NCERT Solutions for Class 7 Maths Chapter 7 Finding the Unknown

This worksheet provides complete and accurate NCERT Solutions for Class 7 Maths Chapter 7 Finding the Unknown from the textbook Ganita Prakash II. This chapter introduces students to the foundational concept of algebra — using variables and equations to find unknown values. It is an important chapter because it builds the thinking skills students need to move from arithmetic to algebraic reasoning. Whether a student is just beginning to learn equations or needs help revising step-by-step solutions, this worksheet covers every section of the chapter in full detail.

Chapter summary: topics and themes

Chapter 7 Finding the Unknown is entirely concept and problem-based. It does not contain any stories or poems. The chapter begins with a visual and intuitive approach through balanced mobiles and weighing scales, asking students to identify unknown weights using logic and the principle of equality. Students then move to matchstick patterns, where they learn to form and solve equations based on visual sequences. The chapter progresses to solving linear equations systematically, generating equations, identifying and correcting mistakes in equation solving, and applying Brahmagupta's historical formula. It concludes with real-life word problems, a puzzle, and a tangram activity. The central theme throughout the chapter is the idea that both sides of an equation are equal, just like a balanced scale.

What this NCERT chapter covers?

The chapter covers a wide range of skills and concepts that are essential for Class 7 Maths, including understanding the balance principle and how it connects to equations, forming equations from visual situations such as balanced mobiles and weighing scales, solving linear equations with one unknown using systematic methods, finding unknown values using trial and error and algebraic methods, identifying mistakes in solved equations and correcting them, working with matchstick patterns and deriving formulas, applying Brahmagupta's ancient formula for solving equations of the form Ax + B = Cx + D, and solving real-life word problems using equations.

How to use these NCERT solutions?

Students should first attempt each question on their own before referring to the solutions in this worksheet. This helps build independent thinking and problem-solving habits. Parents and teachers can use these solutions to verify answers and explain the correct method step by step. All solutions follow the exact order and structure of the NCERT textbook, so students can easily match each question with its solution. This worksheet is especially useful during revision, as it covers every exercise set including Figure It Out Exercise Set 1, Exercise Set 2, Exercise Set 3, the Mind the Mistake section, and the Brahmagupta history section. Regular reference to these solutions will help students understand where they go wrong and how to improve.

Student tips and learning tricks

Always read each equation carefully before solving — identify what is on the left-hand side and what is on the right-hand side. Remember that whatever operation you perform on one side of the equation must also be done on the other side to keep the equation balanced. A very common mistake is adding a number to the right side when it should be subtracted — for example, in 4x + 6 = 10, the correct step is 4x = 10 – 6, not 4x = 10 + 6. When solving equations with brackets, always expand the bracket fully before performing any other operation. For word problems, always define your variable clearly at the beginning — for example, "let the unknown number = x" — before forming the equation. Check your answer at the end by substituting the value back into the original equation to verify that both sides are equal.

Why NCERT solutions are important?

NCERT-aligned solutions ensure that students learn concepts exactly as prescribed by the curriculum, which is essential for performing well in school assessments and building a strong mathematical foundation. Chapter 7 Finding the Unknown is a stepping stone to higher algebra in Classes 8, 9, and 10, so getting the basics right at this stage is very important. These solutions build confidence because students can verify their own work, understand the correct method for each type of question, and identify common errors before they appear in exams. Clear and structured solutions also make revision faster and more effective.

Complete answer key – NCERT solutions

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: Trial and error — try n = 49: 2 × 49 + 1 = 99. Both methods give n = 49.
Another way: Subtract 1 from 99 to get 98, then divide by 2 to get 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: 2e = 6 → e = 3
Fig. 7.7: 4 + 2y = 16 → y = 6
Fig. 7.8: 10 = 4 + b → b = 6
Fig. 7.9: s + 2 = 10 + 2 → s = 10
Fig. 7.10: 2s = s + 14 → s = 14
Fig. 7.11: 4s = 21 → s = 21/4

Solving equations systematically

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

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

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 = 45 → x = 15
Check: 3(15) – 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 → 2m = –20 → m = –10
Check: 4(–10 + 6) – 8 = –24; 2(–10) – 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?):
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 = –5 → x = –1
c) –53w = –15 → w = 15/53
d) 13 – z = 8 → z = 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
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 = 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 = 11
e) –k – 2 = –3 – 2 = –5

Mind the mistake, mend the mistake

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

Problem 2: 7 – 8z = 5
Mistake: 8z = 7 – 5 (sign of 8z was not correctly transposed)
Correction: –8z = 5 – 7 → –8z = –2 → z = 1/4

Problem 3: 2v – 4 = 6
Mistake: v – 4 = 6 – 2 (operation was mixed up)
Correction: 2v = 6 + 4 → 2v = 10 → v = 5

Problem 4: 5z + 2 = 3z – 4
Mistake: Signs were wrong on both sides when transposing.
Correction: 5z – 3z = –4 – 2 → 2z = –6 → z = –3

Problem 5: 15w – 4w = 26
Mistake: 4w was moved to RHS as +4w; the equation was already simplified.
Correction: 11w = 26 → w = 26/11

Problem 6: 3x + 1 = –12
This solution is correct. No mistake.
x + 1 = –4 → x = –5

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

Problem 8: –2(3 – 4x) = 14
Mistake: –2 × (–4x) = +8x, not –8x; also incorrect variable used.
Correction: –6 + 8x = 14 → 8x = 20 → x = 20/8 = 5/2

Problem 9: 3(7y + 4) = 9 + 5y
Mistake: When dividing both sides by 3, 5y on the RHS was not divided.
Correction: 21y + 12 = 9 + 5y → 16y = –3 → y = –3/16

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 – ___ = 2 → ___ = 31

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

2) Let number of ₹50 notes = n; number of ₹100 notes = n
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
Output = 43: 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 → ×3 (top path) and +3 (bottom path) → subtracted → output
Output = 63: 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
x/9 = 5 → x = 45. Input = 45

Machine 2: input → –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: gap = 4 cm
(Students should verify from the actual picture in the textbook.)

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

10) Given 28p – 36 = 98:
28p = 134 → 14p = 67 → 14p – 19 = 48
28p – 38 = 98 – 2 = 96

11a) 6x + 9 = 66
Mistake: 6x + 9 = 66 was treated as 6(x + 9) = 66, giving x + 9 = 11. This is wrong since 6 only multiplies x, not 9.
Correction: 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: wrote 4x = 9x + 8 – 5; should be 4x = 9x + 8 + 5.
Correction: 4x – 9x = 13 → –5x = 13 → x = –13/5

12) Triangle 1 (angles: y, y, y + 15):
y + y + y + 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 = x; second = 2x; third = 6x; fourth = 24x
x + 2x + 6x + 24x = 132 → 33x = 132 → x = 4
First person receives 4.

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

16) Figure 1 (Arrow/Pentagon shape):
Position 1: 6 sticks, Position 2: 11 sticks, Position 3: 16 sticks
Pattern: 5n + 1 sticks at position n. Number of squares = n at position n.
a) Squares at position 11: 11 squares
b) Sticks at position 11: 5(11) + 1 = 56 sticks
c) 5n + 1 = 85 → n = 84/5 = 16.8 (not a whole number). No, cannot make arrangement with exactly 85 sticks.
d) 5n + 1 = 150 → n = 149/5 = 29.8 (not a whole number). No, cannot make arrangement with exactly 150 sticks.

Figure 2 (Plus/cross shape growing):
Position 1: 4 sticks, Position 2: 10 sticks, Position 3: 18 sticks, Position 4: 28 sticks
(Students should count sticks in each position from their actual figure and derive the formula. Answers for a, b, c, d to be determined from the derived formula.)

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) → –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 → 4x = 0 → x = 0
f) 3 – 7s = 7 – 3s → –4 = 4s → s = –1
g) 2x + 1 = 6 – (2x – 3) → 4x = 8 → x = 2
h) 10 – 5x = 3(x – 4) – 2(x – 7) → 8 = 6x → x = 4/3

19) Maze – 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: START → x = 4 → x = 3 → m = –10 → k = 4 → END
(Students should trace the correct path through the maze by matching solved values to the adjacent number tiles as shown in the textbook figure.)

20) Children and donkeys on a beach:
Let donkeys = d; children = c
d + c = 28 (heads); 4d + 2c = 80 (feet)
c = 28 – d → 4d + 2(28 – d) = 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 = 5; Add 3 = 8.
The answer is always 8, regardless of the starting number. The operations cancel out the original number x, always leaving 8.

Tangram (Page 191)

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 and rearrange all 7 pieces to form different shapes such as a square, a triangle, or any other figure of your choice.

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