NCERT Solutions Class 7 Maths Chapter 13 Connecting the Dots

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 7 · M AT H S

NCERT Solutions

Chapter 13: Connecting the
Dots…

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 98 – 134 24 86 English

Solutions, notes, sample papers & more at 106 pages

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 13: Connecting the Dots…
Chapter 5 of Ganita Prakash Grade 7 Part II is about reading data honestly. It starts with a guess — is the 6-
foot-tall friend a man? — and then builds the tools that turn guesses into claims you can defend: the mean,
the median, outliers, dot plots and clustered bar graphs. The habit the chapter teaches matters more than
any formula: say exactly how much the data lets you say, and no more.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 98 – 134

SECTIONS QUESTIONS

24 86

MEDIUM

English

In-text Questions — Page 98
Section 5.1 Of Questions and Statements

Q1 Which of the following are statistical questions? (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?

A question is a statistical question if you must collect data to answer it, and if the answer is
expected to vary from case to case.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

QUESTION STATISTICAL? REASON

(a) Price of a tennis ball in Yes The price is not the same in every shop or every city. You
India would collect prices and then summarise them.

(b) How old are the dogs Yes The dogs are of different ages. You collect the ages and
on this street? describe them.

(c) What fraction of the Yes You must ask every student and count. The answer is a
class likes walking up a proportion found from data.
hill?

(d) Do you like reading? No This is asked of one person and has one answer, yes or
no. Nothing varies, so no data is collected.

(e) Approximately how No One particular wall has one fixed number of bricks.
many bricks are in this Estimating is not the same as collecting data that varies.
wall?

(f) Who was the best Yes To answer it you must look at wickets, runs given and
bowler in the match overs bowled for each bowler — that is data, and people
yesterday? may still disagree on how to combine it.

(g) What was the rainfall Yes Rainfall changes from month to month. You need the
pattern in Barmer last year? whole year's readings to describe a pattern.

Why it happens: the test is not whether the question is hard. It is whether the
answer varies. "Do you like reading?" has one answer for one person. But "What
fraction of the class likes reading?" varies from class to class, so it becomes
statistical.

Careful with (f): "best bowler" is not defined by the data alone. Two people may
collect the same figures and still choose different bowlers. It is a statistical question,
but it does not have one single correct answer.

Math Talk — Pages 98 – 99

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Section 5.2 Representative Values

MATH TALK

Q1 The runs scored by Shubman and Yashasvi in a cricket series are given in the table
below. Who do you think performed better?

MATCH 1 MATCH 2 MATCH 3 MATCH 4

Shubman 0 17 21 90

Yashasvi 67 55 18 35

Runs scored by the two players in the first series — page 98.

Yashasvi performed better in this series, but the honest answer is that it depends on what
you count — and the children in the book each count something different.

WAY OF COMPARING SHUBMAN YASHASVI BETTER

Highest score 90 67 Shubman

Total runs 0+17+21+90 = 128 67+55+18+35 = 175 Yashasvi

Mean per match 128 ÷ 4 = 32 175 ÷ 4 = 43.75 Yashasvi

Range (max − min) 90 − 0 = 90 67 − 18 = 49 Yashasvi (steadier)

Matches won head-to-head Matches 3, 4 Matches 1, 2 Equal

Why it happens: both played the same number of matches, so the total is a fair
comparison here, and Yashasvi's total is larger. He is also more consistent — his
scores stay between 18 and 67, while Shubman swings from 0 to 90. Vaishnavi is
right that Shubman hit the single biggest score, but one big innings does not make a
whole series.

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Class 7 Maths Chapter 13 Connecting the Dots…
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Tip: Shreyas's first remark — "Yashasvi scored more in the first two, Shubman in the

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Runs scored by the two players in the second series — page 99.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q3 Vaishnavi says, “Here, Shubman performed better since his total is 110 runs, while
Yashasvi's total is 96 runs”. What do you think of Vaishnavi's statement?

Her arithmetic is right, but her conclusion is not safe. The two totals are not comparable,
because they come from different numbers of matches.

110 runs from 5 matches → 110 ÷ 5 = 22 per match

96 runs from 4 matches → 96 ÷ 4 = 24 per match

Why it happens: a total grows simply by playing more. Shreyas spots this at once —
"But Yashasvi made 96 runs in 4 matches and Shubman made 110 in 5." Whenever
two groups are of different sizes, compare a per-item value (the mean), not the raw
total.

Tip: the same trap appears everywhere. A shop that sold ₹50,000 of goods in 10
days did not do better than one that sold ₹30,000 in 5 days — that is ₹5,000 a day
against ₹6,000 a day.

Q4 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.

Yes — one number can represent a group, but it can never replace it.

The total is one such number, and it works when the groups are the same size.
The mean (average) works even when the groups differ in size, because it balances the highs
and the lows: Shubman 22, Yashasvi 24.
The highest score and the range are also single numbers, but each tells only one thing.

Why it happens: the mean is a levelling value. If Shubman had scored 22 in every
one of his five matches, his total would still be 110. So 22 is the score that, repeated,
gives the same total — it stands in for the whole group.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

But be careful: a single number hides everything else. Shubman's mean of 22 does
not tell you that one of his scores was 52 and another was 7. That is why we also
look at the minimum, the maximum and the spread.

In-text Questions — Page 100
Average as Fair-Share

Q1 Shreyas and 4 of his friends have collected the following numbers of guavas: 3, 8,
10, 5, and 4. Parag and 5 of his friends have collected the following numbers of
guavas: 5, 4, 6, 3, 4, and 8. Each group will share their guavas equally amongst
themselves. In which group will each member get a bigger share of guavas?

Each member of Shreyas's group gets more — 6 guavas each, against 5 each in Parag's group.

Shreyas's group: 3 + 8 + 10 + 5 + 4 = 30 guavas among 5 people

Each member gets 30 ÷ 5 = 6 guavas

Parag's group: 5 + 4 + 6 + 3 + 4 + 8 = 30 guavas among 6 people

Each member gets 30 ÷ 6 = 5 guavas

Difference = 6 − 5 = 1 guava more each in Shreyas's group

Why it happens: both groups collected exactly the same total, 30 guavas. But
Shreyas's group has 5 people and Parag's has 6. The same total shared among more
people gives each person less. This is the fair-share meaning of the average: total ÷
number of people.

Tip: notice that "who collected more?" and "who gets more each?" have different
answers here. The totals tie at 30, but the shares do not.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q2 Vaishnavi tracks the number of Hibiscus flowers blooming in her garden each day.
The data for the last few days' is 2, 7, 9, 4, 3. What is the average number of Hibiscus
flowers blooming per day in Vaishnavi's garden?

5 flowers per day.

Average = (total number of flowers bloomed) ÷ (number of days)

= (2 + 7 + 9 + 4 + 3) ÷ 5

= 25 ÷ 5

=5

Why it happens: on no single day did exactly 5 flowers bloom. The average tells us
what the daily count would have been if the same number had bloomed every day.
The two high days (7 and 9) make up for the two low days (2 and 3).

Check it yourself: 5 flowers a day for 5 days is 25 flowers — the same total as the
real data. That is the test of a correct mean.

Figure it Out — Page 101
Section 5.2 Representative Values

TRY THIS

Q1 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.

5 bounces per attempt.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Total bounces = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5

=8+9+5+4+6+3+5

= 40
Number of attempts = 8

Average = 40 ÷ 8 = 5 bounces

Why it happens: Shreyas's best attempt was 9 and his worst was 2. The mean of 5
sits between them and balances them out — the 4 extra bounces above 5 in his best
attempt pay for the 3 missing bounces in his worst.

Tip: the median here is also 5. Sorted, the data is 2, 3, 4, 5, 5, 6, 6, 9, and the two
middle values are 5 and 5. Mean and median agree because there is no outlier.

Q2 Try the activity above on your own. Collect data for 7 or more attempts and find the
average.

This is an activity to do yourself. Here is the method, and then one worked sample.

Bounce the ball on the bat and count until it drops. That is one attempt.
Write the count down immediately, before the next attempt.
Do at least 7 attempts. Do not throw away a bad attempt — a 1 or a 0 is real data.
Add all the counts, then divide by the number of attempts.

Sample answer: suppose your 8 attempts give 3, 5, 1, 7, 4, 6, 4, 2.

Total = 3 + 5 + 1 + 7 + 4 + 6 + 4 + 2 = 32

Average = 32 ÷ 8 = 4 bounces per attempt

Why it happens: the more attempts you record, the steadier the average becomes.
With only 2 attempts one lucky try changes everything; with 10 attempts one lucky
try moves the average very little.

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Class 7 Maths Chapter 13 Connecting the Dots…
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Try This: record your attempts again after a week of practice. If your average has

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gone up, you now have evidence that you improved — not just a feeling.

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Q3 Identify a flowering plant in your neighbourhood. Track the number of flowers that

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This is a data-collection activity. Here is how to do it well, and a sample.
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Two friends are training to run a 100 m race. Their runningse times over the past
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Q4

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Neither. Their averages are exactly the same — about 17.43 seconds each.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Nikhil's total = 17 + 18 + 17 + 16 + 19 + 17 + 18 = 122 s

Nikhil's mean = 122 ÷ 7 ≈ 17.43 s

Sunil's total = 20 + 18 + 18 + 17 + 16 + 16 + 17 = 122 s

Sunil's mean = 122 ÷ 7 ≈ 17.43 s

Why it happens: the two lists are made of almost the same numbers in a different
order, so their totals are identical. Since both ran the same number of times,
identical totals must give identical means.

The means tie, so look at the spread — and there the two runners are not the same.

FASTEST SLOWEST RANGE MEDIAN

Nikhil 16 s 19 s 3s 17 s

Sunil 16 s 20 s 4s 17 s

The strongest thing you can say: on average they are equally quick, and their
medians are equal too. But Nikhil is more consistent — his times stay inside a 3-
second band, while Sunil's stretch over 4 seconds. Sunil's slowest run (20 s) is worse
than anything Nikhil produced. For a one-off race that hardly matters; for reliable
performance, Nikhil looks the safer pick.

Q5 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.

Mean enrolment = 1859 students.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Total = 1555 + 1670 + 1750 + 2013 + 2040 + 2126

1555 + 1670 = 3225

3225 + 1750 = 4975
4975 + 2013 = 6988

6988 + 2040 = 9028

9028 + 2126 = 11154

Mean = 11154 ÷ 6 = 1859 students

Why it happens: 1859 lies between the smallest year (1555) and the largest (2126),
as every mean must. It is the number of students the school would have had each
year if the same number had come every year.

Check it yourself: 1859 × 6 = 11154 — the same total we started with. A quick sanity
check: the six numbers all lie between 1500 and 2200, so the mean had to land in
that band.

Math Talk — Pages 101 – 102

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Know Your Onions!

MATH TALK

Q1 The table shows the monthly price of onions, in rupees per kilogram (kg), at two
towns. Where are onions costlier, according to you?

MONTH YAHAPUR WAHAPUR

Jan 25 19

Feb 24 17

Mar 26 23

Apr 28 30

May 30 38

Jun 35 35

Jul 39 42

Aug 43 39

Sep 49 53

Oct 56 60

Nov 59 52

Dec 44 42

Monthly price of onions, in rupees per kilogram, at the two towns — page 101.

Taken over the whole year, onions are slightly costlier in Yahapur — but the two towns are so
close that the answer depends on which measure you use. That is the real point of the question.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

MEASURE YAHAPUR WAHAPUR COSTLIER

Total of the 12 prices ₹458 ₹450 Yahapur

Mean price 458 ÷ 12 ≈ ₹38.17 450 ÷ 12 = ₹37.50 Yahapur

Median price (35 + 39) ÷ 2 = ₹37 (38 + 39) ÷ 2 = ₹38.50 Wahapur

Highest price ₹59 (Nov) ₹60 (Oct) Wahapur

Lowest price ₹24 (Feb) ₹17 (Feb) Yahapur (never as cheap)

Range (max − min) 59 − 24 = ₹35 60 − 17 = ₹43 Wahapur swings more

Months with the higher price 6 months 5 months Yahapur (1 month equal)

Why it happens: Yahapur wins on total, mean and number of dearer months;
Wahapur wins on median and on the single highest price. Wahapur's prices are
more spread out — very cheap in February (₹17), very dear in October (₹60) — so its
low months pull its total down even though its peaks are higher.

Judging the children's claims: Khushboo ("Wahapur, it has the highest price ₹60")
uses one month out of twelve. Vishal ("Wahapur has 3 numbers in the 50s") counts
only part of the data. Jithin's remark about the range is not about costliness at all —
it is about variability. Nafisa's totals and Sampat's month-by-month count are the
two fairest comparisons, and both point mildly to Yahapur.

Q2 Can you think of any other ways to compare the data?

Yes. Beyond the minimum, maximum, mean, total and range, here are ways a Grade 7 student
can actually carry out.

Median — the middle price of each town: ₹37 for Yahapur, ₹38.50 for Wahapur.
Month-by-month differences — subtract the two prices each month and look at the
pattern. In Jan–Mar Yahapur is dearer by ₹3–₹6; in Apr–Jul Wahapur is dearer.
Count how many months each town crossed ₹50: Yahapur 2 (Oct, Nov), Wahapur 3 (Sep,
Oct, Nov).
Group the prices into bands — ₹11–20, ₹21–30, ₹31–40, ₹41–50, ₹51–60 — and count how
many months fall into each band. This is what a dot plot shows at a glance.

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Class 7 Maths Chapter 13 Connecting the Dots…
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Season by season — compare the mean of Jan–Jun with the mean of Jul–Dec for each town.
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Both towns are far dearer in the second half of the year.
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Cost of a fixed purchase — if a family buys 2 kg every month, work out the year's onion bill:
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Yahapur, ₹900 in Wahapur.
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dearer on the whole?" needs the mean or the total. "Which town is riskier for a

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shopkeeper?" needs the range. "When should we buy?" needs the month-by-month ag
pattern.
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In-text Questions — Page 103
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Dot Plots

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Yahapur

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10 20 30 40 50 60

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10 20 30 40 50 60

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The dot plot on page 103. Green circles are Yahapur’s twelve monthly prices, purple
diamonds are Wahapur’s.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

INFORMATION IN THE TABLE? IN THE DOT PLOT?

All 24 price values Yes Yes — one dot each

Which town each price belongs to Yes Yes — by colour

How many times a price repeats Hard to see Yes — stacked dots

Which month a price belongs to Yes No — lost

The order of the months Yes No — lost

Why it happens: a dot plot sorts the values along a number line. Sorting is what
makes clusters, gaps and the highest and lowest values jump out — but sorting is
exactly what destroys the original month-wise order.

Tip: every picture of data is a trade. You gain one kind of clarity and lose another.
That is why the chapter later draws the same onion data again as a clustered bar
graph, where the months come back.

Q2 Looking at it, can we tell the price of onions in Yahapur in the month of January?

No, we cannot. The dot plot shows that Yahapur had a price of ₹25 at some point in the year,
but nothing on it says when.

Why it happens: the horizontal line of the dot plot carries price, not time. Once the
dots are arranged by price, the month label has nowhere left to live. We can say "one
of Yahapur's twelve monthly prices was ₹25" — we cannot say "January's price was
₹25" from this picture alone.

Check it yourself: from the original table, January in Yahapur was indeed ₹25. But
that came from the table, not from the dot plot. Keeping straight what your
evidence actually shows is the whole habit this chapter is teaching.

Math Talk — Page 104

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Know Your Onions!

MATH TALK

Q1 Find the average price of onions at Yahapur and Wahapur.

Yahapur ≈ ₹38.17 per kg; Wahapur = ₹37.50 per kg.

Yahapur: 25 + 24 + 26 + 28 + 30 + 35 + 39 + 43 + 49 + 56 + 59 + 44

= 49 + 26 + 28 + 30 + 35 + 39 + 43 + 49 + 56 + 59 + 44

= 458

Average = 458 ÷ 12 = ₹38.17 (to two decimal places)

Wahapur: 19 + 17 + 23 + 30 + 38 + 35 + 42 + 39 + 53 + 60 + 52 + 42
= 450

Average = 450 ÷ 12 = ₹37.50

Why it happens: both towns have 12 months of data, so here the totals and the
averages point the same way. The gap is tiny — 67 paise per kilogram over a whole
year. On the strength of the averages alone you should say the two towns are nearly
the same, with Yahapur very slightly dearer.

Tip: 458 ÷ 12 = 38 remainder 2, so the answer is 38 and 2/12 = 38.1666… ≈ ₹38.17.
Never round to "about ₹38" and then claim Yahapur is dearer — after rounding, the
two are equal.

Q2 What else do you wonder about?

This is an open question — good data always raises more questions than it answers. Here are
questions that this particular table really can spark, in the spirit of the ones printed in the book.

Both towns are cheapest in February and dearest in October–November. Is that because
the new crop arrives in winter and the old stock runs out before it?

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Yahapur's prices climb steadily month after month; Wahapur's jump about. Does Wahapur
have fewer shops, or does it get its onions from further away?
In April and May, Wahapur is suddenly dearer than Yahapur by ₹2 and ₹8. Was there a
transport problem, or a local festival?
What happens to a family's monthly budget when the price goes from ₹17 to ₹60 — more
than three times — inside one year?
Does the farmer who grew the onions receive three times as much in October as in February,
or does the extra money stay with traders?
Would potatoes, tomatoes or dal show the same shape over the year?

Why it happens: a single statement like "onions cost ₹35 a kilo" invites no further
thought. A table of twelve varying values does, because every rise and dip asks to be
explained. Curiosity is the first step of statistics, not the last.

Tip: pick one of your questions and decide what data you would need to answer it.
That turns a wondering into a statistical question.

Math Talk — Pages 105 – 106
Outliers and Medians

MATH TALK

Q1 The heights of the family members of Yaangba and Poovizhi are as follows:
Yaangba's family: 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm. Poovizhi's family:
170 cm, 173 cm, 165 cm, 118 cm, 175 cm. Find the average height of each family. Can
we say that Yaangba's family is taller than Poovizhi's family?

The averages say Yaangba's family is taller — but that conclusion is not safe, because one very
short child pulls Poovizhi's average down.

Yaangba: 169 + 173 + 155 + 165 + 160 + 164 = 986 cm, 6 members

Average = 986 ÷ 6 ≈ 164.3 cm

Poovizhi: 170 + 173 + 165 + 118 + 175 = 801 cm, 5 members

Average = 801 ÷ 5 = 160.2 cm

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Now look at the actual heights instead of the averages.

YAANGBA'S FAMILY (SORTED) 155 160 164 165 169 173

POOVIZHI'S FAMILY (SORTED) 118 165 170 173 175 —

Why it happens: four of Poovizhi's five members are 165 cm or taller. In Yaangba's
family only two members reach 165 cm. Poovizhi's average is lower only because of
the 118 cm child — a value far below all the others. The average of 160.2 cm is
shorter than 4 of the 5 people it is supposed to represent. When that happens, the
average is not doing its job.

The strongest thing you can say: Poovizhi's family has the lower mean height, but
the taller members. Saying "Yaangba's family is taller" would be reading the mean
without looking at the data behind it.

Q2 Can you think of any other number that can represent the data better?

Yes — sort the data and take the middle value. That number is called the Median.

Poovizhi's family sorted: 118, 165, 170, 173, 175

Middle value → Median = 170 cm

Yaangba's family sorted: 155, 160, 164, 165, 169, 173

Even number of values, so take the average of the two middle ones

Median = (164 + 165) ÷ 2 = 164.5 cm

Why it happens: the median only cares about position, not size. Changing 118 to
100, or to 150, would not move the median at all — it would still be the third value in
the sorted list. The mean, by contrast, adds every value, so one extreme number
changes it a lot.

Page 18 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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co m
m.
Tip: with an even number of values there is no single middle, so we average the two

as e
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middle values. The median then has an equal number of values below it and above
it, exactly.as a g
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it should.

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In this case, does the median represent the heights of the families better than the
.
Q3

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average?

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Yes, clearly — for Poovizhi's family. For Yaangba's family it makes almost no difference.
co m
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FAMILY MEAN MEDIAN GAP

se m
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Yaangba 0.2 cm

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Poovizhi 160.2 cm 170 cm 9.8 cm 118 cm

m a
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Why it happens: the median 170 cm sits right among the heights 165, 170, 173, 175
a g
— it looks like a typical member of Poovizhi's family. The mean 160.2 cm matches
nobody. In Yaangba's family, with no outlier, the mean and median land within 0.2

co m
cm of each other, so either would do.
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a s em The rule to carry away: when the mean and the median are close, the data is fairly

agl balanced and the mean is safe. When they are far apart, look for an outlier and
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prefer the median.

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In-text Questions — Pages

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Are you a bookworm?

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Q1 Find the mean and median in Poovizhi's data without the outlier value 118. What

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Without the outlier the mean jumps by more than 10 cm, while the median barely moves.

m .
as e
a g l

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Remaining heights: 165, 170, 173, 175

Sum = 165 + 170 + 173 + 175 = 683 cm, 4 values

New mean = 683 ÷ 4 = 170.75 cm

Sorted: 165, 170, 173, 175 (even count)

New median = (170 + 173) ÷ 2 = 171.5 cm

WITH 118 CM WITHOUT 118 CM CHANGE

Mean 160.2 cm 170.75 cm +10.55 cm

Median 170 cm 171.5 cm +1.5 cm

Why it happens: removing 118 takes a very small number out of the sum, so the
mean shoots up. The median only shifts by one position in the sorted list, so it hardly
changes. This is the clearest possible demonstration that the median resists outliers
and the mean does not.

Tip: notice that the new mean (170.75) and the new median (171.5) are now close
together — the sign of data with no outlier left in it.

Page 20 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q2 After the summer vacation, a class teacher asked his class how many short stories
they had read. Each student answered the number of stories read on a piece of
paper, as shown below. Find the mean and median number of short stories read.
Before calculating them, can you guess whether the mean will be less than or
greater than the median?

6 8 15
5 0 8
0 40
3 2 12 1
7 5
10

The fifteen slips of paper handed in by the class — page 107.

The guess: the mean will be greater than the median. One student read 40 stories while
everybody else read 15 or fewer — a high outlier drags the mean up but leaves the median
where it is.
Now the calculation. There are 15 papers: 6, 8, 5, 15, 40, 0, 8, 3, 0, 2, 7, 12, 5, 1, 10.

Sum = 6 + 8 + 5 + 15 + 40 + 0 + 8 + 3 + 0 + 2 + 7 + 12 + 5 + 1 + 10

= 14 + 5 + 15 + 40 + 0 + 8 + 3 + 0 + 2 + 7 + 12 + 5 + 1 + 10

= 122

Mean = 122 ÷ 15 ≈ 8.13 stories

Sorted: 0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15, 40

15 values, so the middle one is the 8th → Median = 6 stories

Why it happens: the mean 8.13 is bigger than 11 of the 15 values. The single
reading of 40 adds 40 to the sum all by itself — a third of the entire total came from
one student. The median simply counts positions, so that one student is just "the
last one in the line" and moves nothing.

Page 21 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Check it yourself: the book states that the median is 6 and that half the class read 6
or more stories. Count them: 6, 7, 8, 8, 10, 12, 15, 40 — that is 8 students at or above
6, out of 15. Correct.

Q3 Mark the data, the mean, and the median on the dot plot below.

0 5 10 15 20 25

Page 108 — the empty dot plot printed with this question, with its scale running from 0
to 25.

The number line runs from 0 to 40. Place one dot for each value, stacking repeats. Then draw a
solid line at the mean 8.13 and a dashed line at the median 6.

VALUE 0 1 2 3 5 6 7 8 10 12 15 40

NUMBER OF DOTS 2 1 1 1 2 1 1 2 1 1 1 1

Mean = 8.13
Median = 6

0 5 10 15 20 25 30 35 40

Dot plot of the number of short stories read. The dots crowd between 0 and 15; the lone dot at 40 is
the outlier, and it pulls the mean (solid line) to the right of the median (dashed line).

Page 22 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Why it happens: the picture shows in one glance what the two numbers only hint at
— fourteen dots huddle in the first quarter of the line and one sits far away at the
end. That gap is the outlier.

Q4 Which of the values would you consider an outlier?

40 is the outlier.

All the other values: 0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15

Next highest value after 40 is 15

Gap between them = 40 − 15 = 25 stories

Gap between any other neighbouring pair ≤ 3

Why it happens: an outlier is a value that significantly deviates from the rest. Here
the whole class sits inside a band of 15 stories, and then there is an empty stretch of
25 before the single reading of 40. That empty stretch is the evidence.

Careful: 15 is the second highest value, but it is not an outlier — it is only 3 above
12, which is only 2 above 10. The values step up smoothly until 40.

Q5 Find the mean and median in the absence of the outlier. What change do you
notice?

Removing 40 pulls the mean down by more than 2 stories; the median drops by only half a
story.

Page 23 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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co m
em.
New sum = 122 − 40 = 82, and 14 values remain
m l as
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New mean = 82 ÷ 14 ≈ 5.86 stories
m a g
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aSorted: 0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15

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Even count, so New median = (5 + 6) ÷ 2 = 5.5 stories
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WITH 40 a WITHOUT 40 CHANGE

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Mean 8.13 5.86 −2.27

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Median
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6 5.5
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a Why it happens: the mean changed more than four times as much as the median.

s
Notice also that mean and median have now swapped places — before removal
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mean > median (8.13 > 6); afterwards they are almost equal (5.86 and 5.5). Once the

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high outlier is gone, the data is balanced again.

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The rule: a very high outlier lifts the mean, so mean > median. A very low outlier
. co
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(like 118 cm in Poovizhi's family) drops the mean, so mean < median.

m as e
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sem
g l a
a In-text Questions — Page 108
se m
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Are We on the Same Page?

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agl
Q1 Do you read newspapers? Have you noticed how many pages a newspaper has on
different days of the week — is it the same or different?

com
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com is 16, 18, 20, 22, 26, 16, 10. g l
The number of pages is not the same every day. The data given for one newspaper, Monday to
. a
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Sunday,
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DAY Mon Tue Wed Thu Fri Sat Sun

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PAGES 16 18 20 22 26 16 10

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Total = 16 + 18 + 20 + 22 + 26 + 16 + 10 = 128 pages

Mean = 128 ÷ 7 ≈ 18.29 pages per day

Sorted: 10, 16, 16, 18, 20, 22, 26

Median = 4th value = 18 pages

Why it happens: the page count climbs from Monday to Friday and then falls
sharply. Friday is the biggest (26 pages) because that is when most advertisements
are booked; Sunday is the smallest (10 pages) here. So the answer is "different" —
and knowing how it differs is more useful than knowing that it differs.

Tip: mean 18.29 is only a little above the median 18. This data has no real outlier —
10 is low, but not far away from 16.

Q2 Mark the data, the mean, and the median on the dot plot below.

The number line runs from 0 to 25 and a little beyond. Place the seven dots, stacking the two
16s, then draw the mean at 18.29 and the median at 18.

Mean = 18.29
Median = 18

0 5 10 15 20 25

Newspaper pages, Monday to Sunday. The mean (solid) and the median (dashed) almost coincide,
because the data has no strong outlier.

Why it happens: the dots spread fairly evenly from 10 to 26, with a small pile at 16.
When data is spread out like this instead of being bunched at one end, the mean
and median land almost on top of each other.

Page 25 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Math Talk — Pages 108 – 109
Measures of Central Tendency

MATH TALK

Q1 In the three examples we considered — the heights, short-stories, and newspaper
pages — observe the variability in data when: (a) the mean and median are close to
each other (b) the mean and median are comparatively far apart, with mean <
median (c) the mean and median are comparatively far apart, with mean > median

Each of the three examples illustrates exactly one of the three cases.

CASE EXAMPLE MEAN MEDIAN WHAT THE DATA LOOKS
LIKE

(a) close Newspaper pages 18.29 18 Values spread out evenly, no
together 10, 16, 16, 18, 20, 22, 26 value far from the rest

(b) mean < Poovizhi's family 160.2 170 One value far below the rest —
median 118, 165, 170, 173, 175 a low outlier

(c) mean > Short stories 8.13 6 One value far above the rest —
median 0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, a high outlier
10, 12, 15, 40

Why it happens: the median is anchored to the middle position and cannot be
dragged. The mean is a balance point, so it slides towards whichever side the
extreme value is on. A low value drags it down and it falls below the median; a high
value drags it up and it rises above the median. When nothing extreme exists, there
is nothing to drag it and the two sit together.

Tip: you can use this backwards. If you are told only that a set of exam marks has
mean 42 and median 55, you already know that a few very low marks are pulling the
mean down — without seeing a single mark.

Page 26 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q2 Discuss the effect on the mean and median when outliers are present on both sides.
You may take some example data to examine and explain this.

When there is a low outlier and a high outlier, the two pulls act in opposite directions. Whether
the mean ends up above or below the median depends on which outlier is further from the
middle.
Case 1 — the two outliers balance. Take 2, 18, 19, 20, 21, 22, 38.

Sum = 2 + 18 + 19 + 20 + 21 + 22 + 38 = 140

Mean = 140 ÷ 7 = 20 Median = 4th value = 20

2 is 18 below the middle; 38 is 18 above it — the pulls cancel

Case 2 — the high outlier is further away. Take 2, 18, 19, 20, 21, 22, 60.

Sum = 162, Mean = 162 ÷ 7 ≈ 23.14 Median = 20

Mean > median, because 60 is 40 above the middle while 2 is only 18 below

Case 3 — the low outlier is further away. Take 0, 18, 19, 20, 21, 22, 26.

Sum = 126, Mean = 126 ÷ 7 = 18 Median = 20

Mean < median, because 0 is 20 below the middle while 26 is only 6 above

Why it happens: the mean adds up distances, so it responds to how far each outlier
is, not just to how many there are. The median counts positions, so both outliers are
simply "first in the line" and "last in the line" — and removing or moving them
further out changes the median not at all.

The warning to carry away: in Case 1 the mean and median are equal, which
normally suggests balanced data — but the data is not balanced, it has two wild
values that happen to cancel. So mean = median is good evidence of balance, not
proof of it. Always look at the dot plot as well.

In-text Questions — Pages 109 – 110

Page 27 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Of Ends and the Essence — How Tall is Your Class?

Q1 Suppose you are asked the question, “How tall is your class?” What would you say?
[Boys: 147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 142, 140, 141, 144, 145, 150.
Girls: 143, 136, 150, 144, 154, 140, 145, 148, 156, 150, 150.]

No single number answers this well. A good answer gives a typical height and the two ends.

Boys (17 values): sum = 2430, mean = 2430 ÷ 17 ≈ 142.94 cm

Boys sorted: 128, 130, 134, 135, 135, 140, 141, 142, 144, 145, 146, 146, 147, 150, 154, 155,

158

Median = 9th value = 144 cm

Girls (11 values): sum = 1616, mean = 1616 ÷ 11 ≈ 146.9 cm

Girls sorted: 136, 140, 143, 144, 145, 148, 150, 150, 150, 154, 156

Median = 6th value = 148 cm

Whole class (28 students): sum = 2430 + 1616 = 4046

Mean = 4046 ÷ 28 = 144.5 cm Median = 145 cm

So a full answer is: the class is about 145 cm tall on average, with everybody between 128
cm and 158 cm.

Why it happens: "How tall is your class?" is a statistical question — the answer
varies from student to student. Giving only the mean hides the 30 cm spread
between the shortest and the tallest. Giving the mean, the median and the two ends
describes the class honestly.

Note on the book's figure: the printed dot plot gives the whole-class mean as 144.4
cm. Adding the 28 heights gives 4046, and 4046 ÷ 28 = 144.5 exactly. The correct
mean is 144.5 cm. Nothing else in the discussion changes.

Page 28 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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co m
m.
What can we infer from the dot plots and the central tendency measures?
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Q2

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aglthings can be said, and the book states them all.
Four

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BOYS (17) GIRLS (11) WHOLE CLASS (28)

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Shortest 128 cm
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136 128 cm

Tallest 158 cm
a 156 cm 158 cm

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Range 30 cm 20 cm 30 cm

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Mean 146.9 cm 144.5 cm

l a se
a g Median 144 cm 148 cm 145 cm

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The boys' heights are more spread out, from 128 to 158 cm. The girls lie in a narrower
band, 136 to 156 cm.
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Both the tallest and the shortest student in the class are boys.
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Yet the boys' mean is below the class mean and below the girls' mean. In this class the girls
are taller on average.

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For both groups mean < median (142.94 < 144 and 146.9 < 148), which shows a small pull
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15 students out of 28 are taller than the class average of 144.5 cm. agl
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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

All 28 heights sorted:

128, 130, 134, 135, 135, 136, 140, 140, 141, 142, 143, 144, 144,

145, 145, 146, 146, 147, 148, 150, 150, 150, 150, 154, 154, 155, 156, 158

Class mean = 144.5 cm

Heights above 144.5: 145, 145, 146, 146, 147, 148, 150, 150, 150, 150, 154, 154, 155, 156,

158

Count = 15 students

Why it happens: 15 out of 28 is a little over half. That is quite normal — the mean
does not have to split a group into two equal halves. Only the median does that.
Here the median is 145 cm, and 13 students are below it, 13 above, with 2 students
at exactly 145.

Tip: using the book's printed mean of 144.4 cm gives the same count of 15, because
no student is exactly 144.4 or 144.5 cm tall.

Q4 How many boys are taller than the class' average height?

8 boys are taller than 144.5 cm.

Boys sorted: 128, 130, 134, 135, 135, 140, 141, 142, 144,

145, 146, 146, 147, 150, 154, 155, 158

Boys above 144.5 cm: 145, 146, 146, 147, 150, 154, 155, 158 → 8 boys

Boys at or below 144.5 cm: 17 − 8 = 9 boys

Why it happens: only 8 of the 17 boys (less than half) reach the class average, while
7 of the 11 girls do (145, 148, 150, 150, 150, 154, 156). That is another way of seeing
the same fact — the girls in this class are on the whole taller.

Page 30 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Check it yourself: 8 boys + 7 girls = 15 students above the class average. That
matches the previous answer exactly.

Math Talk — Page 110
How long is a minute?

MATH TALK

Q1 Discuss how well both the groups fared at this activity. Describe and compare the
variability in data and their central tendency. [Group A: Mean = 58.21, Median = 60.
Group B: Mean = 59.28, Median = 59.5.]

Both groups did well, and Group B did slightly better — its mean and median are both closer
to the target of 60 seconds and closer to each other.

GROUP A GROUP B

Mean 58.21 s 59.28 s

Median 60 s 59.5 s

Distance of mean from 60 s 1.79 s 0.72 s

Gap between mean and median 1.79 s 0.22 s

Spread of the dots about 42 s to 70 s about 44 s to 71 s

Central tendency. Group A's median is exactly 60 — half the children in Group A opened their
eyes at or after the full minute. Group B's mean is nearer 60, so as a whole Group B was closer.
Variability. In both groups the dots pile up between about 55 and 65 seconds, with a few
children well short (in the low 40s) and a few well over (around 70). Group B's dots are more
tightly stacked just under 60; Group A has a longer tail of early openers, which is exactly why its
mean (58.21) falls below its median (60).

Why it happens: mean < median for Group A tells us, without counting a single dot,
that the low values in Group A are further from the middle than the high values are.
The children who opened too early were further out than the children who waited
too long.

Page 31 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

The strongest thing you can say: most children in both groups can judge a minute
to within about 5 seconds, and the typical estimate is a little under a minute rather
than over. But with only two groups you cannot claim that children in general
underestimate a minute — that would need far more data.

In-text Questions — Page 111
Zero Median Runs Scored! / Zero vs No value

Q1 In a cricket match, can a team's median runs scored by a player be 0 but the team's
total score be 407/10?

Yes, it can. The median only depends on the middle value of the sorted scores, so it says
nothing about the total.

A team has 11 players, so the median is the 6th score in sorted order.

If 6 or more players score 0, that 6th score is 0 → median = 0.

Example: 0, 0, 0, 0, 0, 0, 12, 41, 96, 108, 131

Median = 6th value = 0

Runs off the bat = 0×6 + 12 + 41 + 96 + 108 + 131 = 388

Extras = 19
Team total = 388 + 19 = 407

The book's own figure follows from this:

Average runs scored by a player = (407 − 19) ÷ 11

= 388 ÷ 11 = 35.27 runs

Why it happens: the extras (19 runs) are not credited to any batter, so they are
subtracted before dividing by 11. And the median ignores how big the big scores are
— six ducks are enough to fix the median at 0 no matter whether the other five
players score 5 runs or 150 runs each.

Page 32 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Zero vs no value: a 0 is a real score and must be counted. A match a player did not
play is not a value at all. So for a player with scores 57, 13, 0, 84, —, 51, 27 the
average is (57 + 13 + 0 + 84 + 51 + 27) ÷ 6 = 232 ÷ 6 ≈ 38.67, dividing by 6 and not by
7. In the same way, Sita's mango tree gave 0, 0, 8, 24, 41, 16, 5, 0, 0, 0, 0, 0 mangoes
— but only the summer months when mangoes are expected should be used to find
a meaningful mean.

Figure it Out — Pages 112 – 113
Section 5.2 Representative Values

MATH TALK

Q1 Find the median of onion prices in Yahapur and Wahapur.

Yahapur: ₹37. Wahapur: ₹38.50.

Yahapur sorted: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59

12 values, so average the 6th and the 7th
Median = (35 + 39) ÷ 2 = 74 ÷ 2 = ₹37

Wahapur sorted: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60

Median = (38 + 39) ÷ 2 = 77 ÷ 2 = ₹38.50

Why it happens: both towns have 12 monthly prices — an even number — so there
is no single middle value and we average the two middle ones.

Compare with the means: Yahapur mean ₹38.17 > median ₹37, while Wahapur
mean ₹37.50 < median ₹38.50. So Yahapur's few very high months (56, 59) pull its
mean above its median, while Wahapur's very cheap months (17, 19) pull its mean
below. By the median, Wahapur is the dearer town; by the mean, Yahapur is. The two
towns are genuinely close, and you should say so rather than pick a winner.

Page 33 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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co m
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Sanskruti asked her class how many domestic animals and pets each had at home.
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Q2

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Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4,
g you describe this
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Mean = 3.6 animals; Median = 2 animals.

l a se
ag "no pets". That leaves 20 values.
First, the two dashes are absent students. They gave no value at all, so they are not counted —
unlike a 0, which is a real answer meaning

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Values: 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4
em.
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The mean of 3.6 is bigger than the value of 16 of the 20 students, so

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Tip: the student with 25 animals probably keeps livestock — cows, goats, hens. That
is not a mistake in the data, just a genuinely different kind of home. Outliers are
often real and interesting; do not delete them, report them.

Q3 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?

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80

Page 112 — the empty dot plot printed with this question, with its scale running from 0
to 80 feet.

Mean ≈ 55.9 feet, Median = 56 feet, and 13 trees are shorter than the average. There are 29
trees.
The dot plot. The printed number line runs from 0 to 80 in steps of 5, but every tree lies
between 43 and 67, so all the dots sit in the middle of the line.

HEIGHT 43 44 45 46 49 50 51 52 54 55 56 57 58 59 60 61 62 63 65 66 67
(FT)

DOTS 1 1 1 1 1 1 1 2 2 2 2 1 1 1 4 2 1 1 1 1 1

Mean ≈ 55.9
Median = 56

40 45 50 55 60 65

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Heights of the 29 date-palm trees. The tallest stack is at 60 feet. The mean (solid) and median
(dashed) fall almost on the same spot.

The mean, the ordinary way:

Sum = 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

= 1621 feet, over 29 trees

Mean = 1621 ÷ 29 = 55.9 feet (to one decimal place)

A quicker way — the assumed mean. All the heights are near 55, so measure everything from
55 instead of from 0.

Write each height as 55 + something:

−5, −10, −12, −3, +6, +8, −9, 0, +5, 0, +4, +1, +1, −6, −1,

+10, +11, −4, −11, +3, +5, −1, −3, +2, +6, +7, +5, +5, +12

Sum of the pluses = 6+8+5+4+1+1+10+11+3+5+2+6+7+5+5+12 = 90

Sum of the minuses = 5+10+12+3+9+6+1+4+11+1+3 = 64

Total = 90 − 64 = +26

Mean = 55 + 26 ÷ 29 = 55 + 0.9 = 55.9 feet

The median. Sorted, the 15th of the 29 heights is the middle one.

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

Median = 15th value = 56 feet

How many trees are shorter than the average, 55.9 feet?

43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55 → 13 trees
(the 14th tree, at 56 feet, is already taller than 55.9)

Page 36 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Describing the heights: the trees stand between 43 and 67 feet, a range of 24 feet. They
cluster strongly in the 50s and low 60s, with the commonest height being 60 feet (4 trees). There
is no outlier, and the mean (55.9) and median (56) agree almost exactly — the sign of a well-
balanced set of data. Roughly half the farm is under 56 feet and half over.

Why the assumed-mean trick works: if you subtract 55 from every value you
subtract 55 × 29 from the total, so the mean also goes down by exactly 55. Adding 55
back at the end restores it. The numbers you actually add up are small, so the
arithmetic is far easier.

Q4 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?

(a) No. Neither the mean nor the median can lie between 25 and 30, because the largest
reading in the whole data is only 20.5 litres.

Data sorted: 3.09, 5.6, 6.5, 7.4, 8, 11.3, 12.1, 12.9, 20.5

Minimum = 3.09 Maximum = 20.5

Sum = 5.6 + 8 + 3.09 + 12.9 + 6.5 + 12.1 + 11.3 + 20.5 + 7.4 = 87.39

Mean = 87.39 ÷ 9 = 9.71 litres

Median = 5th of 9 values = 8 litres

The justification.

Mean. The mean is a fair share of the total. If every one of the 9 days had used the
maximum, 20.5 litres, the total would be 9 × 20.5 = 184.5 and the mean would be 20.5. Since
no day used more than 20.5, the total cannot exceed 184.5 and the mean cannot exceed
20.5. And 20.5 is already below 25.
Median. The median is one of the values themselves (or the average of two of them). Since
every value lies between 3.09 and 20.5, the median must too.

(b) No — never. The mean and the median always lie between the minimum and the
maximum of the data (they may equal one of them, but cannot go outside).

Page 37 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Why it happens: for the mean, think of the fair share. Sharing the total equally
cannot give everyone more than the biggest single value, nor less than the smallest
— if it did, the shares would not add back up to the same total. For the median, the
reason is even simpler: it is chosen from the middle of the sorted list, so there is
always at least one value at or below it and at least one at or above it.

Check it yourself: here the mean 9.71 and the median 8 both sit comfortably inside
3.09 to 20.5. Note also that mean > median, which warns us of the high reading of
20.5 litres — perhaps a washing day.

Q5 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

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

Page 113 — the empty dot plot printed with this question, with its scale running from 0
to 4.5 kg.

Six boys and six girls. The printed line runs from 0 to 4.5 in steps of 0.5, so all twelve dots sit in
its right-hand part.

Page 38 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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Why it happens: with only 6 babies in each group, a difference of 0.05 kg is far too
small to mean anything. Moving one baby's weight by 0.3 kg would flip the result.
The strongest honest claim is that these boys and girls weighed about the
same at birth — roughly 3.4 kg each, between 2.5 kg and 4.1 kg.

Tip: whenever two averages differ by less than the natural variation inside each
group, say so. Claiming "boys are heavier" from a gap of 0.08 kg would be reading
far more into the data than it can support.

Q6 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? Compare the heights of the two sections.
Share your observations.

Mean = 141.21
Whole class Median = 142.5

125 130 135 140 145 150 155

Mean = 142.05
Boys
Median = 143

125 130 135 140 145 150 155

Mean = 140.14
Girls
Median = 140

125 130 135 140 145 150 155

Section 2 dot plots of height in cm, page 113. The solid line marks the mean and the
dashed line the median.

Note on the printed figure: in the book the ‘Whole class’ dot plot on page
113 is the Section 1 plot from page 109 printed again — its dots do not
match the mean 141.21 and median 142.5 given beside it. The boys’ and
girls’ plots are right, and taken together they do give a median of 142.5
and a mean of about 141.2.

Observations about this second section

Page 40 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

The girls' heights are far more spread out — their dots run from about 126 cm to about
158 cm. The boys are packed between about 130 cm and 148 cm.
So in this section both the tallest and the shortest student are girls.
The boys' dots pile up heavily between 141 and 146 cm — that thick stack is why the boys'
mean (142.05) sits above the girls' (140.14) even though the girls own both extremes.
For every group, mean < median (141.21 < 142.5; 142.05 < 143; 140.14 < 140 is the one
exception — here mean > median by a tiny 0.14 cm). So in the whole class and among the
boys, a few short students pull the mean slightly below the middle value.
In this section, the boys are taller on average than the girls.

Comparing the two sections

SECTION 1 (PAGE 109) SECTION 2 DIFFERENCE

Whole class mean 144.5 cm 141.21 cm Section 1 taller by ≈ 3.3 cm

Whole class median 145 cm 142.5 cm Section 1 taller by 2.5 cm

Boys' mean 142.94 cm 142.05 cm Almost the same

Girls' mean 146.9 cm 140.14 cm Section 1 girls taller by ≈ 6.8 cm

Taller group Girls Boys Reversed!

Widest spread Boys Girls Reversed!

Why it matters: the boys of the two sections are practically the same height on
average. Almost the whole 3.3 cm gap between the sections comes from the girls.
And notice that the answer to "are boys or girls taller?" flips from one section to the
other in the same school and the same grade.

The strongest thing you can say: in Section 1 the girls are taller on average, in
Section 2 the boys are. Since two classes of the same grade in the same school give
opposite answers, this data cannot be used to say anything about boys and girls
in general. A class of about 30 children is simply too small a group to settle a
question about everybody.

Page 41 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q7 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?

About 5.7 times — roughly 6 times heavier.

Sumo wrestlers (5 people):
295.2 + 250.7 + 234.1 + 221.0 + 200.9 = 1201.9 kg

Mean = 1201.9 ÷ 5 = 240.38 kg

Ballet dancers (6 people):

40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2 = 254.5 kg

Mean = 254.5 ÷ 6 ≈ 42.42 kg

How many times heavier = 240.38 ÷ 42.42 ≈ 5.67

Why we must use the means: the two groups have different sizes — 5 wrestlers
and 6 dancers — so their totals (1201.9 kg and 254.5 kg) cannot be compared
directly. Dividing each by its own count puts one wrestler against one dancer, which
is what the question asks.

A quick check: the lightest wrestler (200.9 kg) is still more than 4 times the heaviest
dancer (48.2 kg), and the heaviest wrestler (295.2 kg) is more than 7 times the
lightest dancer (37.6 kg). So the true ratio for any pair lies between about 4 and 7 —
and 5.7 sits sensibly in the middle. Notice too that neither group has an outlier, so
the means are trustworthy here.

In-text Questions — Pages 114 – 115

Page 42 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Section 5.3 Visualising Data — Clubbing the Columns

Q1 The two graphs can also be combined into a single graph. We just draw the bars side by
the data in the table matches the graph below.

JAN FEB MAR APR MAY JUN JUL AUG SEP OCT

Yahapur 25 24 26 28 30 35 39 43 49 56

Wahapur 19 17 23 30 38 35 42 39 53 60

Monthly onion prices in Yahapur and Wahapur
60

50

40
Price (in ₹)

30

20

10

0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct

Yahapur Wahapur

The table of monthly onion prices and the clustered column graph drawn from it — page

Yes, it matches. Check the two bars of each cluster against the two rows of the table, month by
month.

Page 43 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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MONTH Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

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YAHAPUR 25 24 26 28 30 35 39 43 49 59 44
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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Gridlines at 0, 10, 20, 30, 40, 50, 60

Gap between two gridlines = ₹10

Highest gridline needed = 60, because the largest price is ₹60 (Wahapur, October)

Why this scale: the prices run from ₹17 to ₹60. A scale of 1 unit = ₹10 needs only 6
gridlines, which keeps the graph short, and still lets you read a bar to the nearest
rupee or two by eye. A scale of 1 unit = ₹1 would need 60 lines; a scale of 1 unit = ₹50
would squash every bar into one or two units and hide all the differences.

Tip: always state the scale before you read any value off a bar graph. A bar reaching
halfway between the 30 and 40 lines means ₹35 on this scale — but ₹350 if the scale
were 1 unit = ₹100.

Q3 Is it now easier to compare month-wise prices in both places?

Yes — much easier.

The two bars of a month touch each other, so which town was dearer that month is settled
without measuring anything.
The size of the gap between the two bars shows by how much. In May the gap is large (₹8); in
June there is no gap at all.
Reading left to right, you can follow how the difference itself changes over the year: Yahapur
ahead in Jan–Mar, Wahapur ahead through most of the middle of the year, Yahapur ahead
again in November.

Why it happens: with two separate graphs your eye has to remember a bar height
while it travels to the other graph — and memory for heights is poor. Putting the
bars next to each other turns a memory task into a direct comparison. This is why
the arrangement is called a clustered column graph, or a double column graph
when there are two bars in each cluster.

What it still does not tell you: the graph shows each month separately, so the
yearly totals and averages are not visible. To answer "which town is dearer overall?"
you must go back to the numbers.

Page 45 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

In-text Questions — Pages 116 – 118
10…9…8…7…6…5…4…3…2…1…Take Off!

Q1 Look at the graph below showing the number of worldwide rocket launches by
different organisations. Share your observations (you may take the teacher's help
to identify the countries these organisations belong to).

Number of worldwide rocket launches by companies/space agencies

SpaceX

CASC

Roscosmos

Arianespace

Rocket Lab

United Launch Alliance

ISRO

Galactic Energy
2021
Expace 2022
2023
Other

0 20 40 60 80 100

Worldwide rocket launches by organisation for 2021, 2022 and 2023 — page 116.

The graph shows rocket launches for 2021, 2022 and 2023, with the bars drawn horizontally —
so the length of a bar, not its height, gives the number.

Page 46 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

ORGANISATION COUNTRY 2021 2022 2023

SpaceX USA ≈ 31 ≈ 61 ≈ 96

CASC China ≈ 48 ≈ 35 ≈ 45

Roscosmos Russia ≈ 16 ≈ 21 ≈ 18

Arianespace France ≈ 15 ≈6 ≈3

Rocket Lab USA ≈6 ≈6 ≈9

United Launch Alliance USA ≈5 ≈8 ≈3

ISRO India ≈2 ≈5 ≈7

Galactic Energy China ≈1 ≈2 ≈7

Expace China ≈4 ≈4 ≈6

Other many ≈ 18 ≈ 17 ≈ 26

SpaceX (USA) is far ahead of everyone, and it is growing fastest — its 2023 bar is longer
than the whole of the CASC cluster put together.
The organisations are already arranged from longest to shortest, so the ranking can be
read straight off.
China has three entries — CASC, Galactic Energy and Expace — and the USA has three as
well: SpaceX, Rocket Lab and United Launch Alliance.
ISRO (India) is the only Indian entry and its launches rose every year, from about 2 to about
7.
The Other bars are not small: about 26 launches in 2023, more than any single organisation
except SpaceX and CASC.

Careful: every number above is read off a bar, so it is an estimate, not an exact
count. The graph itself prints no numbers. Say "about 31" and not "exactly 31".

Page 47 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q2 Step 1: Identify what is given. Notice how the graph is organised, what scale is used,
and what patterns the data shows.

Number of worldwide rocket launches by companies/space agencies

SpaceX

CASC

Roscosmos

Arianespace

Rocket Lab

United Launch Alliance

ISRO

Galactic Energy
2021
Expace 2022
2023
Other

0 20 40 60 80 100

Worldwide rocket launches by organisation for 2021, 2022 and 2023 — page 116.

How it is organised. Each organisation gets three adjacent bars — one for 2021, one for 2022
and one for 2023 — shaded from dark to light. The organisations run down the left side, each
with its country's flag beside it.
The scale. 1 unit length = 20 rockets. The marks along the bottom read 0, 20, 40, 60, 80, 100.
The 'Other' category. It clubs together many smaller organisations from around the world, so
that the graph does not become impossibly long.
Patterns worth noticing.

Increasing year on year: SpaceX — its 2022 bar is almost double its 2021 bar.
Decreasing year on year: Arianespace — its 2023 bar is about half its 2022 bar.
Increase and then decrease: United Launch Alliance — up from 2021 to 2022, then down
sharply in 2023.
SpaceX's 2022 bar lands between the 60 and 80 marks, close to 60. Perhaps 61.

Why the order does not matter here: in the onion-price graph the months had to
stay in order — January to December — because the whole point was the change
over time. Here the organisations could be listed in any order and the meaning
would not change. They have simply been sorted by size to make reading easier.

Page 48 of 106

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Class 7 Maths Chapter 13 Connecting the Dots…
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The boy in the book has a fair complaint: with gridlines only every 20 rockets,

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small bars like ISRO's and Galactic Energy's are hard to read. Lines at 5, 10 and 15
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Why the two steps are kept separate: Step 1 is only about what your eyes can see
— bar lengths, order, scale. Step 2 is where you say what it means. Mixing them is
how people end up "seeing" things the graph never showed.

Q4 Identify which of the following statements can be justified using this data. (a) All
organisations launched more rockets than the previous years. (b) Only an
organisation from the USA launched more than 50 rockets in a single year. (c) The
total number of rockets launched by France in all 3 years is less than 40. (d) The
average number of rockets launched by CASC in these 3 years is around 40. (e) ISRO
launched more rockets than Galactic Energy in these 3 years. (f) Russia launched
more than 60 rockets in these 3 years.

STATEMENT JUSTIFIED? THE EVIDENCE IN THE GRAPH

(a) All organisations No Arianespace fell every year; CASC fell in 2022; United Launch
increased every year Alliance and Roscosmos fell in 2023. One counter-example is
enough to break a claim that says all.

(b) Only a USA Yes SpaceX (USA) passed 50 twice — about 61 in 2022 and about
organisation crossed 50 96 in 2023. The next longest bar anywhere is CASC's 2021 bar
in one year at about 48, which does not reach the halfway point between
the 40 and 60 marks.

(c) France's total over 3 Yes France's only entry is Arianespace: about 15 + 6 + 3 = 24,
years is less than 40 comfortably under 40.

(d) CASC averages Yes About 48 + 35 + 45 = 128, and 128 ÷ 3 ≈ 42.7 — which is
around 40 "around 40".

(e) ISRO launched more Yes ISRO ≈ 2 + 5 + 7 = 14; Galactic Energy ≈ 1 + 2 + 7 = 10. Their
than Galactic Energy 2023 bars are nearly equal, but ISRO's 2021 and 2022 bars are
clearly longer.

(f) Russia launched No Russia's only entry is Roscosmos: about 16 + 21 + 18 = 55,
more than 60 in 3 years which is below 60.

Why (f) needs care: 55 and 60 are close, and these are estimates read off bars with
gridlines 20 apart. Even so, all three Roscosmos bars end well short of the 20-mark
region needed to reach a total of 60, so the graph does not support the statement.
Note also that if any Russian launches are hidden inside the 'Other' bars, the graph
simply cannot tell us — another reason the claim is not justified.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Tip: statements (a) and (f) fail for different reasons. (a) is disproved by the graph —
we can point at Arianespace. (f) is merely unsupported — the graph gives about 55,
and it cannot rule out extra Russian launches sitting in 'Other'. "False" and "not
shown" are not the same verdict.

Q5 List the organisations that have consistently launched more rockets every year.

Reading the three bars of each cluster from dark to light, the ones that get longer every single
time are:

SpaceX — about 31 → 61 → 96
ISRO — about 2 → 5 → 7
Galactic Energy — about 1 → 2 → 7

ORGANISATION PATTERN ACROSS 2021 → CONSISTENT RISE?
2022 → 2023

SpaceX up, up Yes

ISRO up, up Yes

Galactic Energy up, up Yes

Expace level, then up No — 2021 and 2022 look equal

Rocket Lab about level, then up No — the 2021 and 2022 bars are too
close to call

Other slightly down, then up No

CASC, Roscosmos, up then down, or down then up No
ULA

Arianespace down, down No — it fell every year

Why "consistently" is a strong word: it means every step must be an increase, not
just the first and the last. Rocket Lab ended higher in 2023 than in 2021, but its
middle bar does not clearly rise, so it does not qualify.

Page 51 of 106

Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Where the graph runs out: Rocket Lab's and Expace's 2021 and 2022 bars are
almost exactly the same length. At this scale the graph honestly cannot separate
them, and the right answer is to say so rather than to guess.

Q6 Estimate the total number of rockets launched worldwide in 2023. (a) less than 200
(b) 200 to 400 (c) 400 to 600 (d) more than 600

(b) 200 to 400. Adding the ten lightest bars gives roughly 220.

SpaceX ≈ 96

CASC ≈ 45

Roscosmos ≈ 18

Arianespace ≈ 3

Rocket Lab ≈ 9

United Launch Alliance ≈ 3

ISRO ≈ 7

Galactic Energy ≈ 7

Expace ≈ 6

Other ≈ 26

Total ≈ 220

Why this is the safe answer: even a generous reading of every bar cannot push the
total past 250, and even a mean reading keeps it above 180. The only option that fits
comfortably is 200 to 400.

A quicker estimate: SpaceX alone is about 96 and CASC about 45 — already about
141. Every other bar is under 30, and there are eight of them, so the rest add at most
about 100 more. That puts the total between 200 and 250 without adding anything
carefully. Remember to include the Other bar — it is easy to skip, and it is one of the
bigger ones.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Q7 What are you curious to know after looking at this graph?

Number of worldwide rocket launches by companies/space agencies

SpaceX

CASC

Roscosmos

Arianespace

Rocket Lab

United Launch Alliance

ISRO

Galactic Energy
2021
Expace 2022
2023
Other

0 20 40 60 80 100

Worldwide rocket launches by organisation for 2021, 2022 and 2023 — page 116.

This is an open question. Here are questions this graph genuinely provokes, and what data each
would need.

Why does the USA launch so many more rockets than anyone else? Is it one company
doing it, or the whole country? (The graph already hints: SpaceX alone is nearly half the
world's launches in 2023.)
What happened before 2021 and after 2023? Was SpaceX always this far ahead, or is this a
recent jump? You would need the same graph for 2015–2020.
Why did CASC dip in 2022 and recover in 2023?
Why did Arianespace fall so sharply? Did it stop using an old rocket before the new one
was ready?
How many rockets does it take to put one satellite in orbit? A single rocket can carry
many satellites, so launches and satellites are not the same count.
Which organisations are inside 'Other'? Together they launched more than ISRO, Rocket
Lab and Expace combined.
How many launches failed? The graph counts launches, not successes.

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Why this matters: the last two questions point at what the graph hides. A category

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Math Talk — Pages 118 – 121
Summer and Winter at the Same Time
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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

City 1, June: 564 ÷ 30 = 18.8 hours per day

City 1, December: 186 ÷ 31 = 6.0 hours per day

City 2, June: 276 ÷ 30 = 9.2 hours per day

City 2, December: 468 ÷ 31 = 15.1 hours per day

Every answer lies between 0 and 24 hours a day — so the data can be about something that
happens for part of each day. Rain or wind would not follow such a smooth yearly curve, but
daylight would.

The clue that settles it: the two cities move in opposite directions. City 1 peaks in
June and bottoms in December; City 2 does exactly the reverse. Only something tied
to the Earth's tilt behaves like that — the two cities must be in opposite
hemispheres.

Tip: notice that dividing the monthly total by the number of days is itself finding an
average — the average daylight per day for that month. That is exactly how the
clustered bar graph in the book was drawn.

Q2 Does this give some idea of where these two cities are located?

Yes. The pattern tells us two things about their positions, before we are told any names.

Opposite hemispheres. City 1 has its longest days in June, so it is in the Northern
Hemisphere. City 2 has its shortest days in June, so it is in the Southern Hemisphere.
Far from the Equator. City 1 swings from about 19 hours of daylight a day in June to about 6
hours in December — a change of some 13 hours. Such a huge swing only happens well
away from the Equator. City 2 swings from about 9 to about 15 hours, a change of 6 hours,
so it is away from the Equator too, but not as far as City 1.

In June, City 1 has daylight for about 18.8 ÷ 24 ≈ 3/4 of the whole day.

In December, only about 6 ÷ 24 = 1/4 of the day.

The book confirms it: City 1 is Helsinki, Finland and City 2 is Wellington, New Zealand.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

Why the seasons flip: in June the Northern Hemisphere is tilted towards the Sun, so
days there are long and it is summer. At the same moment the Southern
Hemisphere is tilted away, so its days are short and it is winter. The Earth's tilt gives
the two hemispheres opposite seasons at the same time.

Did you know? Very close to the poles the tilt is so extreme that in midsummer the
Sun does not set at all — the Midnight Sun, as at Andøya in Norway.

Q3 Is there anything more that you wish to explore?

An open question. Here are directions this data really opens up.

Where does India fit? India lies much closer to the Equator, so its day lengths vary far less
— roughly 10½ to 13½ hours in Delhi. Collect sunrise and sunset times for your own town for
one year and compare the swing with Helsinki's.
What happens exactly at the Equator? Day and night stay close to 12 hours each, all year.
What happens above the Arctic Circle? The Sun never sets in midsummer and never rises
in midwinter — daylight of 24 hours a day, and of 0.
Do the two cities' yearly totals match? City 1: 210 + 257 + 372 + 441 + 536 + 564 + 555 +
465 + 394 + 310 + 222 + 186 = 4512 hours. City 2: 459 + 384 + 381 + 327 + 304 + 276 + 295 +
318 + 369 + 409 + 435 + 468 = 4425 hours. Both are close to half of 8760 hours in a year, as
you would expect.
Does long daylight mean hot weather? Not always — collect temperature data for the
same cities and see.
How does this affect people's lives? School timings, electricity use, farming, sports — all
shift with daylight.

Why the yearly totals are so alike: every place on Earth gets roughly half a year of
daylight in total. Being far from the Equator changes when you get it, not how much
you get.

In-text Questions — Page 122

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

All it Takes is a Minute

Q1 Answer the following questions based on the graph: Can we tell who batted first?
Who won the match?

15
Runs per over

10

5

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Overs

Runs scored in each over by the two teams, page 122. A circle above a bar marks an over
in which a wicket fell.

The graph does not say directly who batted first — but it can be worked out. The blue
team batted first, and the blue team won by 11 runs.
First read the bars, over by over. The scale is 1 unit = 5 runs.

OVER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

BLUE 2 11 8 7 9 6 3 4 9 8 6 15 6 9 4 10 9 13 17 7

RED 10 8 4 2 11 8 3 6 10 9 9 5 16 11 9 7 7 17 — —

Blue total = 2+11+8+7+9+6+3+4+9+8+6+15+6+9+4+10+9+13+17+7 = 163 runs in 20

overs

Red total = 10+8+4+2+11+8+3+6+10+9+9+5+16+11+9+7+7+17 = 152 runs in 18 overs

Who won? The blue team scored more runs, so the blue team won by 163 − 152 = 11 runs.
Who batted first? The graph does not label the innings — the two bars simply stand side by
side for each over. But two facts decide it.

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Class 7 Maths Chapter 13 Connecting the Dots… AglaSem · NCERT Solutions

The red team has no bars at all in overs 19 and 20. Its innings ended after 18 overs, which
in a 20-over match means the side was all out.
The blue team's score after 19 overs was 2+11+8+7+9+6+3+4+9+8+6+15+6+9+4+10+9+13+17
= 156, already more than the red team's 152. Yet the blue team went on to bat over 20. A
team chasing a target stops the moment it passes that target — so the blue team cannot
have been chasing.

The conclusion: the blue team batted first and made 163 in its full 20 overs. The red
team then chased 164, and was bowled out for 152 in 18 overs. The 7 wicket-circles
all sit above blue bars, in overs 1, 3, 6, 13, 18, 19 and 20 — so the blue team lost 7
wickets while making its 163.

Careful: "blue won" comes straight from the totals — that is proved. "Blue batted
first" is an inference from the missing red bars and from blue still batting in over 20.
It is a strong argument, but the graph never states it. Keep the two kinds of claim
apart.

Q2 How many runs did the blue team score in over 12?

15 runs.

Scale: 1 unit length = 5 runs, so the gridlines are at 0, 5, 10 and 15

In over 12 the blue bar reaches exactly the third gridline

Runs = 3 × 5 = 15

Why we can be sure: the top of the bar sits exactly on a printed gridline, so no
estimating is needed. A bar that ended between two gridlines would only give an
approximate value.

Tip: over 12 was the blue team's second-best over. Only over 19 was better, at about
17 runs. And notice that the red team scored just 5 runs in that same over 12 — a big
swing in blue's favour.

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Over

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Over 12 → 5 runs
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Careful: the red team has no bar at all in overs 19 and 20. That is not "0 runs" — the
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No, it is not easy. The graph shows the runs scored in each over, never the running total.

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