NCERT Solutions Class 8 Maths Chapter 13 Algebra Play

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 13: Algebra Play

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 135 – 147 13 43 English

Solutions, notes, sample papers & more at 48 pages

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 13: Algebra Play
Chapter 6 of Ganita Prakash Grade 8 Part-II puts algebra to work on tricks, puzzles and grids. A ‘Think of a
Number’ trick, a date trick, a number pyramid, a calendar square, a divisibility surprise — each one looks like
magic until you write a letter-number in place of the unknown. Then the trick becomes an expression, the
puzzle becomes an equation, and the reason it always works is there in front of you.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) Part II, 135 – 147

SECTIONS QUESTIONS

13 43

MEDIUM

English

In-text Questions — Page 135
6.1 Algebra Play / 6.2 Thinking about ‘Think of a Number’ Tricks

Q1 I predict you get 2. Am I right? Try it out with different starting numbers. Do you
always end up with the same value, 2? Why?

Yes — the answer is 2 every single time. Try three very different starting numbers:

STEP START WITH 7 START WITH 40 START WITH 1/2

Think of a number 7 40 1/2

Double it 14 80 1

Add four 18 84 5

Divide by two 9 42 5/2

Subtract the original number 2 2 2

Trying numbers shows that it works. To see why, do the same five steps to a letter-number x:

Page 1 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Think of a number: x

Double it: 2x

Add four: 2x + 4
Divide by two: x + 2

Subtract the original number: (x + 2) – x = 2

Why it happens: the letter-number x is not one particular number — it stands for
whatever you thought of. So this single line of working is a check on every starting
number at the same time. Doubling makes the amount of x equal to 2x; halving after
adding 4 brings it back to one x and turns the 4 into 2. The last step removes that
one x, and only the 2 is left. Nothing about x survives to the end, so nothing about
the answer can depend on x.

Tip: notice why step 3 must come between doubling and halving. The 4 is added
when there are two x’s, so halving splits the 4 as well and leaves 2. Add the 4 after
halving and the trick would end at 4 instead.

In-text Questions — Page 136
6.2 Thinking about ‘Think of a Number’ Tricks

MATH TALK

Q1 How would you change this game to make the final answer 3? What about 5?

Change only step 3 — the number you add. Add 6 to finish at 3, and add 10 to finish at 5.

Think of a number: x

Double it: 2x

Add 6: 2x + 6

Divide by two: x + 3

Subtract the original number: 3

Page 2 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Add 10 instead: 2x + 10 → x + 5 → 5

Why it happens: the number you add is put in while there are two x’s, and step 4
halves everything. So the constant that survives is always half of what you added. To
finish at any value k, add 2k:
2x → 2x + 2k → x + k → k.

Check it yourself: start with 11 and add 6 — 22, 28, 14, and 14 – 11 = 3. ✓

Q2 Can you come up with more complicated steps that always lead to the same final
value?

Yes. The recipe is simple: write the expression next to every step, and make sure the last step
removes every x that is left.

Think of a number: x

Multiply by 3: 3x

Add 12: 3x + 12

Divide by 3: x + 4

Add 6: x + 10

Subtract the number you thought of: 10

A longer one, with a detour that changes nothing — it still ends at 10:

Page 3 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
e m.
Think of a number: x
m l as
.co
Multiply by 4: 4x
m a g
l a se
g
Add 40: 4x + 40
aHalve it: 2x + 20

co m
. ag
Subtract 20: 2x
e m
Halve it again: x
g l as
Add 10: x + 10 a

co m
m.
Subtract the number you thought of: 10

m as e
.co a g l
a s em
Why it happens: a trick works when the expression just before the last step is (one

a gl x) + (a constant). Whatever you do in between — multiply, add, subtract, divide —
you only need the amount of x to come back to exactly 1 before the final subtraction.

m a s
.co agl
Multiplying by 4 and later dividing by 4 does exactly that, and the 40 that rode along
becomes 10.
se m
g l a
a
Try This: design one where the final subtraction is not the last step, for example:

co m
.
think of x, treble it, add 15, divide by 3, subtract x, then multiply by 2 — you get 10
e m
as
every time.
m l
m .co a g
l a se
ag
m
How did Shubham figure out the date chosen by Mukta?
se
Q3

com g l a
m . a
ase

a gl
He undid the steps with algebra. Let the month be M and the day be D, and follow the
instructions on the letter-numbers:

co m
m .
as e
com6: 5M + 6
Multiply the month by 5: 5M
.Add a g l
se m
g l a
a Multiply by 4: 20M + 24
c
m .
Add 9: 20M + 33
m a s e
Multiply by 5: 100M + 165
e m . co agl
g l as
Add the day: 100M + 165 + D
a

co m
m .
m ase
.co


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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Mukta’s answer was 291, so

100M + 165 + D = 291

100M + D = 291 – 165 = 126 (subtracting 165 from both sides)

So M = 1 and D = 26 → 26 January

Why it happens: the three multipliers are 5, 4 and 5, and 5 × 4 × 5 = 100. So by the
end the month has been pushed into the hundreds place, while the day D — at most
31 — needs only the tens and units places. The two never mix. Once the constant
165 is taken away, the last two digits of what remains are the day and everything
before them is the month.

Did you know? the 165 also comes straight out of the steps: the 6 gets multiplied by
4 and then by 5, giving 6 × 20 = 120, and the 9 gets multiplied by 5, giving 45. And
120 + 45 = 165.

In-text Questions — Page 137
6.2 Thinking about ‘Think of a Number’ Tricks

MATH TALK

Q1 Mukta thinks of another date, follows the same steps, and reports her answer as
1390. What date did Mukta start with this time?

25 December.

100M + 165 + D = 1390

100M + D = 1390 – 165 = 1225

Last two digits → D = 25 ; what comes before → M = 12

25/12 — 25 December

Page 5 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: 1225 = 100 × 12 + 25. Because D is at most 31 it can never reach
the hundreds place, so splitting 1225 after the last two digits is the only possible split
— the month and the day cannot be confused.

Check it yourself: run 25/12 forward. 12 × 5 = 60, + 6 = 66, × 4 = 264, + 9 = 273, × 5 =
1365, + 25 = 1390. ✓

Q2 Find the dates if the final answers are the following: (i) 1269 (ii) 394 (iii) 296

Subtract 165 from each answer, then read the last two digits as the day and the rest as the
month.

FINAL ANSWER ANSWER – 165 MONTH M DAY D DATE

(i) 1269 1104 11 04 4 November

(ii) 394 229 2 29 29 February

(iii) 296 131 1 31 31 January

(i) 1269 – 165 = 1104 = 100 × 11 + 04 → 4/11

(ii) 394 – 165 = 229 = 100 × 2 + 29 → 29/02

(iii) 296 – 165 = 131 = 100 × 1 + 31 → 31/01

Did you know? 29 February only exists in a leap year, so the friend who gave the
answer 394 was born on a leap day — a birthday that comes round only once in four
years.

Why it happens: each answer is 100M + 165 + D. Removing the 165 leaves 100M + D,
and since 0 < D ≤ 31 the day always fits inside the last two digits. Splitting the
remainder there recovers M and D uniquely.

Page 6 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q3 Can you change the steps in this trick and still find the original date? Instead of
subtracting 165 from the final answer, you might have to subtract some other
number.

Yes. Only one thing really matters: the numbers you multiply the month by must have
product 100. Everything you add along the way just piles up into one constant, and that is the
number to subtract at the end.
Here is a shorter trick. Multiply the month by 5, add 3, multiply by 20, then add the day:

5M

5M + 3

(5M + 3) × 20 = 100M + 60

Add the day: 100M + 60 + D

So subtract 60, not 165.

Test it on 26 January: 1 × 5 = 5, + 3 = 8, × 20 = 160, + 26 = 186. Then 186 – 60 = 126 → M = 1, D =
26. ✓

Why it happens: suppose the multipliers are m₁, m₂, m₃ and the numbers added are
k₁, k₂. The month ends up as m₁m₂m₃ × M, so the product must be 100 for the month
to land in the hundreds place. The constant is k₁m₂m₃ + k₂m₃ — in the original trick 6
× 4 × 5 + 9 × 5 = 120 + 45 = 165, and in the new one 3 × 20 = 60.

Try This: multiply by 2, add 3, multiply by 10, add 7, multiply by 5, add the day. The
month becomes 2 × 10 × 5 = 100 times itself, and the constant is 3 × 10 × 5 + 7 × 5 =
150 + 35 = 185. So subtract 185.

Q4 Try to devise your own ‘Think of a Number’ trick.

Design it backwards: decide what expression you want at each stage, then write the instruction
that produces it.
A trick that always ends at 7:

Page 7 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

SAY THIS OUT LOUD WHAT IT DOES TO X

Think of a number x

Add 5 x+5

Multiply by 4 4x + 20

Subtract 8 4x + 12

Divide by 4 x+3

Add 4 x+7

Subtract the number you thought of 7

A trick that hands the number back:

Think of a number: x

Multiply by 6: 6x

Add 18: 6x + 18

Divide by 3: 2x + 6

Subtract 6: 2x

Halve it: x — “You started with that number!”

Why it happens: a ‘predict the answer’ trick needs the x-terms to vanish, so the step
before the last must leave exactly one x plus a constant, and the last step subtracts
that one x. A ‘guess your number’ trick needs the opposite — the constants must
vanish and exactly one x must be left. Writing the expression beside every
instruction is what lets you control this; without it you are only guessing.

Tip: keep divisions honest. If you say “divide by 4”, make sure the expression really
has a multiple of 4 in every term — as 4x + 12 does — otherwise your friend will be
stuck with fractions.

In-text Questions — Page 138

Page 8 of 48

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Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

6.3 Number Pyramids
co m
e m.
m l as
Q1
.co a g
Use the same rule to fill these pyramids: (i) bottom row 6, 2 (ii) bottom row 3, 4, 3
m
l a se
(iii) bottom row 5, 4, 5, 0

a g

co m
. ag
Work upwards, adding each neighbouring pair.
e m
(i) 6 + 2 = 8.
g l as
a

co m
em.
m l as
m .co a g

8
l a se
a g
m a s
m .co agl
l a se
a g

m .co
m 6 a g
2
l as e m . co m

l a se
ag
se m
com g l a
m . a
ase
The two-row pyramid: the top is simply the sum of the pair below.

a gl
(ii) 3 + 4 = 7 and 4 + 3 = 7, then 7 + 7 = 14.

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

14

7 7

3 4 3
Bottom row 3, 4, 3. The middle row is 7, 7 and the top is 14.

(iii) 5 + 4 = 9, 4 + 5 = 9, 5 + 0 = 5; then 9 + 9 = 18 and 9 + 5 = 14; finally 18 + 14 = 32.

32

18 14

9 9 5

5 4 5 0
Bottom row 5, 4, 5, 0 builds up to 32.

Page 10 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: the rule fixes every box above the bottom row, so once the bottom
row is given the whole pyramid is decided — there is nothing left to choose. That is
why a pyramid can be described completely by its bottom row alone.

Q2 How do we fill this pyramid? (top 10; middle row 4, ?; bottom row 1, ?, ?)

Here the bottom row is not given, so read the rule backwards. “A box is the sum of the two
below it” also says “a missing box below is the box above minus its known neighbour”.

Right box of the middle row: 10 – 4 = 6

Middle box of the bottom row: 4 – 1 = 3

Right box of the bottom row: 6 – 3 = 3

So the pyramid is: bottom 1, 3, 3; middle 4, 6; top 10.

10

4 6

1 3 3
The completed pyramid, filled by subtracting instead of adding.

Page 11 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: if p + q = s and you know s and p, then subtracting p from both
sides gives q = s – p. Subtraction is not a new rule — it is the same rule rearranged,
and rearranging is legitimate because taking the same amount from both sides of
an equation keeps it true.

Check it yourself: add upwards again. 1 + 3 = 4 ✓, 3 + 3 = 6 ✓, 4 + 6 = 10 ✓.

Q3 What about filling in the numbers in this pyramid? Where do we start? (top 60;
middle row ?, ?; bottom row 12, ?, 8)

You cannot start by subtracting — no box here has both the numbers it needs. So name the
empty boxes with letter-numbers and let the rule give you equations.

60

a b

12 c 8
The same pyramid with the three empty boxes named a, b and c.

a + b = 60

12 + c = a

c+8=b

Page 12 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Replace a and b in the first equation by what they equal:

(12 + c) + (c + 8) = 60

20 + 2c = 60

2c = 40 (subtracting 20 from both sides)

c = 20 (dividing both sides by 2)

a = 12 + 20 = 32, b = 20 + 8 = 28

60

32 28

12 20 8
The completed pyramid: bottom 12, 20, 8; middle 32, 28; top 60.

Why it happens: substituting 12 + c for a is allowed because a and 12 + c are the
same number — the second equation says so. Subtracting 20 from both sides and
then dividing both sides by 2 keeps the two sides equal at every stage, because
whatever you do to one side you do to the other. That is the only reason we are
allowed to change an equation at all.

Check it yourself: 12 + 20 = 32 ✓, 20 + 8 = 28 ✓, 32 + 28 = 60 ✓.

Page 13 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
m.
In-text Questions — Page 139
m as e
6.3 Number Pyramids
.co a g l
a s em
aQ1gl Fill the following pyramids: (i) four rows — top 50, second row right box 22, bottom
row 4, ?, 6, ? (ii) four rows — second row left box 40, third row right box 9, bottom

co m
row 5, ?, 7, ? (iii) four rows — top 35, third row right box 7, bottom row 3, 5, ?, ?

e m . ag
g l as
a

In each one, call the two missing bottom entries x and y, write down what the given boxes say,
and solve.
co m
em.
(i) Bottom row 4, x, 6, y. The right box of the second row sits above 6 and is built from x, 6, y:
m l as
.co a g
(xse+m
g l a 6) + (6 + y) = 22 → x + y + 12 = 22 → x + y = 10
a Left box of the second row = 50 – 22 = 28

m a s
.co agl
(4 + x) + (x + 6) = 28 → 2x + 10 = 28

se m
l a
2x = 18 (subtracting 10 from both sides)
a g
x = 9 (dividing both sides by 2), so y = 10 – 9 = 1

co m
m .
m ase
.co a g l
se m
g l a 50
a
se m
com g l a
m . a
e
as28 22
agl

c o m
13 15 7 .
a s em
. com ag l
s e m
a
agl 4 9 6 1 c
m .
m a s e
e m . co agl
g l as
a
Bottom 4, 9, 6, 1 → third row 13, 15, 7 → second row 28, 22 → top 50.

co m
m .
m as e
.co


a g l Page 14 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

(ii) Bottom row 5, x, 7, y. The third-row box marked 9 sits on 7 and y:

7 + y = 9 → y = 2 (subtracting 7 from both sides)

40 = (5 + x) + (x + 7) = 2x + 12

2x = 28 → x = 14

70

40 30

19 21 9

5 14 7 2
Bottom 5, 14, 7, 2 → third row 19, 21, 9 → second row 40, 30 → top 70.

(iii) Bottom row 3, 5, x, y. The third-row box marked 7 sits on x and y, and the top of a four-row
pyramid is a + 3b + 3c + d:

x+y=7

3 + 3(5) + 3x + y = 35 → 18 + 3x + y = 35 → 3x + y = 17

Subtract x + y = 7 from 3x + y = 17: 2x = 10 → x = 5

y=7–5=2

Page 15 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

35

18 17

8 10 7

3 5 5 2
Bottom 3, 5, 5, 2 → third row 8, 10, 7 → second row 18, 17 → top 35.

Why it happens: subtracting one equation from another is legitimate for the same
reason as before — x + y and 7 are the same number, so taking away x + y from the
left of 3x + y = 17 and taking away 7 from the right removes equal amounts from
equal sides. What makes it useful is that the y disappears, leaving a single-letter
equation you can finish.

Check it yourself: rebuild (i) from the bottom — 4 + 9 = 13, 9 + 6 = 15, 6 + 1 = 7; 13 +
15 = 28, 15 + 7 = 22; 28 + 22 = 50 ✓.

Q2 What is the relationship between the numbers in the bottom row and the number
at the top?

The top is always a fixed combination of the bottom row — and the multipliers follow a pattern.

Page 16 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

ROWS BOTTOM ROW TOP MULTIPLIERS

2 a, b a+b 1, 1

3 a, b, c a + 2b + c 1, 2, 1

4 a, b, c, d a + 3b + 3c + d 1, 3, 3, 1

5 a, b, c, d, e a + 4b + 6c + 4d + e 1, 4, 6, 4, 1

Why it happens: think of a number in the bottom row climbing to the top. At each
step it can go up-left or up-right, and it is added into every box it reaches. So a
bottom entry is counted once for every path from its box to the top. The two end
entries have only one path each, so their multiplier is always 1; the inner entries
have many. And the list of multipliers for the next size is made by adding neighbours
in the list above — 1, 3, 3, 1 gives 1, 4, 6, 4, 1 — because a pyramid one row taller is
just two copies of the shorter one added together.

Did you know? the triangle of multipliers 1 / 1 1 / 1 2 1 / 1 3 3 1 / 1 4 6 4 1 is the
famous Pascal’s triangle, known in India centuries earlier as the meru-prastāra of
Piṅgala.

In-text Question — Page 140
6.3 Number Pyramids

Q1 What about a pyramid with three rows? Using letter numbers for the bottom row,
we can write an expression for the top row.

With bottom row a, b, c the middle row is a + b and b + c, so the top is their sum:

(a + b) + (b + c) = a + 2b + c

Page 17 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

a + 2b + c

a+b b+c

a b c

Bottom a, b, c → middle a + b, b + c → top a + 2b + c.

Why it happens: b sits under both boxes of the middle row, so it is added twice on
the way up, while a and c each feed only one box. The number in the middle of the
bottom row therefore has twice the pull of the numbers at the ends — change b by 1
and the top changes by 2.

Check it yourself: take a = 1, b = 9, c = 4. The top should be 1 + 18 + 4 = 23 — and
building it gives 10, 13 and then 23 ✓, exactly the pyramid printed on page 137.

Figure it Out — Page 140
6.3 Number Pyramids

Q1 Without building the entire pyramid, find the number in the topmost row given the
bottom row in each of these cases. (i) 4, 13, 8 (ii) 7, 11, 3 (iii) 10, 14, 25

Use the three-row formula top = a + 2b + c, doubling only the middle entry.

(i) 4 + 2(13) + 8 = 4 + 26 + 8 = 38

(ii) 7 + 2(11) + 3 = 7 + 22 + 3 = 32

(iii) 10 + 2(14) + 25 = 10 + 28 + 25 = 63

Page 18 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
m.
BOTTOM ROW MIDDLE ROW (CHECK) TOP

m as e
4, 13, 8
.co
17, 21
a g l 38

se m
g l a
7, 11, 3
a
18, 14 32

10, 14, 25 24, 39 63

co m
m . ag
l a se
Why it happens: the formula is not a shortcut that skips the pyramid — it is the
g all. Because a + 2b + c was derived from letter-
pyramid, written out once andafor
numbers, it holds for every bottom row, so there is nothing left to build.

co m
em.
m l as
m .co a g
s e Write an expression for the topmost row of a pyramid with 4 rows in terms of the
la values in the bottom row.
Q2

ag
m a s
.co agl

With bottom row a, b, c, d:
se m
g l a
a
Third row: a + b, b + c, c + d

co m
Second row: (a + b) + (b + c) = a + 2b + c and (b + c) + (c + d) = b + 2c + d
m .
m as e
.co
Top: (a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d
a g l
se m
g l a
a Why it happens: a and d sit at the ends, with just one path each to the top, so they
se m
m
are counted once. From b there are three ways up, and the same from c — which is
o l a
m .c also that 1 + 3 + 3 + 1 = 8 = 2³: each extra row ag
se row, because every box is copied into the two
why both carry a multiplier of 3. Notice
doubles the total pull of theabottom
ag l
boxes above it.

c o m
Tip: the pattern continues. Five rows give a + 4b + 6c + 4d + e, m .
s e and each new list of

. om
cmultipliers comes from adding neighbours in the previous
a glalist.
a s em
agl c
m .
m a s e
agl
Without building the entire pyramid, find the number in the topmost row given the
co
Q3

m .
bottom row in each of these cases. (i) 8, 19, 21, 13 (ii) 7, 18, 19, 6 (iii) 9, 7, 5, 11

ase
ANSWER a g l

m
Use top = a + 3b + 3c + d.

. co
e m
m l as
.co a g
Page 19 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

(i) 8 + 3(19) + 3(21) + 13 = 8 + 57 + 63 + 13 = 141

(ii) 7 + 3(18) + 3(19) + 6 = 7 + 54 + 57 + 6 = 124

(iii) 9 + 3(7) + 3(5) + 11 = 9 + 21 + 15 + 11 = 56

BOTTOM ROW THIRD ROW SECOND ROW TOP

8, 19, 21, 13 27, 40, 34 67, 74 141

7, 18, 19, 6 25, 37, 25 62, 62 124

9, 7, 5, 11 16, 12, 16 28, 28 56

Check it yourself: in (ii) and (iii) the second row came out with two equal boxes. A
quick shortcut for those: the top is just twice that number — 2 × 62 = 124 and 2 × 28
= 56 ✓.

Q4 If the first three Virahāṅka-Fibonacci numbers are written in the bottom row of a
number pyramid with three rows, fill in the rest of the pyramid. What numbers
appear in the grid? What is the number at the top? Are they all Virahāṅka-Fibonacci
numbers?

The first three Virahāṅka-Fibonacci numbers are 1, 2, 3.

Page 20 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

8

3 5

1 2 3
Bottom 1, 2, 3 → middle 3, 5 → top 8.

Middle row: 1 + 2 = 3, 2 + 3 = 5

Top: 3 + 5 = 8

The numbers appearing are 1, 2, 3, 3, 5, 8. The number at the top is 8. Yes — every one of them
is a Virahāṅka-Fibonacci number, and 8 is the 5th one in the sequence 1, 2, 3, 5, 8.

Why it happens: the pyramid rule and the Virahāṅka-Fibonacci rule are the same
rule. Adding two neighbours in 1, 2, 3, 5, 8, … gives the next term of the sequence.
So the middle row 1 + 2, 2 + 3 is just 3, 5 — the sequence again, but starting two
places later. Adding once more gives 8.

Q5 What can you say about the numbers in the pyramid and the number at the top in
the following cases? (i) The first four Virahāṅka-Fibonacci numbers are written in
the bottom row of a four row pyramid. (ii) The first 29 Virahāṅka-Fibonacci numbers
are written in the bottom row of a 29 row pyramid.

(i) The bottom row is 1, 2, 3, 5.

Page 21 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

21

8 13

3 5 8

1 2 3 5
Bottom 1, 2, 3, 5 → 3, 5, 8 → 8, 13 → top 21.

Every row is again a run of consecutive Virahāṅka-Fibonacci numbers, each row starting two
places later than the row below. The top is 21, the 7th Virahāṅka-Fibonacci number.

ROW (FROM THE BOTTOM) NUMBERS POSITION IN THE SEQUENCE

1 (bottom) 1, 2, 3, 5 1st to 4th

2 3, 5, 8 3rd to 5th

3 8, 13 5th to 6th

4 (top) 21 7th

(ii) The same thing happens, only for longer. Every number in a 29-row pyramid is a Virahāṅka-
Fibonacci number; row r counted from the bottom holds consecutive terms starting at the (2r –
1)th. The top is the (2 × 29 – 1) = 57th Virahāṅka-Fibonacci number, which is

Top = 591 286 729 879

Page 22 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: write the bottom row as F₁, F₂, …, F₂₉. Adding neighbours gives F₁ +
F₂ = F₃, F₂ + F₃ = F₄, and so on, so the second row is F₃, F₄, …, F₃₀. Each row up loses
one entry and starts two places further along. After 28 steps only one entry is left,
and it starts at position 1 + 2 × 28 = 57.

Tip: you never have to build 29 rows to answer this. Following the position in the
sequence instead of the numbers themselves is what makes the question easy.

Q6 If the bottom row of an n row pyramid contains the first n Virahāṅka-Fibonacci
numbers, what can we say about the numbers in the pyramid? What can we say
about the number at the top?

Every number in the pyramid is a Virahāṅka-Fibonacci number, and the top is the (2n –
1)th one.

Bottom row (row 1): F₁, F₂, …, Fn

Row 2: F₃, F₄, …, Fn+1

Row 3: F₅, F₆, …, Fn+2

Row r: F2r–1, F2r, …, Fn+r–1 (that is n – r + 1 numbers)

Row n (the top): the single number F2n–1

Why it happens: the defining rule Fj + Fj+1 = Fj+2 is exactly the pyramid rule. So
adding neighbours along a run of consecutive terms produces another run of
consecutive terms — one entry shorter, and starting two places later, because the
first sum Fj + Fj+1 lands on Fj+2. Going up n – 1 rows moves the starting position from
1 to 1 + 2(n – 1) = 2n – 1.

Page 23 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
m.
N BOTTOM ROW TOP POSITION 2N – 1

m as e
2
.co
1, 2 3 3rd
a g l
a s em1, 2, 3
gl
3
a
8 5th

4 1, 2, 3, 5 21 7th

co m
5 1, 2, 3, 5, 8
e m .
55 9th ag
g l as
a
Check it yourself: for n = 5 the formula a + 4b + 6c + 4d + e gives 1 + 8 + 18 + 20 + 8
= 55, and 55 is indeed the 9th number in 1, 2, 3, 5, 8, 13, 21, 34, 55. ✓
co m
em.
m l as
.co a g
em
ag las Questions — Page 141
In-text
6.4 Fun with Grids — Calendar Magic

om a s
e
. c
m grid from just knowing this sum? agl
s
Can we find the 4 numbers in the
a
Q1

agl

co m
.
Yes — the sum decides all four numbers. Let a be the top-left date of the 2 × 2 square.
e m
m l as
.co a g
eam
A A+1

a s
agl + 7 a+8

se m
4am+ 16
Sum = a + (a + 1) + (a + 7) + (a + 8) .=co g l a
em a
a s
agl
So a = (Sum – 16) ÷ 4

co
The example on the page checks out: for 6, 7, 13, 14 the sum is 40, and (40 – 16) ÷ 4 = 6 — the
m
m .
e
top-left date.
m l as
.co a g
m Why it happens: a calendar row is a week, so the box directly below any date holds
s e
agla a date 7 larger; the box to the right holds one more. That fixed spacing is what lets
.c
m
all four numbers be written in terms of the single letter-number a. Once the four

m a s e
co agl
cells are 4 copies of a plus the fixed extra 0 + 1 + 7 + 8 = 16, the sum determines a,

m .
and a determines the rest.
ase
a g l

co m
m .
m as e
.co


a g l Page 24 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Tip: not every number can be such a sum. Since Sum = 4a + 16, the sum must be a
multiple of 4 — and a must sit in a place where the 2 × 2 square actually fits inside
the month.

Q2 Suppose you are told that the sum is 36. Can you find the 4 numbers in the grid?

Yes. Solve 4a + 16 = 36.

4a + 16 = 36

4a = 20 (subtracting 16 from both sides)

a = 5 (dividing both sides by 4)

The four dates are a, a + 1, a + 7, a + 8 = 5, 6, 12, 13

5 6

12 13

Why it happens: subtracting 16 from both sides is allowed because both sides name
the same number, so removing 16 from each leaves them still equal; the same holds
for sharing both sides into 4 equal parts. Each step strips away one thing that was
done to a, in the reverse order it was done — that is what “solving” means.

Check it yourself: 5 + 6 + 12 + 13 = 36 ✓, and looking at the August 2025 page, 5
and 6 really do sit above 12 and 13.

In-text Questions — Page 142

Page 25 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

6.4 Fun with Grids — Algebra Grids

MATH TALK

Q1 Create your own calendar trick. For instance, choose a grid of a different size and
shape.

Any shape works — you only have to write every cell in terms of one letter-number. Here are
three tricks, all on the August 2025 page.
1. The 3 × 3 square. Let c be the middle date. The nine dates are:

c–8 c–7 c–6

c–1 c c+1

c+6 c+7 c+8

Sum = 9c + (–8 – 7 – 6 – 1 + 0 + 1 + 6 + 7 + 8) = 9c

So: “Tell me the sum, and I divide by 9 to get the middle date.”

Try it on 11, 12, 13 / 18, 19, 20 / 25, 26, 27: the sum is 171, and 171 ÷ 9 = 19, the middle date. ✓
2. The plus-shape (5 cells). With c in the centre the cells are c – 7, c – 1, c, c + 1, c + 7, so the sum
is 5c — divide by 5.
3. A 2 × 3 rectangle. With a at the top-left the cells are a, a + 1, a + 2, a + 7, a + 8, a + 9, so the
sum is 6a + 27 — subtract 27 and divide by 6.

Why it happens: a calendar is a grid with a fixed step of 1 across and 7 down, so
every cell of a chosen shape is c plus a fixed number. Adding them gives (number of
cells) × c + (a fixed total). If your shape is symmetric about its centre, those fixed
numbers cancel in pairs and the constant is 0 — which is why the 3 × 3 and the plus-
shape give such clean tricks.

Try This: the second grid printed on this page is not a calendar — it runs 2 to 50 with
ten numbers in a row. There the step downwards is 10, not 7, so a 2 × 2 square gives
a + (a + 1) + (a + 10) + (a + 11) = 4a + 22. Every trick has to be re-derived for the grid it
lives on.

Page 26 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q2 In the following grid, shapes represent numbers. In each row, the last column is the
sum of the values to its left. How do we find the values of the shapes?

Read each row as an equation and start with the row that has only one kind of shape.

Row 1: ■ + ■ + ■ = 27, so 3■ = 27

■ = 9 (dividing both sides by 3)

Row 2: ● + ● + ■ = 19

● + ● + 9 = 19 (replacing ■ by the number it stands for)

2● = 10 (subtracting 9 from both sides)

● = 5 (dividing both sides by 2)

So the blue square is 9 and the red circle is 5.

Why it happens: a shape here is doing the job of a letter-number — one shape, one
fixed value throughout the grid. Row 1 has three equal shapes, so it is a one-step
equation. Once ■ is known it can be put into row 2, and that turns a row with two
unknowns into a row with one. Solving several equations is usually this: use one
equation to remove an unknown from another.

Check it yourself: 9 + 9 + 9 = 27 ✓ and 5 + 5 + 9 = 19 ✓.

Q3 In the following grids, find the values of the shapes and fill in the empty squares.

Grid 1 — blue square ■ and red circle ●:

Page 27 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Row 1: ■ + ■ + ● = 27 → 2■ + ● = 27 … (1)

Row 2: ● + ● + ■ = 21 → 2● + ■ = 21 … (2)

Add (1) and (2): 3■ + 3● = 48

■ + ● = 16 (dividing both sides by 3)

Put this into (1), written as ■ + (■ + ●) = 27:

■ + 16 = 27 → ■ = 11

● = 16 – 11 = 5

Row 3: ● + ■ + ● = 5 + 11 + 5 = 21

ROW SHAPES IN NUMBERS SUM

1 ■■● 11 + 11 + 5 27 (given)

2 ●●■ 5 + 5 + 11 21 (given)

3 ●■● 5 + 11 + 5 21

Grid 2 — blue circle ● and purple diamond ◆:

Row 1: ● + ◆ + ◆ = 18 → ● + 2◆ = 18 … (1)

Row 2: ◆ + ● + ● = 15 → ◆ + 2● = 15 … (2)

Add (1) and (2): 3● + 3◆ = 33 → ● + ◆ = 11

From (1), written as (● + ◆) + ◆ = 18: 11 + ◆ = 18 → ◆ = 7

● = 11 – 7 = 4

Row 3: ◆ + ● + ● = 7 + 4 + 4 = 15

Row 3 has exactly the same shapes as row 2, so it had to give the same sum, 15. The fourth row
of this grid is left blank — it is yours to fill: put in any three shapes and work out its sum. For
example ◆ ◆ ◆ would give 21, and ● ● ◆ would give 15.

Page 28 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: adding the two equations is legitimate because each is a statement

m l a se
that two numbers are equal; adding equals to equals keeps equality. It is worth
o g
.c because both rows contain the same total number ofashapes,
m
doing here so the sum

l a se out perfectly symmetric — 3■ + 3●. That single step turns a pair of two-
g
comes
aunknown equations into the very simple ■ + ● = 16, and one substitution finishes it.

o m
Check it yourself: Grid 1 — 11 + 11 + m
e
. c ag
s
5 = 27 ✓, 5 + 5 + 11 = 21 ✓. Grid 2 — 4 + 7 + 7 =
a
agl
18 ✓, 7 + 4 + 4 = 15 ✓.

co m
em.
m
In-text Questions — Pages 142 – 143
coProduct g l as
. a
em
6.5 The Largest

a s
agl
MATH TALK

⬜⬜ ×.⬜co,musing each digit once. What is the largest a s
agl
Q1 Fill the digits 2, 3, and 5 in
product possible?
se m
g l a
a

com
.
The largest is 32 × 5 = 160.

e m
m l as
.co
ARRANGEMENT PRODUCT ARRANGEMENT
a g PRODUCT

a s e23m× 5
a gl 115 35 × 2 70

25 × 3 52 × 3
se m
com a
75 156

. a g l
m
ase
32 × 5 160 53 × 2 106

agl
Why it happens: there are only six arrangements, because the multiplier can be any

. c om
of the three digits and the remaining two can go either way round. The winner puts

s e m
the largest digit (5) as the multiplier and the other two in decreasing order (32).

. cThe a glagets multiplied by 10 and
omreason is place value: the tens digit of the multiplicand
a s em by the multiplier, so it should not be the biggest digit — the biggest digit does more
agl work standing outside, where it multiplies everything.
c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q2 How do we find the largest product among these six options?

Compare in pairs instead of testing all six one by one.

Group by multiplier: (35 × 2, 53 × 2) (25 × 3, 52 × 3) (23 × 5, 32 × 5)

In each pair the multiplier is the same, so the bigger multiplicand wins:

53 × 2, 52 × 3, 32 × 5 survive.

52 × 3 = 156 > 106 = 53 × 2, so 53 × 2 is out.

Now expand the last two:

32 × 5 = (3 × 10 × 5) + (2 × 5) = 150 + 10

52 × 3 = (5 × 10 × 3) + (2 × 3) = 150 + 6

The first parts are equal; the second part is larger for 32 × 5.

Largest = 32 × 5 = 160

Why it happens: splitting each product by place value is what makes the
comparison easy. Both 32 × 5 and 52 × 3 contain the term 3 × 5 × 10 — the same two
digits multiplied and scaled by ten. So they differ only in the smaller term, 2 × 5
against 2 × 3. Comparing two things that share most of their structure is far quicker,
and far more convincing, than computing every product.

Q3 In this case, we used the largest digit as the multiplier. The other two digits were
arranged in decreasing order to form the multiplicand. Will this always be the case?
Let us find out using algebra.

Yes — it is always the case. Take any three digits p, q, r with p < q < r. There are six
arrangements; group them by multiplier.

qr × p, rq × p | pr × q, rp × q | pq × r, qp × r

Page 30 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

In each pair the multiplier is the same, and the multiplicand with the larger tens digit is the
larger number. So only three survive:

rq × p, rp × q, qp × r

Step 1 — rp × q beats rq × p. Write both out by place value:

rq × p = (10r + q) × p = 10rp + qp

rp × q = (10r + p) × q = 10rq + pq

The second terms are equal (qp = pq). Since q > p, we get 10rq > 10rp.

So rp × q > rq × p.

Step 2 — qp × r beats rp × q.

qp × r = (10q + p) × r = 10qr + pr

rp × q = (10r + p) × q = 10rq + pq

Again the first terms are equal (10qr = 10rq). Since r > q, we get pr > pq.

So qp × r > rp × q.

The largest product is therefore qp × r = (10q + p) × r — the largest digit outside as the
multiplier, the other two in decreasing order.

Why it happens: in every product one pair of digits gets the ×10 boost and one pair
does not. Writing (10q + p) × r as 10qr + pr shows that the two largest digits q and r
are the ones multiplied together and scaled by 10 — the best possible use of the
boost. Any other arrangement wastes the ×10 on a pair that includes the smallest
digit p. Testing examples could never establish this for all digit sets; the letter-
numbers do it in two lines.

Check it yourself: with p = 2, q = 3, r = 5 the rule gives (10 × 3 + 2) × 5 = 32 × 5 = 160
— exactly the answer found by listing all six. ✓

Figure it Out — Page 144

Page 31 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

6.5 The Largest Product

Q1 Fill the digits 1, 3, and 7 in ⬜⬜ × ⬜ to make the largest product possible.

Here p = 1, q = 3, r = 7. The rule gives multiplier 7 and multiplicand 31.

Largest = qp × r = 31 × 7 = 217

ARRANGEMENT PRODUCT ARRANGEMENT PRODUCT

13 × 7 91 37 × 1 37

17 × 3 51 71 × 3 213

31 × 7 217 73 × 1 73

Tip: 71 × 3 = 213 comes close, and it is the arrangement that tempts most people —
the biggest digit in the tens place looks right. But 31 × 7 = 10 × 3 × 7 + 1 × 7 = 210 + 7,
while 71 × 3 = 10 × 7 × 3 + 1 × 3 = 210 + 3. Same first term; the second decides it.

Q2 Fill the digits 3, 5, and 9 in ⬜⬜ × ⬜ to make the largest product possible.

Here p = 3, q = 5, r = 9. The rule gives multiplier 9 and multiplicand 53.

Largest = qp × r = 53 × 9 = 477

ARRANGEMENT PRODUCT ARRANGEMENT PRODUCT

35 × 9 315 59 × 3 177

39 × 5 195 93 × 5 465

53 × 9 477 95 × 3 285

Page 32 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: 53 × 9 = 10 × 5 × 9 + 3 × 9 = 450 + 27 = 477, while its nearest rival 93
× 5 = 10 × 9 × 5 + 3 × 5 = 450 + 15 = 465. Both give the two big digits the ×10 boost;
the winner is the one whose leftover digit 3 is multiplied by the larger of the two,
namely 9.

In-text Questions — Page 145
6.6 Decoding Divisibility Tricks

Q1 If we choose other 2-digit numbers, and follow the steps, will there always be no
remainder?

Yes — there is never a remainder, for any starting two-digit number.

Let the number be ab, that is 10a + b. Reversed it is ba = 10b + a.

If b > a, the difference is

(10b + a) – (10a + b)

= 10b – b – 10a + a

= 9b – 9a

= 9(b – a)

Since 9(b – a) is 9 times a whole number, dividing it by 9 always leaves remainder 0.

Why it happens: the two numbers are built from the same two digits, just swapped
between the tens place and the units place. Each digit therefore gains 10 of itself in
one number and loses 1 of itself in the other — a change of 9 of itself. Subtracting
leaves nine of each digit, and any combination of nines is a multiple of 9. Notice how
little we assumed: only that a and b are digits. That is why the trick can never fail.

Check it yourself: 47 → 74. 74 – 47 = 27 = 9 × 3, and b – a = 7 – 4 = 3 ✓.

Page 33 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
m.
Can you work out what happens if a > b?
e
Q2

m l as
.co a g
a
s em
agl the same thing — you just subtract the other way round.
Exactly

com
. ag
If a > b then ab > ba, so the difference is
e m
(10a + b) – (10b + a)
g l as
= 10a – a + b – 10b
a

co m
m.
= 9a – 9b

m as e
.co
= 9(a – b)
a g l
a s em
gSol in both cases the difference is 9 × (the difference of the two digits), and dividing by 9
a leaves no remainder.
m a s
m .co DIFFERENCE agl
se
NUMBER REVERSED ÷9

g l a
72 (a > b) 27
a 45 5=7–2

m
.co
38 (b > a) 83 45 5=8–3

se m
com l a
91 (a > b) 19 72 8=9–1

. a g
e m
as Why it happens: the whole trick depends only on how far apart the two digits are,
agl
m
not on which one is bigger. Writing the difference as 9 × |a – b| covers both cases at

m so that the answer is a genuine digit l a se
codigits”,
once. If a = b the difference is 0 — still a multiple of 9 — which is exactly why the
. a g
em
trick asks for a number “of different
a s
agl
from 1 to 8.

co m
m .
Figure it Out — Pages 145 – 147
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 34 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

6.6 Decoding Divisibility Tricks

MATH TALK

Q1 In the trick given above, what is the quotient when you divide by 9? Is there a
relationship between the two numbers and the quotient?

The quotient is the difference of the two digits.

Difference of the numbers = 9 × (larger digit – smaller digit)

Quotient = |a – b|

NUMBER REVERSED DIFFERENCE QUOTIENT DIGITS DIFFER BY

74 47 27 3 7–4=3

52 25 27 3 5–2=3

81 18 63 7 8–1=7

96 69 27 3 9–6=3

Why it happens: the difference of the two numbers is exactly 9(a – b) or 9(b – a), so
dividing by 9 undoes the 9 and hands back the digit gap. There is a nice
consequence: the quotient does not depend on the digits themselves, only on how
far apart they are. That is why 74, 52 and 96 all give the same quotient 3 — and why
the quotient can only ever be one of 1, 2, …, 8.

Try This: ask a friend for the quotient instead of the number. You will know the gap
between their digits at once — a good start for guessing the number itself.

Page 35 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q2 In the trick given above, instead of finding the difference of the two 2-digit
numbers, find their sum. What will happen? For example: We start with 31. After
reversing we get 13. Adding 31 and 13, we get 44. We start with 28. After reversing
we get 82. Adding 28 and 82, we get 110. We start with 12. After reversing we get 21.
Adding 12 and 21, we get 33. Observe that all these numbers are divisible by 11. Is
this always true? Can we justify this claim using algebra?

Yes, it is always true, and the algebra is one line.

(10a + b) + (10b + a)

= 10a + a + 10b + b

= 11a + 11b

= 11(a + b)

So the sum is always a multiple of 11, and the quotient is a + b, the sum of the digits.

NUMBER REVERSED SUM SUM ÷ 11 A+B

31 13 44 4 3+1=4

28 82 110 10 2 + 8 = 10

12 21 33 3 1+2=3

Why it happens: in the two numbers together, each digit occupies the tens place
once and the units place once. So each digit contributes 10 of itself plus 1 of itself,
that is 11 of itself. Subtracting compared the digits and produced 9s; adding pools
them and produces 11s. Both facts come from the same source — the place values
10 and 1 differ by 9 and add to 11.

Tip: here the two digits need not be different, and the digits may even be equal. 55 +
55 = 110 = 11 × 10 ✓.

Page 36 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q3 Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit
numbers from these digits by cycling these digits around, yielding bca and cab. Now
add the three numbers. Using algebra, justify that the sum is always divisible by 37.
Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]

Write all three numbers by place value and add.

abc = 100a + 10b + c

bca = 100b + 10c + a
cab = 100c + 10a + b

─────────────────────

Sum = (100a + a + 10a) + (10b + 100b + b) + (c + 10c + 100c)

= 111a + 111b + 111c

= 111(a + b + c)

Now factorise 111. Looking at multiples of 37 — 37, 74, 111 — we see that

111 = 3 × 37

So Sum = 3 × 37 × (a + b + c)

The sum is therefore always divisible by 37, and always divisible by 3 as well — in fact it is
divisible by 111.

NUMBER THE THREE CYCLES SUM ÷ 37 ÷3

253 253 + 532 + 325 1110 30 370

147 147 + 471 + 714 1332 36 444

908 908 + 089 + 890 1887 51 629

Why it happens: cycling the digits sends each digit through the hundreds, the tens
and the units place exactly once. So over the three numbers every digit is counted
100 + 10 + 1 = 111 times, whatever the digits are. The whole result is decided by that
111 — and since 111 = 3 × 37, both divisibility claims follow at once. Note the
quotient on dividing by 111 is a + b + c, the digit sum.

Page 37 of 48

Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Tip: in the last row 089 is not really a three-digit number, but the arithmetic is
unaffected — 111(9 + 0 + 8) = 111 × 17 = 1887 ✓.

Q4 Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the
digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do
you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11
and 13.]

You get back the original three-digit number abc, every time — and none of the three
divisions leaves a remainder.

abcabc = abc × 1000 + abc = abc × 1001
And 7 × 11 × 13 = 77 × 13 = 1001

So abcabc = abc × 7 × 11 × 13

Dividing by 7, then 11, then 13 peels off exactly those three factors and leaves abc.

START ÷7 ÷ 11 ÷ 13

253253 36179 3289 253

481481 68783 6253 481

907907 129701 11791 907

Why it happens: writing abc twice shifts the first copy three places to the left, which
multiplies it by 1000, and then adds one more copy — so the six-digit number is abc
× (1000 + 1) = abc × 1001. Nothing about the actual digits matters. The surprise is
really a fact about 1001: it happens to be the product of the three consecutive
primes 7, 11 and 13. The order of the divisions does not matter either, since
multiplication can be done in any order.

Try This: repeat a two-digit number instead — abab = ab × 101, and 101 is prime, so
there is no similar chain of three divisions. Repeating a four-digit block gives × 10001
= 73 × 137.

Page 38 of 48

as e
Class 8 Maths Chapter 13 Algebra Play
a g l AglaSem · NCERT Solutions

co m
m.
There are 3 shrines, each with a magical pond in the front. If anyone dips flowers
se
Q5

o m l a
into these magical ponds, the number of flowers doubles. A person has some
g flowers in shrine 1.
m .c He dips them all in the first pond and then places some
flowers. a
l a se he dips the remaining flowers in the second pond and places some flowers in
Next,
a g shrine 2. Finally, he dips the remaining flowers in the third pond and then places
them all in shrine 3. If he placed an equal number of flowers in each shrine, how

co m
. ag
many flowers did he start with? How many flowers did he place in each shrine?

e m
g l as
a
He started with 7 flowers and placed 8 flowers in each shrine.

co m
m.
Let x be the number of flowers he started with and k the number placed at each shrine.

m as e
c o
. 1: 2x → leaves k → carries 2x – k a g l
e m
After pond
s
a glaAfter pond 2: 2(2x – k) = 4x – 2k → leaves k → carries 4x – 3k
s
After pond 3: 2(4x – 3k) = 8x – 6k → leaves all of it at shrine 3
m a
m .co agl
l a se
So 8x – 6k = k
a g
8x = 7k (adding 6k to both sides)

co m
m .
ase
comis x = 7, which gives k = 8. l
x and k are whole numbers, so 7 must divide 8x, and therefore 7 divides x. The smallest
. a g
em
possibility

a s
agl STAGE FLOWERS

se m
com g l a
.
Start 7

m a
ase
agl
Dip in pond 1 14

Leave 8 at shrine 1 6

co m
Dip in pond 2
m .
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12

m l as
.co a g
em
Leave 8 at shrine 2 4

a s
agl Dip in pond 3 8

.c
s e m
m a
Leave all 8 at shrine 3 0

e m . co agl
g l as
a

co m
m .
m as e
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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: the condition 8x = 7k says the answer is not a single pair of
numbers but a whole family — 7 and 8, 14 and 16, 21 and 24, and so on, all in the
ratio 7 : 8. This is what happens when one equation carries two unknowns: it fixes
the ratio, not the sizes. Requiring whole flowers is what pins down the smallest, and
most natural, answer.

Tip: working backwards is quicker if you only want the smallest case. He ends with 0
after leaving k at shrine 3, so before that dip he had k/2; before the second offering,
k/2 + k = 3k/2; before the second dip, 3k/4; and so on. For all these to be whole
numbers k must be a multiple of 8.

Q6 A farm has some horses and hens. The total number of heads of these animals is 55
and the total number of legs is 150. How many horses and how many hens are on
the farm? Can you solve this without letter-numbers? [Hint: If all the 55 animals
were hens, then how many legs would there be? Using the difference between this
number and 150, can you find the number of horses?]

20 horses and 35 hens.
With letter-numbers. Let h be the number of horses and n the number of hens. Each animal
has one head; a horse has 4 legs and a hen has 2.

h + n = 55 … (heads)

4h + 2n = 150 … (legs)

Halve the second equation: 2h + n = 75 (dividing both sides by 2)

Subtract the first from this: (2h + n) – (h + n) = 75 – 55

h = 20

n = 55 – 20 = 35

Without letter-numbers (the hint’s method):

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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

If all 55 animals were hens, legs = 55 × 2 = 110

But there are 150 legs — that is 150 – 110 = 40 legs too few counted

Turning one hen into a horse adds 4 – 2 = 2 legs

Number of horses = 40 ÷ 2 = 20, so hens = 55 – 20 = 35

Why it happens: the two methods are the same argument in different clothing.
“Assume all hens” is exactly the step of halving the leg equation and subtracting the
head equation: 2h + n – (h + n) = h. Each horse counted as a hen loses 2 legs, so the
shortfall of 40 legs, shared 2 at a time, counts the horses.

Check it yourself: 20 + 35 = 55 heads ✓ and 20 × 4 + 35 × 2 = 80 + 70 = 150 legs ✓.

Q7 A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times
her daughter’s age. How old is the daughter now?

The daughter is 6 years old now, and the mother is 30.

Let the daughter’s present age be d. Then the mother’s present age is 5d.
In 6 years: daughter = d + 6, mother = 5d + 6.

The condition says: 5d + 6 = 3(d + 6)

5d + 6 = 3d + 18

2d + 6 = 18 (subtracting 3d from both sides)

2d = 12 (subtracting 6 from both sides)

d = 6 (dividing both sides by 2)

Mother = 5 × 6 = 30

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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Why it happens: the modelling step is the one to watch. “In 6 years” adds 6 to both
ages, so the mother’s future age is 5d + 6 — not 5(d + 6). And the multiple changes: 5
times now, only 3 times later, because the same 6 years is a much bigger share of
the daughter’s life than of the mother’s. Writing the condition as one equation and
then removing 3d from both sides is what turns a puzzle about two people into a
single statement about d.

Check it yourself: now 30 and 6, and 30 = 5 × 6 ✓. In 6 years, 36 and 12, and 36 = 3 ×
12 ✓.

Q8 Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the
road with their cows. Gauri says to Naina, “You have twice as many cows as I do”.
Naina says, “That’s true, but if I gave you three of my cows, we would each have the
same number of cows”. How many cows do Gauri and Naina have?

Gauri has 6 cows and Naina has 12.

Let Gauri have g cows. Naina’s remark that she has twice as many gives Naina = 2g.

If Naina gives 3 cows to Gauri:

Naina then has 2g – 3, Gauri then has g + 3, and these are equal:

2g – 3 = g + 3

g – 3 = 3 (subtracting g from both sides)
g = 6 (adding 3 to both sides)

Naina = 2 × 6 = 12

Why it happens: giving away 3 cows is a double move — Naina goes down by 3 and
Gauri goes up by 3, so the gap between them closes by 6, not 3. Since the gap has to
close completely, the original gap must have been 6. And the original gap is 2g – g =
g, which is why g = 6 falls out so quickly.

Check it yourself: 12 = 2 × 6 ✓. After the gift, Naina has 9 and Gauri has 9 ✓.

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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q9 I run a small dosa cart and my expenses are as follows: Rent for the dosa cart is
₹5000 per day. The cost of making one dosa (including all the ingredients and fuel)
is ₹10. (i) If I can sell 100 dosas a day, what should be the selling price of my dosa to
make a profit of ₹2000? (ii) If my customers are willing to pay only ₹50 for a dosa,
how many dosas should I aim to sell in a day to make a profit of ₹2000?

(i) ₹80 per dosa. (ii) 175 dosas a day.
(i) Let the selling price be ₹p per dosa.

Money taken in = 100p

Money spent = rent + making cost = 5000 + 100 × 10 = ₹6000
Profit = 100p – 6000 = 2000

100p = 8000 (adding 6000 to both sides)

p = ₹80 (dividing both sides by 100)

(ii) Let the number of dosas sold be n, each at ₹50.

Money taken in = 50n Money spent = 5000 + 10n

Profit = 50n – (5000 + 10n) = 2000

40n – 5000 = 2000

40n = 7000 (adding 5000 to both sides)

n = 175 dosas (dividing both sides by 40)

Why it happens: the two costs behave differently, and the algebra makes that
visible. The ₹5000 rent is fixed — it does not change with n — while the ₹10 per dosa
grows with n. So each dosa sold at ₹50 contributes 50 – 10 = ₹40 towards the rent
and the profit. First ₹5000 of that goes on rent; the ₹2000 profit needs 2000 ÷ 40 =
50 more dosas after the 125 that cover the rent. 125 + 50 = 175.

Check it yourself: (i) 100 × 80 = ₹8000 taken in; 5000 + 1000 = ₹6000 spent; profit
₹2000 ✓. (ii) 175 × 50 = ₹8750 taken in; 5000 + 1750 = ₹6750 spent; profit ₹2000 ✓.

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Class 8 Maths Chapter 13 Algebra Play
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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

Q11 Karim and the Genie. Karim was taking a nap under a tree. He had a dream about a
magical lamp and a genie… The genie said, “Do you see the banyan tree over
there? All you have to do is go around it once. The money in your pocket will
double”, and then, “Since I am bringing you great riches, you should share some of
your gains with me. You must give me 8 coins each time you go around the tree.”
He went around the tree once and the number of coins doubled; he gave 8 coins to
the genie. He made another round; again the number doubled and he gave 8 more
coins. He went around the tree for the third time. The number of coins doubled
again, but to his horror, he was left with only 8 coins, exactly the number of coins
he owed the genie! (i) How many coins did Karim initially have? (ii) For what cost
per round should Karim agree to the deal, if he wants to increase the number of
coins he has? (iii) Through its magical powers, the genie knows the number of
coins that Karim has. How should the genie set the cost per round so that it gets
all of Karim’s coins?

(i) Karim started with 7 coins. Let x be the number of coins he began with.

After round 1: 2x, then pay 8 → 2x – 8

After round 2: 2(2x – 8) = 4x – 16, then pay 8 → 4x – 24

After round 3 (the doubling): 2(4x – 24) = 8x – 48

He was left with exactly 8 coins: 8x – 48 = 8

8x = 56 (adding 48 to both sides)
x = 7 (dividing both sides by 8)

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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

STAGE COINS

Start 7

Round 1 doubles 14

Pay the genie 8 6

Round 2 doubles 12

Pay the genie 8 4

Round 3 doubles 8

Pay the genie 8 0

(ii) The cost per round must be less than the number of coins he has. If he holds x coins and
the cost is c, then after a round he holds 2x – c, and

2x – c > x ⟺ c < x (subtracting x from both sides)

In the story Karim had 7 coins, so any cost of 6 coins or less would have made him richer. With c
= 7 he would stay at 7 forever, and with c = 8 he lost everything.

Why it happens: once c < x holds at the start, it keeps holding — after the round he
has 2x – c, which is more than x, so it is still more than c. His pile then grows every
single round, without limit. The number c is a tipping point: below it Karim wins
forever, above it he is ruined, and exactly at it nothing ever changes.

(iii) The genie must set the cost at c = 2ⁿ x ÷ (2ⁿ – 1), where x is what Karim starts with and n is
the number of rounds the genie wants it to take.

Coins after n rounds = 2nx – c(2n–1 + … + 2 + 1) = 2nx – c(2n – 1)

For this to be 0: c(2n – 1) = 2nx

c = 2nx ÷ (2n – 1)

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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

ROUNDS N COST C WITH X = 7

1 2x 14

2 4x/3 28/3 — not a whole number

3 8x/7 8 ✓ — the genie’s choice

Why it happens: the fastest way is simply c = 2x — Karim doubles to 2x and hands
over all of it in one round. To drag it out over n rounds the genie must charge a little
less than double, namely 2x × 2ⁿ/(2ⁿ – 1), and that fraction shrinks towards 2x as n
grows. In the story x = 7 and n = 3 give c = 8 × 7 ÷ 7 = 8, exactly the price the genie
named — it knew Karim’s 7 coins and chose the cost that emptied his pocket in
precisely three rounds.

Tip: the deal is worth taking only when the cost is below your holding, so the genie’s
trap is that 8 was just one coin above Karim’s 7. Had he carried 9 coins, the same deal
would have made him richer round after round.

Chapter at a glance
A letter-number stands for every possible starting value at once. That is why one line of
algebra settles a trick that no amount of trying out numbers can ever settle — testing 100
starting numbers only tests 100 of them.
Solving an equation means doing the same thing to both sides — adding, subtracting,
multiplying or dividing by the same amount. The two sides name one number, so an
operation applied to both keeps them naming one number. That is the whole justification
for every step.
Modelling a word situation means naming the unknown, then translating each sentence
into an equation about it. “The mother is 5 times her daughter’s age” becomes m = 5d; “in 6
years” becomes d + 6 and m + 6.
In a number pyramid each box is the sum of the two below it. Read forwards it is addition;
read backwards it is subtraction — and when no box has both its neighbours known, letter-
numbers finish the job.
The top of an n-row pyramid is a fixed combination of the bottom row: a + b for 2 rows, a +
2b + c for 3, a + 3b + 3c + d for 4. Each bottom entry is counted once for every way of
climbing from it to the top.
Place value is what most of these tricks really use. A two-digit number is 10a + b, so
reversing it and subtracting leaves 9(b – a); adding leaves 11(a + b); cycling the digits of 100a

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Class 8 Maths Chapter 13 Algebra Play AglaSem · NCERT Solutions

+ 10b + c leaves 111(a + b + c) = 3 × 37 × (a + b + c).
The same habit answers optimisation questions. With digits p < q < r, the largest product of
the form ⬜⬜ × ⬜ is always (10q + p) × r — largest digit as the multiplier, the other two in
decreasing order — and algebra proves it for all digit sets, not just the one you tried.

Quick revision

IDEA WHAT THE ALGEBRA SHOWS KEY EXPRESSION WHERE IT
APPEARS

‘Think of a Every step acts on x, so the x-terms cancel 2x + 4, halved, minus x = Page 135
Number’ trick and a constant survives 2

Date trick ×5, ×4, ×5 multiplies the month by 100, so Answer = 100M + 165 + D Pages 136 –
month and day sit in separate place-value 137
slots

Number pyramid Each box is the sum of the two below; read 10 – 4 = 6, 4 – 1 = 3 Page 138
backwards, a missing box is a subtraction

Pyramid in letter- When no box has both neighbours known, 20 + 2c = 60 → c = 20 Pages 138 –
numbers name the unknowns and solve 139

Top of a pyramid A bottom entry is counted once for each path 3 rows: a + 2b + c; 4 rows: Page 140
up to the top a + 3b + 3c + d

Virahāṅka- Fj + Fj+1 = Fj+2, so each row up starts two Top of n rows = the (2n – Page 140
Fibonacci pyramid places later in the sequence 1)th V-F number

Calendar 2 × 2 A week is 7 days, so the box below a date a + (a+1) + (a+7) + (a+8) = Page 141
grid holds 7 more 4a + 16

Algebra grid Each row is one equation; combining rows 2■ + ● = 27, 2● + ■ = 21 → Page 142
removes a shape ■ = 11, ● = 5

Largest product Largest digit as multiplier, other two in For p < q < r the winner is Pages 142 –
decreasing order (10q + p) × r 144

Reversing two 10b + a minus 10a + b leaves nine of each Difference = 9(b – a); Sum Pages 144 –
digits digit = 11(a + b) 145

Cycling three Each digit visits the hundreds, tens and units abc + bca + cab = 111(a + Page 145
digits place exactly once b + c)

Repeating a block Writing abc twice multiplies it by 1001 1001 = 7 × 11 × 13 Page 145

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