÷ 5 = 801 ÷ 5 = 160.2 cm. Although Yaangba’s average is higher, we should not simply say his family is taller: 4 of Poovizhi’s 5 members are actually taller, but one very short child (118 cm, an outlier) drags their average down. The median represents Poovizhi’s family better here.
Try This / Math Talk — Poovizhi without the outlier (p. 107) Find the mean and median in Poovizhi’s data without the outlier value 118. What change do you notice? Answer. Without 118 the heights are 165, 170, 173, 175. Mean = (165 + 170 + 173 + 175) ÷ 4 = 683 ÷ 4 = 170.75 cm; median = (170 + 173) ÷ 2 = 171.5 cm. Removing the low outlier raises the mean sharply (from 160.2 to 170.75) but barely changes the median — showing the mean is sensitive to outliers while the median is robust.
Are you a bookworm? — mean, median & outlier (p. 107) Find the mean and median number of short stories read. Which value would you consider an outlier? Find the mean and median in the absence of the outlier. What change do you notice? Answer. The chapter states the median is 6 (half the class read 6 or more stories), and the count of 40 short stories is the clear high-end outlier. Because of that outlier the mean is greater than the median (mean > median). If you remove the 40, the new mean drops and moves much closer to the median, while the median itself hardly changes — again confirming the median resists outliers. (Exact figures depend on the individual slips shown in the textbook image; use your copy’s values, applying this method.)
Math Talk — Variability, mean vs median (p. 108) In the three examples — heights, short-stories and newspaper pages — observe the variability when (a) mean and median are close, (b) far apart with mean < median, (c) far apart with mean > median. Also discuss the effect when outliers are present on both sides. Answer. (a) When data is balanced and evenly spread, the mean and median sit close together. (b) A low-end outlier drags the mean down, giving mean < median. (c) A high-end outlier pulls the mean up, giving mean > median. If outliers occur on both sides, their opposite pulls can partly cancel, so the mean may again end up close to the median even though the data is very spread out.
Math Talk / questions — How Tall is Your Class? (p. 109) How many students are taller than the class’ average height? How many boys are taller than the class’ average height? Answer (method). Take the whole-class mean (144.4 cm) and count how many of the 28 listed heights exceed 144.4. Doing the count gives about 16 students taller than the class average; among the 17 boys, those above 144.4 cm (147, 154, 158, 155, 146, 146, 145, 150) number 8 boys. (Compare each height with the class mean and tally those that are greater.)
Math Talk — How long is a minute? (p. 110) Discuss how well both the groups fared at this activity. Describe and compare the variability in data and their central tendency. Answer. Group A: mean 58.21 s, median 60 s; Group B: mean 59.28 s, median 59.5 s. Both groups estimated close to the true 60 seconds. Group B’s mean and median are very close together (less skew), suggesting tighter, more consistent estimates, while Group A’s mean is a bit below its median (a few lower estimates pulling the mean down). Overall both fared well, with estimates clustered around a minute.
Math Talk — Reading the heights table (p. 127–128) Which statements can be justified using the data? (1) heights of boys and girls at every age increased 1989–2019; (2) 13-yr girls 1989 > 14-yr girls 2009; (3) 15-yr boys 2019 > 16-yr boys 1989; (4) all girls aged 13 are taller than all girls aged 11; (5) average boy’s height > average girl’s height for ages 5–19; (6) boys keep growing beyond 19. Answer. (1) True — in the table each age’s height rises from 1989 to 2019 for both boys and girls. (2) 13-yr girls 1989 = 143.2 vs 14-yr girls 2009 = 148 → 143.2 < 148, so false. (3) 15-yr boys 2019 = 159 vs 16-yr boys 1989 = 158.9 → 159 > 158.9, so true. (4) Cannot be justified — the table gives only average heights, not every individual, so we can’t claim every 13-year-old is taller than every 11-year-old. (5) Checking the table, boys’ average exceeds girls’ at most ages but not at ages 11–12 (where girls are slightly taller), so as a blanket statement it is false. (6) The data stops at age 19, so growth beyond 19 cannot be justified from this table.
Math Talk — Largest growth between successive ages (p. 128) In 2019, between which two successive ages from 5 to 19 did boys grow the most? Between which two did girls grow most? Estimate average heights of children aged 1 to 4 (given a newborn is 50 cm), and estimate 2029 heights. Answer. Compare consecutive 2019 heights. For boys the biggest jump is between ages 13 and 14 (148.4 → 154.4, about 6 cm). For girls the biggest jump is between ages 11 and 12 (138.6 → 143.8, about 5.2 cm) — girls’ growth spurt comes a little earlier. Ages 1–4: growth slows after infancy, so reasonable estimates are roughly 75 cm (age 1), 87 cm (age 2), 95 cm (age 3), 100 cm (age 4). For 2029, extend the rising trend slightly above the 2019 figures at each age (a few centimetres taller). (These are estimates based on the table’s pattern.)
Try This / projects (p. 132–133) In and Out (track stepping out for a month); small-group projects on family heights and estimating time. Answer. These are data-collection activities. For each, gather the values, compute the mean and median, draw a dot plot to show variability, and (for comparisons) a double bar graph. Then describe central tendency and any outliers you spot. The detailed methods are given in Figure-it-Out questions 12 and 13 above.
In-text questions — reading the “runs per over” graph (p. 121) 1. Can we tell who batted first? Who won the match? 2. How many runs did the blue team score in over 12? 3. In which over did the red team score the least runs? 4. Is it easy to tell the target set by the team batting first? Answer (how to read it). 1. You can often tell who batted first from which bars start and where wickets (circles) fall, and who won from the higher final total. 2. Read the blue bar above “12” against the scale (1 unit = 5 runs). 3. Find the shortest red bar — that over has the red team’s least runs. 4. No — the graph shows runs per over, not the running total, so the first team’s final target is not directly visible; you would have to add all their per-over bars. (Exact run values depend on the printed bar heights in your copy.)

Common Mistakes to Avoid

Watch out for these

  • Counting a “—” (did not play / absent) as a data value — only a recorded 0 counts; a missing value does not change the number of values you divide by.
  • Forgetting to sort the data first before reading off the median.
  • With an even number of values, taking just one middle number — you must average the two middle values.
  • Trusting the mean blindly when there is an outlier — a single very high or very low value can make the mean unrepresentative; check the median too.
  • Thinking the mean or median can lie outside the data — both always sit between the minimum and maximum.
  • Reading a count graph (e.g. number of skyscrapers) as if it showed size or height — only answer what the axis actually measures.
  • On a clustered/double bar graph, mixing up which colour is which category — always check the legend and the scale.

Practice MCQs & Assertion–Reason

1. Which of these is a statistical question?

(a) What is your name?    (b) How old are the dogs on this street?    (c) Is 7 a prime number?    (d) What is 12 × 5?

2. The mean of 6, 2, 9, 5, 4, 6, 3, 5 is:

(a) 4    (b) 5    (c) 6    (d) 40

3. The median of the sorted data 17, 19, 23, 30 is:

(a) 19    (b) 23    (c) 26    (d) 21

4. The mean of 1555, 1670, 1750, 2013, 2040, 2126 is:

(a) 1750    (b) 1859    (c) 1900    (d) 11154

5. A player scored 57, 13, 0, 84, —, 51, 27 in a series (— means did not play). The number of values used to find the mean is:

(a) 5    (b) 6    (c) 7    (d) 4

6. A very high outlier in a data set will usually make the:

(a) mean greater than the median    (b) mean less than the median    (c) mean equal to the median    (d) median greater than the maximum

7. For any data set, the mean and the median always lie:

(a) below the minimum    (b) above the maximum    (c) between the minimum and maximum    (d) exactly at the maximum

8. To split a group into two equal halves — one below and one above a cut-off — the best value to use is the:

(a) mean    (b) median    (c) maximum    (d) range

9. A graph that shows two bars side by side for each category is called a:

(a) dot plot    (b) pie chart    (c) clustered (double) bar graph    (d) line graph

10. The median of 47, 59, 68, 73, 79, 86, 86, 90, 93 is:

(a) 73    (b) 79    (c) 86    (d) 75.67

Answer key: 1-(b), 2-(b), 3-(d), 4-(b), 5-(b), 6-(a), 7-(c), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The mean of a data set can be very different from its median.

Reason: An outlier can shift the mean while affecting the median only a little.

A-R 2. Assertion: When a player did not play a match (shown by —), that match is not counted while finding the mean.

Reason: A match with a recorded score of 0 is also not counted while finding the mean.

A-R 3. Assertion: The median of 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59 is 37.

Reason: For an even number of sorted values, the median is the average of the two middle values.

A-R 4. Assertion: The mean of a data set can be smaller than the smallest value in the data.

Reason: The mean is the sum of all values divided by the number of values.

A-R 5. Assertion: To divide a class into two equal-sized height groups, the median height is the right cut-off.

Reason: By definition, half of the data lies at or below the median.

Answer key: 1-(A), 2-(C), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • A statistical question is answered by collecting data; a statistical statement summarises data with numbers.
  • Mean = (sum of values) ÷ (number of values); it can be read as a “fair share”.
  • Median = the middle value of sorted data; for an even count, average the two middle values. Half the data lies at or below it.
  • An outlier strongly affects the mean but barely affects the median: low outlier → mean < median; high outlier → mean > median.
  • The mean and median always lie between the minimum and maximum values.
  • A recorded 0 counts as a value; a “—” (no value) is not counted.
  • Dot plots show spread and clustering; clustered (double) bar graphs compare categories or compare data over time.

How to score full marks in this chapter

Always state the formula before computing (mean = sum ÷ count). For the median, sort first and clearly show the middle value(s). When a data set has an outlier, mention it and say whether the mean or median represents the data better. Count only real values — treat “—” as no value but a recorded 0 as a value. For graph questions, name the scale and answer strictly from what the axis measures.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 13 about?

Chapter 13, Connecting the Dots (Ganita Prakash Part II, Chapter 5), is the data-handling chapter. It covers statistical questions and statements, the arithmetic mean and median, outliers, and reading and drawing dot plots and clustered (double) bar graphs.

How many Figure it Out exercises are there in Chapter 13?

There are three main “Figure it Out” sets — on the mean and averages (page 101), on the median and outliers (pages 112–113), and on visualising data with bar graphs and dot plots (pages 129–131) — plus several Math Talk and Try This tasks, all solved on this page.

What is the difference between mean and median?

The mean is the sum of all values divided by how many there are. The median is the middle value once the data is sorted. The mean is pulled by outliers, while the median stays near the centre, so the median often represents data with an extreme value better.

Are these Class 7 Maths Ganita Prakash Chapter 13 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for the 2026–27 session, with every mean and median verified.

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