Finding the Unknown Class 7 Free Notes and Mind Map (Free PDF Download)

This chapter teaches you how to solve algebraic equations by finding unknown values (represented by letters). You will learn systematic methods to solve equations, frame equations from real-life situations, and understand the historical development of algebra in ancient India.

Find the Unknowns

Unknown Weights

We have a weighing scale that behaves as follows. The numbers represent same units of weight.

Question: Find the unknown weights in the following cases:

Figure 7.1:

Left side: 16, Right side: 7, unknown
Solution: 16 = 7 + unknown, so unknown = 16 – 7 = 9

Figure 7.2:

Left side: unknown, Right side: 24
Solution: unknown = 24

Figure 7.3:

Left side: 3, unknown, Right side: 8
Solution: 3 + unknown = 8, so unknown = 8 – 3 = 5

Figure 7.4:

Left side: 18, Right side: 5, unknown
Solution: 18 = 5 + unknown, so unknown = 18 – 5 = 13

Figure 7.5:

Left side: 40, Right side: unknown
Solution: unknown = 40

Figure 7.6:

Left side: 2 + 2 + 2 = 6 (three bread slices, each weighing 2)
Right side: Two fried eggs (each weighing unknown)
Solution: Let each egg weigh e, then 2e = 6, so e = 3

Figure 7.7:

Left side: 16, Right side: 4 + two unknown items
Solution: Let each unknown = y, then 4 + 2y = 16, so 2y = 12, y = 6

Figure 7.8:

Left side: unknown = 10, Right side: 4
Solution: unknown = 10 – 4 = 6

Question: Discuss the answers with your classmates. Give reasons why you think your answer is right.

Find the Unknown Weight of the Sack

Figure 7.9:

Left side: 10 kg + 2 kg, Right side: sack + 2 kg
Hint: If we remove equal weights from both the plates, will the weighing scale still be balanced?
Solution: Remove 2 kg from both sides, so 10 kg = sack. Sack weighs 10 kg.

Figure 7.10: All the sacks have the same weight.

Left side: 2 sacks, Right side: 10 kg + 4 kg = 14 kg
Hint: Remove one sack from each plate.
Solution: Let each sack weigh s kg. 2s = 14, so s = 7 kg.

Figure 7.11:

Left side: 1 kg + 3 sacks, Right side: 10 kg + 10 kg = 20 kg
Hint: Can you remove objects so that the sacks are only on one plate?
Solution: Remove 1 kg from left side and balance, 3s = 20 – 1 = 19 kg, so s = 19/3 kg or approximately 6.33 kg.

Matchstick Pattern

Consider this sequence of matchstick arrangements.

Position	1	2	3	4
Matchsticks	3	5	7	9

Pattern:

The arrangement at position 1 has 2 × (1) + 1 = 3 matchsticks.
The arrangement at position 2 has 2 × (2) + 1 = 5 matchsticks.
The arrangement at position 3 has 2 × (3) + 1 = 7 matchsticks.
And so on.
So, the nth position will have 2n + 1 matchsticks.

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

Solution:

We need to find the value of n such that 2n + 1 = 99.
Solving: 2n = 99 – 1 = 98, so n = 98 ÷ 2 = 49.
The arrangement will be at position 49.

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

Question: Is it possible to make a matchstick arrangement that appears in this sequence using exactly 200 sticks?

Solution:

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

Definition of an Equation

A statement of equality between two algebraic expressions is called an equation.
Nowadays, when using symbols, we write an equation as two algebraic expressions with an equal sign ‘=’ between them.

Examples of equations:

3x + 4 = 7
20 = y – 3
(a – 3)/b = 50
2z + 4 = 5z – 14

Definition of Solving an Equation:

The process of finding the value(s) of the letter-numbers for which the equality holds, or for which the value of the Left Hand Side (LHS) of the equation becomes equal to the Right Hand Side (RHS) of the equation, is called solving the equation.

textLeft Hand Side (LHS)    2n + 1 = 99    Right Hand Side (RHS)

Framing Equations from Weighing Scale Problems

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

For Figure 7.6:

Let us denote the weight of one fried egg as e.
Since each slice of bread is 2, we have 2 + 2 + 2 = 6 on one side and e + e on the other side.
Since they are equal, we have: 6 = e + e, or 2e = 6.

For Figure 7.7:

Bread = 4, and we can denote the weight of one unknown as y.
So, we have 16 on one side, and 4 + 2y on the other side.
Thus, we have the equation: 4 + 2y = 16.

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

Question: Frame 5 equations. Find methods to solve them.

Solving Equations Systematically

Trial and Error Method

How did you solve the various equations framed in the previous section?

One way to solve an equation is to substitute different values in place of the letter-number and to check which value makes LHS = RHS.

Example: Consider the equation 2n + 1 = 99.

If we substitute n = 5, we get LHS = 2 × (5) + 1 = 11. It is far away from 99, the RHS.
We can try n = 10. The LHS is 21. It is still not equal to 99.
Can we try n = 30? The LHS is now 61, still much lower than 99.
Let us try n = 40. The LHS is 81. We are getting closer to 99.
When n = 50, the LHS is 101. This is just a bit too high.
When n = 49, the LHS is 99!

Therefore, the solution to the equation 2n + 1 = 99 is n = 49.

Question: Can this equation have any other solution?

Solution: No, there is only one solution to this equation.

This method is called the trial and error method.

The trial and error method can be inefficient.

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

Properties of Equations

Recall that in the case of the weighing scale, we didn’t use the trial and error method! For some problems, finding the solution was straightforward. For others, we used the fact that when we remove equal weights from both the plates of a balanced weighing scale, it remains balanced. Do equations have a similar property?

Question: Consider 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?

Solution:

For example, you can check by adding 10 to both sides: (15 + 8) + 10 = 23 + 10, which gives 33 = 33 ✓

Since the LHS and the RHS of an equation have the same value, performing the same operation on both sides will clearly not change their equality.

In the weighing scale problems, we removed equal weights so that all the unknown weights were on only one plate of the scale. This made the problem easier to solve. We can use this same strategy to solve an equation as well!

Using Inverse Operations

Let us first solve some arithmetic problems.

Example 1: It is known that 14593 – 1459 + 145 – 14 + 88 = 13353. What is the value of 14593 – 1459 + 145 – 14?

Solution:

To find the value, do we need to evaluate 14593 – 1459 + 145 – 14? No, we can get it by subtracting 88 from 13353.
Why can we do this? We can do this because addition and subtraction are inverse operations.
So, the value of the required expression can be found by subtracting 88 from both sides (LHS and RHS), which removes the term 88 and leaves only the expression to be evaluated on the LHS.
14593 – 1459 + 145 – 14 + 88 – 88 = 13353 – 88
14593 – 1459 + 145 – 14 = 13265

Example 2: It is known that 23 × 41 × 11 × 8 × 7 = 5,80,888. What is the value of the expression 23 × 41 × 11 × 8?

Solution:

Using the fact that multiplication and division are inverse operations, we can simply divide 5,80,888 by 7 to get the value of the expression.
23 × 41 × 11 × 8 = 5,80,888 ÷ 7 = 82,984

Question: 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?

Solution: Yes.

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

Solution:

This problem also can be solved using the fact that addition and subtraction are inverse operations.
However, we can use the following method that is perhaps easier to visualise.
To retain only the expression to be evaluated on the LHS, we need to remove the term –(–67).
We can remove this term by adding (–67) to both sides of the equation.
12345 – 5432 + 135 – 24 – (−67) + (−67) = 7091 + (−67)
12345 – 5432 + 135 – 24 = 7091 + (−67) = 7091 – 67 = 7024

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

Solution:

To retain only the expression to be evaluated on the LHS we need to remove the factor (8/9).
We can do that by dividing both sides by (8/9), which is the same as multiplying both sides by (9/8).
(35/113) × 24 × 14 × (8/9) ÷ (8/9) = (94080/1017) ÷ (8/9)
(35/113) × 24 × 14 = (94080/1017) × (9/8) = 11760/113

Solving Equations Step by Step

Question: Let us use these ideas to solve the equation 5x – 4 = 7.

Solution:

What can we do so that 5x is on one side and the equality between the LHS and the RHS still holds?
To retain only 5x on the LHS, we need to remove the term – 4. This can be done by adding 4 to both sides.
Thus, 5x – 4 + 4 = 7 + 4.
Hence, 5x = 11.
To retain only the unknown x on the LHS, we need to remove the factor 5. This can be done by dividing both sides by 5.
Thus, 5x ÷ 5 = 11 ÷ 5
So, x = 11/5

Question: Can we check that x = 11/5 is the correct solution to the equation?

Solution:

We can check this by substituting x with the value 11/5 in the equation 5x – 4 = 7, and checking if the LHS = RHS.
Substituting x with 11/5 in the LHS we get:
LHS = 5(11/5) – 4 = 11 – 4 = 7
This is the same as the RHS.
So, 11/5 is indeed the correct solution to the equation.

Example 5: Solve the equation 11y + (−5) = 61

Solution:

To retain only the unknown term 11y on one side, we need to remove the term –5.
This can be done by subtracting (–5) from both sides, which is the same as adding 5 to both sides.
11y + (−5) – (-5) = 61 – (-5)
11y = 66
We can directly find the value of y, seeing that 11 × 6 = 66.
We can also divide both sides by 11 to find y.
11y ÷ 11 = 66 ÷ 11
So y = 6 is the solution to the equation 11y + (−5) = 61.
Can you check that this solution is correct?

Example 6: Solve 6y + 7 = 4y + 21

Solution:

In this equation, expressions with an unknown are on both sides.
We have seen how to solve equations when the unknown term is on one side. What can be done to bring the unknown terms to the same side?
Subtracting 4y from both sides, we get:
6y + 7 – 4y = 4y + 21 – 4y
2y + 7 = 21
Subtracting 7 from both sides we get:
2y + 7 – 7 = 21 – 7
2y = 14
We can directly find the value of y, seeing that 2 × (7) = 14.
We can also divide both sides by 2 to find y.
2y ÷ 2 = 14 ÷ 2
y = 7
Remember, it is always good to check your solution.

Figure it Out

Question 1: Solve these equations and check the solutions.

(a) 3x – 10 = 35

Solution:

3x = 35 + 10
3x = 45
x = 45 ÷ 3
x = 15

(b) 5s = 3s

Solution:

5s – 3s = 0
2s = 0
s = 0

(c) 3u – 7 = 2u + 3

Solution:

3u – 2u = 3 + 7
u = 10

(d) 4(m + 6) – 8 = 2m – 4

Solution:

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

(e) u/15 = 6

Solution:

u = 6 × 15
u = 90

Question 2: Frame an equation that has no solution.

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

Solution:

Example: x + 4 = x + 5 has no solution because x + 4 can never equal x + 5 for any value of x.

Efficient Procedure for Solving Equations

The procedure to systematically solve an equation can be made efficient if we make an observation.

Consider the equation 11y + (-5) = 61, which we solved.

As the first step, we subtracted – 5 from both sides to remove the term –5.

11y + (-5) – (-5) = 61 – (– 5)

Since this action removes the term – 5 from the LHS, we could have written this step as:

11y = 61 – (-5)

Note that we could also have arrived at this step by seeing addition and subtraction as inverse operations, as in the case of Example 1.

Similarly, when we had 11y = 66, we divided both sides by 11 to remove the factor 11 from the LHS.

11y ÷ 11 = 66 ÷ 11

Since this action removes the factor 11 in the LHS, we can write this step as:

y = 66 ÷ 11

Again, we could have arrived at this by seeing multiplication and division as inverse operations, as in the case of Example 2.

Two Methods of Writing Solutions

Let us write down both these ways of solving an equation.

Method 1 (Detailed):

text11y + (− 5) = 61
11y + (− 5) – (− 5) = 61 – (– 5)
11y = 66
11y ÷ 11 = 66 ÷ 11
y = 6

Method 2 (Shortcut):

text11y + (-5) = 61
11y = 61 – (-5)
11y = 66
y = 66 ÷ 11
y = 6

Another Example:

text6y + 7 = 4y + 21

Method 1:
6y + 7 – 4y = 4y + 21 – 4y
2y + 7 = 21
2y + 7 – 7 = 21 – 7
2y = 14
2y ÷ 2 = 14 ÷ 2
y = 7

Method 2:
6y + 7 = 4y + 21
6y + 7 - 4y = 21
2y + 7 = 21
2y = 21 – 7
2y = 14
y = 14 ÷ 2
y = 7

Question: What happens in equations like u/15 = 6?

Solution:

Multiplying both sides by 15 leaves only u in the LHS:
u = 6 × 15
u = 90

Observations on Solving Equations

Thus, we can make the following observations:

(a) When a term that is added or subtracted on one side of an equation is removed from that side, its additive inverse should appear as a term on the other side for the equality to hold.

For example, 2y + 7 = 21 becomes 2y = 21 – 7.

(b) If one side of an equation is the product of two or more numbers or expressions, and we remove one of the factors, then the other side should be divided by this factor for the equality to hold.

For example, 2y = 14 becomes y = 14 ÷ 2.

(c) If one side of an equation is the quotient of two numbers or expressions, and we remove the divisor, then the other side should be multiplied by this divisor for the equality to hold.

For example, u/15 = 6 becomes u = 6 × 15.

Solving Problems

Forming and solving equations gives us the ability to find solutions to many problems in our lives. Let us see a few such examples.

Example 7: Ranjana’s Sequence of Arrangements

Question: 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?

Step 1	Step 2	Step 3	Step 4
4 tiles	7 tiles	10 tiles	13 tiles

She can look at the pattern in different ways.

Method 1:

Step 1: 1 + 1 + 1 + 1 = 4
Step 2: 2 + 2 + 2 + 1 = 7
Step 3: 3 + 3 + 3 + 1 = 10
Step 4: 4 + 4 + 4 + 1 = 13
Step k: k + k + k + 1 = 3k + 1

Method 2:

Step 1: 1 + 3 = 4
Step 2: 4 + (4 × 1 + 3) = 4 + 7 = 11 (This doesn’t match the pattern given, so let me recalculate)
Actually looking at the image pattern more carefully:
Step 1: 4 tiles
Step 2: 4 + 3 = 7 tiles
Step 3: 4 + 3 + 3 = 10 tiles
Step 4: 4 + 3 + 3 + 3 = 13 tiles
Pattern: 4 + 3(k-1) = 4 + 3k – 3 = 3k + 1

We have the expression 3k + 1 which gives the number of tiles needed to make an arrangement in Step k.

To check whether an arrangement is possible using 100 tiles at some Step k, we can solve the equation:

3k + 1 = 100

Solution:

3k = 100 – 1
3k = 99
k = 99 ÷ 3
k = 33

So yes, an arrangement using 100 tiles is possible at Step 33.

Example 8: Madhubanti’s Party

Question: 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?

Fatima’s method of solving this problem:

She represented the situation using a rough diagram:
Snack cost: 25 × (number of people)
Delivery cost: + 50
Total: ₹500

Out of ₹500, if we subtract the fixed delivery charge, then Madhubanti is left with ₹450.

So the question becomes “How many plates of snacks, each costing ₹25, can be bought for ₹450?”.

The answer to this is 450 ÷ 25 = 18.

18 plates of snacks can be bought for ₹450. Considering her 5 family members, she can invite 18 – 5 = 13 friends to her party.

Mahesh’s method:

Mahesh represented the unknown quantity of the total number of people who can attend the party, including Madhubanthi and her family members, as p.
Cost incurred = 25p + 50.
Since this cost should be 500, we have the equation:
25p + 50 = 500
Subtracting 50 from both sides, we get:
25p = 500 – 50
25p = 450
Dividing both sides by 25, we get:
p = 450 ÷ 25 = 18
18 people can be at her party, including her 5 family members. That means 13 friends can be invited.

Srikanth’s method:

Srikanth decided to represent the unknown quantity of the total number of friends Madhubanthi can invite as f.
What will be the cost in this case?
Total people = f + 5 (friends + family)
Cost incurred = 25 (f + 5) + 50.
Since Madhubanthi has ₹500, we have the equation:
25 (f + 5) + 50 = 500
Subtracting 50 from both sides:
25 (f + 5) = 450
Dividing both sides by 25:
f + 5 = 18
Subtracting 5 from both sides:
f = 13

Example 9: Two Friends Saving Money

Question: 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?

Solution:

Let m denote the number of months after which their savings are equal.
Jahnavi’s savings after m months = 4000 + 650m
Sunita’s savings after m months = 5050 + 500m
As these amounts are equal after m months, we get the following equation:
4000 + 650m = 5050 + 500m
Subtracting 500m from both sides:
4000 + 650m – 500m = 5050
4000 + 150m = 5050
Subtracting 4000 from both sides:
150m = 5050 – 4000
150m = 1050
Dividing both sides by 150:
m = 1050 ÷ 150
m = 7

So, after 7 months, both will have the same amount of money.

Check the answer:

Jahnavi after 7 months: 4000 + 650 × 7 = 4000 + 4550 = ₹8550
Sunita after 7 months: 5050 + 500 × 7 = 5050 + 3500 = ₹8550 ✓

Example 10: Solve 28(x + 4) + 300 = 1000

Method 1:

Subtracting 300 from both sides, we get:
28 (x + 4) = 1000 – 300
28 (x + 4) = 700
Dividing both sides by 28, we get:
x + 4 = 700 ÷ 28
x + 4 = 25
Subtracting 4 from both sides:
x = 25 – 4
x = 21

Method 2:

Since 28, 300, and 1000 are divisible by 4, we get a simpler equation if we divide both sides by 4.
28(x + 4) + 300 = 1000
Dividing both sides by 4:
[28(x + 4)]/4 + 300/4 = 1000/4
7(x + 4) + 75 = 250
Subtracting 75 from both sides:
7(x + 4) = 175
Expanding:
7x + 28 = 175
Subtracting 28 from both sides:
7x = 147
x = 147 ÷ 7 = 21

Method 3:

Simplifying the LHS:
28x + 112 + 300 = 1000
28x + 412 = 1000
Subtracting 412 from both sides:
28x = 1000 – 412
28x = 588
Dividing both sides by 28:
x = 588 ÷ 28 = 21

Example 11: Riyaz’s Math Trick

Question: Riyaz created a math trick, which he tries out on his friend Akash.

Riyaz asked Akash to perform the following steps without revealing the answer to any of the intermediate steps:

Think of a number.
Subtract 3 from the number.
Multiply the result by 4.
Add 8 to the product.
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.

Solution:

Try the steps using different numbers as the starting number. Do you see any relation between the starting number and final answer?
The answer can be found by algebraically modeling this scenario.
Let the unknown starting number be x.

Steps	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

Since Akash’s final answer was 24, we have the equation:

4x – 4 = 24
4(x – 1) = 24
x − 1 = 6 (Dividing both sides by 4)
x = 7

Thus, Akash thought of the number 7.

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

Solution:

Final answer is 24.
Add 4: 24 + 4 = 28
Divide by 4: 28 ÷ 4 = 7
Rule: (Final answer + 4) ÷ 4 = Starting number

Example 12: Ramesh and Suresh’s Marbles

Question: Ramesh and Suresh have 60 marbles between them. Ramesh has 30 more marbles than Suresh. How many marbles does each boy have?

Solution:

In this problem, we have two unknowns.
If we denote the number of marbles with Ramesh as x and the number of marbles with Suresh as y, then what are the different equations that we have?

Equations:

The total number of marbles is 60: x + y = 60
Ramesh has 30 marbles more than Suresh: x = y + 30

How do we find the unknowns using these equations? So far, we have only developed a strategy to solve an equation with one unknown!

To get such an equation, we can do the following:

Denote the number of marbles with Suresh as y, and the number of marbles with Ramesh as y + 30.
Since the total number of marbles is 60, we have the equation:
y + (y + 30) = 60
2y + 30 = 60
2y = 60 – 30
2y = 30
y = 15

So Suresh has 15 marbles and Ramesh has 15 + 30 = 45 marbles.

Generating Equations

So far, we have solved a given equation to find the value of the letter-number. Can we do the reverse—write equations using a given value of the letter-number?

Question: Write equations whose solution is y = 5. Share the equations you made with each other and discuss the methods used.

Examples:

y + 1 = 6
3y = 15
2y – 3 = 7
y/5 = 1

Chains of Equations

Consider the following chains of equations, where one is obtained from the previous one by performing the same operation on both sides.

Chain 1:

texty + 1 = 6
Multiplying by -1: -y - 1 = -6
Adding y: -1 = -6 + y
Adding 6: 5 = y

Chain 2:

text3y = 15
Adding 6: 3y + 6 = 21
Dividing by 3: y + 2 = 7
Subtracting 2: y = 5

Question: Can you form a chain going from the bottom equation to the top? Compare the operations used when going from the top to the bottom and from the bottom to the top.

Question: Without calculating, can you find the value of the unknown in each equation in the chains above?

Hint: We have seen that the value that satisfies an equation also satisfies the new equation obtained by performing the same operation on both sides of the original equation.

Solution: Yes, all equations in the chain have the solution y = 5.

Example 13: Real-Life Situation from an Equation

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

Solution:

If we think of these numbers as representing money, we can see that the total money is ₹250.
There are two terms that are adding to ₹250. The second term in the LHS, ₹75, is fixed and does not change.
The value of the first term would change based on ‘x‘. So we can think of 100 as the number of units and ‘x‘ as the cost per unit.
For instance, x could represent the cost of a plate of snacks and 75 could be the delivery charge.
Situation: If each plate of snacks costs ₹x and there are 100 plates, plus a delivery charge of ₹75, the total cost is ₹250.

Figure it Out

Question 1: Write 5 equations whose solution is x = −2.

Solution:

x + 5 = 3
2x = -4
x – 3 = -5
3x + 10 = 4
x/2 = -1

Question 2: Find the value of each unknown:

(a) 2y = 60

Solution: y = 30

(b) -8 = 5x – 3

Solution: 5x = -8 + 3 = -5, x = -1

(c) -53w = −15

Solution: w = -15 ÷ (-53) = 15/53

(d) 13 – z = 8

Solution: -z = 8 – 13 = -5, z = 5

(e) k + 8 = 12 – k

Solution: k + k = 12 – 8, 2k = 4, k = 2

(f) 7m = m – 3

Solution: 7m – m = -3, 6m = -3, m = -3/6 = -1/2

(g) 3n = 10 + n

Solution: 3n – n = 10, 2n = 10, n = 5

Question 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?

Solution:

Let unit’s digit = x
Ten’s digit = x – 3
Hundred’s digit = (x – 3) – 3 = x – 6
Sum: x + (x – 3) + (x – 6) = 15
3x – 9 = 15
3x = 24
x = 8
Number: 258 (hundred’s digit = 2, ten’s digit = 5, unit’s digit = 8)

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

Solution:

Let weight = w kg
w = w/2 + 1
2w = w + 2
w = 2 kg

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

Solution:

Let number = n
n/4 + 9 = n
n/4 – n = -9
-3n/4 = -9
n = 12

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

(a) 8k + 2

Solution: 4k = 12, so 8k = 24, 8k + 2 = 26

(b) 4k

Solution: 4k = 12

(c) k

Solution: k = 3

(d) 4k – 1

Solution: 4k – 1 = 12 – 1 = 11

(e) -k – 2

Solution: -k – 2 = -3 – 2 = -5

Mind the Mistake, Mend the Mistake

Question: 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.

Problem 1:

text4x + 6 = 10
4x = 10 + 6    ← MISTAKE
4x = 16
x = 4

Correction:

When removing +6 from LHS, it should become -6 on RHS.
Correct solution:

text4x + 6 = 10
4x = 10 - 6
4x = 4
x = 1

Problem 2:

text7 - 8z = 5
8z = 7 - 5
8z = 2
z = 4    ← MISTAKE (should be z = 2/8 = 1/4)

Correction:

Correct solution:

text7 - 8z = 5
-8z = 5 - 7
-8z = -2
z = -2 ÷ (-8) = 1/4

Problem 3:

text2v - 4 = 6
v - 4 = 6 - 2    ← MISTAKE
v - 4 = 4
v = 8

Correction:

When dividing LHS by 2, must divide RHS by 2 as well.
Correct solution:

text2v - 4 = 6
2v = 6 + 4
2v = 10
v = 5

Problem 4:

text5z + 2 = 3z – 4
5z + 3z = -4 + 2    ← MISTAKE (should be 5z - 3z)
8z = -2
z = -2/8

Correction:

text5z + 2 = 3z – 4
5z - 3z = -4 - 2
2z = -6
z = -3

Problem 5:

text15w - 4w = 26
15w = 26 + 4w    ← MISTAKE
15w = 30
w = 2

Correction:

text15w - 4w = 26
11w = 26
w = 26/11

Problem 6:

text3x + 1 = -12
x + 1 = -12/3
x + 1 = -4
x = -5

This is CORRECT ✓

Problem 7:

text4(4q + 2) = 50
4(4q) = 50 – 2    ← MISTAKE
16q = 48
q = 3

Correction:

Must divide entire RHS by 4, not just subtract 2.

text4(4q + 2) = 50
4q + 2 = 50/4 = 12.5
4q = 12.5 - 2 = 10.5
q = 10.5/4 = 2.625

Or expand first:

text16q + 8 = 50
16q = 42
q = 42/16 = 21/8

Problem 8:

text-2(3 - 4x) = 14
-6 - 8x = 14    ← MISTAKE (should be -6 + 8x)
-8x = 14 + 6
-8x = 20
x = -20/8

Correction:

text-2(3 - 4x) = 14
-6 + 8x = 14
8x = 14 + 6
8x = 20
x = 20/8 = 5/2

Problem 9:

text3(7y + 4) = 9 + 5y
7y + 4 = 9/3 + 5y    ← MISTAKE
7y + 4 = 3 + 5y
7y – 5y + 4 = 3
2y = 4 - 3
y = 1/2    ← MISTAKE (should be y = 1/2 is wrong calculation)

Correction:

text3(7y + 4) = 9 + 5y
21y + 12 = 9 + 5y
21y - 5y = 9 - 12
16y = -3
y = -3/16

A Pinch of History

Forming expressions using symbols and solving equations with such expressions was an important component of mathematical explorations in ancient India. This area of mathematics was termed bījagaņita, also now known as algebra. The word bīja means seed. Just as a tree is hidden inside a seed, the answer to a problem is hidden inside an unknown number. Solving the problem is like helping the tree grow— step by step, we discover what is hidden.

We have seen Brahmagupta’s contributions to different areas of mathematics like integers and fractions. In Chapter 18 of his book Brāhmasphuțasiddhānta (628 CE), he also explained how to add, subtract, and multiply unknown numbers using letters — similar to how we use x or y today. This chapter was one of the earliest known works in algebra in history. As renowned mathematician and Fields Medalist David Mumford remarked, ‘Brahmagupta is the key person in the creation of algebra as we know it’.

In the 8th century, Indian mathematical ideas were translated into Arabic. They influenced a well-known mathematician named Al-Khwarizmi, who lived in present-day Iraq. Around 825 CE, he wrote a book called Hisab al-jabr wal-muqabala, which means ‘calculation by restoring and balancing’.

These ideas spread even further. By the 12th century, Al-Khwarizmi’s book was translated into Latin and brought to Europe, where it became very popular. The word al-jabr from his book gave us the word algebra, which we also still use today.

Similar to how we use letters from the alphabet today to represent unknowns, ancient Indian mathematicians from the time of Brahmagupta used distinct symbols like yā, kā, nī, pī, lo, etc., for different unknowns. The symbol yā was short for yāvat-tāvat (meaning ‘as much as needed’). Others like kā and nī referred to as the first letters of the names of colours – kālaka (black), nīlaka (blue), and so on. In contrast to these unknowns, the known quantities in an expression had a specific form (rūpa) and were denoted by rū.

Ancient Indian Notation Examples

Modern Notation	Ancient Indian Notation	Explanation
2x + 1	yā 2 rū 1	In each term the indication of unknown/known came first
2x – 8	yā 2 rū 8̇	A dot over the number indicated that it was negative
3x + 4 = 2x + 8	yā 3 rū 4 yā 2 rū 8	The two sides of an equation were presented one below the other

Example 16: Bhāskarāchārya’s Problem (1150 CE)

Bījgaņita by Bhāskarāchārya (1150 CE) mentions this problem:

Question: 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.

Solution:

Let price of one horse = ₹x
First man’s wealth: 300 + 6x
Second man’s wealth: 10x – 100
According to the problem:
300 + 6x = 10x – 100
300 + 6x + 100 = 10x
400 + 6x = 10x
400 = 10x – 6x
400 = 4x
400 ÷ 4 = x
100 = x

Therefore, the price of one horse = ₹100.

Brahmagupta’s Formula

Such problems and solutions were well understood in ancient India. In fact, a very systematic way to solve problems with single unknowns was first proposed by Aryabhata (499 CE) and Brahmagupta has outlined it in his Brāhmasphuțasiddhānta (628 CE).

Let us understand his method.

Let us look at a few equations of the following form:

5x + 4 = 3x + 8, or 3x – 6 = 2x + 4

Question: 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?

To get a formula, we can generalise equations of this form by taking the four numbers as A, B, C, and D. That is:

Ax + B = Cx + D

Brahmagupta gives the solution to equations of this form with this formula:

textx = (D - B)/(A - C)

Using this approach, we can find the solution to the equation:

650m + 4000 = 500m + 5050

simply by calculating:

textm = (5050 - 4000)/(650 - 500) = 1050/150 = 7

Question: Using this formula can you solve this equation 2x + 3 = 4x + 5?

Solution:

textx = (5 - 3)/(2 - 4) = 2/(-2) = -1

Ancient Indian mathematicians were excellent at converting complex mathematical ideas into simple procedures so that everyone could use these procedures to solve problems!

Bījagaņita or algebra is the branch of mathematics that uses letter symbols to solve mathematical problems. We have seen some examples in previous pages. By studying algebra, we learn to generalise patterns in numbers, shapes, and situations. We express these generalisations using the language of algebra, which is both precise and powerful.

Bījagaņita also gives us a way to justify mathematical claims (like why the sum of two odd numbers is always even) and to solve problems of many kinds.

The power of algebra was well recognised by ancient Indian mathematicians. We hope you recognise it too — and enjoy using it!

Figure it Out

Question 1: Fill in the blanks with integers.

(a) 5 × ___ = 37

Solution: 37/5 = 7.4 (not an integer, so this might be a trick question or typo)

(b) 37 – (___) = 35

Solution: 2

(c) –3 × (−11 + ___) = 45

Solution: -11 + ___ = 45 ÷ (-3) = -15, so ___ = -15 + 11 = -4

Question 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?

Solution:

Let number of ₹50 notes = x, number of ₹100 notes = x
50x + 100x = 750
150x = 750
x = 750/150 = 5
She has 5 notes of ₹50 and 5 notes of ₹100.

Question 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.

Solution:

Let each blob cover x dots
Let’s say there are n blobs and some visible dots v
Equation: nx + v = 25
(Need to see the image to determine exact values of n and v)

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

(a) Machine: +3, ×4, -5

Example: Input 12 → +3 → 15 → ×4 → 60 → -5 → 55

Find the inputs:

Input ? → +3 → ×4 → -5 → 43

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

Input ? → +3 → ×4 → -5 → 75

Solution: Let input = x
((x + 3) × 4) – 5 = 75
(x + 3) × 4 = 80
x + 3 = 20
x = 17

(b) Machine: ×3, +3

Example: Input 12 → ×3 → 36 → +3 → 39 (not 21 as shown, might be different machine)

Let me recalculate based on the example given:

12 → ×3 → 36, but output shown is 21
This suggests: 12 → +3 → 15 → Maybe different operation

Question 5: What are the inputs to these machines?

? → +3 → +5 → -4 → = -11

Solution: Let input = x
x + 3 + 5 – 4 = -11
x + 4 = -11
x = -15

Question 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.

Solution:

Let distance = d km
800 + 20d = 2200
20d = 2200 – 800
20d = 1400
d = 70 km

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

Solution:

Let smaller number = x
Larger number = 3x
x + 3x = 76
4x = 76
x = 19
Numbers are 19 and 57.

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

Given: Total width = 34 cm, Rod width = 2 cm, End margins = 3 cm each

Solution:

Let gap = g cm
Need to count number of rods and gaps from the diagram
Assuming 5 rods and 4 gaps:
2(3) + 5(2) + 4g = 34
6 + 10 + 4g = 34
4g = 18
g = 4.5 cm

Question 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.

Solution:

Let milkshake cost = ₹m
Fruit juice cost = ₹(m – 15)
4(m – 15) + 7m = 600
4m – 60 + 7m = 600
11m = 660
m = 60
Milkshake = ₹60, Fruit juice = ₹45

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

Solution:

From 28p – 36 = 98:
28p = 98 + 36 = 134
14p – 19 = (28p/2) – 19 = (134/2) – 19 = 67 – 19 = 48
28p – 38 = 134 – 38 = 96

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

(a) 6x + 9 = 66

textx + 9 = 11    ← MISTAKE (should divide entire LHS by 6)
x = 11 - 9
x = 2

Correction:

text6x + 9 = 66
6x = 66 - 9
6x = 57
x = 57/6 = 9.5

(b) 14y + 24 = 36

text7y + 12 = 18    (Correct - divided by 2)
7y = 6
y = 6/7

This is CORRECT ✓

(c) 4x – 5 = 9x + 8

text4x = 9x + 8 - 5
4x = 9x + 3
4x - 9x = 3
-5x = 3
x = -3/5

This is CORRECT ✓

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

Triangle 1: Angles are y, y + 15, x – 10

Solution: y + (y + 15) + (x – 10) = 180
Need more information to solve completely

Triangle 2: Angles are x, x + 10, another angle

Need complete information from diagram

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

Solution:

u – 2 = 4
2u = 12
u + 5 = 11
3u – 10 = 8

Question 14: The Bakhshāli 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?

Solution:

Let first person get = x
Second person = 2x
Third person = 3(2x) = 6x
Fourth person = 4(6x) = 24x
Total: x + 2x + 6x + 24x = 132
33x = 132
x = 4

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

Solution:

Let height = h
h = h/2 + 2.5
2h = h + 5
h = 5 meters

Question 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?

(Need to see the actual figures to provide specific answers)

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

Solution:

Let number = x
x + 36 = 10x
36 = 10x – x
36 = 9x
x = 4

Question 18: Solve these equations:

(a) 5(r + 2) = 10

Solution: r + 2 = 2, r = 0

(b) –3(u + 2) = 2(u – 1)

Solution: -3u – 6 = 2u – 2, -5u = 4, u = -4/5

(c) 2(7 – 2n) = – 6

Solution: 14 – 4n = -6, -4n = -20, n = 5

(d) 2(x – 4) = – 16

Solution: x – 4 = -8, x = -4

(e) 6(x – 1) = 2(x − 1) − 4

Solution: 6x – 6 = 2x – 2 – 4, 4x = 0, x = 0

(f) 3 – 7s = 7 – 3s

Solution: -7s + 3s = 7 – 3, -4s = 4, s = -1

(g) 2x + 1 = 6 – (2x – 3)

Solution: 2x + 1 = 6 – 2x + 3, 4x = 8, x = 2

(h) 10 – 5x = 3(x – 4) – 2(x – 7)

Solution: 10 – 5x = 3x – 12 – 2x + 14, 10 – 5x = x + 2, -6x = -8, x = 4/3

Question 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.

(This is a maze-like puzzle where you solve equations and follow paths based on answers)

Question 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?

Solution:

Let children = c, donkeys = d
Equation 1 (heads): c + d = 28
Equation 2 (feet): 2c + 4d = 80 (children have 2 feet, donkeys have 4 feet)
From equation 1: c = 28 – d
Substitute in equation 2: 2(28 – d) + 4d = 80
56 – 2d + 4d = 80
2d = 24
d = **

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