Class 7 Maths Chapter 7 Finding the Unknown Notes & Solutions | Ganita Prakash

Class 7 Maths Chapter 7 Finding the Unknown notes and book solutions from Ganita Prakash Part 2 with easy explanations, important concepts, examples and complete textbook question answers for revision and exam preparation.

Finding the Unknown

Grade 7 Mathematics – Ganita Prakash Part-II

Easy Notes, Concepts and Complete Book Solutions

Easy Notes and Concepts

What is an unknown?
An unknown is a quantity whose value is not known. We can use a letter such as x, y, z, m or n to represent it.
What is an equation?
An equation is a mathematical statement showing that two expressions are equal. The equal sign = tells us that the value on the Left Hand Side (LHS) is equal to the value on the Right Hand Side (RHS).
Example: 3x + 4 = 10
LHS = 3x + 4
RHS = 10
Solving an equation
Solving an equation means finding the value of the unknown that makes the LHS equal to the RHS.
Important rule
Whatever operation we perform on one side of an equation, we must perform the same operation on the other side.
5x – 4 = 7
5x = 7 + 4
5x = 11
x = 11 ÷ 5
x = 11/5
Inverse operations
  • Addition and subtraction are inverse operations.
  • Multiplication and division are inverse operations.
Useful rules
  • If +5 is removed from one side, subtract 5 from the other side.
  • If -5 is removed from one side, add 5 to the other side.
  • If a number is multiplied with the unknown, divide by that number.
  • If the unknown is divided by a number, multiply by that number.
Trial and error method
We can try different values of the unknown until the LHS becomes equal to the RHS. This is called the trial and error method.
Checking a solution
After finding the unknown, substitute the value back into the original equation. If LHS = RHS, the answer is correct.

7.1 Find the Unknowns

A letter-number can be used to represent an unknown quantity. Equations can then be formed from situations involving unknown quantities.

For example, if two equal unknown weights together weigh 6 kg, we can write: 2e = 6, so e = 3 kg.
Key idea: An equation helps us turn an unknown quantity into something we can calculate.

7.2 Solving Equations Systematically

Step 1: Simplify the equation if needed.
Step 2: Bring unknown terms to one side.
Step 3: Bring constant terms to the other side.
Step 4: Divide or multiply to get the unknown alone.
Step 5: Check the answer.

Example: 6y + 7 = 4y + 21

6y + 7 = 4y + 21
6y – 4y = 21 – 7
2y = 14
y = 7
Answer: y = 7

Example: 28(x + 4) + 300 = 1000

28(x + 4) + 300 = 1000
28(x + 4) = 700
x + 4 = 25
x = 21
Answer: x = 21

Example: Riyaz’s Number Trick

Think of a number, subtract 3, multiply by 4 and add 8. The final answer is 24.

Let the number be x.
4(x – 3) + 8 = 24
4x – 12 + 8 = 24
4x – 4 = 24
4x = 28
x = 7
The starting number is 7.

Figure it Out – Section 7.2

1. 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)

3x – 10 = 35
3x = 45
x = 15

(b)

5s = 3s
2s = 0
s = 0

(c)

3u – 7 = 2u + 3
3u – 2u = 3 + 7
u = 10

(d)

4(m + 6) – 8 = 2m – 4
4m + 24 – 8 = 2m – 4
4m + 16 = 2m – 4
2m = -20
m = -10

(e)

u/15 = 6
u = 6 × 15
u = 90
Answers: (a) x = 15, (b) s = 0, (c) u = 10, (d) m = -10, (e) u = 90.

2. Frame an equation that has no solution.

[Hint: 4 more than a number, and 5 more than a number can never be equal!]

x + 4 = x + 5
Subtracting x from both sides:
4 = 5

This is impossible. Therefore, the equation has no solution.

Example of an equation with no solution: x + 4 = x + 5.

Figure it Out – Section 7.2: Applications

1. Write 5 equations whose solution is x = -2.

Many answers are possible. One set is:

x + 2 = 0
x – 1 = -3
2x = -4
3x + 6 = 0
5x = -10

2. Find 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

(a)

2y = 60
y = 30

(b)

-8 = 5x – 3
-5 = 5x
x = -1

(c)

-53w = -15
w = 15/53

(d)

13 – z = 8
-z = -5
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
Answers: (a) y = 30, (b) x = -1, (c) w = 15/53, (d) z = 5, (e) k = 2, (f) m = -1/2, (g) n = 5.

3. 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?

Let the hundred’s digit be x.

Ten’s digit = x + 3
Unit’s digit = x + 6

x + (x + 3) + (x + 6) = 15
3x + 9 = 15
3x = 6
x = 2

Hundred’s digit = 2, ten’s digit = 5, unit’s digit = 8.

Answer: 258

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

Let the weight be x kg.
x = x/2 + 1
x/2 = 1
x = 2
Answer: 2 kg

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

Let the number be x.
x/4 + 9 = x
9 = 3x/4
36 = 3x
x = 12
Answer: 12

6. Given 4k + 1 = 13, find the values of:

(a) 8k + 2

(b) 4k

(c) k

(d) 4k – 1

(e) -k – 2

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
Answers: (a) 26, (b) 12, (c) 3, (d) 11, (e) -5.

7.3 Mind the Mistake, Mend the Mistake

The following are some equations along with the steps used to solve them to find the value of the letter-number. 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.

1. 4x + 6 = 10

4x = 10 + 6
4x = 16
x = 4

The mistake is in the first step. When +6 is removed, we must subtract 6, not add 6.

4x + 6 = 10
4x = 10 – 6
4x = 4
x = 1
Correct answer: x = 1

2. 7 – 8z = 5

8z = 7 – 5
8z = 2
z = 4

The mistake is that the sign of 8z was changed incorrectly.

7 – 8z = 5
-8z = 5 – 7
-8z = -2
z = 1/4
Correct answer: z = 1/4

3. 2v – 4 = 6

v – 4 = 6 – 2
v – 4 = 4
v = 8

The first step is incorrect because 2 was divided out from the left side but not correctly handled on the right side.

2v – 4 = 6
2v = 10
v = 5
Correct answer: v = 5

4. 5z + 2 = 3z – 4

5z + 3z = -4 + 2
8z = -2
z = -2/8

The terms were moved with incorrect signs.

5z + 2 = 3z – 4
5z – 3z = -4 – 2
2z = -6
z = -3
Correct answer: z = -3

5. 15w – 4w = 26

15w = 26 + 4w
15w = 30
w = 2

The first step is incorrect. The equation already has like terms on the left.

15w – 4w = 26
11w = 26
w = 26/11
Correct answer: w = 26/11

6. 3x + 1 = -12

x + 1 = -12/3
x + 1 = -4
x = -5

The first step divides 3 only from the first term. The +1 must also be handled correctly.

3x + 1 = -12
3x = -13
x = -13/3
Correct answer: x = -13/3

7.4 A Pinch of History

The study of algebra in ancient India was called bījagaṇita. The word bīja means seed. The idea is that just as a tree is hidden inside a seed, the answer to a problem can be hidden inside an unknown number.
Brahmagupta made important contributions to mathematics. His work included methods involving integers, fractions and unknown quantities. Ancient Indian mathematicians used symbols for unknown quantities long before modern algebraic notation became common.
A useful general form of an equation is: Ax + B = Cx + D
x = (D – B)/(A – C)
For example:
2x + 3 = 4x + 5
2x – 4x = 5 – 3
-2x = 2
x = -1

Figure it Out – Final Section

1. Fill in the blanks with integers.

(a) 5 × ___ – 8 = 37

(b) 37 – (33 – ____ ) = 35

(c) -3 × (-11 + ____ ) = 45

(a)

5x – 8 = 37
5x = 45
x = 9

(b)

37 – (33 – x) = 35
37 – 33 + x = 35
4 + x = 35
x = 31

(c)

-3(-11 + x) = 45
-11 + x = -15
x = -4
Answers: (a) 9, (b) 31, (c) -4.

2. Ranju 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?

Let the number of each type of note be x.

50x + 100x = 750
150x = 750
x = 5
Answer: 5 notes of ₹50 and 5 notes of ₹100.

3. In 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.

25 dots

There are 3 black blobs and 4 visible blue dots.

Let the number of dots hidden by one blob be x.
3x + 4 = 25
3x = 21
x = 7
Answer: One blob covers 7 dots.

4. Here are machines that take an input, perform an operation on it and send out the result as an output.

(a) Find the inputs in the following cases:

Machine rule: Add 3, multiply by 4, then subtract 5.

Case 1: Output = 43

Case 2: Output = 75

For output 43:

4(x + 3) – 5 = 43
4x + 12 – 5 = 43
4x + 7 = 43
4x = 36
x = 9

For output 75:

4(x + 3) – 5 = 75
4x + 7 = 75
4x = 68
x = 17
Answers: 9 and 17.

(b) Find the inputs in the following cases:

The machine shown performs multiplication by 3 on one branch and addition of 3 on the other branch, and then subtracts the lower result from the upper result.

Case 1: Output = 63

Case 2: Output = 227

The machine rule is:

3x – (x + 3) = output
2x – 3 = output

For output 63:

2x – 3 = 63
2x = 66
x = 33

For output 227:

2x – 3 = 227
2x = 230
x = 115
Answers: 33 and 115.

5. What are the inputs to these machines?

Machine 1: Divide by 3, then divide by 3, output +5.

Machine 2: Subtract 4, then subtract 4, output -11.

Machine 1:

x ÷ 3 ÷ 3 = 5
x ÷ 9 = 5
x = 45

Machine 2:

x – 4 – 4 = -11
x – 8 = -11
x = -3
Answers: 45 and -3.

6. A 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.

Let the distance be x km.
800 + 20x = 2200
20x = 1400
x = 70
Answer: 70 kilometres.

7. The sum of two numbers is 76. One number is three times the other number. What are the numbers?

Let the smaller number be x.

x + 3x = 76
4x = 76
x = 19

The other number is:

3 × 19 = 57
Answer: 19 and 57.

8. The figure shows the diagram for a window with a grill. What is the gap between two rods in the grill?

3 cm 2 cm 34 cm

Using the measurements shown in the figure, let the required gap between successive rods be x. There are five rods, each 2 cm thick, and the top and bottom gaps are 3 cm each.

34 = 3 + 3 + 5(2) + 4x
34 = 16 + 4x
4x = 18
x = 4.5
Answer: The gap between two rods is 4.5 cm.

9. In 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.

Let the cost of one chocolate milkshake be ₹x.

Fruit juice = ₹(x – 15)
4(x – 15) + 7x = 600
4x – 60 + 7x = 600
11x = 660
x = 60

Milkshake = ₹60.

Fruit juice = ₹60 – ₹15 = ₹45.

Answer: Fruit juice = ₹45; chocolate milkshake = ₹60.

10. Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.

28p – 36 = 98
28p = 134

Therefore:

14p = 67
14p – 19 = 67 – 19 = 48
28p – 38 = 134 – 38 = 96
Answers: 14p – 19 = 48 and 28p – 38 = 96.

11. The steps to solve three equations are shown below. Identify and correct any mistakes.

(a)

6x + 9 = 66
x + 9 = 11
x = 11 – 9
x = 2

(b)

14y + 24 = 36
7y + 12 = 18
7y = 6
y = 6/7

(c)

4x – 5 = 9x + 8
4x = 9x + 8 – 5
4x = 9x + 3
4x – 9x = 3
-5x = 3
x = -5/3

(a) Mistake: The first step divides 6 from 6x but not from 9.

6x + 9 = 66
6x = 57
x = 57/6
x = 19/2

(b) Correct.

14y + 24 = 36
7y + 12 = 18
7y = 6
y = 6/7

(c) Mistake: When -5 is removed from the LHS, 5 must be added to the RHS.

4x – 5 = 9x + 8
4x = 9x + 13
4x – 9x = 13
-5x = 13
x = -13/5
Answers: (a) x = 19/2, (b) y = 6/7, (c) x = -13/5.

12. Find the measures of the angles of these triangles.

y y + 15 y + 15 x x – 10 x + 10

First triangle:

The two sloping sides are equal, so the two base angles are equal.

y + (y + 15) + (y + 15) = 180
3y + 30 = 180
3y = 150
y = 50

Angles are:

50°, 65°, 65°

Second triangle:

x + (x – 10) + (x + 10) = 180
3x = 180
x = 60

Angles are:

60°, 50°, 70°
Answers: First triangle: 50°, 65°, 65°. Second triangle: 60°, 50°, 70°.

13. Write 4 equations whose solution is u = 6.

Many answers are possible. For example:

u + 4 = 10
u – 2 = 4
2u = 12
3u = 18
Each of these equations has u = 6 as its solution.

14. The 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?

Let the first person receive x.

First = x
Second = 2x
Third = 3(2x) = 6x
Fourth = 4(6x) = 24x
x + 2x + 6x + 24x = 132
33x = 132
x = 4
Answer: The first person receives 4.

15. The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?

Let height = x metres.
x = x/2 + 2.5
x/2 = 2.5
x = 5
Answer: 5 metres.

16. Two 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

Position 1 has 1 square, position 2 has 2 squares, and position 3 has 3 squares. So position n has n squares.

Number of squares = n
Number of sticks = 3n + 3

For position 11:

Squares = 11
Sticks = 3(11) + 3 = 36

For 85 sticks:

3n + 3 = 85
3n = 82

82 is not divisible by 3, so no such position exists.

For 150 sticks:

3n + 3 = 150
3n = 147
n = 49
Figure 1:
(a) 11 squares
(b) 36 sticks
(c) 85 sticks: No
(d) 150 sticks: Position 49

Figure 2

The number of squares increases by 3 in every new position. The first position has 3 squares, the second has 6, and the third has 9. Therefore:

Number of squares = 3n
Number of sticks = 9n + 1

For position 11:

Squares = 3(11) = 33
Sticks = 9(11) + 1 = 100

For 85 sticks:

9n + 1 = 85
9n = 84
n = 84/9

This is not a whole number, so no such position exists.

For 150 sticks:

9n + 1 = 150
9n = 149

This is not a whole-number position.

Figure 2:
(a) 33 squares
(b) 100 sticks
(c) 85 sticks: No
(d) 150 sticks: No

17. A number increased by 36 is equal to ten times itself. What is the number?

Let the number be x.
x + 36 = 10x
36 = 9x
x = 4
Answer: 4

18. Solve these equations:

(a) 5(r + 2) = 10

(b) -3(u + 2) = 2(u – 1)

(c) 2(7 – 2n) = -6

(d) 2(x – 4) = -16

(e) 6(x – 1) = 2(x – 1) – 4

(f) 3 – 7s = 7 – 3s

(g) 2x + 1 = 6 – (2x – 3)

(h) 10 – 5x = 3(x – 4) – 2(x – 7)

(a)

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

(b)

-3(u + 2) = 2(u – 1)
-3u – 6 = 2u – 2
-5u = 4
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 – 6
4x = 0
x = 0

(f)

3 – 7s = 7 – 3s
-4 = 4s
s = -1

(g)

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
8 = 6x
x = 4/3
Answers:
(a) r = 0
(b) u = -4/5
(c) n = 5
(d) x = -4
(e) x = 0
(f) s = -1
(g) x = 2
(h) x = 4/3

19. Solve 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.

START 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 END

The path is followed by solving each equation and choosing the arrow carrying the same answer.

Start:
8x = 20 + 3x
5x = 20
x = 4

Follow 4.

2x – 9 = -3
2x = 6
x = 3

Follow 3.

-2x = -42
x = 21

Follow 21.

15 = 19 – 4x
4x = 4
x = 1

Follow 1.

2x + 3 = x + 5
x = 2

Follow 2.

2x + 5 = 3(x – 1)
2x + 5 = 3x – 3
x = 8

Follow 8.

2(x + 1) – 10 = 18
2x + 2 – 10 = 18
2x = 26
x = 13

Follow 13.

8m + 8 = -72
8m = -80
m = -10

Follow -10.

-4 = 16 – 5k
-20 = -5k
k = 4

Follow +4.

2x – 9 = 3 – x
3x = 12
x = 4

Follow 4 to reach the final part of the path.

Path values: 4 → 3 → 21 → 1 → 2 → 8 → 13 → -10 → 4 → 4

20. There 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?

Let the number of donkeys be d.

Then the number of children is 28 – d.

Every donkey has 4 feet and every child has 2 feet.

4d + 2(28 – d) = 80
4d + 56 – 2d = 80
2d = 24
d = 12

Children:

28 – 12 = 16
Answer: 12 donkeys and 16 children.

Chapter Summary

  • An algebraic equation shows equality between two algebraic expressions.
  • The two sides of an equation are called the LHS and RHS.
  • The value that makes LHS equal to RHS is called the solution.
  • The same operation must be performed on both sides of an equation.
  • Addition and subtraction are inverse operations.
  • Multiplication and division are inverse operations.
  • Trial and error can be used to find an unknown.
  • Equations can be used to solve real-life problems.
  • Patterns can often be represented using algebraic expressions.
  • Ancient Indian mathematicians made important contributions to algebra.

A Magic Trick

Think of any number.

Now multiply it by 2.

Add 10.

Divide by 2.

Now subtract the original number you thought of.

Finally, add 3.

The final answer is always 8.

Let the original number be 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
Therefore, the trick always gives 8.

Prepare for your exams with this easy and student-friendly worksheet based on Class 7 Maths Chapter 7 from Ganita Prakash Part 2. Practice all the questions carefully to revise the chapter, strengthen your concepts, and improve your mathematical understanding.

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CLASS 7 • MATHEMATICS PART 2

Class 7 Maths Part 2 Chapter 7 Worksheet

Practice worksheet for Class 7 Mathematics Part 2 Chapter 7 with important questions, extra practice and answer key.

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