NCERT Solutions Class 8 Maths Chapter 12 Tales By Dots and Lines

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 8 · M AT H S

NCERT Solutions

Chapter 12: Tales by Dots and
Lines

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 103 – 133 20 80 English

Solutions, notes, sample papers & more at 78 pages

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 12: Tales by Dots and Lines
Chapter 5 of Ganita Prakash Grade 8 Part II looks at the mean and the median from a new angle — the mean
as the balance point of a dot plot, where the total distance on the left equals the total distance on the right. It
then turns to line graphs, infographics and activity strips, and asks what a picture of data honestly lets you
conclude.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) Part II, 103 – 133

SECTIONS QUESTIONS

20 80

MEDIUM

English

In-text Questions — Page 103
Section 5.1 The Balancing Act

Q1 Consider any 2 numbers. Find their average/arithmetic mean. Repeat this by taking
other pairs. What do you observe?

For two numbers the mean always lands exactly halfway between them.

3 and 7 → (3 + 7) ÷ 2 = 5, and 5 is 2 away from each

8 and 9 → (8 + 9) ÷ 2 = 8.5, and 8.5 is 0.5 away from each

40 and 100 → (40 + 100) ÷ 2 = 70, and 70 is 30 away from each

Why it happens: take any two numbers p and q with p < q. Their mean is (p + q) ÷ 2.
Its distance from p is (p + q)/2 − p = (q − p)/2, and its distance from q is q − (p + q)/2 =
(q − p)/2. The two distances are the same expression, so the mean must sit at the
midpoint.

Tip: this is the first hint of the big idea of this section — the mean is the point that
balances the data. With two numbers, balancing simply means standing midway.

Page 1 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q2 Calculate and mark the mean of each collection of data below.

Reading the four dot plots on page 103 and computing each mean:

DOT PLOT VALUES SUM MEAN

Top left (blue) 6, 7, 8 21 21 ÷ 3 = 7

Top right (blue) 3, 6, 9 18 18 ÷ 3 = 6

Bottom left (yellow) 2, 4, 9 15 15 ÷ 3 = 5

Bottom right (yellow) 4, 11, 15 30 30 ÷ 3 = 10

Mark each mean with a small cross on the number line of that plot: at 7, 6, 5 and 10.

Check it yourself: in the first plot the mean 7 is also the middle dot, but in the third
plot the mean 5 is not one of the dots at all, and it is not the midpoint of 2 and 9
either. So ‘centre’ here must mean something other than ‘middle dot’ — that is
exactly what page 104 sorts out.

In-text Questions — Page 104
Section 5.1 The Balancing Act

MATH TALK

Q1 Can you explain how the mean is the centre of each collection?

Measure how far each value is from the mean, and add the distances up on each side. The two
totals come out equal.

Page 2 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

COLLECTION MEAN DISTANCES ON THE LEFT DISTANCES ON THE RIGHT

6, 7, 8 7 1 (from 6) 1 (from 8)

3, 6, 9 6 3 (from 3) 3 (from 9)

2, 4, 9 5 3+1=4 4 (from 9)

4, 11, 15 10 6 (from 4) 1+5=6

Why it happens: think of the number line as a see-saw with a dot of equal weight at
each value. The mean is the point where the see-saw balances, because the pull of
everything on the left exactly cancels the pull of everything on the right. In the third
collection, 2 and 4 are close in but there are two of them (3 + 1 = 4), while the single
value 9 is far out (4) — small distances in bulk balance one large distance.

Q2 Mark the mean for the collections below.

Reading the four dot plots and computing:

DOT PLOT VALUES SUM MEAN

Top left (orange) 11, 13, 17, 19 60 60 ÷ 4 = 15

Top right (orange) 5, 6, 15, 16 42 42 ÷ 4 = 10.5

Bottom left (green) 10, 10, 11, 17 48 48 ÷ 4 = 12

Bottom right (green) 3, 5, 10, 12 30 30 ÷ 4 = 7.5

Mark a cross at 15, 10.5, 12 and 7.5. Notice that in two of these the mean falls between the
marked numbers, not on one of them.

Q3 Can you explain how the mean is the centre of each collection?

Again, add up the distances on either side of the mean:

Page 3 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co m
em.
11, 13, 17, 19 with mean 15 → LHS = 4 + 2 = 6, RHS = 2 + 4 = 6 ✓
m l as
.co
5, 6, 15, 16 with mean 10.5 → LHS = 5.5 + 4.5 = 10, RHS = 4.5 + 5.5 = 10 ✓
m a g
l a se
g
10, 10, 11, 17 with mean 12 → LHS = 2 + 2 + 1 = 5, RHS = 5 ✓
a3, 5, 10, 12 with mean 7.5 → LHS = 4.5 + 2.5 = 7, RHS = 2.5 + 4.5 = 7 ✓

co m
m . ag
l a se
Why it happens: the third collection is the interesting one. Its mean, 12, is nowhere
ag 10 and 17 (that would be 13.5). It sits low because
near the midpoint of the extremes
three of the four values are down near 10 and only one is out at 17. The balance is 1

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+ 2 + 2 on the left against 5 on the right — three short pulls against one long one.

m as e
.co a g l
a s em
a glQ4 Verify that this holds for all the collections of data shown earlier.
m a s
.co agl

se m
l a
It holds for every one of the eight collections on pages 103 and 104:
g
a
COLLECTION MEAN LHS TOTAL RHS TOTAL

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6, 7, 8 7 1 1

3, 6, .9co
m l a
ag 3
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6 3

a 2, 4, 9
agl 5 3+1=4 4

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4, 11, 15 10 6 1+5=6

. a g
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11, 13, 17, 19 15 4+2=6 2+4=6

5, 6, 15, 16 agl10.5 5.5 + 4.5 = 10 4.5 + 5.5 = 10

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10, 10, 11, 17 12 2+2+1=5 5

e m
m l as
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7.5
a
3, 5, 10, 12 4.5 + 2.5 = 7 2.5 + 4.5 = 7

se m
g l a
a Why it always works: let the mean be a. Then x1 + x2 + … + xn = na, so (x1 − a) + (x2
c
m .
− a) + … + (xn − a) = na − na = 0. The values above a give positive terms, the values
m a s e
m . co
below a give negative terms, and since the whole lot adds to zero the two groups
e agl
l as
must be equal in size. That is precisely ‘LHS total = RHS total’.
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a

co m
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m as e
.co


a g l Page 4 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q5 Can there be more than one such ‘centre’? In other words, is there any other value
such that the sum of the distances to the values lower than it and the values higher
than it will still be equal?

No — there is exactly one such point, and it is the mean.

Why it happens: suppose you slide the balance point from the mean a to a + d, with
d > 0. Every value below the new point is now d further away, and every value above
it is d closer. So the left total goes up and the right total goes down — they can no
longer be equal. Sliding to a − d does the opposite. Since moving either way spoils
the balance, only one point can balance the data.

Tip: in algebra: (x1 − c) + (x2 − c) + … + (xn − c) = na − nc, which is zero only when c = a.

Q6 In the case of the collection 10, 10, 11, and 17 whose mean is 12, suppose there is a
different centre larger than 12.

2
2
1 5

Mean = 12

9 11 13 15 17

LHS = 2 + 2 + 1 = 5 RHS = 5

The dot plot for this collection on page 104, with the distances marked on each side of
the mean.

Take the trial centre to be 13 instead of 12 and measure again.

At 12 → LHS = 2 + 2 + 1 = 5, RHS = 5 ✓ balanced

At 13 → LHS = 3 + 3 + 2 = 8, RHS = 4 ✗ not balanced
At 11 → LHS = 1 + 1 + 0 = 2, RHS = 6 ✗ not balanced

Page 5 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Moving up by 1 made all three left-hand distances grow by 1 each (a gain of 3) and the single
right-hand distance shrink by 1. The totals part company at once.

Why it happens: when you move the trial centre up by d, the left total grows by
(number of values on the left) × d and the right total falls by (number of values on
the right) × d. They can only stay equal if d = 0. So 12 is the only centre.

In-text Questions — Page 105
Section 5.1 The Balancing Act

MATH TALK

Q1 Will including a new value in the data increase or decrease the mean?

It depends entirely on where the new value sits compared with the present mean.

New value greater than the mean → the mean increases.
New value less than the mean → the mean decreases.
New value equal to the mean → the mean does not change.

Old data 4, 6, 8 → mean 6

Include 12 → (4 + 6 + 8 + 12) ÷ 4 = 30 ÷ 4 = 7.5 (up)

Include 2 → (4 + 6 + 8 + 2) ÷ 4 = 20 ÷ 4 = 5 (down)

Why it happens: with n values and mean a, adding x gives a new mean of (na + x) ÷
(n + 1). Subtracting the old mean,
(na + x)/(n + 1) − a = (na + x − na − a)/(n + 1) = (x − a) ÷ (n + 1).
So the shift has the same sign as x − a. In balance language: a value dropped on the
right makes that side heavier, so the balance point has to slide right until the
distances even out again.

Tip: the formula also tells you how much the mean moves — the gap x − a shared out
over n + 1 values. A big class barely notices one new value; a small group feels it a
lot.

Page 6 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q2 What happens to the mean when an existing value is removed? When will the mean
increase, decrease, or stay the same?

Removing a value does the opposite of adding it.

Remove a value greater than the mean → the mean decreases.
Remove a value less than the mean → the mean increases.
Remove a value equal to the mean → the mean stays the same.

Data 4, 6, 8, 12 → mean 7.5

Remove 12 → (4 + 6 + 8) ÷ 3 = 6 (down)

Remove 4 → (6 + 8 + 12) ÷ 3 = 26 ÷ 3 ≈ 8.67 (up)

Why it happens: with n values and mean a, removing x leaves (na − x) ÷ (n − 1). The
change works out to (a − x) ÷ (n − 1) — the mirror image of the ‘adding’ formula.
Taking away weight from the heavy side lets the balance point slide back towards
the other side.

Q3 What happens to the mean if a value equal to the mean is included or removed? Try
to explain this using the fair-share interpretation of mean that we studied last year.

Nothing changes — the mean stays exactly where it was.

Data 4, 6, 8, 12, mean 7.5

Include 7.5 → (30 + 7.5) ÷ 5 = 37.5 ÷ 5 = 7.5

Remove 7.5 from that new list → 30 ÷ 4 = 7.5

Page 7 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Fair-share explanation: the mean is what each person would get if everything were
pooled and shared equally. Suppose five friends pool their money and each ends up
with ₹7.5. Now a sixth friend joins carrying exactly ₹7.5 — she already has her fair
share, so nothing has to be passed to her and nothing has to be taken from her. The
equal share stays ₹7.5. In the balance picture, a dot placed exactly at the balance
point has zero distance on either side, so it tips nothing.

In-text Questions — Pages 105–106
Unchanging Mean!

Q1 Explore if it is possible to include or remove 2 values such that the mean is
unchanged. You may use the following data to experiment with.

Yes. Reading the dot plot on page 106, the data is

2.5, 5, 6.5, 7, 7.5, 8, 8, 8, 8, 9, 10, 10.5, 11, 12, 12, 13, 15
Number of values = 17, sum = 153, mean = 153 ÷ 17 = 9

To keep the mean at 9 while adding two values, the two must be the same distance below and
above 9 — that is, they must add up to 18.

Include 6 and 12 → new sum = 153 + 18 = 171, new count = 19

New mean = 171 ÷ 19 = 9 ✓

Include 4 and 14, or 8 and 10, or 9 and 9 → all give 9 again

Removing works the same way: take out 8 and 10 (they add to 18), leaving a sum of 135 over 15
values, and 135 ÷ 15 = 9.

Why it happens: two new values x and y change the total by x + y and the count by
2. The mean survives only if the extra total is exactly two average-sized shares, i.e. x
+ y = 2 × 9 = 18. In balance language, one dot pulls left and the other pulls right by
an equal amount, so the see-saw does not tip.

Page 8 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co m
m.
How about including or removing 3 values without changing the mean? Is it
e
Q2

m l as
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possible?

a g
sem
g l a
a

Yes — three new values keep the mean at 9 as long as they add up to 3 × 9 = 27.

. c om ag
s e m count = 20
Include 7, 9 and 11 → sum = 153 + 27 = 180,

New mean = 180 ÷ 20 = 9 ✓ agl
a

m
Include 2, 10 and 15 → 2 + 10 + 15 = 27 → mean still 9 ✓
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se m.
o m l a
g share of 9 on
Why it.chappens: each extra value must bring exactly its ownafair
m
se They need not each equal 9 — only their total must equal 27, because then
g l a
average.
a the extra total (27) and the extra count (3) are in the same ratio as the old total to the

s
old count.
m a
m .co agl
l a se
ag
Tip: the general rule is now clear — including k values leaves the mean unchanged
exactly when those k values add up to k × (the mean).

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m .
m as e
Q3 .co
a g l
em that the mean remains the same?
Can we include 2 values less than the mean and 1 value greater than the mean, so

l a s
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Yes. The two small values pull the mean down;
m
the one large value must pull it up by the same
a
total amount.
l a se
ag
m
Take 5 and 6 (below 9): shortfalls are 9 − 5 = 4 and 9 − 6 = 3, total 7 short

. co
So the third value must be 7 above the mean → 9 + 7 = 16
em
m l as
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Check: 5 + 6 + 16 = 27 = 3 × 9 ✓
m New mean = (153 + 27) ÷ 20 = 180 ÷ 20 = 9 ✓ a g
l a se
ag
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s e m
. com a gla
Why it happens: what matters is not how many values sit on each side, but the total
distance. Two values 4 ande3mbelow the mean can be balanced by a single value 7
g l as rule from page 104 applied to the newcomers alone.
a
above it — exactly the balance

co m
m .
m ase
.co


a g l Page 9 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

The book’s own example: the green dots added in the figure are at 7, 7 and 13. The
two 7s are each 2 below the mean (total 4) and 13 is 4 above it, so the pulls cancel —
and 7 + 7 + 13 = 27 = 3 × 9, exactly as the rule requires.

Q4 Try to include 2 values greater than the mean and 1 value less than the mean, so
that the mean stays the same.

Mirror the previous idea: the two large values must be balanced by one small one.

Take 11 and 13 (above 9): excesses are 2 and 4, total 6 extra

So the third value must be 6 below the mean → 9 − 6 = 3

Check: 11 + 13 + 3 = 27 = 3 × 9 ✓

New mean = (153 + 27) ÷ 20 = 9 ✓

Another set that works: 10 and 12 (excesses 1 and 3, total 4) together with 9 − 4 = 5, since 10 +
12 + 5 = 27.

Check it yourself: pick any two numbers above 9, add up how far above they are,
and go that far below 9 for the third. It will work every time.

In-text Questions — Pages 106–107
Relatively Unchanged!

Q1 We saw what happens to the mean when values are included or removed from the
collection. What happens to the mean if every value in the collection increases by
some fixed number?

The mean increases by that same fixed number.

Data 2, 4, 6 → mean 4

Add 5 to each → 7, 9, 11 → mean = 27 ÷ 3 = 9 = 4 + 5 ✓

Page 10 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Why it happens: the whole dot plot slides bodily along the number line. The gaps
between the dots are untouched, so the point that balanced them before still
balances them — it has just moved along with everything else. Nothing about the
shape of the data changed, only its position.

Q2 Consider the data: 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Calculate its mean.

Sum = 8 + 3 + 10 + 13 + 4 + 6 + 7 + 7 + 8 + 8 + 5

= 11 + 10 + 13 + 4 + 6 + 7 + 7 + 8 + 8 + 5

= 79

Number of values = 11

Mean = 79 ÷ 11 = 7.18 (to two decimal places)

Tip: 79 ÷ 11 = 7.1818…, a repeating decimal. The book rounds it to 7.18, which is
what the dot plot on page 106 shows.

Q3 Now, consider this data with every value increased by 10: 18, 13, 20, 23, 14, 16, 17, 17,
18, 18, 15. What is its mean? Is there a quicker way to find out?

The mean is 17.18, and yes — there is a much quicker way.

Quick way: every value went up by 10, so the mean goes up by 10.

New mean = 7.18 + 10 = 17.18

Long way (as a check): new sum = 79 + (11 × 10) = 79 + 110 = 189

New mean = 189 ÷ 11 = 17.18 ✓

Page 11 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Why it happens: adding 10 to each of 11 values adds 11 × 10 = 110 to the total.
Dividing that extra 110 among the same 11 values gives each an extra 10. On the dot
plot the whole pattern has shifted 10 places to the right, and so has the balance
point — the mean sits in exactly the same relative position among the dots.

In-text Questions — Page 107
Relatively Unchanged!

TRY THIS

Q1 Try to explain, using algebra, what the average is when a fixed number, e.g., 2 is
subtracted from every value in the collection.

The average goes down by 2.

Let the values be x1, x2, …, xn with

(x1 + x2 + … + xn) ÷ n = a

New average = [(x1 − 2) + (x2 − 2) + … + (xn − 2)] ÷ n

= (x1 + x2 + … + xn − 2n) ÷ n

= (x1 + x2 + … + xn) ÷ n − 2n ÷ n

=a−2

In general, subtracting a fixed number c from every value lowers the average by exactly c.

Check it yourself: the data 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5 has mean 7.18. Reduce every
value by 1 and the mean becomes 6.18 — which is the same as the shift the book
shows when the +10 data (mean 17.18) is reduced by 1 to give 16.18.

Page 12 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q2 Try to explain this using the fair-share interpretation of average that you learnt last
year.

Picture the values as amounts of money held by n friends, sharing fairly to get ₹a each.

If everyone pays ₹2 to a shopkeeper, every purse is ₹2 lighter. Pooling and re-sharing now
gives ₹(a − 2) each.
If instead everyone receives ₹3, the fair share rises to ₹(a + 3).

Why it happens: a fair share is fair for everybody at once. Treating all n friends
identically cannot create any new inequality — it just shifts the whole pool up or
down by the same amount per person, so the share shifts by that amount too.

Q3 What happens to the average if every value in the collection is doubled?

The average is doubled too.

Data 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5 → mean 7.18

Doubled: 16, 6, 20, 26, 8, 12, 14, 14, 16, 16, 10

New sum = 2 × 79 = 158, new mean = 158 ÷ 11 = 14.36 = 2 × 7.18 ✓

In algebra, for a multiplier of 5:

[(5x1) + (5x2) + … + (5xn)] ÷ n

= 5(x1 + x2 + … + xn) ÷ n (distributive property)

= 5 × [(x1 + x2 + … + xn) ÷ n] = 5a

Why it happens: multiplying stretches the dot plot away from 0 rather than sliding
it. Every distance from 0 is scaled by the same factor, so the balance point is scaled
by that factor as well. This is the rule you use whenever you change units — heights
in centimetres to millimetres, prices in rupees to paise.

Page 13 of 78

as e
Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co m
m.
In-text Questions — Page 108
m as e
Tinkering with Median
.co a g l
a s em
aQ1gl Will including a new value to the data increase or decrease the median?

co m
. ag

e m
l as
The data shown in the dot plot is 4, 7, 8, 12, 14, whose median is the middle value, 8.
g
a
m
Include 11 (a value greater than 8):
co
em.
as
Sorted: 4, 7, 8, 11, 12, 14 → six values, so the median is the average of the two middle ones
m+ 11) ÷ 2 = 9.5 (it increased)
Median.c=o(8 g l
em a
a s
agl
Include 5 instead (a value less than 8):
m a s
Sorted: 4, 5, 7, 8, 12, 14
m .co agl
l a se
g
Median = (5 + 7) ÷ 2 = 6 (it decreased)
a
So a new value greater than the median pushes the median up, and one less than the median
co m
m .
e
pushes it down — but only by a little, and it can never jump past the neighbouring value.

m l as
.co a g
a s em Why it happens: the median only cares about position. Adding a value above 8 puts

agl one more number on the high side, so 8 is no longer in the middle — the middle
m
slides one place towards the larger numbers. Notice the crucial difference from the

a se
com l
mean: it does not matter whether you add 11 or 1100, the median moves to the
. a g
m
ase
same place, because the median never asks how far a value is, only which side it is on.

agl
Tip: that is why the median is the safer summary when a data set has one wild value

co m
.
— a single crorepati in a village changes the mean income a lot and the median

em
as
income hardly at all.
m l
m.co a g
l a se
ag
In-text Questions — Page 109 .c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 14 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Finding the Unknown

Q1 Coach Balwan noted down the weights of the kushti players (wrestlers) and the
mean as shown. But one value that was written down got smudged. Can you find
out the missing value?

The recorded weights are 42, 40, 39, 33, 48, 38, 42, 35, 32 and one smudged value w, and the
mean of all 10 is 39.2 kg.

Sum of the nine known weights

= 42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 = 349

(349 + w) ÷ 10 = 39.2

349 + w = 39.2 × 10 = 392

w = 392 − 349 = 43

The missing weight is 43 kg.

Why it works: the mean tells you the total in disguise. If ten players average 39.2
kg, they must weigh 392 kg between them. Subtract the nine weights you can read
and only the smudged one is left. Working backwards from the mean to the total is
the key move in every ‘find the unknown’ problem of this kind.

Q2 Venkayya keeps track of the coconut harvest in his farm. He calculates the average
harvest per tree as 25.6. His son verifies the counts and finds that one tree’s harvest
count is incorrectly noted as 3 more than the actual number. Can you find the
correct average if the number of trees is 15?

The correct average is 25.4 coconuts per tree.

Wrong total = average × number of trees = 25.6 × 15 = 384

One tree was counted 3 too many, so the correct total = 384 − 3 = 381

Correct average = 381 ÷ 15 = 25.4

Page 15 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Why it works: you never need the individual tree counts. The average multiplied by
the number of trees recovers the total, and an error of 3 in one tree is an error of 3
in the total. Spreading that mistake of 3 over 15 trees changes the average by 3 ÷ 15
= 0.2, which is exactly the drop from 25.6 to 25.4.

Tip: this is the ‘adding a value’ formula in another guise — an error of e in the total
shifts the mean by e ÷ n. The more trees, the less one miscount matters.

In-text Questions — Page 110
Mean and Median with Frequencies

Q1 What is the average family size of students in your class? How would you find this
out?

Collect the data first, then use a frequency table rather than a long list.

1. Ask every student how many members there are in their family and note each answer.
2. Make a table of family size against how many students gave that answer — this is the
frequency.
3. Multiply each family size by its frequency, add these products to get the total number of
family members counted.
4. Divide by the total number of students.

Average family size = (sum of all the values) ÷ (number of values)

= Σ(family size × frequency) ÷ Σ(frequency)

Tip: the frequency table is not just tidier — it is what stops you from adding each
family size only once when several students gave the same answer.

Q2 What is the average family size of this class?

5.22 members — not 6.5.

Page 16 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

FAMILY SIZE 3 4 5 6 7 8 9 10

FREQUENCY 3 11 9 7 3 1 1 1

SIZE × FREQUENCY 9 44 45 42 21 8 9 10

Total number of students = 3 + 11 + 9 + 7 + 3 + 1 + 1 + 1 = 36

Total of all family sizes = 9 + 44 + 45 + 42 + 21 + 8 + 9 + 10 = 188

Average = 188 ÷ 36 = 5.22

Why 6.5 is wrong: (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) ÷ 8 = 6.5 averages the eight different
sizes, as though one student had each. But eleven students have a family of 4 and
only one has a family of 10. The mean must count 4 eleven times over. Because the
common sizes are the small ones, the true average is pulled well below 6.5.

Q3 What is the median family size of this class?

5 members.

There are 36 values, so the median is the average of the 18th and 19th values in order.

Running (cumulative) totals:

up to 3: 3 → positions 1–3

up to 4: 3 + 11 = 14 → positions 4–14

up to 5: 14 + 9 = 23 → positions 15–23

The 18th and 19th values both fall in that block, so both are 5.

Median = (5 + 5) ÷ 2 = 5

Mean 5.22 but median 5: the mean is dragged slightly up by the few large families
(8, 9 and 10 members), while the median simply reports where the middle student
stands and ignores how large those families are.

Page 17 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q4 Do we need to write all the 36 numbers in order? Is there a quicker way to find out?

No — the frequency table already holds the order. Add the frequencies from the smallest
value upwards until you pass the middle position.

FAMILY SIZE FREQUENCY CUMULATIVE FREQUENCY POSITIONS IT OCCUPIES

3 3 3 1 to 3

4 11 14 4 to 14

5 9 23 15 to 23

6 7 30 24 to 30

7 3 33 31 to 33

8 1 34 34

9 1 35 35

10 1 36 36

Positions 18 and 19 lie in the row for 5, so the median is 5 — found without writing a single one
of the 36 numbers.

Tip: the cumulative frequency column answers “how many students have a family of
this size or smaller?”. Once you can read positions off it, finding a median in data of
any size becomes a matter of one column of additions.

In-text Questions — Pages 112–113
Spreadsheets

Q1 Can you tell which cell has the marks obtained by Farooq in Mathematics?

Cell E5.

Page 18 of 78

as e
Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co m
e m.
Columns: A = Name, B = Odia, C = Telugu, D = English, E = Maths, F = Social Science, G
m l as
= Science
m .co a g
l a se
g
Rows: 1 = the headings, 2 = Ratna, 3 = Nagesh, 4 = Ashwin, 5 = Farooq
aColumn E, row 5 → E5, which holds 42

com
m . ag
l a se
Tip: a cell name is always column letter first, then row number. Read across to find
agnumber, and the two meet at the cell you want.
the letter and down to find the

co m
em.
m l as
.co a g
Q2 Can you tell what data is in column B7?

a s em
a l
gANSWER
Cell B7 holds Gowri’s marks in Odia, which are 27.
m a s
m.co agl
l a se
g
Column B = Odia
a
Row 7 → row 2 Ratna, 3 Nagesh, 4 Ashwin, 5 Farooq, 6 Mrinal, 7 Gowri

co m
.
So B7 = Gowri’s Odia marks = 27
e m
m l as
.co a g
a s em
a gl Q3 In which subjects has Ashwin scored more than 30 marks?
se m
com g l a
m . a
ase
agl
Ashwin’s row (row 4) reads:

ODIA TELUGU ENGLISH MATHS SOCIAL SCIENCE SCIENCE

co m
m .
ase
29 31 33 34 30 28

c o m gl
. a
a s em
He scored more than 30 in Telugu (31), English (33) and Mathematics (34).

agl c
Careful: Social Science is exactly 30, which is not more than 30, so it does not belong
m .
m a s e
co agl
in the list.

m .
as e
a g l

co m
m .
m ase
.co


a g l Page 19 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q4 What formula would you type to find out the class average marks in Science?

Science is column G, and the 22 students occupy rows 2 to 23. So type

=AVERAGE(G2:G23)

Working it out by hand as a check: the Science marks add up to 729, and 729 ÷ 22 = 33.14 (to
two decimal places).

Why the range matters: G2:G23 names the first and last cell of the block you want.
If you typed G1:G23 the heading ‘Science’ would be included; if you typed G2:G22
you would silently leave Jyothi out. A spreadsheet will not warn you — it simply
averages whatever you point it at.

Q5 Find out if the class average marks in Odia is greater than the class average marks
in Telugu.

No — Odia’s average is lower than Telugu’s.

=AVERAGE(B2:B23) → Odia total 687, average = 687 ÷ 22 = 31.23
=AVERAGE(C2:C23) → Telugu total 739, average = 739 ÷ 22 = 33.59

31.23 < 33.59, so the Telugu average is higher by about 2.36 marks

Tip: you can let the spreadsheet answer the comparison itself by typing
=AVERAGE(B2:B23)>AVERAGE(C2:C23), which returns FALSE.

Q6 Show the average marks in other subjects after the last row by typing the
appropriate formulae.

The last student is in row 23, so put the averages in row 24. In B24 type =AVERAGE(B2:B23) and
drag it across to G24 — the column letter updates by itself.

Page 20 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

SUBJECT FORMULA IN ROW 24 TOTAL OF 22 MARKS CLASS AVERAGE

Odia =AVERAGE(B2:B23) 687 31.23

Telugu =AVERAGE(C2:C23) 739 33.59

English =AVERAGE(D2:D23) 718 32.64

Maths =AVERAGE(E2:E23) 751 34.14

Social Science =AVERAGE(F2:F23) 690 31.36

Science =AVERAGE(G2:G23) 729 33.14

The class did best in Mathematics (34.14) and weakest in Odia (31.23), though the six subject
averages are all within about three marks of one another.

Q7 Get the total scores of each student by typing the appropriate formulae.

Use a new column H headed ‘Total’. In H2 type =SUM(B2:G2) and drag it down to H23.

STUDENT TOTAL STUDENT TOTAL STUDENT TOTAL

Ratna 200 Aishwarya 273 Shanker 232

Nagesh 250 Hari 147 Vyshnavi 197

Ashwin 185 Trupti 188 Govind 103

Farooq 266 Veeresh 148 Shiva 177

Mrinal 194 Vidhya 213 Tarun 246

Gowri 183 Sanskruti 242 Jyothi 183

Pankaj 112 Jaya 248 Ganesh 225

Shravan 102

The highest total is Aishwarya’s 273 out of 300 and the lowest is Shravan’s 102.

Page 21 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Why a spreadsheet is worth learning here: that is 22 sums of six numbers each,
plus six averages of 22 numbers — around 150 additions by hand. Typing two
formulae and dragging them does the lot, and if a single mark is corrected every
total and every average updates on its own.

Figure it Out — Pages 113–116
Section 5.1 The Balancing Act

MATH TALK

Q1 Find the mean of the following data and share your observations: (i) The first 50
natural numbers. (ii) The first 50 odd numbers. (iii) The first 50 multiples of 4.

All three lists are evenly spaced, so in each case the mean is simply the average of the first and
last terms.

(i) 1, 2, 3, …, 50

Sum = 50 × 51 ÷ 2 = 1275

Mean = 1275 ÷ 50 = 25.5 (= (1 + 50) ÷ 2)

(ii) 1, 3, 5, …, 99

Sum of the first 50 odd numbers = 502 = 2500
Mean = 2500 ÷ 50 = 50 (= (1 + 99) ÷ 2)

(iii) 4, 8, 12, …, 200

Sum = 4 × (1 + 2 + … + 50) = 4 × 1275 = 5100

Mean = 5100 ÷ 50 = 102 (= (4 + 200) ÷ 2)

Observations

For every one of these lists the mean is the midpoint of the smallest and largest value,
because the dots are spread symmetrically about the middle.
The mean of the first n odd numbers is n itself — here 50.

Page 22 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

(iii) is (i) with every value multiplied by 4, and sure enough its mean is 4 × 25.5 = 102 — the
scaling rule from page 108.
In (i) and (iii) the mean is not even a member of its own list — 25.5 is not a natural number
and 102 is not a multiple of 4. A mean need not be one of the data values.

Tip: for any evenly spaced list you can pair the terms from the outside in — 1 with
50, 2 with 49, and so on. Every pair has the same sum, so the average of the whole
list is the average of any one pair.

Q2 The dot plot below shows a collection of data and its average; but one dot is
missing. Mark the missing value so that the mean is 9 (as shown below).

0 2 4 6 8 10 12 14 16

The dot plot printed on page 113. The vertical line marks the average of the collection.

The missing value is 16.
Reading the ten dots that are printed: one at 4, one at 7, two at 8, five at 9 and one at 11.

Sum of the printed dots = 4 + 7 + (8 × 2) + (9 × 5) + 11

= 4 + 7 + 16 + 45 + 11 = 83

With the missing dot there are 11 values, and the mean must be 9, so

Total needed = 11 × 9 = 99

Missing value = 99 − 83 = 16

Page 23 of 78

as e
Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co m
m.
Balance check: measure the distances from 9. Below: 5 (from 4), 2 (from 7), 1 + 1

m must sit 7 above the mean: 9 + 7 = 16. ✓ as e
l
(from the two 8s) → total 9. Above: 2 (from 11) → total 2. The left side is 7 heavier, so
. codot a g
em
the missing
a s
agl

co m
ag
Sudhakar, the class teacher, asks Shreyas to measure the heights of all 24 students
.
Q3

e m
as
in his class and calculate the average height. Shreyas informs the teacher that the

a g l
average height is 150.2 cm. Sudhakar discovers that the students were wearing
uniform shoes when the measurements were taken and the shoes add 1 cm to the

com
height. (i) Should the teacher get all the heights measured again without the shoes

.
to find the correct average height? Or is there a simpler way? (ii) What is the correct
m 149.2 cm (e)
average height of the class? (a) 174.2 cm (b) 126.2 cm (c) 150.2 cme(d)
m a s
. co cm (f) None of the above (g) Insufficient informationagl
em
151.2

g l as
a ANSWER
m
(i) There is no need to measure anyone again. Every single reading is exactly 1 cm too large,
a s
m .co
because every student wore the same shoes. Subtracting a fixed number from every value
agl
l a se
lowers the mean by that same number, so simply take 1 cm off the average.

ag
m
Correct average = 150.2 − 1 = 149.2 cm
. co
e m
m l as
m .co
The long way, as a check:
a g
a s eMeasured
agl total = 150.2 × 24 = 3604.8 cm

Total of the 24 shoe-thicknesses = 24 × 1 = 24 cm
se m
com g l a
.
Correct total = 3604.8 − 24 = 3580.8 cm
m a
Correct average = 3580.8 ÷ 24a=se149.2 cm ✓
agl
(ii) Option (d), 149.2 cm.
co m
m .
m as e
.co g l
Why it happens: this is the ‘subtract a constant’ rule of page 107 doing real work.
a
a s em The error is systematic — the same for everybody — so it shifts the whole dot plot 1

agl cm to the right without changing its shape, and the balance point shifts with it. Had
.c
m
the shoes been of different thicknesses, this shortcut would not be available and the

m a s e
co agl
heights really would need measuring again.

m .
as e
a g l

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m .
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.co


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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q4 The three dot plots below show the lengths, in minutes, of songs of different
albums. Which of these has a mean of 5.57 minutes? Explain how you arrived at the
answer.

Album A.
Reading the dots off the three plots (the scale is marked in steps of 0.5, and dots also sit halfway
between two marks):

A: 5, 5, 5.25, 5.5, 5.75, 6, 6.5 → 7 songs

Sum = 5 + 5 + 5.25 + 5.5 + 5.75 + 6 + 6.5 = 39

Mean = 39 ÷ 7 = 5.571… ≈ 5.57 minutes ✓

B: 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5 → 8 songs

Sum = 19.25, mean = 19.25 ÷ 8 = 2.41 minutes

C: 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5 → 8 songs

Sum = 31.25, mean = 31.25 ÷ 8 = 3.91 minutes

A quicker way to reach the same answer: the mean must lie inside the range of
the data. Every dot in B sits between 0.5 and 5, and every dot in C between 3.5 and
4.5, so neither can have a mean as large as 5.57. Only A has dots spread around 5.5,
so A is the only candidate — then one calculation confirms it.

Q5 Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. (i) If we
include one value to the data (in the given list) without affecting the median, what
could that value be? (ii) If we include two values to the data without affecting the
median what could the two values be? (iii) If we remove one value from the data
without affecting the median what could the value be?

The list is already in order and has 16 values, so the median is the average of the 8th and 9th.

Page 25 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

8th value = 41, 9th value = 41

Median = (41 + 41) ÷ 2 = 41

Note that there are seven values below 41, two 41s, and seven values above 41. That symmetry
is the key to all three parts.
(i) Any value at all. With 17 values the median becomes the 9th.

Add something less than 41 (say 20): now eight values are below, so the 9th is 41. ✓
Add something greater than 41 (say 60): seven below, then 41, 41 — the 9th is 41. ✓
Add 41 itself: three 41s at positions 8, 9, 10 — the 9th is 41. ✓

(ii) One value 41 or less together with one value 41 or more — for example 20 and 60, or 41
and 100. With 18 values the median is the average of the 9th and 10th, and only a value added
on each side keeps 41 in both those places. Two values both below 41 would push a smaller
number into 9th place; two values both above would push a larger number into 10th place.
(iii) Any value at all. With 15 values the median is the 8th.

Remove one from below (say 8): six values below, then 41, 41 — the 8th is 41. ✓
Remove one from above (say 92): seven below, then 41 — the 8th is 41. ✓
Remove one 41: seven below, then the remaining 41 in 8th place. ✓

Why this data set is so stubborn: the two middle values are equal, and there is a
matching count of seven values on either side. That gives the median a cushion — a
single value added or removed nudges the middle position by one place, and 41 is
still sitting there.

Q6 Examine the statements below and justify if the statement is always true,
sometimes true, or never true. (i) Removing a value less than the median will
decrease the median. (ii) Including a value less than the mean will decrease the
mean. (iii) Including any 4 values will not affect the median. (iv) Including 4 values
less than the median will increase the median.

(i) Never true. Removing a value from below the middle leaves fewer numbers on the low side,
so the middle position moves towards the larger values. The median can rise or stay put, but it
cannot fall.

1, 2, 3, 4, 5 → median 3. Remove 1 → 2, 3, 4, 5 → median 3.5 (rose)

1, 3, 3, 3, 5 → median 3. Remove 1 → 3, 3, 3, 5 → median 3 (unchanged)

Page 26 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

(ii) Always true. From the formula on page 105, the shift in the mean is (x − a) ÷ (n + 1). If x < a
this is negative, whatever the data.

4, 6, 8 → mean 6. Include 2 → 20 ÷ 4 = 5 (fell by 1 = (2 − 6) ÷ 4)

(iii) Sometimes true. It depends on where the four values land.

1, 2, 3 → median 2. Include 1, 1, 3, 3 → 1, 1, 1, 2, 3, 3, 3 → median 2 ✓ unchanged

1, 2, 3 → median 2. Include 0, 0, 0, 0 → 0, 0, 0, 0, 1, 2, 3 → median 0 ✗ changed

(iv) Never true. Four values added below the median make the low side heavier in count, so the
middle position slides down the sorted list. The median can fall or stay the same, never rise.

1, 2, 3, 4, 5 → median 3. Include 0, 0, 0, 0 → nine values, median = 5th = 1 (fell)

2, 2, 2, 2, 2 → median 2. Include 1, 1, 1, 1 → nine values, median = 5th = 2 (unchanged)

The pattern behind all four: the mean responds to size, so a rule about “less than
the mean” settles its direction completely — statement (ii) is the only ‘always’. The
median responds to counts, so adding or removing on one side moves the middle
position in a fixed direction, but it may land on an equal value and appear not to
move at all. That is why the median statements come out as ‘never’ or ‘sometimes’,
never ‘always’.

Page 27 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q7 The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.

There are 8 numbers, so the total must be

8 × 10.375 = 83

Sum of the seven known numbers

= 8 + 13 + 10 + 4 + 5 + 20 + 10 = 70

y = 83 − 70 = 13

Check: 8 + 13 + 10 + 4 + 5 + 20 + 13 + 10 = 83, and 83 ÷ 8 = 10.375. ✓

Tip: 10.375 = 10 + 3/8, which is a clue that the denominator is 8 — a decimal that
ends in .375 comes from eighths.

Q8 The mean of a set of data with 15 values is 134. Find the sum of the data.

Mean = sum ÷ number of values

So sum = mean × number of values

Sum = 134 × 15 = 2010

Why it works: the mean is the fair share each of the 15 values would get if the total
were divided equally. Fifteen shares of 134 rebuild the whole total: 134 × 15 = 2010.
Notice that this tells you nothing about the individual values — thousands of
different data sets have 15 values adding to 2010.

Page 28 of 78

as e
Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co m
m.
Consider the data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of the following
e
Q9

om30 l as
number(s) could be p if the median of this data is 29? (i) 10 (ii) 25 (iii) 40 (iv) 100 (v) 29
(vi) 47.c(vii) g
em a
a s
a gl

m
p can be 40, 100, 29, 47 or 30 — that is options (iii), (iv), (v), (vi) and (vii).
. co ag
e m
g l as
a
Sort the ten known values:

8, 8, 12, 18, 25, 29, 35, 39, 47, 73

co m
m.
With p there are 11 values, so the median is the 6th value.

o m l a se
Five ofm a g is 29 only when p does
.cknown values (8, 8, 12, 18, 25) are below 29. So the 6th value
a e
ssqueeze
the

ag l
not in ahead of it.

a s
com agl
P SORTED ORDER AROUND THE MIDDLE 6TH VALUE WORKS?

m .
ase
10 8, 8, 10, 12, 18, 25, 29, … 25 No

25 8, 8, 12, 18, 25, 25, 29, … agl 25 No

co m Yes
.
29 8, 8, 12, 18, 25, 29, 29, … 29

e m
c o m8, 8, 12, 18, 25, 29, 30, … g l as
.
Yes
a
30 29

a s e40m
gl
8, 8, 12, 18, 25, 29, 35, … 29 Yes
a
47 8, 8, 12, 18, 25, 29, 35, … 29 Yes
se m
com g l a
m . a
e
100 8, 8, 12, 18, 25, 29, 35, … 29 Yes

s
a gla
The general rule: the median is 29 exactly when p ≥ 29. Any p below 29 gives the
6th place to 25; a p between 26 and 28 would itself take the 6th place. And once p is
co m
m .
e
29 or more it does not matter how much more — 30 and 100 give the same median,
m l as
.co g
which is the median’s whole character.
m a
l a se
ag
.c
s e m
m a
em . co agl
g l as
a

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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q10 The number of times students rode their cycles in a week is shown in the dot plot
below.

0 2 4 6 8 10

The dot plot printed on page 115 — one dot for each student.

Four students rode their cycles twice in that week. (i) Find the average number of
times students rode their cycles. (ii) Find the median number of times students
rode their cycles. (iii) Which of the following statements are valid? Why? (a)
Everyone used their cycle at least once. (b) Almost everyone used their cycle a few
times. (c) There are some students who cycled more than once on some days. (d)
Exactly 5 students have used their cycles more than once on some days. (e) The
following week, if all of them cycled 1 more time than they did the previous week,
what would be the average and median of the next week’s data?

Counting the dots in each column of the plot:

TIMES CYCLED 0 1 2 3 4 5 6 7 8 10 TOTAL

NO. OF STUDENTS 3 1 4 7 7 5 4 6 3 2 42

The column above 2 has four dots, which matches the hint in the question.
(i)

Total rides = (0×3) + (1×1) + (2×4) + (3×7) + (4×7) + (5×5) + (6×4) + (7×6) + (8×3) +

(10×2)

= 0 + 1 + 8 + 21 + 28 + 25 + 24 + 42 + 24 + 20 = 193

Average = 193 ÷ 42 = 4.6 rides (to one decimal place)

Page 30 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

(ii) With 42 students the median is the average of the 21st and 22nd values.

Cumulative counts: 0 → 3, 1 → 4, 2 → 8, 3 → 15, 4 → 22

Positions 16 to 22 all hold the value 4, so the 21st and 22nd are both 4

Median = 4 rides

(iii)

(a) Not valid. Three dots sit above 0, so three students did not ride at all.
(b) Valid. 38 of the 42 students rode at least twice, and most cluster between 3 and 7 rides —
‘a few times’ describes the bulk of the class well.
(c) Valid. A week has only 7 days, so the three students who rode 8 times and the two who
rode 10 times must have ridden more than once on some day.
(d) Not valid. We can be sure about those 5 students, but nothing rules out a student who
rode 6 times riding twice on one day and not at all on two others. The plot gives weekly
totals, not day-by-day counts, so ‘exactly 5’ cannot be justified.
(e) If everyone rides once more, every value goes up by 1, so both measures go up by 1:

New average = 4.6 + 1 = 5.6 (check: 235 ÷ 42 = 5.595)

New median = 4 + 1 = 5

Why the median beats the mean here: the average of 4.6 is dragged up by the two
students who rode 10 times. The median of 4 says something more useful about a
typical student — and note that 4.6 is not a possible number of rides for anyone,
while 4 is.

Page 31 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q11 A dart-throwing competition was organised in a school. The number of throws
participants took to hit the bull’s eye (the centre circle) is given in the table below.

No. of trials 1 2 3 4 5 6 7 8 9 10

No. of students 1 0 0 1 4 9 12 15 10 10

The table printed on page 116.

Describe the data using its minimum, maximum, mean and median.

NO. OF TRIALS 1 2 3 4 5 6 7 8 9 10

NO. OF STUDENTS 1 0 0 1 4 9 12 15 10 10

CUMULATIVE 1 1 1 2 6 15 27 42 52 62

Minimum = 1 trial (one lucky student hit it on the first throw)
Maximum = 10 trials

Number of students = 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62

Total trials = (1×1) + (4×1) + (5×4) + (6×9) + (7×12) + (8×15) + (9×10) + (10×10)

= 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473

Mean = 473 ÷ 62 = 7.63 trials

Median = average of the 31st and 32nd values.

The cumulative count reaches 27 at 7 trials and 42 at 8 trials, so positions 28 to 42 all hold 8.

Median = 8 trials

Describing the data: hitting the bull’s eye was hard. The trials ran from 1 to 10, but only 2 of
the 62 students managed it in fewer than 5 throws, while 47 needed 7 or more. Both the mean
(7.63) and the median (8) sit near the top of the range, and the most common result was 8 trials.
The mean is slightly below the median because that single 1-trial student pulls the average
down a little; the median is unmoved by that one lucky throw.

Page 32 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Tip: when the mean is a little below the median, the data has a tail stretching to the
left — a few unusually small values. Here that tail is the pair of students who hit the
target in 1 and 4 trials.

In-text Questions — Pages 116–119

Page 33 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

Section 5.2 Visualising and Interpreting Data — Line Graphs: Temperature
co m
em.
m as
MATH TALK

.co a g l
a s em observe the following graph. Do both these graphs represent the same
gl
Now,
a
Q1
information?

co m
em .
Graph 1 — clustered-column graph ag
g l as
a
Monthly Maximum Temperature
Kerala Punjab

40
co m
e m.
m 30 l as
Temperature (in Celsius)

m .co a g
l a se
ag 20

m a s
.co agl
10

a s em
0
Jan gl Apr May
Feb aMar Jun Jul Aug Sep Oct Nov Dec

co m
m .
e
Graph 2 — line graph
m l as
m.co Monthly Maximum Temperature
a g
l a se Kerala Punjab

a g 40

se m
com l a
Temperature (in Celsius)

30
. a g
m
ase
20
agl

co m
.
10

se m
m l a
.co Jul agAug
0

m
ase
Jan Feb Mar Apr May Jun Sep Oct Nov Dec

agl c
The clustered-column graph printed on page 116 and the line graph printed on page 117.
m .
m a s e
e m . co agl
g l as
a

Yes — exactly the same numbers, drawn in two different ways. Both show the monthly

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m .
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.co


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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

maximum temperature in Kerala and Punjab through 2023, with the months along the
horizontal axis and the temperature in °C up the vertical axis.

CLUSTERED-COLUMN GRAPH LINE GRAPH

How a value is Height of a bar Height of a marked point
shown

What stands out Comparing the two states month by How each state changes through the
month year

Reading a trend Eye has to hop from bar top to bar top The joining line shows the trend
directly

Why the line graph reads better here: the data is a sequence in time, and the
joining segments turn ‘rising’, ‘flat’ and ‘falling’ into shapes you can take in at a
glance. Punjab’s hump and Kerala’s almost-flat line are obvious in the line graph; in
the column graph the same fact has to be assembled from 24 separate bar heights.

Tip: the line between two months is only a guide for the eye. There was no single
‘temperature’ midway between January and February — the graph gives one figure
per month, and the segment merely links them.

Q2 How do we get the maximum temperature over a month in a state?

A state is large, so no one thermometer speaks for it. A few weather stations spread across the
state record the local temperature regularly. For a given month, the state’s monthly maximum is
taken as the largest of all the values recorded at those stations during that month.

Why knowing this matters: once you know how a number was produced you can
judge what it can and cannot tell you. This figure is a peak, not a typical day —
Punjab’s 38 °C in June does not mean every June day was 38 °C, or that every part of
Punjab reached it. It also means the answer depends on where the stations happen
to be: more stations in the hottest district would push the recorded maximum up.
Asking how data was collected is how you spot such limitations and bias.

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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q3 What thoughts or questions occur to you?

Some questions worth carrying further, of the kind the chapter is inviting:

Why are the two trends so different? Punjab is far inland and far north, so it swings from
about 19 °C in January to 38 °C in June. Kerala sits near the equator with the sea on one side,
and the sea heats and cools slowly, so its monthly maxima stay between about 29 °C and 33
°C all year.
What decides a region’s temperature? Latitude, distance from the sea, height above sea
level, and the monsoon — Kerala’s dip to about 29 °C comes in July, its wettest month, when
cloud and rain keep the days cool.
Which other states would look like Punjab? Try Rajasthan, Haryana, Delhi and Uttar
Pradesh — all inland and northern.
What would the monthly minimum temperatures look like? Punjab’s winter minima
would drop very low, so the gap between its two curves would be much wider than Kerala’s.
Is the difference between the two states growing over the years? That needs data from
several years, not one.

Try This: pick your own district, find its monthly maximum temperatures for last
year, and draw them on the same axes. Does your line behave more like Kerala’s or
like Punjab’s?

In-text Questions — Pages 119–120
Space Jam: A Traffic Problem in the Future?

MATH TALK

Q1 What could be the possible method used to derive this data? Discuss.

Every launch that reaches orbit is tracked and catalogued. Space agencies and defence radar
networks record each object put into orbit — satellites, probes, landers, crewed spacecraft and
space-station parts — and international registers keep a running list with its launch date and
launching country. Counting the entries for each year gives the worldwide figure; sorting them
by country gives the separate lines for the USA, China and Russia.

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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Why this matters for reading the graph: the counts come from an official register,
so an object that never reached orbit, or one that was never declared, would not
appear. The graph also counts objects, not launches — a single rocket carrying 60
small satellites adds 60 to the count. That alone explains much of the steep climb
after 2019, when large satellite constellations began going up.

Q2 Which of the following statements are valid inferences? • From 2012 till 2024, the
worldwide count of space object launches increased every year. • USA is a major
contributor in the years 2022 – 24, launching about 3/4th of the worldwide count. •
Nepal did not launch any object in the period 2012 – 24. • The combined count of
object launches by China and Russia in 2024 is about 400.

“Increased every year” — not valid. The overall climb is steep, but the world line dips in a
few places: it falls slightly from 2014 to 2015 and again to 2016, dips from 2017 to 2018, and
falls from about 2900 in 2023 to about 2800 in 2024. ‘Every year’ is a much stronger claim
than ‘overall’.
“USA launched about 3/4th in 2022–24” — valid. Reading the two lines: in 2022 about 1950
out of 2500; in 2023 about 2250 out of 2900; in 2024 about 2280 out of 2800. Those fractions
are roughly 0.78, 0.78 and 0.81 — close to three-quarters.
“Nepal did not launch any object” — not valid. The graph has only four lines: World,
United States, China and Russia. Nepal is simply not shown. A graph that does not mention
something says nothing about it, and you cannot turn silence into evidence.
“China and Russia together about 400 in 2024” — valid. Reading the two lower lines at
2024 gives roughly 300 for China and roughly 110 for Russia, a total of about 410 — ‘about
400’ is a fair reading.

The habit to build: before accepting a statement, ask which line and which points
on the graph support it. The Nepal claim fails not because it is false in the world but
because this graph cannot decide it — the commonest mistake in reading data.

Q3 Identify two consecutive years where the worldwide count increased by 2 times or
more.

Two such jumps show up on the world line:

Page 37 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

2016 → 2017: about 245 → about 505

505 ÷ 245 ≈ 2.1 times ✓

2019 → 2020: about 610 → about 1280

1280 ÷ 610 ≈ 2.1 times ✓

No other consecutive pair comes close: even the big rise from 2021 to 2022 is roughly 1810 to
2500, only about 1.4 times.

Careful: ‘increased by 2 times or more’ is about the ratio, not the gap. From 2021 to
2022 the count rose by about 690 objects, a larger gap than the 260 gained between
2016 and 2017 — yet only the smaller gap counts here, because 260 on top of 245 is
a doubling and 690 on top of 1810 is not. On a graph, steepness shows the gap; you
have to compare the two heights to see the ratio.

In-text Questions — Pages 120–121
Catch the (Pattern in) Rain

MATH TALK

Q1 What could be the possible method to compile this data?

Rain gauges in each city measure the rainfall every day. The daily figures are added to give a
monthly total, and this is repeated for several years. The totals for the same month across those
years are then averaged — so ‘average rainfall in June at Kovalam’ means the mean of the June
totals over all the years recorded.

Why an average is used: a single year’s rainfall can be freakish — one cloudburst,
or one failed monsoon. Averaging over years smooths out those accidents and
leaves the underlying pattern of the seasons, which is what the graph is trying to
show. The price is that the graph can no longer tell you about any particular year.

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Class 8 Maths Chapter 12 Tales by Dots and Lines
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co m
m.
Mark these cities on a map of India. What is common to how they are grouped in
e
Q2

m l as
.co
the graphs? Share your observations and inferences about the graphs.

a g
se m
g l a
a

The six cities split neatly into two coasts, and that is exactly how the two graphs are grouped.

co m
m . ag
se Along the west coast (Kerala, Karnataka, Maharashtra)
GROUP CITIES WHERE THEY ARE

l a
First graph Kovalam, Udupi, Mumbai
ag

co m
Second graph Rameswaram, Chennai, Puri Along the east coast (Tamil Nadu, Tamil Nadu, Odisha)

se m.
o m l a
Observations: the west-coast cities show one tall, narrow peak in the middle of the year, and

m .c is much heavier — Udupi’s peak runs to several hundredagmillimetres a month. The
se cities have flatter, later peaks and receive less rain overall.
their rainfall

l a
ag
east-coast

The inference: the west coast faces the Arabian Sea and the Western Ghats stand
m a s
.co
right behind it, so the monsoon winds are forced up and drop their moisture at once
em agl
a s
agtol wait for the returning winds later in the year.
— a lot of rain, in a short season. The east coast lies in the rain shadow of those hills
for the same winds and has

co m
m .
m ase
Q3
.co
Identify the peak months and low months of rainfall for each city.
a g l
a s em
agl ANSWER
se m
CITY COAST PEAK MONTHS
com
LOW MONTHS
g l a
m . a
Kovalam West
gl ase
June – August January – March

Udupi West
aJune – August January – March

co m
Mumbai West
m .
January – March

se January – March
June – August

m l a
.co ag
em
Puri East July – September

a s
agl Chennai East Rain from June, peaking in November, January – March

.c
m
continuing into December

m a s e
. co agl
Rameswaram East October – December January – September (very

e m
las
little rain)

ag

co m
m .
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.co


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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

January to March are dry for every one of the six cities. Rameswaram is the extreme case — it
gets almost nothing for nine months and then nearly all of its rain in the last three.

Q4 Read about the south-west monsoon and north-east monsoon and which regions
come under the influence of these and when.

India has two monsoons, and the graphs are a picture of both.

SOUTH-WEST MONSOON NORTH-EAST MONSOON

When June to September October to December

Wind From the south-west, off the Arabian Sea From the north-east, back across the Bay
direction and Bay of Bengal of Bengal

Who gets Most of India — especially the west coast, Tamil Nadu, coastal Andhra Pradesh,
the rain the Western Ghats and the north-east south Karnataka, Kerala’s coast,
Puducherry

How the graphs show it: Kovalam, Udupi, Mumbai and Puri all peak between June
and September — the south-west monsoon. Rameswaram and Chennai peak
between October and December — the north-east monsoon, sometimes called the
retreating monsoon. Chennai shows both: some rain from June onwards and its real
peak in November. This is why a ‘monsoon holiday’ falls at very different times in
Mumbai and in Chennai.

Did you know? The north-east monsoon carries winds that have crossed the Bay of
Bengal and picked up moisture there, which is why they bring rain to the eastern
coast rather than to the dry interior they started from.

Figure it Out — Pages 122–123

Page 40 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Section 5.2 Visualising and Interpreting Data

Q1 The average number of customers visiting a shop and the average number of
customers actually purchasing items over different days of the week is shown in
the table below.

Mon Tue Wed Thu Fri Sat Sun

Visiting 16 19 10 14 20 22 35

Purchasing 10 8 7 11 12 16 26

The table printed on page 122.

Visualise this data on a line graph.

DAY MON TUE WED THU FRI SAT SUN

VISITING 16 19 10 14 20 22 35

PURCHASING 10 8 7 11 12 16 26

Put the days along the horizontal axis and the number of customers up the vertical axis. A scale
of 1 cm to 5 customers, running from 0 to 35, holds every value comfortably. Plot both series on
the same axes and use different marks so the two can be told apart even in black and white.

Page 41 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

35

28

21

14

7

0
Mon Tue Wed Thu Fri Sat Sun

Visiting Purchasing

Customers visiting and purchasing at the shop across the week. Both lines dip on Wednesday and
climb steeply into Sunday.

What the graph says: both lines rise towards the weekend and peak sharply on Sunday.
Wednesday is the quietest day. The gap between the lines is the number who look but do not
buy — widest on Tuesday (19 visiting, only 8 purchasing) and narrowest on Monday and
Thursday.

Page 42 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q2 The average number of days of rainfall in each month for a few cities is shown in the ta
below:

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov De

Mangaluru 0.1 0 0.1 1.8 6.2 24.1 27.7 24.5 14 8.8 3.9 0.

New Delhi

Port Blair 2.4 1.3 0.9 3.3 15.5 18.7 17.3 18.8 16.8 14.1 11.3 5.

Rameswaram 2.6 1.3 1.9 3.4 2.5 0.4 1 1 1.9 8.1 10.4 7.

The table printed on page 122; the row for New Delhi is left blank in the book.

(i) What could be the possible method to compile this data? (ii) Mark the data for
Mangaluru, Port Blair, and Rameswaram in the line graph shown below. You can round
the values to the nearest integer. (iii) Based on the line for New Delhi in the graph fill th
data in the table. (iv) Which city among these receives the most number of days of rain
per year? Which city gets the least number of days of rainfall per year? (v) Looking at th
table, when is the rainy season in New Delhi and Rameswaram?

30

25
Number of days

20

15

10

5

0

The line graph printed on page 122. It already carries the line for New Delhi. The twelve tick
along the horizontal axis are the months January to December of the table. (Source: weather
and-climate.com)

Page 43 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

(i) For each city, note on how many days of a given month rain was recorded at the weather
co m
se
station. Do this for several years, then average the counts for that month. So ‘24.1 days in June
m.
o m g l a
.c — no single June has 24.1 rainy days.
at Mangaluru’ means that on average it rained on about 24 days of June. That is why the figures

m a
se each figure to the nearest whole number and plot the points, joining each city’s
come with decimals

l a
ag
(ii) Round
points with its own line:

. comJUN ag
m
CITY JAN FEB MAR APR MAY JUL AUG SEP OCT NOV DEC

l a se6
ag
Mangaluru 0 0 0 2 24 28 25 14 9 4 1

Port Blair 2 1 1 3 16 19 17 19 17 14 11 5

. com10
se m
Rameswaram 3 1 2 3 3 0 1 1 2 8 8

o m l a
g value fits.
m
The vertical
.c scale of the printed graph runs from 0 to 30, so every rounded
a
(iii) s
a e
a g l Reading the New Delhi line off the graph, month by month:

s
MONTH JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC

m a
m .co 4 agl
se
New 1 2 2 1 2 10 10 4 1 0 1
Delhi

g l a
a
Read to the nearest whole number, as the question allows. Reading a graph is never exact — a
classmate might read July as 9 or 10, and either is a fair reading of the printed line.
co m
m .
(iv) Add each row across:
m as e
.co a g l
s m
eMangaluru
g l a = 0.1 + 0 + 0.1 + 1.8 + 6.2 + 24.1 + 27.7 + 24.5 + 14 + 8.8 + 3.9 + 0.9 = 112.1
a
m
days

a se
.com
Port Blair = 2.4 + 1.3 + 0.9 + 3.3 + 15.5 + 18.7 + 17.3 + 18.8 + 16.8 + 14.1 + 11.3 + 5.4 =
a g l
m
ase
agl
125.8 days
Rameswaram = 2.6 + 1.3 + 1.9 + 3.4 + 2.5 + 0.4 + 1 + 1 + 1.9 + 8.1 + 10.4 + 7.8 = 42.3 days

co m
.
New Delhi (from the graph) ≈ 1 + 2 + 2 + 1 + 2 + 4 + 10 + 10 + 4 + 1 + 0 + 1 = 38 days

em
m l as
m .co a
Port Blair gets the most rainy days (about 126 a year) and New Delhi the fewest (about 38). g
l a se
ag
.c
Worth noticing: Mangaluru has fewer rainy days than Port Blair yet is far wetter
overall — its rain arrives in three furious months. ‘Number of rainy days’ and
s e m
m a
. co agl
‘amount of rain’ are different measurements, and this table only records the first.
e m
g l as
a

co m
m .
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.co


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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

(v) New Delhi’s rainy season is July and August, with the build-up in June and the tail in
September — the south-west monsoon. Rameswaram’s rainy season is October to December,
with November the wettest — the north-east monsoon. The two cities get their rain at almost
opposite times of the year.

Q3 The following line graph shows the number of births in every month in India over a
time period:

2M

1.5M
Live births

1M

0.5M
Jul 2017 Jan 2018 Jul 2018 Jan 2019 Jul 2019 Jan 2020
Time

The line graph printed on page 123.

(i) What are your observations? (ii) What was the approximate number of births in
July 2017? (iii) What time period does the graph capture? (iv) Compare the number
of births in the month of January in the years 2018, 2019, and 2020. (v) Estimate the
number of births in the year 2019.

(The book prints this question as ‘2’ again; it is the third question of this set.)
(i) Observations. The line never falls below about 1.4 million or rises above about 2 million, so
births each month stay in a fairly narrow band. Inside that band there is a clear repeating yearly
pattern: a peak around August–October every year and a dip around February and April. The
pattern repeats three times, once for each year on the graph, and the whole band drifts very
slightly upwards over the three years.
(ii) Reading the point above July 2017: about 1.75 million births (roughly 17.5 lakh).
(iii) The first point sits about three months before the July 2017 gridline and the last about three
months after January 2020, so the graph runs from about April 2017 to March 2020 — three
years, 36 monthly points.
(iv) Reading the three January points:

Page 45 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

MONTH APPROXIMATE BIRTHS

January 2018 about 1.68 million

January 2019 about 1.76 million

January 2020 about 1.77 million

January births rose noticeably from 2018 to 2019 and then stayed almost level in 2020. All three
are close to one another, and all three are well below that year’s autumn peak — January is a
middling month for births, not a low one.
(v) Add the twelve monthly readings for 2019:

1.76 + 1.57 + 1.77 + 1.51 + 1.64 + 1.64 + 1.85 + 1.99 + 1.96 + 1.99 + 1.93 + 1.87

≈ 21.5 million births in 2019 (about 2.15 crore)

Quicker estimate: the 2019 points hover around 1.8 million a month, and 12 × 1.8 =
21.6 million — close to the careful total. When every value in a graph sits near one
level, multiplying that level by the number of points is a good estimate, and it is a
use of the mean in reverse.

In-text Questions — Pages 123–124
Infographics

Q1 Share your observations. Based on this infographic, answer the following: (i) The
value of Karnataka is hidden. Can you guess what it could be? (ii) Which are the top
5 states where rice is the most popular? (iii) Which are the top 5 states where wheat
is the most popular? (iv) List a few states where the preference between rice and
wheat is more or less balanced.

Observations. The scale runs from −100 (mostly wheat) through 0 (both equally) to +100
(mostly rice). The red line drawn across the map splits India almost cleanly in two: the north and
west are negative (wheat) and the east, north-east and south are positive (rice). The values
printed on the map are:

Page 46 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

WHEAT SIDE (NEGATIVE) VALUE RICE SIDE (POSITIVE) VALUE

Rajasthan −93 Manipur +100

Haryana −81 Nagaland +99

Punjab −78 Mizoram +97

Madhya Pradesh −60 Tripura +96

Delhi −45 Sikkim +86

Gujarat −37 Lakshadweep / Puducherry +80 each

Uttar Pradesh −30 Kerala +79

Himachal −19 Andaman & Nicobar +65

Uttarakhand −18 Goa +57

Maharashtra −15 Jammu & Kashmir (undivided) +41

Jharkhand +45

Bihar +3

(i) Karnataka sits in the southern rice belt, hemmed in by Goa (+57), Kerala (+79) and the
strongly rice-eating states of the far south, with only Maharashtra (−15) leaning the other way.
Its shade matches the moderate-rice group, so a sensible guess is somewhere around +50, say
between +40 and +60 — clearly a rice state, but not as one-sided as Kerala or the north-east.
(ii) Top 5 rice states (of the values printed): Manipur (+100), Nagaland (+99), Mizoram (+97),
Tripura (+96), Sikkim (+86) — all in the north-east.
(iii) Top 5 wheat states: Rajasthan (−93), Haryana (−81), Punjab (−78), Madhya Pradesh (−60),
Delhi (−45) — all in the north-west.
(iv) More or less balanced (values close to 0): Bihar (+3), Maharashtra (−15), Uttarakhand
(−18) and Himachal Pradesh (−19). In these states people eat a good deal of both grains.

Reading the scale carefully: +100 does not mean the state eats no wheat at all. It
means the difference between rice and wheat eaten per person is the largest on this
scale. The infographic measures a gap, not a total — a point worth remembering
whenever a map colours states by a ‘score’.

Page 47 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Careful: the printed map carries numbers for only some states; for the rest
(Karnataka, Tamil Nadu, Odisha, West Bengal, Assam and others) you have to judge
from the shade against the colour scale, so different readers may put the ‘top 5’ in a
slightly different order.

In-text Questions — Page 125
What can a Strip Say?

Q1 Look at the three coloured strips carefully. (i) What activity does each colour stand
for? (ii) The three strips correspond to the days Friday – Sunday in some order.
Which day do you think each strip represents? (iii) On one of these days, he went
out with friends to watch a long movie. When do you think this happened? (iv) At
what time does his school break for lunch? (v) What more can the strips tell us?

Each strip is one day cut into 48 half-hour boxes, from midnight to midnight, with the hours
numbered down the side.
(i) What each colour stands for

COLOUR ACTIVITY THE CLUE ON THE STRIPS

Light Sleeping The long unbroken block through the night on all
blue three strips

Purple Showering and getting dressed, yoga An hour or so right after waking, every day
or exercise

Green Eating Three short half-hour blocks a day, morning,
midday and evening

Grey Travelling Half an hour immediately before and after the long
daytime blocks

Yellow Attending classes, studying and The long block that starts right after the morning
homework travel

Orange Meeting friends, hobbies, media, time The evening block, and the whole afternoon on the
with family day with no school

(ii) Which day is which

Page 48 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

co mDAY
m.
STRIP WHAT THE DAYTIME LOOKS LIKE

m as e
Middle
. c o
Travel at 8:30, classes 9:00–12:30 and 1:30–4:30, travel home at 4:30 — a full
a g l Friday

sem
school day

gla
aRight Travel at 8:30, classes 9:00–12:30 only, then home and a nap — a half day Saturday

m
.co ag
Left No classes at all; a long afternoon out, only about two hours of study Sunday

la sem
g
(iii) On the left-hand strip (Sunday). After lunch at 12:30 there is half an hour of travel at 1:00,
a
then a single orange block from 1:30 to 4:30, then travel again from 4:30 to 5:00. Travel on

m
both sides of a three-hour block means he went somewhere and came back — long enough for

co
m.
a film.

o m l a se
(iv) On the school-day strip the classes stop at 12:30, there is eating from 12:30 to 1:00 and free
g for lunch at 12:30 pm,
time from.c1:00 to 1:30, and classes start again at 1:30. So school breaks
a
for s
m
e hour.
l a
ag(v) What more the strips tell us. Count the boxes and compare:
one

m a s
ACTIVITY SUNDAY (LEFT)
m.co FRIDAY (MIDDLE) SATURDAY (RIGHT)
agl
l a se
Sleeping
ag
10½ h 10½ h 10½ h

m
Eating 1½ h 1½ h 1½ h

. co
Classes / study
se m
com a
2h 7h 4 h

. ag l
e m
as
Friends, hobbies, family 7½ h 3h 5½ h

agl Travelling 1h 1h 1½ h

se m
Washing, dressing, exercise
com g l a
.
1½ h 1h 1h

e m a
g l as on every one of the three days — but not in one stretch.
Manoj sleeps exactly 10½ hours
a
On Sunday he takes a half-hour nap at 5 pm and on Saturday a full hour at 2 pm, and each

m
time he goes to bed correspondingly later.

. co
m
He eats 1½ hours a day, always in three half-hour meals.

o m l a se
Study time and free time trade places: 7 hours of class on Friday against 3 hours with friends,
g is roughly 9½ hours on each
.cand almost exactly the reverse on Sunday. The total of theatwo
se m day.
g l a
a c
.
Travel only ever appears in half-hour pieces at the edges of an outing, so nothing he does is
far from home.
s e m
m a
m . co agl
l a se
Careful: the text says he recorded ‘five types of activities’ but six are listed, and six
agthe strips. Count the colours, not the sentence.
colours do appear on

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m .
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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

In-text Questions — Page 126
What can a Strip Say?

TRY THIS MATH TALK

Q1 What would your strip for a weekday look like? How similar or different is it to
Manoj’s?

Make your own strip of 48 boxes and colour it in for yesterday, using Manoj’s six colours. Here is
a sample weekday, which you should replace with your own:

TIME ACTIVITY BOXES

10:00 pm – 6:00 am Sleeping 16 (8 h)

6:00 – 7:00 am Washing, dressing, exercise 2 (1 h)

7:00 – 7:30 am Eating 1

7:30 – 8:00 am Travelling 1

8:00 am – 2:00 pm Classes (with a lunch break) 12 (6 h)

2:00 – 2:30 pm Travelling 1

2:30 pm – 10:00 pm Eating, homework, family, play 15 (7½ h)

Comparing with Manoj’s Friday: look at three things — when your sleep block starts and ends,
how many boxes your classes take, and how many boxes are left for yourself. Many students will
find they sleep less than Manoj’s 10½ hours and travel more than his 1 hour. What matters is not
who has ‘more’ of anything, but that the 48 boxes are the same for everybody: every extra box
for one activity is a box taken from another.

Q2 What would a strip of your typical day during your vacation look like? How
similar/different would it look?

The 48 boxes stay the same; only their colours move around.

The whole yellow block of classes disappears, freeing about 12 boxes (6 hours).
Those boxes usually turn orange — play, friends, screen time, family — and sometimes light
blue, because sleep often starts later and ends later.

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Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

The two grey travel boxes vanish on most days, but a holiday trip can produce a strip with a
huge grey block instead.
Meal times spread out and become less regular, so the green boxes shift about.

Try This: colour a vacation strip and a school-day strip on the same sheet, one under
the other. Line up the same hour on both and you can see at a glance exactly which
hours changed and which stayed put.

Q3 What would a strip for any of the adults in your family look like? Make a strip of a
day for any adult at home. Compare your strip with theirs. What do you find
interesting?

Record an adult’s day in the same 48 boxes and lay the two strips side by side. Differences that
usually show up:

A GRADE 8 STUDENT AN ADULT AT HOME

Sleep About 9–10 hours Usually about 6–8 hours

Longest daytime Classes Work, or housework and care of the
block family

Free-time boxes Several Often only in the evening, and fewer

Shape of the day One long block, fixed by the school Often broken into many short blocks
timetable

What is interesting: the adult’s strip is usually far more fragmented — many small blocks
rather than a few long ones — because household work interrupts everything else. Work at
home very often fills more boxes than paid work outside, and it rarely appears on any timetable.
The strip makes that visible in a way a list never does, which is exactly what a good picture of
data should do.

In-text Questions — Pages 126–127

Page 51 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Data Story: Sleepy-Deepy

Q1 Share your observations on this graph.

Night sleep across ages

12

10
Sleep (hours/day)

8

6

4

2

0
10 30 50 70
Age (years)

Page 52 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Night sleep across ages

12.0
11.5
11.0
Sleep (hours/day)

10.5
10.0
9.5
9.0
8.5
8.0
7.5
7.0
10 30 50 70
Age (years)
The two line graphs printed on page 126 — the same data drawn on two different
vertical scales.

What do you find interesting?

The graph shows the typical daily sleep of Indians from age 6 to age 75. The second picture is
the same data with the vertical axis stretched: it runs from 7.0 to 12.0 hours in steps of half an
hour, instead of starting at 0.

A 6-year-old sleeps about 9.5 to 9.7 hours a day — the highest point on the curve.
Sleep falls steeply through the teenage years, then more gently, reaching its lowest point of
about 8 hours between ages 35 and 40, and staying near 8 hours from about 30 to 50.
After 50 it climbs again — about 8.5 hours by age 70 and a shade under 9 hours by 75.
So the curve is a shallow valley — steep down, flat, then a slow climb — and the whole range
from highest to lowest is under 2 hours.

Why the curve looks smooth: it holds about 80 points, one for each age, packed
close together, so the eye reads them as a curve rather than a chain of segments.
Drawn as a column graph the same data would need 70 columns and the gentle
valley would be lost in the clutter — this is a case where a line graph is plainly the
right choice.

Page 53 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

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a g l Page 54 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Section 5.2 Visualising and Interpreting Data

MATH TALK

Q1 Mean Grids: (i) Fill the grid with 9 distinct numbers such that the average along
each row, column, and diagonal is 10. (ii) Can we fill the grid by changing a few
numbers and still get 10 as the average in all directions?

The empty grid printed on page 127.

(i) Each line holds 3 numbers, so an average of 10 means every row, column and diagonal must
add up to 30. Take the ordinary 3 × 3 magic square (which sums to 15 in every direction) and
add 5 to each entry:

Page 55 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

13 6 11

8 10 12

9 14 7

Rows: 13 + 6 + 11 = 30, 8 + 10 + 12 = 30, 9 + 14 + 7 = 30

Columns: 13 + 8 + 9 = 30, 6 + 10 + 14 = 30, 11 + 12 + 7 = 30

Diagonals: 13 + 10 + 7 = 30, 11 + 10 + 9 = 30

All nine numbers 6, 7, 8, 9, 10, 11, 12, 13, 14 are distinct ✓

Why adding 5 works: every line of the magic square has 3 entries, so adding 5 to
each entry adds 15 to every line total — 15 + 15 = 30. This is the ‘add a constant’ rule
of page 107, used nine times at once.

(ii) Yes, and there are endless ways. The centre is forced to be 10, but everything else is free.
Fill the grid like this, choosing any two numbers a and b:

10 + a 10 − a − b 10 + b

10 − a + b 10 10 + a − b

10 − b 10 + a + b 10 − a

Every row, column and diagonal here adds to 30 whatever a and b are — the a’s and b’s cancel
out. Taking a = 3, b = 1 gives the grid above; taking a = 4, b = 1 gives a fresh one:

14 5 11

7 10 13

9 15 6

Why the centre must be 10: add the middle row, the middle column and both
diagonals. That is 4 lines, total 4 × 30 = 120. Every cell of the grid is used once in that
sum except the centre, which is used four times. So (sum of all 9 cells) + 3 × (centre) =
120. But the three rows show the sum of all 9 cells is 90, so 3 × (centre) = 30 and the
centre is 10 — it can never be anything else.

Page 56 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q2 Give two examples of data that satisfy each of the following conditions: (i) 3
numbers whose mean is 8. (ii) 4 numbers whose median is 15.5. (iii) 5 numbers
whose mean is 13.6. (iv) 6 numbers whose mean = median. (v) 6 numbers whose
mean > median.

CONDITION EXAMPLE 1 EXAMPLE 2 HOW IT WAS BUILT

(i) 3 numbers, mean 8 6, 8, 10 1, 8, 15 Total must be 3 × 8 = 24

(ii) 4 numbers, median 10, 15, 16, 20 1, 15, 16, 100 The two middle values must average
15.5 15.5, so 15 + 16 = 31

(iii) 5 numbers, mean 10, 12, 14, 16, 13, 13, 14, 14, Total must be 5 × 13.6 = 68
13.6 16 14

(iv) 6 numbers, mean = 1, 2, 3, 4, 5, 6 2, 4, 6, 8, 10, Evenly spaced data: both come out at
median 12 3.5 and 7

(v) 6 numbers, mean > 1, 2, 3, 4, 5, 1, 1, 2, 2, 3, 20 Put one very large value far out to the
median 100 right

Checks

(i) 6 + 8 + 10 = 24, 24 ÷ 3 = 8 ✓ 1 + 8 + 15 = 24 ✓

(ii) (15 + 16) ÷ 2 = 15.5 ✓ in both

(iii) 10 + 12 + 14 + 16 + 16 = 68, 68 ÷ 5 = 13.6 ✓ 13 + 13 + 14 + 14 + 14 = 68 ✓

(iv) 1+2+3+4+5+6 = 21, mean 3.5; median = (3 + 4) ÷ 2 = 3.5 ✓

(v) 1+2+3+4+5+100 = 115, mean ≈ 19.2; median = (3 + 4) ÷ 2 = 3.5, and 19.2 > 3.5 ✓

What (iv) and (v) are really about: when the data is spread symmetrically the mean
and median land in the same place. Pull one value far out to the right and the mean
chases it — because the mean measures distance — while the median stays put,
because it only counts positions. That gap between mean and median is a signal
that the data has an extreme value.

Page 57 of 78

Class 8 Maths Chapter 12 Tales by Dots and Lines AglaSem · NCERT Solutions

Q3 Fill in the blanks such that the median of the collection is 13: 5, 21, 14, _____, ______,
______. How many possibilities exist if only counting numbers are allowed?

There will be six numbers, so the median is the average of the 3rd and 4th when sorted.

Median = 13 → (3rd value) + (4th value) = 2 × 13 = 26

The fixed values sorted are 5, 14, 21, and 13 lies between 5 and 14. Only two pairs of whole
numbers can occupy the two middle places:

12 and 14. Put 12 in one blank; then one blank must be 12 or less and the other 14 or more,
so 12 is 3rd and 14 is 4th.

Example: 5, 21, 14, 12, 3, 40 → sorted 3, 5, 12, 14, 21, 40 → median (12 + 14) ÷ 2 = 13

✓

13 and 13. Put 13 in two blanks; the third must be 13 or less.

Example: 5, 21, 14, 13, 13, 2 → sorted 2, 5, 13, 13, 14, 21 → median 13 ✓

How many possibilities? Infinitely many. In the first family the largest blank may be any
counting number from 14 upwards — 14, 15, 16, …, 1000, … — and each choice gives a fresh
answer with the median still 13.

Why any pair other than 12&14 or 13&13 fails: suppose you tried 11 and 15 (they
also add to 26). The value 14 is already in the collection and lies between 11 and 15,
so 14 would have to take one of the two middle places itself. The middle pair must
therefore fit into the gap between 5 and 14, or use 14 as its upper member — which
leaves exactly the two families above.

Page 58 of 78

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Class 8 Maths Chapter 12 Tales by Dots and Lines
a g l AglaSem · NCERT Solutions

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a g l Page 59 of 78