NCERT Solutions Class 7 Maths Chapter 15 Finding the Unknown

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

C L A S S 7 · M AT H S

NCERT Solutions

Chapter 15: Finding the
Unknown

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 164 – 190 20 89 English

Solutions, notes, sample papers & more at 95 pages

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 15: Finding the Unknown
Chapter 7 of Ganita Prakash Grade 7 Part II begins with a weighing scale and ends with algebra. Once you
notice that removing the same weight from both pans never disturbs a balance, you have the one rule that
solves every equation: do the same thing to both sides. The chapter builds from picture-puzzles to
equations like 6y + 7 = 4y + 21, to word problems, and finally to Brahmagupta's own formula for solving them.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 164 – 190

SECTIONS QUESTIONS

20 89

MEDIUM

English

In-text Questions — Pages 164 – 165

Page 1 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

7.1 Find the Unknowns — Unknown Weights

Q1 Find the unknown weights in the following cases:

16

=3 = =

Fig. 7.1 — the scale is marked 16 and the beam hangs level. A leaf weighs 3.

Bud = 2 and flower = 8.
First read what the picture is telling us. The number in the circle at the top is the total weight
hanging from the scale. The beam is level, so the two strings carry equal weight.

Page 2 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Total = 16, and the two sides are equal

So each side carries 16 ÷ 2 = 8

Left string: leaf + bud + leaf = 8

3+b+3=8

b+6=8

b=8−6=2

Right string: flower = 8, so flower = 8

16

3

f
b

3

3+b+3=8 f=8
Total 16 splits equally into 8 and 8, because the beam is level.

Why it happens: the sample pictures at the start of the section show the rule. A
scale marked 4 with 2 and 2 below it hangs level; a scale marked 7 with 4 and 3
below it tilts. So the circled number is the sum of the two sides, and a level beam
means the two sides are equal. Once you know one side is 8, the left string becomes
a small puzzle: two leaves already use 6, so the bud must make up the last 2.

Page 3 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: 3 + 2 + 3 = 8 on the left, 8 on the right, and 8 + 8 = 16 — the

m as e
l
number in the circle. Everything fits.

m .co a g
l a se
a g
Q2 Find the unknown weights in the following cases:

co m
m . ag
l a se
ag 24

co m
em.
m l as
m .co a g
l a se
a g
m a s
m.co agl
l a se
ag

co m
m .
m l a s=e
.co ag
=2 =
se m
g l a
a
m
Fig. 7.2 — the scale is marked 24 and the beam hangs level. A starfish weighs 2.

a se
. com a g l
m
ase
agl
Striped fish = 8 and grey fish = 4.

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

com
m .
m ase
.co


a g l Page 4 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Total = 24, beam is level, so each side = 24 ÷ 2 = 12

Left string: starfish + striped fish + starfish = 12
2 + f + 2 = 12

f + 4 = 12

f = 8 (the striped fish)

Right string: striped fish + grey fish = 12

8 + g = 12

g = 12 − 8 = 4 (the grey fish)

Why it happens: notice the order of work. The left string had only one unknown, so
it could be solved straight away. That answer then unlocked the right string, which
had two unknowns to start with. When a puzzle has several unknowns, always look
first for the part that has just one.

Check it yourself: left = 2 + 8 + 2 = 12, right = 8 + 4 = 12, total = 24. ✓

Page 5 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q3 Find the unknown weights in the following cases:

8

= =

Fig. 7.3 — the scale is marked 8. A book hangs on the left; on the right a second beam
carries two identical pencil boxes. Both beams hang level.

Book = 4 and each pencil box = 2.

Top beam is level, total = 8

So each side of the top beam = 8 ÷ 2 = 4

Left side is only the book, so book = 4

Right side = the small beam with its two boxes = 4

The small beam is level too, so its two sides are equal

box + box = 4

2 × box = 4

box = 4 ÷ 2 = 2

Page 6 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why it happens: a scale hanging from a scale is just the same rule used twice.
Whatever hangs from a beam is shared equally between its two strings when the
beam is level. So 8 splits into 4 and 4, and then the 4 on the right splits again into 2
and 2.

Tip: work from the top downwards in these hanging puzzles. The top number is the
biggest piece of information you have.

Q4 Find the unknown weights in the following cases:

18

= =5 =

Fig. 7.4 — the scale is marked 18 and the beam hangs level. The sun weighs 5.

Cloud = 1 and lightning bolt = 3.

Page 7 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Total = 18, beam is level, so each side = 18 ÷ 2 = 9

Left string: sun + 4 clouds = 9
5 + 4c = 9

4c = 9 − 5 = 4

c = 4 ÷ 4 = 1 (one cloud)

Right string: 3 lightning bolts = 9

3L = 9

L = 9 ÷ 3 = 3 (one bolt)

Why it happens: here the same object appears several times, so we can group. Four
clouds weigh 4c together. Taking away the sun's 5 from both sides leaves 4c = 4, and
sharing 4 equally among 4 clouds gives 1 each. This is exactly the two moves you will
use again and again in this chapter: remove a term, then remove a factor.

Check it yourself: 5 + 1 + 1 + 1 + 1 = 9 and 3 + 3 + 3 = 9, and 9 + 9 = 18. ✓

Page 8 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Find the unknown weights in the following cases:
e
Q5

m l as
m .co a g
l a se 40
a g

co m
e m . ag
g l as
a

co m
em.
m l as
m.co a g
l a se
ag
m a s
m .co agl
l a se
ag
= =

co m
m .
as e
com l
Fig. 7.5 — the scale is marked 40 and the beam hangs level.

. a g
m
ase
agl ANSWER
se m
Crown = 5 and gem = 3.
com g l a
m . a
ase
agl

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 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Total = 40, beam is level, so each side = 40 ÷ 2 = 20

Right string first — it has only one kind of object:
4 crowns = 20

4k = 20, so k = 20 ÷ 4 = 5 (one crown)

Left string: crown + 5 gems = 20

5 + 5g = 20

5g = 20 − 5 = 15

g = 15 ÷ 5 = 3 (one gem)

Why it happens: the right string is the easy door into this puzzle — four equal
crowns sharing 20 must be 5 each. Neither string could be solved on its own if we
had started on the left, because the left has two different unknowns. Choosing the
right starting point saves all the work.

Check it yourself: left = 5 + 3 + 3 + 3 + 3 + 3 = 20, right = 5 + 5 + 5 + 5 = 20, total 40. ✓

Page 10 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q6 Find the unknown weights in the following cases:

=2 =

Fig. 7.6 — the balance hangs level. One slice of bread weighs 2.

One fried egg = 3.
This picture has no number at the top. It is a plain balance: the beam is level, so the two arms
carry equal weight.

Left arm: 3 slices of bread = 2 + 2 + 2 = 6

Right arm: 2 fried eggs = e + e = 2e

The arms balance, so

2e = 6

e=6÷2=3

Why it happens: a level beam is a statement of equality — exactly what an equation
is. The book writes this same picture as 6 = e + e, or 2e = 6. From here on, every
balance picture in this chapter can be turned into an equation in one line.

Page 11 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Check it yourself: two eggs weigh 3 + 3 = 6, the same as three slices of bread. The
beam stays level. ✓

Q7 Find the unknown weights in the following cases:

=4 =

Fig. 7.7 — the balance hangs level. One star weighs 4.

One doughnut = 6.

Left arm: 4 stars = 4 × 4 = 16

Right arm: doughnut + star + doughnut = y + 4 + y = 4 + 2y

The beam is level, so

4 + 2y = 16

2y = 16 − 4 (remove the term 4 from the left side)

2y = 12

y = 12 ÷ 2 = 6

Page 12 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why it happens: there is a star on both arms. Taking that one star off both arms at
the same time does not disturb the balance — it leaves 3 stars = 12 on the left and 2
doughnuts on the right, so 2y = 12. Removing equal weights from both pans is the
whole idea of this chapter, and it is the same as the algebra step "subtract 4 from
both sides".

Tip: the book frames this picture as the equation 4 + 2y = 16. Compare it with your
own working — they are the same steps, written in symbols.

Q8 Find the unknown weights in the following cases:

= 10 =4 =

Fig. 7.8 — the balance hangs level. The watermelon slice weighs 10 and the orange
weighs 4.

One banana = 3.

Page 13 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
e m.
Left arm: watermelon = 10
m l as
.co
Right arm: banana + orange + banana = b + 4 + b = 2b + 4
m a g
l a se
g
aThe beam is level, so

co m
. ag
2b + 4 = 10
e m
g l as
2b = 10 − 4 (remove the orange from the right arm and 4 from the left)

2b = 6 a

co m
m.
b=6÷2=3

m as e
.co a g l
a s em
Why it happens: think of it as taking the orange off the right pan. To keep the

a gl balance you must take away the same 4 units from the left pan too, leaving 6 units
against two bananas. Each banana is therefore 3.

m a s
m .co agl
l a se
Check it yourself: 3 + 4 + 3 = 10, which is exactly the watermelon. ✓
ag

co m
m .
e
Math Talk — Page 165
m l as
.co a g
7.1 Find the Unknowns

a s emTALK
gl
MATH

a
se m
Q1
com
Discuss the answers with your classmates. Give reasons why you think your answer
g l a
is right.
m . a
ase
ANSWER agl
m
A reason is not the same as an answer. Say why each step is allowed, not just what the number

. co
m
is. Here is the kind of reasoning to bring to the discussion.

m as e
.co a g l
Why the top number splits in half: the sample pictures show a scale marked 4 carrying 2

a s em and 2 hanging level, and a scale marked 7 carrying 4 and 3 tilting. So the circled number is

agl the total, and a level beam means the two sides are equal. Equal parts of 16 are 8 and 8.

.c
m
Why a level beam gives an equation: "the left side weighs the same as the right side" is

m a s e
co agl
exactly what the sign '=' means.

m .
se
Why removing the same weight from both sides is safe: if two things were equal and you

l a
g and the orange in Fig. 7.8.
take away the same amount from each, what is left must still be equal. That is how the star
was cancelled in Fig.a
7.7

co m
m .
m ase
.co


a g l Page 14 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why the answer must be the only one: in Fig. 7.4, once 4c = 4, no number other than 1 can
work for c, because 4 × 1 is the only product of 4 and a single value that gives 4.

Try this in class: two friends may have used different routes — one may have
started from the left string, another from the right. Compare them. Different
methods that reach the same answer are a good sign that the answer is right.

In-text Questions — Pages 165 – 166
7.1 Find the Unknowns — Unknown Weights

Q1 Find the unknown weight of the sack in the following cases.

2kg 10 kg 2kg

Fig. 7.9 — the weighing scale is balanced.

[Hint: If we remove equal weights from both the plates, will the weighing scale still
be balanced?]

The sack weighs 10 kg.

Page 15 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Left pan = sack + 2 kg Right pan = 10 kg + 2 kg = 12 kg

The scale is balanced, so

s + 2 = 12
s + 2 − 2 = 12 − 2 (take the 2 kg weight off both pans)

s = 10 kg

Why it happens: there is a 2 kg weight sitting on each pan. Lifting both off at the
same moment removes the same amount from each side, so the scale stays
balanced. What is left is one sack on the left and 10 kg on the right — and now the
answer can simply be read off.

Tip: the aim in every one of these puzzles is to get the sacks alone on one pan. Every
weight you can cancel off both pans brings you closer.

Page 16 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q2 Find the unknown weight of the sack in the following cases. In Fig. 7.10, all the
sacks have the same weight.

10 kg 4kg

Fig. 7.10 — the weighing scale is balanced. All the sacks have the same weight.

[Hint: Remove one sack from each plate for Fig. 7.10.]

Each sack weighs 14 kg.

Left pan = 2 sacks = 2s Right pan = 10 + 4 + one sack = 14 + s

The scale is balanced, so

2s = 14 + s

2s − s = 14 + s − s (remove one sack from each pan)

s = 14 kg

Page 17 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

sack + sack 14 kg + sack

strike out one sack on each pan
what is left: 1 sack = 14 kg

Removing one sack from each pan keeps the balance and leaves the answer in plain sight.

Why it happens: all the sacks weigh the same, so one sack on the left and one sack
on the right are equal weights. Taking one off each pan removes the same amount
from both sides, so the scale is still balanced. The right pan then holds only the fixed
weights, 10 + 4 = 14 kg, and the left holds a single sack.

Check it yourself: two sacks weigh 28 kg. The other pan holds 10 + 4 + 14 = 28 kg.
The scale balances. ✓

Page 18 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Find the unknown weight of the sack in the following cases.
e
Q3

m l as
m .co a g
l a se
a g

co m
e m . ag
l as
10 kg 10 kg

g
1kg

a

co m
se m.
o m l a
c 7.11 — the weighing scale is balanced. All the sacks have atheg same weight.
m .Fig.
l a se
a g [Hint: Can you remove objects so that the sacks are only on one plate?]

m a s
m .co agl
se

g l a
a
Each sack weighs 7 kg.

co m
.
Left pan = 5 sacks = 5s

e m
o m l as
Right pan = 1 + 10 + 10 + 2 sacks = 21 + 2s
. c a g
s e m scale is balanced, so
The
a
agl 5s = 21 + 2s

se m
com a
5s − 2s = 21 + 2s − 2s (remove two sacks from each pan)
. a g l
m
ase
3s = 21

s = 21 ÷ 3 = 7 kg
agl

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 19 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

5 sacks 21 kg + 2 sacks

take 2 sacks off take 2 sacks off

3 sacks = 21 kg, so 1 sack = 7 kg

Two sacks are common to both pans, so both can go.

Why it happens: the hint asks you to get the sacks onto one pan only. Two sacks
appear on each side, so removing two from each side is removing the same weight
from both — the balance survives. Three sacks are left facing 21 kg, and sharing 21
equally among 3 gives 7.

Check it yourself: five sacks weigh 5 × 7 = 35 kg. The right pan holds 21 + 7 + 7 = 35
kg. ✓

Page 20 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q4 Find the unknown weight of the sack in the following cases.

500 kg
50 kg

90 sacks in all 60 sacks in all

Fig. 7.12 — the weighing scale is balanced. All the sacks have the same weight.

Each sack weighs 15 kg.

Left pan = 90 sacks + 50 kg = 90s + 50

Right pan = 60 sacks + 500 kg = 60s + 500

The scale is balanced, so

90s + 50 = 60s + 500

Remove 60 sacks from each pan:

30s + 50 = 500

Remove 50 kg from each pan:

30s = 500 − 50 = 450

s = 450 ÷ 30 = 15 kg

Page 21 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why it happens: two removals are needed here, but each one is the same move as
before. First the 60 sacks that sit on both pans go, leaving 30 sacks on the left. Then
the 50 kg that the left pan carries is taken off both pans, leaving 30 sacks against
450 kg. Notice that we never had to know the weight of a sack to make either move
— that is what makes the method powerful.

Check it yourself: 90 × 15 + 50 = 1350 + 50 = 1400 kg. And 60 × 15 + 500 = 900 + 500
= 1400 kg. The pans match. ✓

In-text Questions — Pages 166 – 167
Matchstick Pattern

MATH TALK

Q1 Jasmine decides to make a matchstick arrangement that appears in this sequence,
using exactly 99 sticks. What will be the position number of this arrangement in the
sequence?

1 2 3 4

The first four arrangements in the matchstick sequence, with their position numbers.

Position 49.
First find the rule for the number of sticks. Look at the arrangements one by one.

Position 1 has 2 × (1) + 1 = 3 matchsticks

Position 2 has 2 × (2) + 1 = 5 matchsticks

Position 3 has 2 × (3) + 1 = 7 matchsticks

So position n has 2n + 1 matchsticks

Page 22 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Now Jasmine wants 99 sticks, so we need the value of n for which 2n + 1 is 99.

2n + 1 = 99

2n = 99 − 1 = 98

n = 98 ÷ 2 = 49

Why it happens: the first triangle uses 3 sticks and every new triangle after that
adds 2 more (the new triangle shares one stick with the one before it). So the count
is "1 stick to start, then 2 for each triangle" — which is 2n + 1. Writing 2n + 1 = 99
turns Jasmine's question into a one-line equation.

Check it yourself: 2 × 49 + 1 = 98 + 1 = 99. ✓

Q2 Can you find ways to get the value of n, such that 2n + 1 = 99?

Yes — here are three different ways, and all three give n = 49.

1. Undo the steps. The expression 2n + 1 was built by doubling n and then adding 1. Undo it in
the reverse order: from 99 subtract 1 to get 98, then halve 98 to get 49.
2. Same operation on both sides. Subtract 1 from both sides: 2n + 1 − 1 = 99 − 1, so 2n = 98.
Now divide both sides by 2: n = 98 ÷ 2 = 49.
3. Trial and error. Try n = 40 → LHS = 81, too small. Try n = 50 → LHS = 101, a bit too big. Try n =
49 → LHS = 99. Correct.

Why it happens: 2n + 1 grows steadily as n grows — every time n goes up by 1, the
LHS goes up by 2. So the LHS passes 99 exactly once, and there is only one value of n
that works. That is why all three methods must land on the same answer.

Tip: trial and error is fine for small numbers, but here you might have tested a dozen
values. The second method reaches the answer in two lines, and it works just as fast
for 2n + 1 = 9999.

Page 23 of 95

ase
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Is it possible to make a matchstick arrangement that appears in this sequence
e
Q3

m l as
.co
using exactly 200 sticks?

a g
se m
g l a
a
o m
1 2 3 m.c
e 4 ag
ag las

co m
m.
Page 166 — the first four matchstick arrangements in the sequence, with their position

m
numbers.
as e
.co a g l
se m
g l a
a ANSWER
m a s
agl
No, it is not possible.

m.co
l a se
We would need 2n + 1 = 200
ag
2n = 200 − 1 = 199

co m
n = 199 ÷ 2 = 99.5
m .
o m l a se
g
.c number must be a whole number — there is no 99.5thaarrangement.
s e m
A position So no
aarrangement
agl
in this sequence uses exactly 200 sticks.

se m
coofmsticks: 3, 5, 7, 9, 11, … . 200 is even, so it can
Why it happens: 2n is always even, so 2n + 1 is always odd. Every arrangement in
g l a
. a
em
this sequence uses an odd number

l a s
g
never appear in that list. The arrangements jump from 199 sticks (position 99)
a
straight to 201 sticks (position 100).

c o m
.
mposition 100 needs
s e
Check it yourself: position 99 needs 2 × 99 + 1 = 199 sticks and

. omThere is no way to stop at 200.
c201. a gla
as em
agl c
m .
In-text Questions — Page 168
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


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Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

7.1 Find the Unknowns

Q1 For the weighing scale problems in figures 7.6, 7.7, 7.8, 7.9, 7.10, and 7.11, frame
equations by using letter-numbers to denote the unknown weight.

Name the unknown with a letter, write what each side weighs, and join the two sides with '='
because the scale is balanced.

FIGURE UNKNOWN LEFT SIDE RIGHT SIDE

Fig. 7.6 e = weight of one fried egg 2+2+2=6 2e, so 2e = 6

Fig. 7.7 y = weight of one doughnut 4 stars = 16 4 + 2y, so 4 + 2y = 16

Fig. 7.8 b = weight of one banana watermelon = 10 2b + 4, so 2b + 4 = 10

Fig. 7.9 s = weight of the sack s+2 12, so s + 2 = 12

Fig. 7.10 s = weight of one sack 2s 14 + s, so 2s = 14 + s

Fig. 7.11 s = weight of one sack 5s 21 + 2s, so 5s = 21 + 2s

Why it happens: a picture and an equation say the same thing in two languages.
"Three slices of bread balance two eggs" and "6 = 2e" carry exactly the same
information. Once it is written in symbols you no longer have to look at the picture
— you can work on the symbols alone.

Tip: always write down in words what your letter stands for ("let e be the weight of
one fried egg"). An equation with an unexplained letter is hard to check later.

Q2 Solve the equations that you frame and check if you get the same value for the
unknown weight as you got previously.

Yes — the equations give exactly the same answers as the pictures did.

Page 25 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

EQUATION WORKING ANSWER MATCHES THE PICTURE?

2e = 6 e=6÷2 e=3 Yes, Fig. 7.6

4 + 2y = 16 2y = 16 − 4 = 12; y = 12 ÷ 2 y=6 Yes, Fig. 7.7

2b + 4 = 10 2b = 10 − 4 = 6; b = 6 ÷ 2 b=3 Yes, Fig. 7.8

s + 2 = 12 s = 12 − 2 s = 10 Yes, Fig. 7.9

2s = 14 + s 2s − s = 14 s = 14 Yes, Fig. 7.10

5s = 21 + 2s 5s − 2s = 21; 3s = 21; s = 21 ÷ 3 s=7 Yes, Fig. 7.11

Why it happens: every step you took with the picture had a twin in the algebra. "Lift
the same weight off both pans" became "subtract the same number from both
sides". "Share the weight equally among 3 sacks" became "divide both sides by 3".
The pictures were never a different method — they were this method, drawn.

Check it yourself: put each answer back in. For 5s = 21 + 2s with s = 7: LHS = 35, RHS
= 21 + 14 = 35. ✓

Q3 Frame 5 equations. Find methods to solve them.

Here are five equations of five different shapes, each with the method that suits it. Yours may be
different — compare them with your classmates'.

EQUATION SHAPE METHOD SOLUTION

x + 12 = 30 a term added Remove 12: x = 30 − 12 x = 18

7m = 91 a factor Remove the factor 7: m = 91 ÷ 7 m = 13

t÷6=9 a divisor Remove the divisor 6: t = 9 × 6 t = 54

4a − 5 = 23 term and factor 4a = 23 + 5 = 28, then a = 28 ÷ 4 a=7

5p + 8 = 3p + 20 unknown on both sides 2p + 8 = 20, so 2p = 12, p = 12 ÷ 2 p=6

Page 26 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why it happens: there are really only three moves in the whole chapter — remove a
term, remove a factor, remove a divisor — and every equation is some combination
of them. Deal with the added or subtracted term first, and the multiplying factor last;
that keeps the numbers whole for as long as possible.

Try This: make one equation whose answer is a negative number, such as 3x + 10 = 1
(giving x = −3). Swap equations with a friend and check each other's answers by
substituting.

Math Talk — Page 168
7.2 Solving Equations Systematically

MATH TALK

Q1 Can this equation have any other solution? [The equation 2n + 1 = 99, solved by trial
and error to give n = 49.]

No. n = 49 is the only solution.

If n = 48, LHS = 2 × 48 + 1 = 97 (too small)

If n = 49, LHS = 2 × 49 + 1 = 99 (correct)

If n = 50, LHS = 2 × 50 + 1 = 101 (too big)

Why it happens: every time n increases by 1, the LHS increases by exactly 2 — it
never goes back down. So the LHS passes through 99 once and only once. Solving it
systematically shows the same thing: 2n + 1 = 99 forces 2n = 98, and 98 has only one
half, namely 49.

Tip: not every equation behaves like this. x + 4 = x + 5 has no solution, and 5s = 3s
has exactly one (s = 0). Equations you will meet later can have more than one. It is
always worth asking the question.

Page 27 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q2 Try solving 5x – 4 = 7 using trial and error.

Trial and error works, but this equation shows why it is a poor tool — the answer is not a whole
number.

TRY X = LHS = 5X − 4 COMPARE WITH RHS = 7

1 5−4=1 too small

2 10 − 4 = 6 a little too small

3 15 − 4 = 11 too big

2.1 10.5 − 4 = 6.5 still small

2.2 11 − 4 = 7 correct

So x = 2.2 = 11/5

Check: 5 × (11/5) − 4 = 11 − 4 = 7 ✓

Why it happens: the answer lies between 2 and 3, so whole-number guessing can
never reach it. You have to start guessing decimals, and there are infinitely many of
those. The systematic method has no such trouble: 5x − 4 = 7 gives 5x = 11 and then
x = 11/5 straight away.

Tip: use trial and error to understand an equation and to check an answer. Use the
balance method to find the answer.

In-text Questions — Pages 169 – 170

Page 28 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

7.2 Solving Equations Systematically
co m
em.
m as
MATH TALK

.co a g l
a s em an equation 15 + 8 = 23. If we add, subtract, multiply or divide the same
gl
Consider
a
Q1
number on both sides, will it still preserve the equality of LHS and RHS? For
example, you can check by adding 10 to both sides.

com
m . ag
l a se
ag operations, as long as you do the same thing to both
Yes — the equality survives all four
sides.

. c om
a s emEQUAL?
com
OPERATION NEW LHS NEW RHS STILL

m . agl
ase
Add 10 15 + 8 + 10 = 33 23 + 10 = 33 Yes

aglSubtract 7 15 + 8 − 7 = 16 23 − 7 = 16 Yes

m a s
.co agl
Multiply by 4 (15 + 8) × 4 = 92 23 × 4 = 92 Yes

se m
Divide by 23
g l a
(15 + 8) ÷ 23 = 1 23 ÷ 23 = 1 Yes
a
Why it happens: the LHS and the RHS are not two different things — they are two
co m
m .
names for the same number, 23. If you do the same thing to a number, you get the
m as e
.co
same answer, whichever name you started from. So the two sides can never drift
a g l
s e m This is precisely the weighing-scale rule: remove (or add) equal weights on
apart.

agla both pans and the scale stays balanced.
se m
m is dividing by 0 — that is not allowed for
cowith g l a
.
Tip: the one operation to be careful
m a
any number. Everything else sisesafe.
a
agl

co m
m .
e
Example 1: It is known that 14593 – 1459 + 145 – 14 + 88 = 13353. What is the value of
as
Q2

c o m14593 – 1459 + 145 – 14? g l
. a
a s emANSWER
agl
.c
13265. And we do not have to evaluate the long expression at all.
s e m
m a
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 29 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

14593 − 1459 + 145 − 14 + 88 = 13353

Subtract 88 from both sides:

14593 − 1459 + 145 − 14 + 88 − 88 = 13353 − 88
14593 − 1459 + 145 − 14 = 13265

Why it happens: addition and subtraction are inverse operations. The term + 88 on
the LHS is cancelled by − 88, and that leaves exactly the expression we were asked
about. Because we subtracted 88 from the RHS as well, the equality is untouched.

Check it yourself: 14593 − 1459 = 13134; 13134 + 145 = 13279; 13279 − 14 = 13265.
And 13265 + 88 = 13353. ✓

Q3 Example 2: It is known that 23 × 41 × 11 × 8 × 7 = 5,80,888. What is the value of the
expression 23 × 41 × 11 × 8? Is this the same as dividing both sides by 7, which
removes the factor 7 and leaves only the expression to be evaluated on the LHS?

82,984 — and yes, dividing 5,80,888 by 7 is exactly the same as dividing both sides by 7.

23 × 41 × 11 × 8 × 7 = 5,80,888

Divide both sides by 7:

(23 × 41 × 11 × 8 × 7) ÷ 7 = 5,80,888 ÷ 7

23 × 41 × 11 × 8 = 82,984

Why it happens: multiplication and division are inverse operations, so the factor 7
on the LHS is removed by dividing by 7. Doing the division "only on the right" and
doing it "on both sides" are the same act described in two ways — on the left the
division simply cancels the 7 and leaves nothing to compute.

Check it yourself: 82984 × 7 = 580888. ✓ (Also 23 × 41 = 943, 943 × 11 = 10373,
10373 × 8 = 82984.)

Page 30 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q4 Example 3: It is known that 12345 – 5432 + 135 – 24 – (–67) = 7091. What is the value
of the expression 12345 – 5432 + 135 – 24?

7024.

12345 − 5432 + 135 − 24 − (−67) = 7091

Add (−67) to both sides — this is easier to picture than subtracting:

12345 − 5432 + 135 − 24 − (−67) + (−67) = 7091 + (−67)

The terms − (−67) and + (−67) cancel, so

12345 − 5432 + 135 − 24 = 7091 − 67 = 7024

Why it happens: − (−67) means "add 67", so its additive inverse is "add (−67)", that
is, subtract 67. Putting the two together gives zero and the unwanted term
disappears from the LHS. The RHS then loses 67 as well, keeping the equality.

Note: the book's working line prints 132 in place of 135. That is a slip — the figure
given in the question is 135, and 12345 − 5432 + 135 − 24 = 7024, which is the
answer the book itself states.

Check it yourself: 12345 − 5432 = 6913; 6913 + 135 = 7048; 7048 − 24 = 7024. And
7024 − (−67) = 7024 + 67 = 7091. ✓

Q5 Example 4: It is known that (35/113) × 24 × 14 × (8/9) = 94080/1017. What is the value
of the expression (35/113) × 24 × 14?

11760/113.

Page 31 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

To keep only the wanted expression on the LHS we must remove the factor 8/9.

Divide both sides by 8/9, that is, multiply both sides by 9/8:

(35/113) × 24 × 14 × (8/9) × (9/8) = (94080/1017) × (9/8)

(8/9) × (9/8) = 1, so the LHS is just the expression we want

RHS = (94080 × 9) ÷ (1017 × 8) = 846720 ÷ 8136 = 11760/113

Why it happens: dividing by a fraction is the same as multiplying by its reciprocal.
The pair (8/9) and (9/8) multiply to 1, so the factor vanishes from the LHS — exactly
like taking a weight off a pan. The same multiplication done on the RHS keeps the
two sides equal.

Check it yourself: (35/113) × 24 × 14 = (35 × 24 × 14)/113 = 11760/113. And
(11760/113) × (8/9) = 94080/1017. ✓

In-text Questions — Pages 170 – 172
7.2 Solving Equations Systematically

Q1 Let us use these ideas to solve the equation 5x – 4 = 7. What can we do so that 5x is
on one side and the equality between the LHS and the RHS still holds?

Add 4 to both sides. Then x = 11/5.

Page 32 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

5x − 4 = 7

To keep only 5x on the LHS, remove the term − 4 by adding 4 to both sides:

5x − 4 + 4 = 7 + 4

5x = 11

To keep only x on the LHS, remove the factor 5 by dividing both sides by 5:

5x ÷ 5 = 11 ÷ 5

x = 11/5

Why it happens: two different kinds of clutter sit around the x. The − 4 is a term, so
it is removed by adding its additive inverse, + 4. The 5 is a factor, so it is removed by
dividing. Deal with the term first: if you divide by 5 at the start you get x − 4/5 = 7/5,
which works but brings in fractions sooner.

Tip: "add 4 to both sides" and "take the − 4 across and change its sign" are the same
move. The first version explains why the second is allowed.

Q2 Can we check that x = 11/5 is the correct solution to the equation 5x – 4 = 7?

Yes — substitute 11/5 for x and see that LHS = RHS.

LHS = 5 × (11/5) − 4

= 11 − 4 (the 5 and the 5 cancel)

=7

RHS = 7

LHS = RHS, so x = 11/5 is indeed the solution.

Page 33 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: solving an equation is a chain of steps, and any one of them could

as e
coitmcatches mistakes that re-reading would miss. l
hide a slip. Substituting the answer back tests the original equation, not your
working, .so a g
a s em
agl
Tip: make checking a habit. It takes one line and it is the only way to be sure.

com
m . ag
l a se
Q3 ag 11y + (–5) = 61. Can you check that this solution is
Example 5: Solve the equation
correct?

co m
em.
omcheck confirms it. l as

y = 6, and.cthe a g
a s em
agl 11y + (−5) = 61

m a s
.co agl
Remove the term (−5) by subtracting (−5) from both sides:

se m
11y + (−5) − (−5) = 61 − (−5)
g l a
11y = 61 + 5 = 66 a

co m
m .
m as e
.co l
Now remove the factor 11 by dividing both sides by 11:
a g
a s em
11y ÷ 11 = 66 ÷ 11

agl y=6

se m
com g l a
m . a
ase
Check: LHS = 11 × 6 + (−5) = 66 − 5 = 61 = RHS ✓

agl
Why it happens: subtracting (−5) is the same as adding 5, because the additive
co m
m .
e
inverse of −5 is +5. That is why the 61 grows to 66 rather than shrinking. Once 11y =
m l as
.co g
66, you can either divide by 11 or simply recall that 11 × 6 = 66 — both are the same

em a
s
step.
l a
ag c
m .
Tip: when a negative number appears in brackets, write the subtraction out fully
m a s e
. co agl
before simplifying. 61 − (−5) = 61 + 5 is where most slips happen.
e m
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q4 Example 6: Solve 6y + 7 = 4y + 21. In this equation, expressions with an unknown are
on both sides. What can be done to bring the unknown terms to the same side?

Subtract 4y from both sides. Then y = 7.

6y + 7 = 4y + 21

Subtract 4y from both sides:

6y + 7 − 4y = 4y + 21 − 4y

2y + 7 = 21

Subtract 7 from both sides:

2y + 7 − 7 = 21 − 7

2y = 14

Divide both sides by 2:

2y ÷ 2 = 14 ÷ 2

y=7

Check: LHS = 6 × 7 + 7 = 49. RHS = 4 × 7 + 21 = 49. ✓

Why it happens: this is Fig. 7.10 all over again — sacks on both pans. Taking 4y off
each side removes the same amount from both, so the equality holds, and it leaves
the unknown on one side only. After that the equation has the familiar shape 2y + 7
= 21, which we already know how to finish.

Tip: subtract the smaller multiple of y (here 4y, not 6y). That keeps the coefficient of y
positive and avoids negative numbers.

Figure it Out — Page 172

Page 35 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

7.2 Solving Equations Systematically

MATH TALK

Q1 Solve these equations and check the solutions. (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 =
2u + 3 (d) 4 (m + 6) – 8 = 2m – 4 (e) u/15 = 6

(a) x = 15 (b) s = 0 (c) u = 10 (d) m = −10 (e) u = 90

(a) 3x − 10 = 35

3x = 35 + 10 = 45 (remove the term −10)

x = 45 ÷ 3 = 15

Check: 3 × 15 − 10 = 45 − 10 = 35 ✓

(b) 5s = 3s

5s − 3s = 3s − 3s (subtract 3s from both sides)

2s = 0

s=0÷2=0

Check: LHS = 5 × 0 = 0, RHS = 3 × 0 = 0 ✓

(c) 3u − 7 = 2u + 3

3u − 7 − 2u = 2u + 3 − 2u (subtract 2u from both sides)

u−7=3

u = 3 + 7 = 10

Check: 3 × 10 − 7 = 23 and 2 × 10 + 3 = 23 ✓

Page 36 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

(d) 4(m + 6) − 8 = 2m − 4

4m + 24 − 8 = 2m − 4 (open the bracket)

4m + 16 = 2m − 4

4m + 16 − 2m = 2m − 4 − 2m

2m + 16 = −4

2m = −4 − 16 = −20

m = −20 ÷ 2 = −10

Check: LHS = 4(−10 + 6) − 8 = 4 × (−4) − 8 = −24. RHS = 2 × (−10) − 4 = −24 ✓

(e) u ÷ 15 = 6

Remove the divisor 15 by multiplying both sides by 15:

u = 6 × 15 = 90

Check: 90 ÷ 15 = 6 ✓

Why it happens: notice how part (b) surprises people. "5s = 3s" looks impossible
until you remember that s = 0 makes both sides zero. And in (d) the bracket must be
opened first, because 4 multiplies the whole of (m + 6), not just the m.

Tip: in (e) the unknown is being divided, so you undo it by multiplying — the
opposite of what you do when the unknown is multiplied.

Q2 Frame an equation that has no solution. [Hint: 4 more than a number, and 5 more
than a number can never be equal!]

One such equation is x + 4 = x + 5.

Page 37 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

x+4=x+5

Subtract x from both sides:

x+4−x=x+5−x
4 = 5 — which is false

No value of x can make this true, so the equation has no solution.

Some more equations with no solution:

2y + 1 = 2y + 9 (subtracting 2y gives 1 = 9)
3(t + 2) = 3t + 10 (this becomes 3t + 6 = 3t + 10, so 6 = 10)
5 − p = 8 − p (adding p gives 5 = 8)

Why it happens: in every one of these the unknown cancels completely, leaving a
statement about numbers alone. If that statement is false, no value of the unknown
can rescue it. Picture it as a balance: both pans hold the same sack, but one pan also
has 4 kg and the other 5 kg. Whatever is in the sack, that scale can never balance.

Try This: what if the leftover statement is true? Try 2(x + 3) = 2x + 6. Cancelling gives
6 = 6, which is always true — so every number is a solution.

In-text Questions — Page 173
7.2 Solving Equations Systematically

Q1 What happens in cases like u/15 = 6?

The 15 is a divisor, so it is removed by multiplying both sides by 15.

u ÷ 15 = 6

(u ÷ 15) × 15 = 6 × 15
u = 90

This gives the third of the chapter's three observations. Put together, they are:

Page 38 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
WHAT SITS WITH THE HOW TO REMOVE IT EXAMPLE BECOMES
UNKNOWN
m as e
. c o a g l
s em
A term added or subtracted
a
Its additive inverse appears on the 2y + 7 = 21 2y = 21 − 7

agl
other side

A factor Divide the other side by it 2y = 14 y = 14 ÷ 2

co m
e m . ag
as
A divisor Multiply the other side by it u/15 = 6 u = 6 × 15

a g l
Why it happens: each row is one pair of inverse operations. Addition is undone by

c o m
subtraction, multiplication by division, division by multiplication. In every case you
are still doing the same thing to both sides — the table is just a shortm .
s e way of writing

. com
that down.
a gla
a s em
a gl Check it yourself: 90 ÷ 15 = 6. ✓

m a s
em
.co agl
a s
a l – 175
In-text Questions — Pagesg174
Solving Problems

co m
m .
m l a se
Example 7: Ranjana creates a sequence of arrangements with square tiles as shown
o ag
Q1

.cbelow. Can she extend the sequence and make an arrangement
m yes, which step in the sequence will it be?
using 100 tiles? If

l a se
ag
se m
com2 g l a
Step 1 .
Step Step 3 a
a sem
agl

com
m .
m ase
.co a g l
se m
g l a
a c
m .
The first three steps of Ranjana’s sequence of square tiles.
m a s e
em . co agl
g l as
ANSWER a
m
Yes — it will be Step 33.

. co
e m
m l as
.co a g
Page 39 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

First get the rule. The book shows two ways of seeing the same pattern.

STEP METHOD 1: THREE ARMS + METHOD 2: K TILES + (2K + 1) TILES
CENTRE TILES

1 1+1+1+1 1+3 4

2 2+2+2+1 2+5 7

3 3+3+3+1 3+7 10

4 4+4+4+1 4+9 13

k k+k+k+1 k + (2k + 1) 3k + 1

For 100 tiles: 3k + 1 = 100

3k = 100 − 1 = 99

k = 99 ÷ 3 = 33

Why it happens: both ways of looking at the picture give the same expression, 3k +
1, which is a good sign that the rule is right. Since 33 is a whole number, Step 33
really exists, so the arrangement can be made. Had the division not come out exact,
the answer would have been "no".

Check it yourself: 3 × 33 + 1 = 99 + 1 = 100. ✓

Q2 To check whether an arrangement is possible using 100 tiles at some Step k, we can
solve the equation: 3k + 1 = 100. Find the value of k.

k = 33.

Page 40 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

3k + 1 = 100

3k + 1 − 1 = 100 − 1 (remove the term 1 from both sides)

3k = 99

3k ÷ 3 = 99 ÷ 3 (remove the factor 3)

k = 33

Why it happens: the same two moves as always — first the term, then the factor.
The answer being a whole number is what makes the arrangement possible. If we
had asked about 101 tiles we would get 3k = 100, and 100 is not divisible by 3, so no
step of the sequence uses 101 tiles.

Try This: which nearby numbers of tiles are possible? 3k + 1 gives 4, 7, 10, 13, … —
every number that leaves remainder 1 on division by 3.

In-text Questions — Pages 175 – 177
Solving Problems

Q1 Example 8: Madhubanti wants to organise a party. She decides to buy snacks for the
party from the chaat shop in town. Each plate of snacks costs ₹25. The shop charges
an additional fixed amount of ₹50 to deliver the snacks to Madhubanti's house.
There are 5 members in Madhubanti's family, including herself. Her parents tell her
she can spend ₹500 on this party. How many friends can she invite to the party if
she wants to give a plate of snacks to each person, including her family and
friends?

She can invite 13 friends.
Let p be the total number of people at the party, family and friends together.

Page 41 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Cost = 25p + 50 (₹25 per plate, plus ₹50 delivery)

This must be ₹500, so

25p + 50 = 500
25p = 500 − 50 = 450

p = 450 ÷ 25 = 18 people

Of these 18, five are her family, so

friends = 18 − 5 = 13

Fatima reached the same answer without algebra: take the ₹50 delivery out of ₹500 first, leaving
₹450 for snacks; ₹450 ÷ ₹25 = 18 plates; 18 − 5 = 13 friends.

Why it happens: the ₹50 is a fixed charge — it is paid once, however many plates are
bought. The ₹25 is a rate — it is paid for each plate. So the total cost is "rate ×
number + fixed", which is 25p + 50. Removing the fixed part from both sides is what
makes the rest a simple division.

Check it yourself: 18 plates cost 18 × 25 = ₹450, plus ₹50 delivery = ₹500 exactly. ✓

Q2 Srikanth decided to represent the unknown quantity of the total number of friends
Madhubanthi can invite as f. What will be the cost in this case?

Cost = 25 (f + 5), and solving gives f = 13 — the same answer.

If f friends come, the number of plates is f + 5 (the 5 family members too).

Cost of snacks = 25 (f + 5)

Madhubanti has ₹450 for snacks after the ₹50 delivery is paid, so

25 (f + 5) = 450

f + 5 = 450 ÷ 25 = 18 (divide both sides by 25)

f = 18 − 5 = 13

Page 42 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why it happens: Mahesh named the total number of people and Srikanth named
the number of friends. Different unknowns give different-looking equations — 25p +
50 = 500 and 25(f + 5) = 450 — but they describe the same party, so they must agree.
Note that p and f are linked by p = f + 5, and 18 = 13 + 5. ✓

Tip: when a bracket is multiplied by a number that divides the other side exactly,
divide first. Here dividing by 25 was far quicker than opening 25(f + 5) into 25f + 125.

Q3 Example 9: Two friends want to save money. Jahnavi starts with an initial amount of
₹4000, and in addition, saves ₹650 per month. Sunita starts with ₹5050 and saves
₹500 per month. After how many months will they have the same amount of
money?

After 7 months.
Let m be the number of months.

Jahnavi's savings after m months = 4000 + 650m

Sunita's savings after m months = 5050 + 500m

They are equal, so

4000 + 650m = 5050 + 500m

4000 + 650m − 500m = 5050 (subtract 500m from both sides)

4000 + 150m = 5050

150m = 5050 − 4000 (subtract 4000 from both sides)

150m = 1050

m = 1050 ÷ 150 = 7

Why it happens: Sunita begins ₹1050 ahead, but Jahnavi saves ₹150 more every
month. So Jahnavi closes the gap at ₹150 a month, and 1050 ÷ 150 = 7 months to
close it completely. The algebra does exactly this arithmetic, but it finds the two
numbers 1050 and 150 for you.

Page 43 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Tip: subtracting 500m rather than 650m keeps the coefficient of m positive, which

m as e
l
makes the last division easier.

m .co a g
l a se
a g
Q4 Check the answer. [For Example 9, m = 7 months.]

com
m . ag
l a se
ag₹8550 after 7 months.
The check works out — both have

co m
m.
Jahnavi = 4000 + 650 × 7 = 4000 + 4550 = ₹8550

m as e
.co
Sunita = 5050 + 500 × 7 = 5050 + 3500 = ₹8550
a g l
a s em
l
The two amounts are equal, so m = 7 is correct.
a g
m a s
.co agl
MONTH JAHNAVI (₹) SUNITA (₹) GAP (₹)

se m
Start 4000
g l a 5050 1050

1 4650
a 5550 900

m
co750
5300 6050
m .
e
2

m l as
.co a g
e7m
6 7900 8050 150

a s
agl 8550 8550 0

se m
com g l a
. a
Why it happens: the gap shrinks by exactly ₹150 each month — 1050, 900, 750, … ,
m
ase
150, 0. It reaches zero for the first time in month 7, and after that Jahnavi is ahead.

agl
The table and the equation tell the same story.

co m
m .
In-textmQuestions — Pages 177 – 179 as e
.co a g l
a s em
Solving Problems

agl
.c
Q1 Example 10: Solve 28 (x + 4) + 300 = 1000.
s e m
m a
m . co agl
l a se
ag routes; all three end at the same place.
x = 21. The book gives three

co m
m .
m ase
.co


a g l Page 44 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Way 1 — remove the term, then the factor

28(x + 4) + 300 = 1000

28(x + 4) = 1000 − 300 = 700

x + 4 = 700 ÷ 28 = 25

x = 25 − 4 = 21

Way 2 — divide the whole equation by 4 first

28, 300 and 1000 are all divisible by 4, so divide both sides by 4:

7(x + 4) + 75 = 250

7(x + 4) = 250 − 75 = 175

7x + 28 = 175

7x = 147, so x = 147 ÷ 7 = 21

Way 3 — open the bracket first

28x + 112 + 300 = 1000

28x + 412 = 1000

28x = 1000 − 412 = 588

x = 588 ÷ 28 = 21

Check: 28(21 + 4) + 300 = 28 × 25 + 300 = 700 + 300 = 1000 ✓

Why it happens: every route uses only the balance rule, so every route must land on
21. Which one is quickest depends on the numbers. Way 1 is shortest here because
700 ÷ 28 is exact. Way 2 is handy when a common factor makes the numbers smaller.
Way 3 always works but produces the biggest arithmetic.

Tip: look at the numbers before you start. Spotting that 700 ÷ 28 = 25 saves several
lines.

Page 45 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q2 Example 11: Riyaz asked Akash to perform the following steps: 1. Think of a number.
2. Subtract 3 from the number. 3. Multiply the result by 4. 4. Add 8 to the product. 5.
Reveal the final answer. The final answer revealed by Akash was 24. Using this,
Riyaz correctly figured out the starting number that Akash had thought of. Find this
number.

Akash thought of 7.
Let the starting number be x, and follow the steps in symbols.

STEP EXPRESSION

Think of a number x

Subtract 3 from the number x−3

Multiply the result by 4 4(x − 3) = 4x − 12

Add 8 to the product 4x − 12 + 8 = 4x − 4

The final answer was 24, so

4x − 4 = 24

4(x − 1) = 24

x − 1 = 24 ÷ 4 = 6 (divide both sides by 4)

x=6+1=7

Check: 7 → 7 − 3 = 4 → 4 × 4 = 16 → 16 + 8 = 24 ✓

Why it happens: Riyaz never sees the middle steps, but he does not need to. All four
steps together do just one thing to the starting number: they turn x into 4x − 4.
Knowing that single expression is enough to undo the whole trick.

Page 46 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q3 Try the steps using different numbers as the starting number. Do you see any
relation between the starting number and final answer?

Yes — the final answer is always 4 less than 4 times the starting number.

START X X−3 ×4 + 8 (FINAL ANSWER) 4X − 4

1 −2 −8 0 4(1) − 4 = 0 ✓

5 2 8 16 4(5) − 4 = 16 ✓

7 4 16 24 4(7) − 4 = 24 ✓

10 7 28 36 4(10) − 4 = 36 ✓

12 9 36 44 4(12) − 4 = 44 ✓

Another way to say it: each time the starting number goes up by 1, the final answer goes up by
4.

Why it happens: the algebra shows it in one line — the four steps together give 4x −
12 + 8, which is 4x − 4. The "− 3" and the "+ 8" do not disappear; they combine into a
single "− 4" after the multiplication by 4 has stretched the − 3 into − 12.

Q4 Can you think of a simple rule that you can use to get the starting number from the
final answer?

Add 4 to the final answer and divide by 4. In short: starting number = (final answer ÷ 4) + 1.

final answer = 4x − 4

4x = final answer + 4

x = (final answer + 4) ÷ 4

Since 4x − 4 = 4(x − 1), the same rule can be written as

x = (final answer ÷ 4) + 1

Page 47 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

FINAL ANSWER ÷4 +1 STARTING NUMBER

24 6 6+1 7

36 9 9+1 10

0 0 0+1 1

Why it happens: the trick multiplies by 4 and then shifts by −4, so undoing it means
shifting back by +4 and then dividing by 4. Doing the inverse operations in the
reverse order is what unlocks any such trick.

Try This: invent your own. "Think of a number, add 5, double it, subtract 4" gives 2x +
6, so your rule to undo it is (answer − 6) ÷ 2.

Q5 Example 12: Ramesh and Suresh have 60 marbles between them. Ramesh has 30
more marbles than Suresh. How many marbles does each boy have? Use this to find
both the unknowns. [Hint: denote Suresh's marbles as y and Ramesh's as y + 30,
giving y + (y + 30) = 60.]

Suresh has 15 marbles and Ramesh has 45 marbles.

Let Suresh have y marbles. Then Ramesh has y + 30.

Total is 60, so
y + (y + 30) = 60

2y + 30 = 60

2y = 60 − 30 = 30

y = 30 ÷ 2 = 15 (Suresh)

Ramesh = y + 30 = 15 + 30 = 45

Check: 15 + 45 = 60 marbles in all ✓ and 45 − 15 = 30 more for Ramesh ✓

Page 48 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: writing x + y = 60 and x = y + 30 gives two unknowns, and so far we

m l a se
can only solve equations with one. The trick is to use the second fact at once: since
o
m .c has 30 more than Suresh, we can call his count ya+g30 instead of x.
Ramesh always
a e y is unknown, and one equation is enough.
sonly
ag l
Now

o m
c ag
Tip: think of it as the balance again. Set aside Ramesh's extra 30 marbles; the
.
m the two boys, 15 each. Ramesh then
s e
remaining 30 marbles split equally between
a
agl
takes his 30 back, giving 45.

co m
e m.
com
Math Talk — Pages 179 – 180
g l as
. a
em
Generating Equations

a s
agl
MATH TALK

om a s
. c agl
Q1 Write equations whose solution is y = 5. Share the equations you made with each
other and discuss the methodsm
a s e used.

agl

co m
.
Start from y = 5 and do the same thing to both sides — every step gives a new equation with the

e m
as
same solution.
m l
m .co a g
l a se
OPERATION APPLIED TO Y = 5 EQUATION YOU GET CHECK WITH Y = 5

a g ✓
m
Add 1 y+1=6 5+1=6

a se
Multiply by 3
. com
3y = 15 15 = 15 ✓
a g l
m
ase
agl
Multiply by 3, then add 6 3y + 6 = 21 15 + 6 = 21 ✓
Subtract from 12 12 − y = 7 12 − 5 = 7 ✓
co m✓
m .
as e
Divide by 5 y÷5=1 5÷5=1

com by 4, then subtract y
.Multiply a g l
sem
4y = 3y + 5 20 = 15 + 5 ✓
a
agl c
Why it happens: this is solving run backwards. When you solve, you strip operations
m .
m a s e
co agl
away until only y is left. When you generate, you build operations on until the

m .
e
equation looks interesting. Because every step is done to both sides, the value y = 5
as
a g l
keeps satisfying every equation in the chain.

co m
m .
m ase
.co


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Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Try This: make one whose unknown appears on both sides, such as 4y = 3y + 5, and
give it to a friend to solve.

Q2 Can you form a chain going from the bottom equation to the top? [The chains
shown are: y + 1 = 6 → (multiplying by −1) −y − 1 = −6 → (adding y) −1 = −6 + y →
(adding 6) 5 = y; and 3y = 15 → (adding 6) 3y + 6 = 21 → (dividing by 3) y + 2 = 7 →
(subtracting 2) y = 5.]

Yes — climb back up by applying the inverse operation at each step, taken in the reverse order.

Second chain, bottom to top

y=5

Add 2 to both sides: y + 2 = 7

Multiply both sides by 3: 3y + 6 = 21

Subtract 6 from both sides: 3y = 15 ✓

First chain, bottom to top

5=y

Subtract 6 from both sides: −1 = y − 6, that is, −1 = −6 + y

Subtract y from both sides: −1 − y = −6, that is, −y − 1 = −6
Multiply both sides by −1: y + 1 = 6 ✓

Why it happens: going down the chain you added 6; to come back up you subtract
6. Going down you divided by 3; to come back up you multiply by 3. And the order
must be reversed, because the last thing you did on the way down is the first thing
you must undo on the way up — just like taking off shoes and socks.

Tip: multiplying by −1 is its own inverse. Doing it twice brings you back to where you
started.

Page 50 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q3 Compare the operations used when going from the top to the bottom and from the
bottom to the top.

They are inverse operations, in the opposite order.

CHAIN TOP → BOTTOM BOTTOM → TOP

3y = 15 … y = 5 add 6, divide by 3, subtract 2 add 2, multiply by 3, subtract 6

y+1=6…5=y multiply by −1, add y, add 6 subtract 6, subtract y, multiply by −1

"Add 6" is undone by "subtract 6"; "divide by 3" is undone by "multiply by 3".
The list is read backwards, because the last step down is the first step up.
Multiplying by −1 is special: it undoes itself.

Why it happens: every operation in these chains is reversible, and reversing a
sequence of reversible steps means reversing each step and the order. That is exactly
what happens when you solve an equation — solving is nothing but climbing back
up a chain that somebody else built.

Q4 Without calculating, can you find the value of the unknown in each equation in the
chains above? [Hint: We have seen that the value that satisfies an equation also
satisfies the new equation obtained by performing the same operation on both
sides of the original equation.]

Yes — every equation in both chains has the same solution, y = 5.

Page 51 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

EQUATION PUT Y = 5 TRUE?

y+1=6 5+1=6 Yes

−y − 1 = −6 −5 − 1 = −6 Yes

−1 = −6 + y −6 + 5 = −1 Yes

5=y 5=5 Yes

3y = 15 3 × 5 = 15 Yes

3y + 6 = 21 15 + 6 = 21 Yes

y+2=7 5+2=7 Yes

y=5 5=5 Yes

Why it happens: each equation in a chain was built from the one above it by doing
the same operation to both sides. If y = 5 made the two sides equal before, it must
make them equal after — the operation treated both sides alike. So you can read the
answer off the bottom of the chain and know it fits every line above.

Tip: this is why writing a chain of equations is a safe way to solve. As long as every
step is done to both sides, the answer never changes.

In-text Questions — Page 180
Generating Equations

Q1 Example 13: Can you give a real-life situation that can be modelled as the equation,
100x + 75 = 250?

Yes. Read the numbers as rupees: ₹75 is a fixed charge, x is the cost of one item, 100 is the
number of items, and ₹250 is the total bill.

100x + 75 = 250

100x = 250 − 75 = 175

x = 175 ÷ 100 = ₹1.75 per item

Page 52 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Some situations that fit:

A shop sells 100 pencils at ₹x each and charges ₹75 for packing. The bill is ₹250. What does
one pencil cost?
A tent for a school function is booked for 100 chairs at ₹x a chair, with a fixed ₹75 for
transport, and the total is ₹250.
An internet plan costs ₹75 a month plus ₹x for each of 100 units used; the bill comes to ₹250.

Why it happens: the shape of the equation tells you what story fits. A term with the
unknown in it is something that changes with the quantity — a rate. A plain number
added to it is something fixed. The number on the other side is the total.
Madhubanti's party, the taxi fare and the savings problems in this chapter all share
that shape.

Check it yourself: 100 × 1.75 = 175, and 175 + 75 = 250. ✓

Figure it Out — Page 181
7.2 Solving Equations Systematically

Q1 Write 5 equations whose solution is x = – 2.

Build each one by doing the same thing to both sides of x = − 2.

EQUATION HOW IT WAS BUILT CHECK WITH X = − 2

x+2=0 Add 2 to both sides −2+2=0 ✓
3x = − 6 Multiply both sides by 3 3 × (− 2) = − 6 ✓
5x + 4 = − 6 Multiply by 5, then add 4 − 10 + 4 = − 6 ✓
x ÷ 4 = − 0.5 Divide both sides by 4 − 2 ÷ 4 = − 0.5 ✓
7x + 9 = 4x + Multiply by 7 and add 9; the same value also gives 4x − 14 + 9 = − 5 and − 8 + 3 = − 5
3 +3 ✓

Why it happens: the value x = − 2 satisfies x = − 2, and every operation applied to
both sides carries that value forward. So you can make as many equations as you
like, and they will all have − 2 as their solution.

Page 53 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
m.
Tip: to make the last kind, choose any two expressions that are equal at x = − 2. Here

m as e
l
7x + 9 and 4x + 3 both give − 5.

m .co a g
l a se
a g
Q2 Find the value of each unknown: (a) 2y = 60 (b) – 8 = 5x – 3 (c) – 53w = –15 (d) 13 – z = 8

com
ag
(e) k + 8 = 12 – k (f) 7m = m – 3 (g) 3n = 10 + n

m .
ase
a g l
(a) y = 30 (b) x = − 1 (c) w = 15/53 (d) z = 5 (e) k = 2 (f) m = − 1/2 (g) n = 5

co m
em.
m as
(a) 2y = 60 → y = 60 ÷ 2 = 30 Check: 2 × 30 = 60 ✓
.co a g l
se m
g l a
a (b) − 8 = 5x − 3

om a s
agl
− 8 + 3 = 5x → − 5 = 5x → x = − 5 ÷ 5 = − 1

Check: 5(− 1) − 3 = − 5 − 3 = − 8 ✓m . c
l a se
ag
(c) − 53w = − 15
co m
m .
as e
com− 53 × 15/53 = − 15 ✓
w = (− 15) ÷ (− 53) = 15/53
. a g l
m
ase
Check:

agl
se m
(d) 13 − z = 8
com g l a
m . a
ase
13 − 8 = z → z = 5 Check: 13 − 5 = 8 ✓

agl
(e) k + 8 = 12 − k
co m
m .
m
k + k + 8 = 12 (add k to both sides)
o l a se
c = 12 − 8 = 4 → k = 2 Check: 2 + 8 = 10 and 12 − 2 = a10g✓
.2k
se m
g l a
a c
m .
a s e
om agl
(f) 7m = m − 3
. c
7m − m = − 3 → 6m = − 3 → m
s e m= − 3 ÷ 6 = − 1/2
la and − 1/2 − 3 = − 3.5 ✓
−g3.5
Check: 7 × (− 1/2) = a

co m
m .
m ase
.co


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Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

(g) 3n = 10 + n

3n − n = 10 → 2n = 10 → n = 5 Check: 15 = 10 + 5 ✓

Why it happens: parts (e), (f) and (g) all have the unknown on both sides, so the first
move is always the same — bring the unknown terms together by adding or
subtracting one of them from both sides. In (b) the unknown happens to sit on the
right; that changes nothing, since an equation reads equally well in either direction.

Tip: in (c) the answer is a fraction and that is perfectly fine. Dividing a negative by a
negative gives a positive.

Q3 I am a 3-digit number. My hundred's digit is 3 less than my ten's digit. My ten's digit
is 3 less than my unit's digit. The sum of all the three digits is 15. Who am I?

The number is 258.
Let the units digit be u.

Tens digit = u − 3

Hundreds digit = (tens digit) − 3 = u − 3 − 3 = u − 6

The three digits add up to 15:
(u − 6) + (u − 3) + u = 15

3u − 9 = 15

3u = 15 + 9 = 24

u = 24 ÷ 3 = 8

Tens digit = 8 − 3 = 5 Hundreds digit = 8 − 6 = 2

Check: 2 + 5 + 8 = 15 ✓; 2 is 3 less than 5 ✓; 5 is 3 less than 8 ✓

Page 55 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Why it happens: naming the units digit is the smart choice, because the other two
are described in terms of it. Had we named the hundreds digit h, we would get h, h +
3, h + 6 and the same equation 3h + 9 = 15, giving h = 2 — the same number.

Tip: always check that the digits are single digits between 0 and 9, and that the
hundreds digit is not 0. Here 2, 5 and 8 pass both tests.

Q4 The weight of a brick is 1 kg more than half its weight. What is the weight of the
brick?

The brick weighs 2 kg.

Let the weight of the brick be w kg.

"1 kg more than half its weight" is w/2 + 1, and that equals w:

w = w/2 + 1

w − w/2 = 1 (subtract w/2 from both sides)

w/2 = 1

w = 1 × 2 = 2 kg

Check: half of 2 kg is 1 kg, and 1 kg more than that is 2 kg. ✓

Why it happens: a whole brick minus half a brick is half a brick. So that missing half
must be exactly the 1 kg mentioned in the question — and a brick is two halves, so 2
kg. Picture a balance: on one pan the whole brick, on the other half a brick and a 1
kg weight. Remove half a brick from each pan and you are left with half a brick
against 1 kg.

Tip: read such sentences slowly. "1 kg more than half its weight" is w/2 + 1, not (w +
1)/2.

Page 56 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

Q5 One quarter of a number increased by 9 gives the same number. What is the
number?

The number is 12.

Let the number be x.

x/4 + 9 = x

9 = x − x/4 (subtract x/4 from both sides)

9 = 3x/4 (because x − x/4 = 4x/4 − x/4 = 3x/4)

9 × 4 = 3x → 36 = 3x

x = 36 ÷ 3 = 12

Check: one quarter of 12 is 3, and 3 + 9 = 12 ✓

Why it happens: the number is made up of four quarters. Taking one quarter away
leaves three quarters, and those three quarters are worth the 9 that was added. So
one quarter is 3, and the whole number is 4 × 3 = 12.

Q6 Given 4k + 1 = 13, find the values of: (a) 8k + 2 (b) 4k (c) k (d) 4k – 1 (e) – k – 2

(a) 26 (b) 12 (c) 3 (d) 11 (e) − 5

From 4k + 1 = 13: 4k = 13 − 1 = 12, so k = 12 ÷ 4 = 3

Page 57 of 95

Class 7 Maths Chapter 15 Finding the Unknown AglaSem · NCERT Solutions

EXPRESSION QUICK WAY VALUE

(a) 8k + 2 = 2 × (4k + 1) = 2 × 13 26

(b) 4k = 13 − 1 12

(c) k = 12 ÷ 4 3

(d) 4k − 1 = 12 − 1 11

(e) − k − 2 =−3−2 −5

Why it happens: part (a) needs no value of k at all. 8k + 2 is exactly twice 4k + 1, so it
is twice 13. Spotting a relationship like this is faster than solving, and it is the same
idea used in Examples 1 and 2 — using what is given instead of recomputing.

Check it yourself: with k = 3, 8k + 2 = 24 + 2 = 26 ✓, 4k − 1 = 12 − 1 = 11 ✓, − k − 2 = −
3−2=−5✓

Mind the Mistake, Mend the Mistake — Pages 181 – 182
7.3 Mind the Mistake, Mend the Mistake

Q1 Go through each solution and decide whether the steps are correct. If there is a
mistake, describe the mistake, correct it and solve the equation. 4x + 6 = 10 / 4x = 10
+ 6 / 4x = 16 / x = 4

There is a mistake in the first step. The correct answer is x = 1.
The mistake: the term + 6 was removed from the LHS but was written as + 6 on the RHS. When
a term is removed from one side, its additive inverse must appear on the other side — so it
should be − 6.

Correct solution

4x + 6 = 10

4x = 10 − 6 = 4

x=4÷4=1

Page 58 of 95

as e
Class 7 Maths Chapter 15 Finding the Unknown
a g l AglaSem · NCERT Solutions

co m
e m.
Check: 4 × 1 + 6 = 4 + 6 = 10 ✓ (the wrong answer 4 gives 4 × 4 + 6 = 22, not 10)
m l as
m .co a g
l a se
g
Why it happens: think of the balance. To take a 6 kg weight off the left pan you
amust also take 6 kg off the right pan, not add it. That is why 10 becomes 10 − 6.

co m
e m . ag
g l as
a
Q2 7 – 8z = 5 / 8z = 7 – 5 / 8z = 2 / z = 4

. com

m a
The first two steps are fine; the last step is wrong. The correct answer is
s em
z = 1/4.

. co from 8z = 2 the division was done the wrong way roundag—l 8 was divided by 2
emof 2 being divided by 8.
The mistake:

a s
agl
instead

m a s
.co agl
Correct solution

7 − 8z = 5
a s em
aglis correct)
7 − 5 = 8z (this step in the book

m
8z = 2
. co
e m
as
z = 2 ÷ 8 = 1/4
m l
.co a g
a s em
agl Check: 7 − 8 × (1/4) = 7 − 2 = 5 ✓ (z = 4 would give 7 − 32 = − 25)
se m
com g l a
m . a
e
Why it happens: to remove the factor 8 from the LHS you divide both sides by 8. The
8.sDividing 8 by 2 is not the same operation at all.
RHS is 2, so it becomes 2 ÷la
a g

co m
.
Tip: a quick sanity check helps. Since 8z = 2 and 2 is smaller than 8, z must be

em
as
smaller than 1.
m l
m .co a g
l a se
ag 2v – 4 = 6 / v – 4 = 6 – 2 / v – 4 = 4 / v = 8
.c
m
Q3

m a s e
em . co agl
as

a g l
The first step is wrong. The correct answer is v = 5.

co m
m .
m ase
.co


a g l Page 59 of 95