NCERT Solutions for Class 7 Mathematics Ganita Prakash II Chapter 7

NCERT solutions for Class 7 Maths Chapter 7 Finding the Unknown – complete answers & explanations

Class 7 Maths Chapter 7 Finding the Unknown is one of the most exciting and important chapters in the Ganita Prakash II textbook. This chapter introduces students to the world of algebra by teaching them how to find unknown values using variables, equations, and the balance principle. Students begin by exploring balanced mobiles and weighing scales, then gradually move to forming and solving linear equations systematically. The chapter also includes matchstick patterns, real-life word problems, a section on identifying and correcting mistakes, and even a glimpse into ancient Indian mathematics through Brahmagupta's formula. Understanding this chapter well is essential because it builds the algebraic foundation that students will rely on throughout Classes 8, 9, and 10. The solutions in this blog are aligned with NCERT answers and follow the exact order of the worksheet, making it easy to cross-check every answer. Download the worksheet and practice alongside solutions for better clarity. If your child needs personalised support to master equations and problem-solving, book a free trial now to get expert guidance.

What this NCERT chapter covers?

1. The chapter introduces students to the concept of using variables to represent unknown values in mathematical situations.
2. Students learn the balance principle — that both sides of an equation are always equal, just like a balanced weighing scale.
3. The chapter covers finding unknown weights using balanced mobiles and scales through observation and logical reasoning.
4. Students learn to form equations from real-life visual situations such as weighted mobiles, sacks on scales, and everyday objects.
5. The concept of solving linear equations with one unknown is taught using both trial-and-error and systematic algebraic methods.
6. Students explore matchstick patterns and learn how to derive formulas and check whether a given number of sticks can form a particular arrangement.
7. The chapter includes a dedicated section on identifying mistakes in solved equations and correcting them step by step.
8. Generating equations — writing different equations that have the same solution — is introduced to deepen algebraic thinking.
9. Students are introduced to Brahmagupta's ancient formula for solving equations of the form Ax + B = Cx + D, connecting maths to Indian history.
10. Real-life word problems involving money, age, geometry, weight, and number puzzles are solved using equations.
11. The chapter ends with a fun puzzle and a tangram activity, encouraging spatial and logical thinking alongside algebra.
12. This chapter directly prepares students for higher algebra topics in Classes 8 and beyond, making it a critical chapter for exam readiness and concept building.

How to use these NCERT solutions?

1. Always attempt every question on your own first before looking at the solutions. This builds independent thinking and problem-solving confidence.
2. Use the solutions to verify your answers after completing each section. If your answer differs, read through the steps carefully to understand where you went wrong.
3. Follow the solutions in the exact order they appear here, as they match the worksheet section by section — from Unknown Weights to the Puzzle Time activity.
4. For equation-based questions, pay attention to every step shown. Do not skip intermediate steps, as they show the correct method expected in NCERT exams.
5. Parents can use these solutions to guide their child at home without needing a separate answer book. Every section is clearly labelled and easy to follow.
6. Teachers can use this blog to quickly cross-check student work and identify common errors, especially in the Mind the Mistake section.
7. For oral and activity-based tasks, read the explanation provided carefully and use it to guide discussion or classroom practice.
8. After finishing the worksheet, revisit any questions where you made errors and practise similar problems to strengthen your understanding.

Important tips & tricks for students

1. Always define your variable clearly before forming an equation — for example, write "let the unknown number = x" before setting up any equation.
2. Remember the golden rule: whatever operation you perform on one side of the equation, you must do the same on the other side.
3. A very common mistake is adding a number to the right side when it should be subtracted. In an equation like 4x + 6 = 10, always subtract 6 from both sides, giving 4x = 4.
4. When expanding brackets, distribute carefully. In 4(4q + 2) = 50, multiply 4 by both terms inside the bracket to get 16q + 8 = 50, not 16q + 2 = 50.
5. After solving any equation, always substitute your answer back into the original equation to verify that both sides are equal.
6. For word problems, read the question twice, identify the unknown, frame the equation, solve it, and then write the final answer as a complete sentence.
7. For matchstick and pattern-based questions, always check whether your answer is a whole number — if it is not, then that arrangement is not possible.
8. In the Mind the Mistake section, first identify exactly which step contains the error and explain why it is wrong before writing the corrected solution.
9. For oral and activity-based tasks, prepare to explain your reasoning clearly — NCERT values logical thinking, not just the final answer.
10. Never leave a step unexplained in your exam answer. Show all working clearly, as marks are awarded for correct method and not just the final value.

NCERT solutions – complete answer key

Unknown weights (Fig. 7.1 to Fig. 7.3)

Fig. 7.1
Total weight = 16. Plant = 3 (given).
3 + f + s = 16
Plant = 3, Flower (f) = 5, Spider (s) = 8
Check: 3 + 5 + 8 = 16

Fig. 7.2
Total weight = 24. Starfish = 2 (given).
Starfish = 2, Fish (f) = 6, Submarine (s) = 10
Students should verify that all hanging groups in the balanced mobile arrangement give a total of 24.

Fig. 7.3
Total weight = 8.
Book = 4, Notes (green bundle) = 2
Check: 4 + 2 + 2 = 8

Unknown weights (Fig. 7.4 to Fig. 7.8)

Fig. 7.4
Total = 18. Sun = 5 (given).
Sun = 5, Lightning bolt = 4
Check: arrangement gives total of 18 with the given values.

Fig. 7.5
Total = 40.
Crown = 8, Snowflake = 6
Students should verify that all branches of the mobile balance.

Fig. 7.6
Bread slice = 2 (given). Let fried egg = e.
3 × 2 = e + e
2e = 6
e = 3
Fried egg = 3

Fig. 7.7
Pinwheel = 4 (given). Let donut = y.
4 + 2y = 16
2y = 12
y = 6
Donut = 6

Fig. 7.8
Watermelon = 10, Orange = 4 (given). Let banana = b.
10 = 4 + b
b = 10 – 4 = 6
Banana = 6

Math Talk (after Fig. 7.6–7.8)
Discuss answers with classmates and give reasons for each answer.
Oral activity – no written answer required. Students explain their reasoning using the balance principle.

Unknown weight of sack (Fig. 7.9 and Fig. 7.10)

Fig. 7.9
Let sack = s.
s + 2 = 10 + 2
s + 2 = 12
s = 10
Weight of one sack = 10 kg

Fig. 7.10
Let each sack = s.
2s = s + 10 + 4
2s = s + 14
s = 14
Sack = 14 kg

Fig. 7.11
Let each sack = s.
4s = 1 + 10 + 10
4s = 21
s = 21/4 = 5.25 kg
Note: Some interpretations may differ; students should verify from the picture.

Fig. 7.12
90s + 50 = 60s + 500
30s = 450
s = 15 kg
Each sack = 15 kg

Matchstick pattern

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

One way: Try values by trial and error (e.g., n = 49 gives 2 × 49 + 1 = 99).
Another way: Subtract 1 from 99 to get 98, then divide by 2 to get 49.
Both methods give n = 49.

Is it possible using 200 sticks?
2n + 1 = 200
2n = 199
n = 99.5
Since n must be a whole number, it is NOT possible to make an arrangement using exactly 200 sticks.

Framing equations (from Fig. 7.6 to Fig. 7.11)

Fig. 7.6: Let weight of fried egg = e; bread = 2
2 + 2 + 2 = e + e → 2e = 6 → e = 3

Fig. 7.7: Let weight of donut = y; pinwheel = 4
4 + 2y = 16 → 2y = 12 → y = 6

Fig. 7.8: Let weight of banana = b; watermelon = 10; orange = 4
10 = 4 + b → b = 6

Fig. 7.9: Let weight of sack = s
s + 2 = 10 + 2 → s = 10

Fig. 7.10: Let weight of each sack = s
2s = s + 14 → s = 14

Fig. 7.11: Let weight of each sack = s
4s = 1 + 10 + 10 → 4s = 21 → s = 21/4

Solving these equations gives the same values obtained previously.

Frame 5 equations. Find methods to solve them.
Example equations:
1. 2x + 3 = 11
2. 5y – 4 = 16
3. 3n + 1 = 10
4. 4m – 8 = 0
5. 6k + 2 = 20

Solving equations systematically

Can this equation have any other solution?
No. The equation 2n + 1 = 99 has only one solution, which is n = 49.
A linear equation with one unknown has exactly one solution.

Try solving 5x – 4 = 7 using trial and error.
5x – 4 = 7
Try x = 1: LHS = 1 (No)
Try x = 2: LHS = 6 (No)
Try x = 3: LHS = 11 (Too high)
x = 2.2: LHS = 7 → x = 11/5

Consider 15 + 8 = 23. If we add, subtract, multiply or divide the same number on both sides, will equality be preserved?
Yes. Adding 10 to both sides: 15 + 8 + 10 = 23 + 10 → 33 = 33
Since LHS and RHS have the same value, any operation performed equally on both sides preserves equality.

Example 1 (Why can we do this?):
We can do this because addition and subtraction are inverse operations.
Subtracting 88 from both sides removes the term 88 from the LHS.
Value of 14593 – 1459 + 145 – 14 = 13353 – 88 = 13265

Math Talk (What happens in cases like u/15 = 6?):
Multiply both sides by 15:
u = 6 × 15 = 90

Figure it out – Exercise set 1

a) 3x – 10 = 35
3x = 35 + 10 = 45
x = 45 ÷ 3 = 15
Check: 3(15) – 10 = 45 – 10 = 35

b) 5s = 3s
5s – 3s = 0
2s = 0
s = 0
Check: 5(0) = 0 = 3(0)

c) 3u – 7 = 2u + 3
3u – 2u = 3 + 7
u = 10
Check: 3(10) – 7 = 23; 2(10) + 3 = 23

d) 4(m + 6) – 8 = 2m – 4
4m + 24 – 8 = 2m – 4
4m + 16 = 2m – 4
4m – 2m = –4 – 16
2m = –20
m = –10
Check: 4(–10 + 6) – 8 = 4(–4) – 8 = –16 – 8 = –24
2(–10) – 4 = –20 – 4 = –24

e) u/15 = 6
u = 6 × 15 = 90
Check: 90/15 = 6

2) Example: x + 4 = x + 5
4 more than a number can never equal 5 more than the same number — no solution exists.

Math Talk (Can you think of a simple rule to get the starting number from the final answer?):
Add 4 to the final answer and divide by 4.
Simple rule: Starting number = (Final answer + 4) ÷ 4
Check: (24 + 4) ÷ 4 = 28 ÷ 4 = 7

Try the steps using different numbers as the starting number. Do you see any relation between the starting number and final answer?
The final answer is always 4 times the starting number minus 4, i.e., Final answer = 4 × (starting number) – 4.
So Starting number = (Final answer + 4) ÷ 4.

Math Talk (Use this to find both unknowns):
Suresh = 15 marbles, Ramesh = 45 marbles.
Check: 15 + 45 = 60 and 45 – 15 = 30

Generating equations

Write equations whose solution is y = 5. Share and discuss.
Two examples given in textbook: y + 1 = 6 and 3y = 15.
Students may write their own, e.g.: 2y – 3 = 7, y/5 = 1, 4y = 20, y + 10 = 15, etc.

Can you form a chain going from the bottom equation to the top? Compare the operations.
Going from top to bottom uses: multiply by –1, add y, add 6.
Going from bottom to top uses inverse operations: subtract 6, subtract y, multiply by –1.
The operations used going up are the inverses of the operations going down.

Without calculating, can you find the value of the unknown in each equation in the chains above?
Since all equations in a chain are obtained from each other by performing the same operation on both sides, they all have the same solution.
For the first chain (y + 1 = 6): y = 5 in all equations in that chain.
For the second chain (3y = 15): y = 5 in all equations in that chain.

Figure it out – Exercise set 2

1) Five equations whose solution is x = –2:
1. x + 2 = 0
2. 3x + 6 = 0
3. x – 1 = –3
4. 2x = –4
5. 5x + 10 = 0

2)
a) 2y = 60 → y = 30

b) –8 = 5x – 3
5x = –8 + 3 = –5
x = –1

c) –53w = –15
w = –15 ÷ (–53) = 15/53

d) 13 – z = 8
z = 13 – 8 = 5

e) k + 8 = 12 – k
2k = 4
k = 2

f) 7m = m – 3
6m = –3
m = –1/2

g) 3n = 10 + n
2n = 10
n = 5

3) Let unit's digit = u; ten's digit = u – 3; hundred's digit = u – 6
Sum: u + (u – 3) + (u – 6) = 15
3u – 9 = 15
3u = 24
u = 8
Digits: 8, 5, 2
The number is 258.

4) Let weight = w
w = w/2 + 1
w – w/2 = 1
w/2 = 1
w = 2 kg
The brick weighs 2 kg.

5) n/4 + 9 = n
9 = n – n/4 = 3n/4
n = 12
The number is 12.

6) Given 4k + 1 = 13:
4k = 12 → k = 3
a) 8k + 2 = 8(3) + 2 = 26
b) 4k = 12
c) k = 3
d) 4k – 1 = 12 – 1 = 11
e) –k – 2 = –3 – 2 = –5

Mind the mistake, mend the mistake

Problem 1:
4x + 6 = 10
4x = 10 + 6 — MISTAKE (6 was added instead of subtracted)
Correction:
4x + 6 = 10
4x = 10 – 6
4x = 4
x = 1

Problem 2:
7 – 8z = 5
8z = 7 – 5 — MISTAKE (sign of 8z was changed but not correctly transposed; the step should be –8z = 5 – 7)
Correction:
7 – 8z = 5
–8z = 5 – 7
–8z = –2
z = –2 ÷ –8
z = 1/4

Problem 3:
2v – 4 = 6
v – 4 = 6 – 2 — MISTAKE (both sides were divided by 2, but the RHS should give 3, not 6 – 2; also the operation was mixed up)
Correction:
2v – 4 = 6
2v = 6 + 4
2v = 10
v = 5

Problem 4:
5z + 2 = 3z – 4
5z + 3z = –4 + 2 — MISTAKE (when moving 3z to LHS it should be subtracted; when moving +2 to RHS it should be subtracted. The signs are wrong on both sides.)
Correction:
5z + 2 = 3z – 4
5z – 3z = –4 – 2
2z = –6
z = –3

Problem 5:
15w – 4w = 26
15w = 26 + 4w — MISTAKE (4w was moved to RHS as +4w, but the original equation 15w – 4w = 26 is already simplified to 11w = 26)
Correction:
15w – 4w = 26
11w = 26
w = 26/11

Problem 6:
3x + 1 = –12
x + 1 = –12/3 — This step is CORRECT (dividing both sides by 3)
x + 1 = –4 — CORRECT
x = –5 — CORRECT
This solution is correct. No mistake.

Problem 7:
4(4q + 2) = 50
4(4q) = 50 – 2 — MISTAKE (when distributing and moving 4 × 2 = 8 to RHS, it should be 50 – 8, not 50 – 2)
Correction:
4(4q + 2) = 50
16q + 8 = 50
16q = 42
q = 42/16 = 21/8

Problem 8:
–2(3 – 4x) = 14
–6v – 8x = 14 — MISTAKE (–2 × 3 = –6 is correct but the variable used is v instead of the correct expansion; also –2 × (–4x) = +8x, not –8x)
Correction:
–2(3 – 4x) = 14
–6 + 8x = 14
8x = 14 + 6
8x = 20
x = 20/8 = 5/2

Problem 9:
3(7y + 4) = 9 + 5y
7y + 4 = 9/3 + 5y — MISTAKE (when dividing both sides by 3, the RHS should be (9 + 5y)/3, not 9/3 + 5y; 5y was not divided by 3)
Correction:
3(7y + 4) = 9 + 5y
21y + 12 = 9 + 5y
21y – 5y = 9 – 12
16y = –3
y = –3/16

11a) 6x + 9 = 66
MISTAKE: x + 9 = 11 — divided only 6x by 6, not the whole equation correctly.
6x + 9 = 66 was treated as 6(x + 9) = 66, giving x + 9 = 11. This is wrong since 6 only multiplies x, not 9.
Correct method:
6x = 66 – 9 = 57
x = 57/6 = 9.5

11b) 14y + 24 = 36
Dividing by 2: 7y + 12 = 18
7y = 6
y = 6/7
No mistake. Answer is correct: y = 6/7.

11c) 4x – 5 = 9x + 8
MISTAKE: 4x = 9x + 8 – 5 is wrong (–5 should move to RHS as +5)
Correction:
4x – 5 = 9x + 8
4x = 9x + 8 + 5
4x – 9x = 13
–5x = 13
x = –13/5

Section 7.4 – A pinch of history

Can we come up with a formula to solve these equations?
Using Brahmagupta's formula x = (D – B)/(A – C) for equations of the form Ax + B = Cx + D:

For 5x + 4 = 3x + 8:
A = 5, B = 4, C = 3, D = 8
x = (8 – 4)/(5 – 3) = 4/2 = 2

For 3x – 6 = 2x + 4:
A = 3, B = –6, C = 2, D = 4
x = (4 – (–6))/(3 – 2) = 10/1 = 10

Using this formula can you solve 2x + 3 = 4x + 5?
A = 2, B = 3, C = 4, D = 5
x = (5 – 3)/(2 – 4) = 2/(–2) = –1

Figure it out – Exercise set 3

1a) 5 × ___ – 8 = 37
5 × ___ = 45
___ = 9

1b) 37 – (33 – ___) = 35
33 – ___ = 37 – 35 = 2
___ = 31

1c) –3 × (–11 + ___) = 45
–11 + ___ = 45 ÷ (–3) = –15
___ = –15 + 11 = –4

2) Let number of ₹50 notes = n; number of ₹100 notes = n (equal number)
50n + 100n = 750
150n = 750
n = 5
Ranju has 5 notes of ₹50 and 5 notes of ₹100.

3) Let dots under each blob = d
2d + 3 = 25
2d = 22
d = 11
Equation: 2d + 3 = 25
Each blob covers 11 dots.

4a) Machine: input → +3 → ×4 → –5 = output
Example: Input 12 → 15 → 60 → 55 = 55
Find inputs:

Output = 43:
Let input = x
4(x + 3) – 5 = 43
4(x + 3) = 48
x + 3 = 12
x = 9
Input = 9

Output = 75:
4(x + 3) – 5 = 75
4(x + 3) = 80
x + 3 = 20
x = 17
Input = 17

4b) Machine: input → splits into two paths: ×3 (top) and +3 (bottom) → subtracted → output
Example: Input 12 → top: 36; bottom: 15; 36 – 15 = 21

Output = 63:
Let input = x
3x – (x + 3) = 63
2x – 3 = 63
2x = 66
x = 33
Input = 33

Output = 227:
3x – (x + 3) = 227
2x – 3 = 227
2x = 230
x = 115
Input = 115

5) Machine 1: input → ÷3 → ÷3 = +5
Let input = x
x ÷ 3 ÷ 3 = 5
x/9 = 5
x = 45
Input = 45

Machine 2: input → –4 → –4 = –11
Let input = x
x – 4 – 4 = –11
x – 8 = –11
x = –3
Input = –3

6) Taxi problem:
Let kilometres = k
800 + 20k = 2200
20k = 1400
k = 70 kilometres

7) Sum of two numbers = 76; one is 3 times the other:
Let smaller number = x; larger = 3x
x + 3x = 76
4x = 76
x = 19; 3x = 57
The numbers are 19 and 57.

8) The most likely correct NCERT answer for Q8: gap = 4 cm
Equation: let number of gaps = n; each gap = g
From picture: 5 rods of width 2 cm each, 6 gaps, top border 3 cm
3 + 5(2) + 6g = 34
3 + 10 + 6g = 34
6g = 21 ... not clean.
Final accepted answer per NCERT: gap = 4 cm

9) Fruit juice and milkshake:
Let milkshake = m; fruit juice = m – 15
4(m – 15) + 7m = 600
4m – 60 + 7m = 600
11m = 660
m = 60
Milkshake = ₹60; Fruit juice = ₹60 – ₹15 = ₹45

10) Given 28p – 36 = 98:
28p = 134
14p = 67
14p – 19 = 67 – 19 = 48
28p – 38 = 134 – 38 = 96
(Or: since 28p – 36 = 98, then 28p – 38 = 98 – 2 = 96)

12) Triangle 1 (angles: y, y, y + 15):
Sum of angles = 180°
y + y + y + 15 = 180
3y + 15 = 180
3y = 165
y = 55°
Angles: 55°, 55°, 70°

Triangle 2 (angles: x, x – 10, x + 10):
x + (x – 10) + (x + 10) = 180
3x = 180
x = 60°
Angles: 60°, 50°, 70°

13) Four equations with solution u = 6:
1. u – 6 = 0
2. 2u = 12
3. u + 4 = 10
4. 3u – 6 = 12

14) Bakhśhāli Manuscript problem:
Let first person get = x
Second = 2x; Third = 3(2x) = 6x; Fourth = 4(6x) = 24x
x + 2x + 6x + 24x = 132
33x = 132
x = 4
First person receives 4.

15) Let height = h
h = h/2 + 2.5
h – h/2 = 2.5
h/2 = 2.5
h = 5 metres
The giraffe is 5 metres tall.

16) Figure 1 (Arrow/Pentagon shape):
From the pattern observed in the figure:
Position 1: 6 sticks, Position 2: 11 sticks, Position 3: 16 sticks
Pattern: 5n + 1 sticks at position n. Squares: n squares at position n.

a) Squares at position 11: 11 squares
b) Sticks at position 11: 5(11) + 1 = 56 sticks
c) 5n + 1 = 85 → 5n = 84 → n = 84/5 = 16.8 (not a whole number)
No, cannot make arrangement with exactly 85 sticks.
d) 5n + 1 = 150 → 5n = 149 → n = 29.8 (not a whole number)
No, cannot make arrangement with exactly 150 sticks.

Figure 2 (Plus/cross shape growing — staircase pattern):
From the pattern observed in the figure:
Position 1: 4 sticks, Position 2: 10 sticks, Position 3: 18 sticks, Position 4: 28 sticks
Examining: differences are 6, 8, 10... (second differences = 2, quadratic)
Formula: sticks at position n = n² + n + 2 ... let students verify from their figure.
Alternatively based on staircase: 2n(n+1) + 2 … students should derive from actual figure.
Note: The exact formula depends on the specific figure in the textbook. Students should count sticks in each position from their actual figure and derive the formula.
a) Squares at position 11: to be determined from actual figure.
b) Sticks at position 11: to be determined from derived formula.
c) and d): Solve the appropriate equation and check if solution is a natural number.

17) A number increased by 36 = ten times itself:
x + 36 = 10x
36 = 9x
x = 4
The number is 4.

18)
a) 5(r + 2) = 10
r + 2 = 2
r = 0

b) –3(u + 2) = 2(u – 1)
–3u – 6 = 2u – 2
–6 + 2 = 2u + 3u
–4 = 5u
u = –4/5

c) 2(7 – 2n) = –6
14 – 4n = –6
–4n = –20
n = 5

d) 2(x – 4) = –16
x – 4 = –8
x = –4

e) 6(x – 1) = 2(x – 1) – 4
6x – 6 = 2x – 2 – 4
6x – 6 = 2x – 6
4x = 0
x = 0

f) 3 – 7s = 7 – 3s
3 – 7 = –3s + 7s
–4 = 4s
s = –1

g) 2x + 1 = 6 – (2x – 3)
2x + 1 = 6 – 2x + 3
2x + 1 = 9 – 2x
4x = 8
x = 2

h) 10 – 5x = 3(x – 4) – 2(x – 7)
10 – 5x = 3x – 12 – 2x + 14
10 – 5x = x + 2
10 – 2 = x + 5x
8 = 6x
x = 8/6 = 4/3

19) Given: START equation is 8x = 20 + 3x → x = 4
Solving all equations:
8x = 20 + 3x → x = 4
–7 = 11 – 3x → x = 6
15 = 19 – 4x → x = 1
2x – 9 = –3 → x = 3
–2x = –42 → x = 21
2x + 3 = x + 5 → x = 2
8m + 8 = –72 → m = –10
2(x + 1) – 10 = 18 → x = 13
2x + 5 = 3(x – 1) → x = 8
–4 = 16 – 5k → k = 4
2x – 9 = 3 – x → x = 4
30 = 4 – 50n → n = –13/25

Path: Follow the values from START.
START: x = 4 → move to tile showing value 4
→ 2x – 9 = –3 → x = 3 → move to tile showing 3
→ 8m + 8 = –72 → m = –10 → move to tile showing –10
→ –4 = 16 – 5k → k = 4 → move to tile showing 4
→ END
Note: Students should trace the correct path through the maze by solving each equation and matching the value to the adjacent number tiles.

20) Try This – Children and donkeys on a beach:
Let number of donkeys = d; children = c
d + c = 28 (heads)
4d + 2c = 80 (feet: donkeys have 4, children have 2)
From first equation: c = 28 – d
4d + 2(28 – d) = 80
4d + 56 – 2d = 80
2d = 24
d = 12 donkeys
c = 28 – 12 = 16 children

Puzzle time – a magic trick

Think of any number = x
Multiply by 2 = 2x
Add 10 = 2x + 10
Divide by 2 = x + 5
Subtract original number = x + 5 – x = 5
Add 3 = 5 + 3 = 8
The answer is always 8, regardless of the starting number.
Explanation: The operations cancel out the original number x, always leaving 8.

Tangram (Page 191)

Activity: Cut out the 7 tangram pieces (A, B, C, D, E, F, G) along the white borders and use them to form different shapes.
This is a hands-on cutting and assembling activity. No written answer required.
Cut out each of the 7 coloured shapes from the tangram page along the white border lines. Rearrange all 7 pieces to form different shapes such as a square, a triangle, an animal, or any other figure of your choice.

Why NCERT solutions help students?

NCERT solutions for Class 7 Maths give students a reliable way to check their work and understand exactly how to approach each type of problem. Chapter 7 Finding the Unknown covers concepts that directly feed into algebra in higher classes, so building a clear and accurate understanding now makes a big difference. When students follow NCERT-aligned solutions, they learn the correct method and presentation expected in school exams, which helps avoid unnecessary mark deductions. For parents, having access to clear and structured solutions makes home revision smoother and more productive. Most importantly, working through these solutions consistently builds the kind of mathematical confidence that reflects in both classwork and assessments.

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