NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown (PDF) – 2027
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NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
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.
NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown – Questions and Answers
All 89 questions of the chapter (textbook pages Part II, 164 – 190) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
Q1Find the unknown weights in the following cases: (The book first shows how the scale behaves: a scale marked 4 with 2 and 2 hanging from it stays level; a scale marked 7 with 4 and 3 hanging from it tilts. The number in the circle is the total weight.)Fig. 7.1 — the scale is marked 16 and the beam hangs level. A leaf weighs 3.
Answer
Bud = 2 and flower = 8.
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
Q2Find the unknown weights in the following cases:Fig. 7.2 — the scale is marked 24 and the beam hangs level. A starfish weighs 2.
Answer
Striped fish = 8 and grey fish = 4.
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)
Q3Find the unknown weights in the following cases: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.
Answer
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
Q4Find the unknown weights in the following cases:Fig. 7.4 — the scale is marked 18 and the beam hangs level. The sun weighs 5.
Answer
Cloud = 1 and lightning bolt = 3.
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)
Q5Find the unknown weights in the following cases:Fig. 7.5 — the scale is marked 40 and the beam hangs level.
Answer
Crown = 5 and gem = 3.
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)
Q6Find the unknown weights in the following cases:Fig. 7.6 — the balance hangs level. One slice of bread weighs 2.
Answer
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
Q7Find the unknown weights in the following cases:Fig. 7.7 — the balance hangs level. One star weighs 4.
Answer
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 (subtract 4 from both sides)
2y = 12
y = 12 ÷ 2 = 6
Q8Find the unknown weights in the following cases:Fig. 7.8 — the balance hangs level. The watermelon slice weighs 10 and the orange weighs 4.
Answer
One banana = 3.
Left arm: watermelon = 10
Right arm: banana + orange + banana = b + 4 + b = 2b + 4
The beam is level, so
2b + 4 = 10
2b = 10 − 4 (subtract 4 from both sides)
2b = 6
b = 6 ÷ 2 = 3
Math Talk — Page 1657.1 Find the Unknowns
Math Talk
Q1Discuss the answers with your classmates. Give reasons why you think your answer is right.
Answer
The number at the top is the total weight, and a level beam means both sides weigh the same. So in Fig. 7.1 each side is 16 ÷ 2 = 8, and the bud is 8 − 3 − 3 = 2. Taking the same weight off both sides keeps the balance (the star in Fig. 7.7). Putting our answers back (3 + 2 + 3 = 8) shows they are right.
Q1Find the unknown weight of the sack in the following cases.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?]
Answer
The sack weighs 10 kg.
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
Q2Find the unknown weight of the sack in the following cases. In Fig. 7.10, all the sacks have the same weight.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.]
Answer
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
Removing one sack from each pan keeps the balance and leaves the answer in plain sight.
Q3Find the unknown weight of the sack in the following cases.Fig. 7.11 — the weighing scale is balanced. All the sacks have the same weight. [Hint: Can you remove objects so that the sacks are only on one plate?]
Answer
Each sack weighs 7 kg.
Left pan = 5 sacks = 5s
Right pan = 1 + 10 + 10 + 2 sacks = 21 + 2s
The scale is balanced, so
5s = 21 + 2s
5s − 2s = 21 + 2s − 2s (remove two sacks from each pan)
3s = 21
s = 21 ÷ 3 = 7 kg
Two sacks are common to both pans, so both can go.
Q4Find the unknown weight of the sack in the following cases.Fig. 7.12 — the weighing scale is balanced. All the sacks have the same weight.
Answer
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
Q1Jasmine 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?The first four arrangements in the matchstick sequence, with their position numbers.
Answer
Position 49.
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
2n + 1 = 99
2n = 99 − 1 = 98
n = 98 ÷ 2 = 49
Q2Can you find ways to get the value of n, such that 2n + 1 = 99?
Answer
Yes. n = 49.
2n + 1 = 99
2n + 1 − 1 = 99 − 1 (subtract 1 from both sides)
2n = 98
n = 98 ÷ 2 = 49
Q3Is it possible to make a matchstick arrangement that appears in this sequence using exactly 200 sticks?
Page 166 — the first four matchstick arrangements in the sequence, with their position numbers.
Answer
No, it is not possible.
We would need 2n + 1 = 200
2n = 200 − 1 = 199
n = 199 ÷ 2 = 99.5
A position number must be a whole number — there is no 99.5th arrangement. So no arrangement in this sequence uses exactly 200 sticks.
In-text Questions — Page 1687.1 Find the Unknowns
Q1For 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.
Fig. 7.6 — the balance hangs level. One slice of bread weighs 2.Fig. 7.7 — the balance hangs level. One star weighs 4.Fig. 7.8 — the balance hangs level. The watermelon slice weighs 10 and the orange weighs 4.Fig. 7.9 — the weighing scale is balanced.Fig. 7.10 — the weighing scale is balanced. All the sacks have the same weight.Fig. 7.11 — the weighing scale is balanced. All the sacks have the same weight.
Answer
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
Q2Solve the equations that you frame and check if you get the same value for the unknown weight as you got previously.
Answer
Yes — the equations give exactly the same answers as the pictures did.
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
Q3Frame 5 equations. Find methods to solve them.
Answer
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
Math Talk — Page 1687.2 Solving Equations Systematically
Math Talk
Q1Can this equation have any other solution? [The equation 2n + 1 = 99, solved by trial and error to give n = 49.]
Answer
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)
Q2Try solving 5x – 4 = 7 using trial and error.
Answer
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 ✓
Q1Consider an equation 15 + 8 = 23. If we add, subtract, multiply or divide the same 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.
Answer
Yes — the equality survives all four operations, as long as you do the same thing to both sides.
Operation
New LHS
New RHS
Still equal?
Add 10
15 + 8 + 10 = 33
23 + 10 = 33
Yes
Subtract 7
15 + 8 − 7 = 16
23 − 7 = 16
Yes
Multiply by 4
(15 + 8) × 4 = 92
23 × 4 = 92
Yes
Divide by 23
(15 + 8) ÷ 23 = 1
23 ÷ 23 = 1
Yes
Q2Example 1: It is known that 14593 – 1459 + 145 – 14 + 88 = 13353. What is the value of 14593 – 1459 + 145 – 14?
Answer
13265. And we do not have to evaluate the long expression at all.
Q3Example 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?
Answer
82,984 — and yes, dividing 5,80,888 by 7 is exactly the same as dividing both sides by 7.
Q1Let 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?
Answer
Add 4 to both sides. Then x = 11/5.
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
Q2Can we check that x = 11/5 is the correct solution to the equation 5x – 4 = 7?
Answer
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.
Q3Example 5: Solve the equation 11y + (–5) = 61. Can you check that this solution is correct?
Answer
y = 6, and the check confirms it.
11y + (−5) = 61
Remove the term (−5) by subtracting (−5) from both sides:
11y + (−5) − (−5) = 61 − (−5)
11y = 61 + 5 = 66
Now remove the factor 11 by dividing both sides by 11:
11y ÷ 11 = 66 ÷ 11
y = 6
Check: LHS = 11 × 6 + (−5) = 66 − 5 = 61 = RHS ✓
Q4Example 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?
Q1Example 7: Ranjana creates a sequence of arrangements with square tiles as shown below. Can she extend the sequence and make an arrangement using 100 tiles? If yes, which step in the sequence will it be?The first three steps of Ranjana’s sequence of square tiles.
Answer
Yes — it will be Step 33.
First get the rule. The book shows two ways of seeing the same pattern.
Q1Example 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?
Answer
She can invite 13 friends.
Let p be the total number of people at the party, family and friends together.
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
Q2Srikanth 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?
Answer
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
Q3Example 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?
Answer
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
Q4Check the answer. [For Example 9, m = 7 months.]
Answer
The check works out — both have ₹8550 after 7 months.
Jahnavi = 4000 + 650 × 7 = 4000 + 4550 = ₹8550
Sunita = 5050 + 500 × 7 = 5050 + 3500 = ₹8550
The two amounts are equal, so m = 7 is correct.
28(x + 4) + 300 = 1000
28(x + 4) = 1000 − 300 = 700 (subtract 300 from both sides)
x + 4 = 700 ÷ 28 = 25 (divide both sides by 28)
x = 25 − 4 = 21
Q2Example 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.
Answer
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
Q3Try the steps using different numbers as the starting number. Do you see any relation between the starting number and final answer?
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.
Q4Can you think of a simple rule that you can use to get the starting number from the final answer?
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
Final answer
÷ 4
+ 1
Starting number
24
6
6 + 1
7
36
9
9 + 1
10
0
0
0 + 1
1
Q5Example 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.]
Answer
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
Math Talk — Pages 179 – 180Generating Equations
Math Talk
Q1Write equations whose solution is y = 5. Share the equations you made with each other and discuss the methods used.
Answer
Start from y = 5 and do the same thing to both sides — every step gives a new equation with the same solution.
Operation applied to y = 5
Equation you get
Check with y = 5
Add 1
y + 1 = 6
5 + 1 = 6 ✓
Multiply by 3
3y = 15
15 = 15 ✓
Multiply by 3, then add 6
3y + 6 = 21
15 + 6 = 21 ✓
Subtract from 12
12 − y = 7
12 − 5 = 7 ✓
Divide by 5
y ÷ 5 = 1
5 ÷ 5 = 1 ✓
Multiply by 4, then subtract y
4y = 3y + 5
20 = 15 + 5 ✓
Q2Can you form a chain going from the bottom equation to the top?
Chain 1
Chain 2
y + 1 = 6
3y = 15
↓ multiplying by −1
↓ adding 6
−y − 1 = −6
3y + 6 = 21
↓ adding y
↓ dividing by 3
−1 = −6 + y
y + 2 = 7
↓ adding 6
↓ subtracting 2
5 = y
y = 5
Answer
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 ✓
Q3Compare the operations used when going from the top to the bottom and from the bottom to the top.
Answer
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.
Q4Without 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.]
Answer
Yes — every equation in both chains has the same solution, y = 5.
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
In-text Questions — Page 180Generating Equations
Q1Example 13: Can you give a real-life situation that can be modelled as the equation, 100x + 75 = 250?
Answer
A shop sells 100 pencils at ₹x each and charges ₹75 for packing. The total bill is ₹250. What does one pencil cost? This is 100x + 75 = 250.
Figure it Out — Page 1817.2 Solving Equations Systematically
Q1Write 5 equations whose solution is x = – 2.
Answer
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 + 3
Multiply by 7 and add 9; the same value also gives 4x + 3
− 14 + 9 = − 5 and − 8 + 3 = − 5 ✓
Q2Find the value of each unknown: (a) 2y = 60 (b) – 8 = 5x – 3 (c) – 53w = –15 (d) 13 – z = 8 (e) k + 8 = 12 – k (f) 7m = m – 3 (g) 3n = 10 + n
Answer
(a) y = 30 (b) x = −1 (c) w = 15/53 (d) z = 5 (e) k = 2 (f) m = −1/2 (g) n = 5
(a) 2y = 60 → y = 60 ÷ 2 = 30 (b) −8 = 5x − 3 → −8 + 3 = 5x → −5 = 5x → x = −1 (c) −53w = −15 → w = (−15) ÷ (−53) = 15/53 (d) 13 − z = 8 → 13 − 8 = z → z = 5 (e) k + 8 = 12 − k → k + k = 12 − 8 → 2k = 4 → k = 2 (f) 7m = m − 3 → 7m − m = −3 → 6m = −3 → m = −1/2 (g) 3n = 10 + n → 3n − n = 10 → 2n = 10 → n = 5
Q3I 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?
Answer
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
Q4The weight of a brick is 1 kg more than half its weight. What is the weight of the brick?
Answer
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
Q5One quarter of a number increased by 9 gives the same number. What is the number?
Answer
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
Q6Given 4k + 1 = 13, find the values of: (a) 8k + 2 (b) 4k (c) k (d) 4k – 1 (e) – k – 2
Answer
(a) 26 (b) 12 (c) 3 (d) 11 (e) − 5
From 4k + 1 = 13: 4k = 13 − 1 = 12, so k = 12 ÷ 4 = 3
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
Mind the Mistake, Mend the Mistake — Pages 181 – 1827.3 Mind the Mistake, Mend the Mistake
Q1Go 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
Answer
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.
The first two steps are fine; the last step is wrong. The correct answer is z = 1/4.
The mistake: from 8z = 2 the division was done the wrong way round — 8 was divided by 2 instead of 2 being divided by 8.
Correct solution
7 − 8z = 5
7 − 5 = 8z (this step in the book is correct)
8z = 2
z = 2 ÷ 8 = 1/4
Q32v – 4 = 6 v – 4 = 6 – 2 v – 4 = 4 v = 8
Answer
The first step is wrong. The correct answer is v = 5.
The mistake: 2v is a product, so the 2 is a factor, not a term. It cannot be moved across as a subtraction. To remove a factor you divide the whole of the other side by it.
Correct solution
2v − 4 = 6
2v = 6 + 4 = 10 (remove the term − 4 first)
v = 10 ÷ 2 = 5
Both signs in the first step are wrong. The correct answer is z = − 3.
The mistake: 3z was moved to the LHS but its sign was not changed (it should become − 3z), and + 2 was moved to the RHS but also kept its sign (it should become − 2).
The mistake is in the third step. The correct answer is w = 26/11.
The mistake: 15w = 26 + 4w is allowed (4w was added to both sides), but then 26 + 4w was written as 30. 26 and 4w are not like terms, so they cannot be added. It is simpler to collect the like terms on the LHS first.
Correct solution
15w − 4w = 26
11w = 26 (collect the like terms on the LHS)
w = 26/11
Q63x + 1 = – 12 x + 1 = – 12/3 x + 1 = – 4 x = – 5
Answer
The first step is wrong. The correct answer is x = − 13/3.
The mistake: both sides were divided by 3, but on the LHS only the 3x was divided. The 1 was left untouched. Dividing a side means dividing every term of it.
In-text Questions — Pages 183 – 1847.4 A Pinch of History
Q1Example 16: Bījgaṇita by Bhāskarāchārya (1150 CE) mentions this problem. One man has ₹300 rupees and 6 horses. Another man has 10 horses and a debt of ₹100. If they are equally rich and the price of each horse is the same, tell me the price of one horse.
Answer
One horse costs ₹100.
Let the price of one horse be ₹x.
First man’s wealth = 300 + 6x
Second man’s wealth = 10x − 100 (a debt of ₹100 is − 100)
They are equally rich, so
300 + 6x = 10x − 100
300 + 6x + 100 = 10x (add 100 to both sides)
400 + 6x = 10x
400 = 10x − 6x (subtract 6x from both sides)
400 = 4x
x = 400 ÷ 4 = ₹100
Q2Let us look at a few equations of the following form: 5x + 4 = 3x + 8, or 3x – 6 = 2x + 4. Can we come up with a formula to solve these equations? That is, for the first equation, can we perform some operations using 5, 4, 3, and 8 that will directly give us the solution? Using a similar method, can you solve the second equation using the numbers 3, – 6, 2 and 4?
Answer
Yes. Both equations have the form Ax + B = Cx + D, and Brahmagupta’s rule gives the solution straight from the four numbers.
Ax + B = Cx + D
Ax − Cx = D − B (subtract Cx and subtract B from both sides)
(A − C)x = D − B x = (D − B) ÷ (A − C)
Equation
A, B, C, D
x = (D − B) ÷ (A − C)
Solution
5x + 4 = 3x + 8
5, 4, 3, 8
(8 − 4) ÷ (5 − 3) = 4 ÷ 2
x = 2
3x − 6 = 2x + 4
3, − 6, 2, 4
(4 − (− 6)) ÷ (3 − 2) = 10 ÷ 1
x = 10
Q3Using this formula can you solve this equation 2x + 3 = 4x + 5?
Answer
x = − 1.
Here A = 2, B = 3, C = 4, D = 5.
x = (D − B) ÷ (A − C)
x = (5 − 3) ÷ (2 − 4)
x = 2 ÷ (− 2) = − 1
Figure it Out — Pages 185 – 1897.4 A Pinch of History
Q2Ranju is a daily wage labourer. She earns ₹ 750 a day. Her employer pays her in 50 and 100 rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have?
Answer
5 notes of ₹50 and 5 notes of ₹100.
Let n be the number of notes of each kind.
Value of the ₹50 notes = 50n
Value of the ₹100 notes = 100n
Total = ₹750, so
50n + 100n = 750
150n = 750
n = 750 ÷ 150 = 5
Q3In the given picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? Write an equation to describe this problem.Three black blobs and the blue dots that are not covered; altogether there are 25 dots.
Answer
Each blob hides 7 dots. The equation is 3b + 4 = 25.
Let b be the number of dots hidden by one blob.
Dots hidden by 3 blobs = 3b
Dots you can see = 4
3b + 4 = 25
3b = 25 − 4 = 21
b = 21 ÷ 3 = 7
Q4Here are machines that take an input, perform an operation on it and send out the result as an output. (a)Machine (a): +3, then ×4, then −5. The first row is the book’s worked example (12 → 15 → 60 → 55). Find the inputs when the output is 43 and when it is 75.(b)Machine (b): the input goes into both ×3 (top) and +3 (bottom); the bottom result is subtracted from the top result. Worked example: 12 gives 36 and 15, and 36 − 15 = 21. Find the inputs when the output is 63 and when it is 227.
Q5What are the inputs to these machines?Machine 1: ÷3, then ÷3, gives +5. Machine 2: −4, then −4, gives −11.
Answer
Machine 1: input 45. Machine 2: input −3.
Machine 1: (x ÷ 3) ÷ 3 = 5 → x ÷ 9 = 5 → x = 5 × 9 = 45 Machine 2: x − 4 − 4 = −11 → x − 8 = −11 → x = −11 + 8 = −3
Q6A taxi driver charges a fixed fee of ₹800 per day plus ₹20 for each kilometer traveled. If the total cost for a taxi ride is ₹2200, determine the number of kilometres traveled.
Answer
70 kilometres.
Let k be the number of kilometres travelled.
Cost = 800 + 20k
800 + 20k = 2200
20k = 2200 − 800 = 1400
k = 1400 ÷ 20 = 70 km
Q7The sum of two numbers is 76. One number is three times the other number. What are the numbers?
Answer
The numbers are 19 and 57.
Let the smaller number be x. Then the other is 3x.
x + 3x = 76
4x = 76
x = 76 ÷ 4 = 19
The other number = 3 × 19 = 57
Q8The figure shows the diagram for a window with a grill. What is the gap between two rods in the grill?The window grill: it is 34 cm from top to bottom, the band at each end is 3 cm wide and each rod between them is 2 cm wide.
Answer
The gap between two rods is 3 cm.
Read the grill from top to bottom. Between the 3 cm band at the top and the 3 cm band at the bottom there are 5 rods, each 2 cm wide, and 6 equal gaps.
Let the gap be g cm.
Total height = top band + bottom band + 5 rods + 6 gaps
34 = 3 + 3 + 5 × 2 + 6g
34 = 6 + 10 + 6g
34 = 16 + 6g
6g = 34 − 16 = 18
g = 18 ÷ 6 = 3 cm
Q9In a restaurant, a fruit juice costs ₹15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹600, find the cost of the fruit juice and milkshake.
Answer
A fruit juice costs ₹45 and a chocolate milkshake costs ₹60.
Let the milkshake cost ₹m. Then a juice costs ₹(m − 15).
(a) has a mistake (x = 19/2); (b) is correct (y = 6/7); (c) has two mistakes (x = −13/5).
(a) Mistake: both sides were divided by 6, but the 9 was not divided.
6x + 9 = 66 → 6x = 66 − 9 = 57 → x = 57 ÷ 6 = 19/2
(b) No mistake. Dividing every term by 2 gives 7y + 12 = 18, so 7y = 6 and y = 6/7.
(c) Mistakes: −5 moved across should become +5 (not −5); and −5x = 3 gives x = 3 ÷ (−5), not −5/3.
4x − 5 = 9x + 8 → 4x = 9x + 13 → 4x − 9x = 13 → −5x = 13 → x = −13/5
Q12Find the measures of the angles of these triangles.Two triangles. In the first, the two slanting sides carry equal tick marks; the top angle is y and the angle at the bottom right is y + 15. In the second, the top angle is x and the bottom angles are x – 10 and x + 10.
Answer
First triangle: 50°, 65°, 65°. Second triangle: 60°, 50°, 70°.
First triangle: the two marked sides are equal, so both base angles are y + 15.
y + (y + 15) + (y + 15) = 180 → 3y + 30 = 180 → 3y = 150 → y = 50
Angles: 50°, 65°, 65°
Second triangle: x + (x − 10) + (x + 10) = 180 → 3x = 180 → x = 60
Angles: 60°, 50°, 70°
Q13Write 4 equations whose solution is u = 6.
Answer
Build them from u = 6 by doing the same thing to both sides.
Equation
How it was built
Check with u = 6
u − 6 = 0
Subtract 6 from both sides
6 − 6 = 0 ✓
2u = 12
Multiply both sides by 2
12 = 12 ✓
u ÷ 3 = 2
Divide both sides by 3
6 ÷ 3 = 2 ✓
5u + 1 = 31
Multiply by 5, then add 1
30 + 1 = 31 ✓
4u − 7 = 3u − 1
Two expressions that both give 17 at u = 6
24 − 7 = 17 and 18 − 1 = 17 ✓
Q14The Bakhśhāli Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second; and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?
Answer
The first person is given 4.
Let the first person get x.
Second = 2 × x = 2x
Third = 3 × 2x = 6x
Fourth = 4 × 6x = 24x
Total = 132, so
x + 2x + 6x + 24x = 132
33x = 132
x = 132 ÷ 33 = 4
Person
Expression
Amount
First
x
4
Second
2x
8
Third
6x
24
Fourth
24x
96
Total
33x
132
Q15The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?
Answer
The giraffe is 5 metres tall.
Let the height be h metres.
h = h/2 + 2.5
h − h/2 = 2.5 (subtract h/2 from both sides)
h/2 = 2.5
h = 2.5 × 2 = 5 m
Q16Two separate figures are given below. Each figure shows the first few positions in a sequence of arrangements made with sticks. Identify the pattern and answer the following questions for each figure: (a) How many squares are in position number 11 of the sequence? (b) How many sticks are needed to make the arrangement in position number 11 of the sequence? (c) Can an arrangement in this sequence be made using exactly 85 sticks? If yes, which position number will it correspond to? (d) Can an arrangement in this sequence be made using exactly 150 sticks? If yes, which position number will it correspond to?
Figure 1
Figure 2
Figure 1 shows positions 1 to 3; Figure 2 shows positions 1 to 4.
Answer
Figure 1: (a) 11 squares (b) 36 sticks (c) No (d) Yes, position 49. Figure 2: (a) 34 squares (b) 103 sticks (c) Yes, position 9 (d) No.
Figure 1 — sticks: 6, 9, 12, … so position n has n squares and 3n + 3 sticks.
(a) 11 squares (b) 3 × 11 + 3 = 36 sticks
(c) 3n + 3 = 85 → 3n = 82; 82 is not divisible by 3, so no.
(d) 3n + 3 = 150 → 3n = 147 → n = 49, so yes, position 49.
Q19Solve the equations to find a path from Start to the End. Show your work in the given boxes provided and colour your path as you proceed.
Page 189 — the puzzle grid: twelve boxes, each holding an equation, joined by numbered arrows. The Start box at the top left is worked out for you.
Answer
Solve every box, then follow the arrow marked with its answer.
Q20There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?
Answer
There are 12 donkeys and 16 children.
Let there be d donkeys. Since every child and every donkey has one head,
number of children = 28 − d
Feet: a child has 2, a donkey has 4.
4d + 2(28 − d) = 80
4d + 56 − 2d = 80
2d + 56 = 80
2d = 80 − 56 = 24
d = 24 ÷ 2 = 12 donkeys
Children = 28 − 12 = 16
A Magic Trick — Page 190Puzzle page
Try This
Q1Think of any number. Now multiply it by 2. Add 10. Divide by 2. Now subtract the original number you thought of. Finally, add 3. I predict that you now have 8. Am I correct? Can you explain why the trick works? [Hint: Denote the first number thought of by x.]
Answer
Yes, the prediction is always correct — the answer is 8 whatever number you start with.
Let the number you think of be x, and follow the steps in algebra.
Step
Expression
If x = 7
If x = 100
Think of a number
x
7
100
Multiply it by 2
2x
14
200
Add 10
2x + 10
24
210
Divide by 2
x + 5
12
105
Subtract the original number
x + 5 − x = 5
5
5
Add 3
5 + 3 = 8
8
8
Q2Try the trick on your friends and family! Can you make your own such tricks?
Answer
Yes. Here is my own trick — the x cancels at the end, so the answer is always 10.
My trick
Think of a number → x
Multiply by 3 → 3x
Add 12 → 3x + 12
Divide by 3 → x + 4
Subtract the number you thought of → 4
Add 6 → 10
Prediction: the answer is always 10.