---
title: "NCERT Solutions Class 7 Maths Expressions using Letter-Numbers"
url: https://www.swavid.com/maths/class/7/chapter/expressions-using-letter-numbers/ncert-solutions
dateModified: 2026-10-07T16:58:15+00:00
---

# NCERT Solutions Class 7 Maths Expressions using Letter-Numbers

This chapter's questions cover foundational concepts of algebraic expressions, formulas, simplification, like and unlike terms, and identifying patterns using letter-numbers.

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## Page No. 81

### Question 1

*2 marks · Very short answer*

Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be?

**Solution**

1. Shabnam is 3 years older than Aftab.
2. When Aftab's age is 18 years, Shabnam's age will be $18 + 3 = 21$ years.

**Answer:** 21 years

> Common mistake: Subtracting 3 instead of adding 3.

### Question 2

*2 marks · Very short answer*

Given Aftab’s age, how will you find out Shabnam’s age?

**Solution**

1. We add 3 to Aftab's age to get Shabnam's age.

**Answer:** Add 3 to Aftab's age.

> Common mistake: Multiplying by 3 instead of adding.

### Question 3

*2 marks · Very short answer*

Can we write this as an expression?

**Solution**

1. Yes, Shabnam's age can be written as an algebraic expression.
2. Using $a$ for Aftab's age and $s$ for Shabnam's age, the expression is $s = a + 3$.

**Answer:** Yes, $s = a + 3$

> Common mistake: Writing $a = s + 3$ incorrectly.

### Question 4

*2 marks · Very short answer*

If $a$ is 23 (Aftab’s age in years), then what is Shabnam’s age?

**Solution**

1. Replace $a$ by 23 in the expression $a + 3$.
2. We get $23 + 3 = 26$ years.

**Answer:** 26 years

> Common mistake: Multiplying 23 by 3 instead of substituting.

## Page No. 82

### Question 1

*2 marks · Very short answer*

Given the age of Shabnam, write an expression to find Aftab’s age.

**Solution**

1. We know that Aftab is 3 years younger than Shabnam, so Aftab's age is 3 less than Shabnam's age.
2. Using $a$ for Aftab's age and $s$ for Shabnam's age, the algebraic expression is $a = s - 3$.

**Answer:** $a = s - 3$

> Common mistake: Writing $s - a = 3$ or $s + 3$ instead of $a = s - 3$.

### Question 2

*2 marks · Very short answer*

Use this expression to find Aftab’s age if Shabnam’s age is 20.

**Solution**

1. Substitute $s = 20$ in the expression $a = s - 3$.
2. We get $a = 20 - 3 = 17$ years.

**Answer:** $17$ years

> Common mistake: Adding 3 instead of subtracting 3.

## Page No. 83

### Question 1

*2 marks · Very short answer*

How much should she pay if she buys 10 coconuts and 5 kg jaggery?

**Solution**

1. The cost of 10 coconuts is $10 \times 35 = 350$ rupees.
2. The cost of 5 kg jaggery is $5 \times 60 = 300$ rupees.
3. The total amount to be paid is $350 + 300 = 650$ rupees.

**Answer:** ₹650

> Common mistake: Multiplying the quantities with wrong unit prices.

### Question 2

*2 marks · Very short answer*

How much should she pay if she buys 8 coconuts and 9 kg jaggery?

**Solution**

1. The cost of 8 coconuts is $8 \times 35 = 280$ rupees.
2. The cost of 9 kg jaggery is $9 \times 60 = 540$ rupees.
3. The total amount to be paid is $280 + 540 = 820$ rupees.

**Answer:** ₹820

> Common mistake: Arithmetic errors while adding the total cost.

### Question 3

*2 marks · Very short answer*

Write an algebraic expression to find the total amount to be paid for a given number of coconuts and quantity of jaggery.

**Solution**

1. Let $c$ represent the number of coconuts and $j$ represent the quantity of jaggery in kg.
2. The cost of $c$ coconuts is $c \times 35$ and the cost of $j$ kg jaggery is $j \times 60$.
3. The total algebraic expression for the amount to be paid is $c \times 35 + j \times 60$ or $35c + 60j$.

**Answer:** 35c + 60j

> Common mistake: Confusing the price of coconuts and jaggery.

### Question 4

*2 marks · Very short answer*

Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery.

**Solution**

1. Substitute $c = 7$ and $j = 4$ in the expression $c \times 35 + j \times 60$.
2. The cost for 7 coconuts is $7 \times 35 = 245$ rupees.
3. The cost for 4 kg jaggery is $4 \times 60 = 240$ rupees.
4. The total amount to be paid is $245 + 240 = 485$ rupees.

**Answer:** ₹485

> Common mistake: Wrong substitution of values for coconuts and jaggery.

### Question 5

*2 marks · Very short answer*

What is the perimeter of a square with sidelength 7 cm? Use the expression to find out.

**Solution**

1. The perimeter of a square is given by the expression $4 \times q$, where $q$ is the sidelength.
2. Substitute $q = 7$ cm into the expression.
3. The perimeter is $4 \times 7 = 28$ cm.

**Answer:** 28 cm

> Common mistake: Using the formula for area instead of perimeter.

## Figure it Out

### Question 1

*3 marks · Short answer*

Write formulas for the perimeter of: (a) triangle with all sides equal. (b) a regular pentagon (c) a regular hexagon

**Part (a)**

1. Let the length of each side of the equilateral triangle be $a$ units.
2. The perimeter of a triangle with all sides equal is the sum of its 3 sides, which is $3a$ units.

Answer (a): $3a$ units

**Part (b)**

1. Let the length of each side of the regular pentagon be $a$ units.
2. A regular pentagon has 5 equal sides, so its perimeter is $5a$ units.

Answer (b): $5a$ units

**Part (c)**

1. Let the length of each side of the regular hexagon be $a$ units.
2. A regular hexagon has 6 equal sides, so its perimeter is $6a$ units.

Answer (c): $6a$ units

**Answer:** Perimeters are $3a$ units, $5a$ units, and $6a$ units respectively.

> Common mistake: Confusing the number of sides of a pentagon and hexagon.

### Question 2

*3 marks · Short answer*

Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.

**Solution**

1. The length of the first pipe is given as $20\text{ m}$.
2. The length of the other joined pipe is denoted by the letter-number $k$ meters.
3. The combined length of the pipe is the sum of both lengths, which is $(20 + k)\text{ meters}$.

**Answer:** $(20 + k)\text{ meters}$

> Common mistake: Multiplying the lengths instead of adding them.

### Question 3

*3 marks · Short answer*

What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:

**Part (a)**

1. For 8 notes of ₹100, 4 notes of ₹20, and $z$ notes of ₹5, the total amount expression is formed by multiplying the note value by its count.
2. The expression and total amount is $8 \times 100 + 4 \times 20 + z \times 5 = 880 + 5z$.

Answer (a): $880 + 5z$

**Part (b)**

1. For $x$ notes of ₹100, $y$ notes of ₹20, and $z$ notes of ₹5, the total amount expression is formed similarly.
2. The expression and total amount is $x \times 100 + y \times 20 + z \times 5 = 100x + 20y + 5z$.

Answer (b): $100x + 20y + 5z$

**Answer:** The completed table expressions are $8 \times 100 + 4 \times 20 + z \times 5$ and $x \times 100 + y \times 20 + z \times 5$.

> Common mistake: Swapping the coefficients of the variables incorrectly.

### Question 4

*1 mark · MCQ*

Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially?

- $10 + 8 + y$
- $(10 + 8) \times y$
- $10 \times 8 \times y$
- $10 + 8 \times y$
- $10 \times y + 8$

**Solution**

1. The machine takes a fixed $10\text{ seconds}$ to start running initially.
2. Grinding $y\text{ kg}$ of grain takes $8\text{ seconds}$ per kg, which is represented as $8 \times y$.
3. Adding the start-up time and the grinding time gives the total time expression $10 + 8 \times y$.

**Answer:** (d) $10 + 8 \times y$

> Common mistake: Adding the fixed start-up time and per-kg time before multiplying by $y$.

### Question 5

*3 marks · Short answer*

Write algebraic expressions using letters of your choice. (a) 5 more than a number (b) 4 less than a number (c) 2 less than 13 times a number (d) 13 less than 2 times a number

**Part (a)**

1. Let the number be represented by the letter $d$.
2. 5 more than the number is obtained by adding 5 to $d$.

Answer (a): $d + 5$

**Part (b)**

1. Let the number be represented by the letter $d$.
2. 4 less than the number is obtained by subtracting 4 from $d$.

Answer (b): $d - 4$

**Part (c)**

1. Let the number be represented by the letter $d$.
2. 13 times the number is $13d$, and 2 less than that is $13d - 2$.

Answer (c): $13d - 2$

**Part (d)**

1. Let the number be represented by the letter $d$.
2. 2 times the number is $2d$, and 13 less than that is $2d - 13$.

Answer (d): $2d - 13$

**Answer:** The expressions are (a) $d + 5$, (b) $d - 4$, (c) $13d - 2$, (d) $2d - 13$.

> Common mistake: Writing subtraction in the reverse order for 'less than' phrases.

### Question 6

*3 marks · Short answer*

Describe situations corresponding to the following algebraic expressions: (a) $8 \times x + 3 \times y$ (b) $15 \times j - 2 \times k$

**Part (a)**

1. Consider two items priced at ₹$x$ and ₹$y$ respectively.
2. A shopkeeper sells a pen for ₹$x$ and a notebook for ₹$y$. Abha buys 8 pens and 3 notebooks, making the total cost $8x + 3y$.

Answer (a): A shopkeeper sells a pen for ₹$x$ and a notebook for ₹$y$. Abha buys 8 pens and 3 notebooks. The total cost is $8x + 3y$.

**Part (b)**

1. Consider a production process with a daily output and some losses over days.
2. A factory makes 15 chairs every day for $j$ days, but 2 chairs break daily for $k$ days. The remaining good chairs can be represented by $15j - 2k$.

Answer (b): A factory makes 15 chairs a day for $j$ days, and 2 chairs break daily for $k$ days. The remaining chairs are given by $15j - 2k$.

**Answer:** Situations describing the given algebraic expressions are provided in the parts.

> Common mistake: Failing to assign clear meanings to the variables and their coefficients.

### Question 7

*3 marks · Short answer*

In a calendar month, if any $2 \times 3$ grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

**Solution**

1. Let the bottom middle cell of the $2 \times 3$ grid have the date $w$.
2. The cell immediately above $w$ is in the previous week, so its date is $w - 7$.
3. The cells to the left and right in the bottom row are $w - 1$ and $w + 1$, and in the top row are $w - 8$ and $w - 6$.

**Answer:** Top row: $w - 8$, $w - 7$, $w - 6$; Bottom row: $w - 1$, $w$, $w + 1$

> Common mistake: Subtracting 1 instead of 7 for the cell directly above in the calendar grid.

## Page No. 86

### Question 1

*2 marks · Very short answer*

Find the value that the expression $5m + 3$ takes when $m = 2$.

**Solution**

1. Substitute $m = 2$ in the expression $5m + 3$.
2. $5 \times 2 + 3 = 10 + 3 = 13$.

**Answer:** 13

> Common mistake: Adding 5 and 3 first before multiplying by $m$.

## Mind the Mistake, Mend the Mistake (Page 87)

### Question 1

*3 marks · Short answer*

Some simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained. Observe each of them, identify if there is a mistake, explain what might have gone wrong, and then correct it and give the value.

**Part (i)**

1. Given expression is $10 - a$ and $a = -4$.
2. Substituting $a = -4$ gives $10 - (-4) = 10 + 4$.
3. Simplifying the expression gives $14$.

Answer (i): $14$

**Answer:** The mistake in evaluating $10 - a$ for $a = -4$ is that subtracting a negative number was treated as subtracting a positive number.

> Common mistake: Forgetting that subtracting a negative number is equivalent to addition.

## Page No. 88

### Question 1

*2 marks · Very short answer*

If $c = ₹50$, find the total amount earned by the scale of pencils.

**Solution**

1. The total amount earned by selling pencils is given by the expression $18c$.
2. Substitute $c = 50$ into the expression to get $18 \times 50 = ₹900$.

**Answer:** ₹900

> Common mistake: Multiplying incorrectly or forgetting the unit.

### Question 2

*3 marks · Short answer*

Write the expression for the total money earned by selling erasers. Then, simplify the expression.

**Solution**

1. The number of erasers sold on Day 1, Day 2, and Day 3 are 4, 6, and 1 respectively, with price $d$ per eraser.
2. The expression for the total money earned by selling erasers is $4 \times d + 6 \times d + 1 \times d$.
3. Using the distributive property, simplify by adding like terms: $(4 + 6 + 1)d = 11d$.

**Answer:** $11d$

> Common mistake: Adding the coefficients incorrectly.

### Question 3

*2 marks · Very short answer*

Check that both expressions take the same value when $c$ is replaced by different numbers.

**Solution**

1. The unsimplified expression is $5c + 3c + 10c$ and the simplified expression is $18c$.
2. Substitute $c = 10$ to get $5(10) + 3(10) + 10(10) = 50 + 30 + 100 = 180$, and $18(10) = 180$, showing both expressions take the same value.

**Answer:** Both expressions yield the same value for any value of $c$.

> Common mistake: Failing to check with multiple test values.

## Page No. 90

### Question 1

*2 marks · Very short answer*

Write an expression for the total number of rupees paid if $x$ chairs and $y$ tables are rented.

**Solution**

1. The amount paid for $x$ chairs at the beginning is $40x$ rupees.
2. The amount paid for $y$ tables at the beginning is $75y$ rupees.
3. The total amount paid at the beginning is $40x + 75y$, and the total amount returned is $6x + 10y$.
4. The expression for the total number of rupees paid is $(40x + 75y) - (6x + 10y)$.

**Answer:** $(40x + 75y) - (6x + 10y)$

> Common mistake: Writing the refund as a positive addition instead of subtraction.

### Question 2

*2 marks · Very short answer*

Describe the procedure to get these amounts.

**Solution**

1. Find the total amount paid initially by multiplying the number of chairs by 40 and tables by 75, then adding them to get $40x + 75y$.
2. Find the total amount returned by multiplying the number of returned chairs by 6 and tables by 10, then adding them to get $6x + 10y$.
3. Subtract the returned amount from the initial amount paid to get the final rupees paid.

**Answer:** Subtract the total returned amount from the total initial amount paid.

> Common mistake: Confusing initial payment with the amount returned.

### Question 3

*2 marks · Very short answer*

Can we simplify this expression? If yes, how? If not, why not?

**Solution**

1. Yes, the expression can be simplified by opening the brackets.
2. Remove the brackets to get $40x + 75y - 6x - 10y$.
3. Group the like terms together to get $(40 - 6)x + (75 - 10)y$.
4. Subtract the coefficients to obtain the simplified expression $34x + 65y$.

**Answer:** Yes, by opening brackets and grouping like terms to get $34x + 65y$.

> Common mistake: Making sign errors while opening brackets with a negative sign outside.

## Page No. 91

### Question 1

*2 marks · Very short answer*

Could we have written the initial expression as $(40x + 75y) + (-6x - 10y)$?

**Solution**

1. Yes, subtracting an expression is the same as adding its negative terms.
2. Therefore, $(40x + 75y) - (6x + 10y)$ can be written as $(40x + 75y) + (-6x - 10y)$.

**Answer:** Yes, it can be written as $(40x + 75y) + (-6x - 10y)$.

> Common mistake: Forgetting to change the signs of all terms inside the bracket when removing the negative sign.

### Question 2

*2 marks · Very short answer*

What do each of the expressions mean?

**Solution**

1. The expressions $7p - 3q$, $8p - 4q$, and $6p - 2q$ represent Charu's scores in the first, second, and third rounds respectively.
2. Here, $p$ is the score for a correct answer and $q$ is the penalty for an incorrect answer.

**Answer:** They represent Charu's scores in the three rounds of the quiz.

> Common mistake: Confusing the score for correct answers with the penalty for wrong answers.

### Question 3

*2 marks · Very short answer*

If the score for a correct answer is 4 ($p = 4$) and the penalty for a wrong answer is 1 ($q = 1$), find Charu’s score in the first round.

**Solution**

1. Charu's score in the first round is given by the expression $7p - 3q$.
2. Substituting $p = 4$ and $q = 1$ into the expression, we get $7 \times 4 - 3 \times 1$.
3. Evaluating the products gives $28 - 3 = 25$.

**Answer:** $25$

> Common mistake: Multiplying incorrectly or not following the order of operations by subtracting before multiplying.

### Question 4

*3 marks · Short answer*

What are her scores in the second and third rounds?

**Solution**

1. Charu's score in the second round is given by the expression $8p - 4q$.
2. Charu's score in the third round is given by the expression $6p - 2q$.
3. Thus, her scores in the second and third rounds are $8p - 4q$ and $6p - 2q$ respectively.

**Answer:** Second round score is $8p - 4q$ and third round score is $6p - 2q$.

> Common mistake: Writing the scores as numerical values without considering the given expressions for each round.

### Question 5

*2 marks · Very short answer*

What if there is no penalty? What will be the value of $q$ in that situation?

**Solution**

1. If there is no penalty for incorrect answers, then the penalty value $q$ is zero.
2. Therefore, the value of $q$ will be $0$.

**Answer:** $q = 0$

> Common mistake: Assuming $q = 1$ when there is no penalty.

### Question 6

*2 marks · Very short answer*

What is her final score after the three rounds?

**Solution**

1. Her final score is the sum of scores in all three rounds: $(7p - 3q) + (8p - 4q) + (6p - 2q)$.
2. Grouping like terms gives $(7 + 8 + 6)p + (-3 - 4 - 2)q = 21p - 9q$.

**Answer:** $21p - 9q$

> Common mistake: Incorrectly adding the negative coefficients of $q$.

## Page No. 92

### Question 1

*2 marks · Very short answer*

Give some possible scores for Krishita in the three rounds so that they add up to give $23p - 7q$.

**Solution**

1. Any three round scores whose coefficients of $p$ add up to 23 and coefficients of $q$ add up to $-7$ are valid.
2. For example, $8p - 2q$, $9p - 3q$, and $6p - 2q$ are possible scores for the three rounds.

**Answer:** Possible round scores are $8p - 2q$, $9p - 3q$, and $6p - 2q$.

> Common mistake: Adding coefficients incorrectly.

### Question 2

*2 marks · Very short answer*

Can we say who scored more? Can you explain why?

**Solution**

1. We cannot say who scored more without knowing the numerical values of $p$ and $q$.
2. The scores depend on the points for a correct answer and the penalty for a wrong answer.

**Answer:** We cannot say who scored more because the final scores depend on the values of $p$ and $q$.

> Common mistake: Assuming letter-numbers have fixed values.

### Question 3

*2 marks · Very short answer*

Simplify this expression further.

**Solution**

1. Write the expression for the difference between Krishita's and Charu's scores: $23p - 7q - (21p - 9q)$.
2. Remove brackets and group like terms: $23p - 7q - 21p + 9q = (23 - 21)p + (-7 + 9)q = 2p + 2q$.

**Answer:** $2p + 2q$

> Common mistake: Making sign errors when opening the brackets with a negative sign outside.

### Question 4

*3 marks · Short answer*

Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that $5u$ and $5 + u$ take.

**Solution**

1. Substitute $u = 2, 5, 8, 11$ into the expression $5u$ to get $10, 25, 40, 55$.
2. Substitute the same values into the expression $5 + u$ to get $7, 10, 13, 16$.
3. Comparing the values, $5u$ and $5 + u$ are not equal for any given value of $u$ except when $u = 1.25$.

**Answer:** For $u = 2, 5, 8, 11$, $5u$ gives $10, 25, 40, 55$ and $5 + u$ gives $7, 10, 13, 16$, showing the two expressions are different.

> Common mistake: Confusing multiplication with addition during substitution.

## Page No. 93

### Question 1

*2 marks · Very short answer*

After filling in the two diagrams, do you think the two expressions are equal?

**Solution**

1. The expression $10y - 3$ means 3 less than 10 times $y$.
2. The expression $10(y - 3)$ means 10 times (3 less than $y$), which evaluates to $10y - 30$.
3. Since the computed values for different values of $y$ in the diagrams are different, the two expressions are not equal.

**Answer:** No, the two expressions $10y - 3$ and $10(y - 3)$ are not equal.

> Common mistake: Confusing $10y - 3$ with $10(y - 3)$ by forgetting to distribute the 10.

## Figure it Out (Page 93-94)

### Question 1

*3 marks · Short answer*

Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing.

**Part (a)**

1. Method 1 (Group by like terms): $5y + x + x + 5y - 6 + 2 = (5y + 5y) + (x + x) + (-6 + 2) = 10y + 2x - 4$.
2. Method 2 (Group by rows): Top row sum is $5y - 6 + x$ and bottom row sum is $x + 2 + 5y$, which combines to $10y + 2x - 4$.
3. Thus, both methods give the same simplified expression: $10y + 2x - 4$.

Answer (a): $10y + 2x - 4$

**Part (b)**

1. Method 1 (Group by term type): Count $2p$ four times ($8p$), $3q$ four times ($12q$), and numbers $3 - 2 - 2 + 3 = 2$, giving $8p + 12q + 2$.
2. Method 2 (Group by columns): Sum each column from left to right to get $(2p + 3q) + (3q + 2p) + (1 + 2p + 3q) + (1 + 3q + 2p) = 8p + 12q + 2$.
3. Thus, both methods give the same simplified expression: $8p + 12q + 2$.

Answer (b): $8p + 12q + 2$

**Part (c)**

1. Method 1 (Group by term type): Count $-5g$ four times ($-20g$) and $5k$ twelve times ($60k$), giving $-20g + 60k$.
2. Method 2 (Group by columns): Sum the columns to get $(-10g + 10k) + 40k + (-10g + 10k) = -20g + 60k$.
3. Thus, both methods give the same simplified expression: $-20g + 60k$.

Answer (c): $-20g + 60k$

**Answer:** The expressions are simplified by grouping like terms or by rows and columns, giving the same result.

> Common mistake: Mixing up unlike terms or forgetting to account for negative signs when opening brackets.

### Question 2

*3 marks · Short answer*

Simplify each of the following expressions: (a) $p + p + p + p$, $p + p + p + q$, $p + q + p - q$ (b) $p - q + p - q$, $p + q - p + q$ (c) $p + q - (p + q)$, $p - q - p - q$ (d) $2d - d - d - d$, $2d - d - d - c$ (e) $2d - d - (d - c)$, $2d - (d - d) - c$ (f) $2d - d - c - c$

**Part (a)**

1. For $p + p + p + p$, combine like terms to get $4p$.
2. For $p + p + p + q$, combine like terms to get $3p + q$.
3. For $p + q + p - q$, group like terms as $(p + p) + (q - q)$ to get $2p$.

Answer (a): $4p$, $3p + q$, $2p$

**Part (b)**

1. For $p - q + p - q$, group like terms as $(p + p) + (-q - q)$ to get $2p - 2q$.
2. For $p + q - p + q$, group like terms as $(p - p) + (q + q)$ to get $2q$.

Answer (b): $2p - 2q$, $2q$

**Part (c)**

1. For $p + q - (p + q)$, open the bracket to get $p + q - p - q = 0$.
2. For $p - q - p - q$, group like terms as $(p - p) + (-q - q)$ to get $-2q$.

Answer (c): $0$, $-2q$

**Part (d)**

1. For $2d - d - d - d$, combine like terms to get $-d$.
2. For $2d - d - d - c$, combine like terms to get $-c$.

Answer (d): $-d$, $-c$

**Part (e)**

1. For $2d - d - (d - c)$, open the bracket to get $2d - d - d + c = c$.
2. For $2d - (d - d) - c$, solve inside the bracket first to get $2d - 0 - c = 2d - c$.

Answer (e): $c$, $2d - c$

**Part (f)**

1. For $2d - d - c - c$, combine like terms for $d$ and $c$ to get $d - 2c$.

Answer (f): $d - 2c$

**Answer:** Simplified expressions for each given set.

> Common mistake: Incorrectly handling signs when removing parentheses.

## Mind the Mistake, Mend the Mistake (Page 94-95)

### Question 1

*5 marks · Long answer*

Observe each simplification, see if there is a mistake, explain what might have gone wrong, and simplify it correctly.

**Part (1)**

1. Mistake: The terms $3a$ and $2b$ are unlike terms and cannot be added together to get $5$.
2. Correction: The expression $3a + 2b$ is already in its simplest form as it contains unlike terms.

Answer (1): $3a + 2b$

**Part (2)**

1. Mistake: No mistake. Like terms are correctly combined: $3b - 2b - b = 0$.
2. Correction: $0$

Answer (2): $0$

**Part (3)**

1. Mistake: The number outside the bracket was not multiplied by the second term inside the bracket ($2$).
2. Correction: Using the distributive property, $6(p + 2) = 6 \times p + 6 \times 2 = 6p + 12$.

Answer (3): $6p + 12$

**Part (4)**

1. Mistake: The negative sign was not correctly distributed to all terms inside the second bracket.
2. Correction: $(4x + 3y) - (3x + 4y) = 4x + 3y - 3x - 4y = (4x - 3x) + (3y - 4y) = x - y$.

Answer (4): $x - y$

**Part (5)**

1. Mistake: The negative sign outside the bracket was not distributed to $-6z$, making it $-6z$ instead of $+6z$.
2. Correction: $5 - (2 - 6z) = 5 - 2 + 6z = 3 + 6z$.

Answer (5): $3 + 6z$

**Answer:** The expressions are simplified by correctly removing brackets using the distributive property and grouping like terms, rather than adding unlike terms or incomplete multiplications.

> Common mistake: Forgetting to distribute a negative sign or a multiplier to all terms inside a bracket.

### Question 2

*2 marks · Very short answer*

Is there any relation between the number of terms and the number of letter-numbers these expressions have?

**Solution**

1. By observing the corrected simplest forms of the algebraic expressions, we find that the number of letter-numbers is less than or equal to the number of terms.
2. That is, the number of letters is less than or equal to the number of terms in the simplified algebraic expression.

**Answer:** The number of letter-numbers is less than or equal to the number of terms in the simplified expression.

> Common mistake: Assuming the number of letter-numbers is always strictly equal to the number of terms, missing cases where a term contains no letters or multiple terms contain the same letter.

## Formula Detective (Page 95-96)

### Question 1

*3 marks · Short answer*

Find out the formula of this number machine.

**Solution**

1. Observe the input pairs and the output: $(5, 2) \rightarrow 8$, $(8, 1) \rightarrow 15$, $(9, 11) \rightarrow 7$, $(10, 10) \rightarrow 10$, and $(6, 4) \rightarrow 8$.
2. Test the relation $2a - b$ for the first pair: $2 \times 5 - 2 = 10 - 2 = 8$.
3. Verify for the second pair: $2 \times 8 - 1 = 16 - 1 = 15$.
4. The formula for the number machine is $2a - b$, where $a$ is the first input and $b$ is the second input.

**Answer:** The formula is $2a - b$.

> Common mistake: Confusing the order of inputs or failing to verify the formula against all given sets.

### Question 2

*3 marks · Short answer*

Find the formulas of the number machines below and write the expression for each set of inputs.

**Part (i)**

1. Observe the inputs and outputs: $(5, 2) \rightarrow 5$, $(8, 1) \rightarrow 7$, $(9, 11) \rightarrow 18$, $(10, 10) \rightarrow 18$.
2. Test the relation $a + b - 2$: $5 + 2 - 2 = 5$, $8 + 1 - 2 = 7$, $9 + 11 - 2 = 18$, $10 + 10 - 2 = 18$.
3. The formula is $a + b - 2$.

Answer (i): Formula: $a + b - 2$

**Part (ii)**

1. Observe the inputs and outputs: $(4, 1) \rightarrow 5$, $(6, 0) \rightarrow 1$, $(3, 2) \rightarrow 7$, $(10, 3) \rightarrow 31$.
2. Test the relation $ab + 1$: $4 \times 1 + 1 = 5$, $6 \times 0 + 1 = 1$, $3 \times 2 + 1 = 7$, $10 \times 3 + 1 = 31$.
3. The formula is $ab + 1$.

Answer (ii): Formula: $ab + 1$

**Answer:** The formulas for the two machines are $a + b - 2$ and $ab + 1$.

> Common mistake: Forgetting to add 1 in the second machine or miscalculating the constant term.

### Question 3

*3 marks · Short answer*

Now, make a formula on your own. Write a few number machines as examples using that formula. Challenge your classmates to figure it out!

**Solution**

1. Choose a formula: $3a + b$.
2. Create example 1: If $a = 2$ and $b = 3$, the output is $3 \times 2 + 3 = 9$.
3. Create example 2: If $a = 4$ and $b = 1$, the output is $3 \times 4 + 1 = 13$.

**Answer:** Formula: $3a + b$; Examples: $(2, 3) \rightarrow 9$ and $(4, 1) \rightarrow 13$.

> Common mistake: Choosing a formula that is too simple or too complex for the level.

## Page No. 96-97

### Question 1

*3 marks · Short answer*

Somjit wonders if there is a way to describe all the positions where the (i) Design A occurs, (ii) Design B occurs, and (iii) Design C occurs.

**Part (i)**

1. Design A appears at positions 1, 4, 7, and so on.
2. The $n^{\text{th}}$ position for Design A is given by the expression $3n - 2$.

Answer (i): $3n - 2$

**Part (ii)**

1. Design B appears at positions 2, 5, 8, and so on.
2. The $n^{\text{th}}$ position for Design B is given by the expression $3n - 1$.

Answer (ii): $3n - 1$

**Part (iii)**

1. Design C appears at positions 3, 6, 9, and so on.
2. The $n^{\text{th}}$ position for Design C is given by the expression $3n$.

Answer (iii): $3n$

**Answer:** The expressions for the $n^{\text{th}}$ positions are $3n - 2$ for Design A, $3n - 1$ for Design B, and $3n$ for Design C.

> Common mistake: Confusing the position number expressions for different designs.

### Question 2

*2 marks · Very short answer*

Where would design C appear for the nth time?

**Solution**

1. Design C appears at positions that are multiples of 3.
2. Therefore, the $n^{\text{th}}$ occurrence of Design C is at position $3n$.

**Answer:** $3n$

> Common mistake: Writing $n + 3$ instead of $3n$.

### Question 3

*2 marks · Very short answer*

Similarly, find the formula that gives the position where the other Designs appear for the nth time.

**Solution**

1. Design B appears at positions that are one less than the multiples of 3, giving $3n - 1$.
2. Design A appears at positions that are two less than the multiples of 3, giving $3n - 2$.

**Answer:** Design B is at position $3n - 1$ and Design A is at position $3n - 2$.

> Common mistake: Mixing up the offsets for Design A and Design B.

### Question 4

*2 marks · Very short answer*

Given a position number can we find out the design that appears there? Which Design appears at Position 122?

**Solution**

1. Divide the position number by 3 to check the remainder.
2. Position 122 divided by 3 gives a quotient of 40 and a remainder of 2, which corresponds to Design B.

**Answer:** Design B appears at Position 122.

> Common mistake: Incorrectly matching the remainder to the design.

### Question 5

*2 marks · Very short answer*

Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below.

**Solution**

1. Yes, the remainder when divided by 3 can be used to identify the design.
2. A remainder of 0 gives Design C, a remainder of 2 gives Design B, and a remainder of 1 gives Design A.

**Answer:** Yes, remainders 0, 1, and 2 correspond to Designs C, A, and B respectively.

> Common mistake: Stating that the quotient determines the design instead of the remainder.

### Question 6

*2 marks · Very short answer*

Use this to find what design appears at positions 99, 122, and 148.

**Solution**

1. Position 99 leaves a remainder 0 on division by 3, so it is Design C.
2. Positions 122 and 148 leave remainders 2 and 1 respectively, giving Design B and Design A.

**Answer:** Design C appears at position 99, Design B at position 122, and Design A at position 148.

> Common mistake: Miscalculating the remainder for position 148.

## Page No. 98-99

### Question 1

*3 marks · Short answer*

Will the diagonal sums be equal in every $2 \times 2$ square in this endless grid? How can we be sure?

**Solution**

1. To be sure, we cannot check all $2 \times 2$ squares because there are an unlimited number of them.
2. Instead, we use algebraic modelling by taking the top left number as $a$.
3. We express the other numbers in terms of $a$ and show that the diagonal sums are equal for any value of $a$.

**Answer:** Yes, diagonal sums are equal in every $2 \times 2$ square, which is proved using algebraic expressions.

> Common mistake: Trying to check individual calendar squares instead of using algebraic variables.

### Question 2

*2 marks · Very short answer*

Given that we know the top left number, how do we find the other numbers in this $2 \times 2$ square?

**Solution**

1. The number to the right of $a$ is $1$ more than it, written as $a + 1$.
2. The number below $a$ is $7$ more than it, written as $a + 7$, and the diagonal number is $a + 8$.

**Answer:** The other numbers are $a + 1$, $a + 7$, and $a + 8$.

> Common mistake: Adding wrong offsets like $1$ or $7$ for the diagonal or vertical positions.

### Question 3

*2 marks · Very short answer*

Verify this expression for diagonal sums by considering any $2 \times 2$ square and taking its top left number to be ‘a’.

**Solution**

1. The first diagonal sum is $a + (a + 8) = 2a + 8$.
2. The second diagonal sum is $(a + 1) + (a + 7) = 2a + 8$, which proves both sums are equal.

**Answer:** Both diagonal sums simplify to $2a + 8$.

> Common mistake: Making errors while opening brackets and combining like terms.

### Question 4

*3 marks · Short answer*

Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?

**Solution**

1. For the cross shape with center $15$, the numbers are $8$, $14$, $15$, $16$, and $22$.
2. Their sum is $8 + 14 + 15 + 16 + 22 = 75$, which is $5 \times 15$.
3. Repeating this for another cross shape, we observe that the total sum is always $5$ times the number in the centre.

**Answer:** The total sum of the numbers in the cross shape is always 5 times the center number.

> Common mistake: Adding the numbers incorrectly or missing one of the arms of the cross.

### Question 5

*2 marks · Very short answer*

Will this always happen? How do you show this?

**Solution**

1. Let the center number be $a$. The numbers in the cross are $a - 7$, $a - 1$, $a$, $a + 1$, and $a + 7$.
2. Adding these gives $(a - 7) + (a - 1) + a + (a + 1) + (a + 7) = 5a$, which proves it always happens.

**Answer:** Yes, by taking the center as $a$, the sum simplifies to $5a$.

> Common mistake: Incorrectly writing the surrounding numbers in terms of $a$.

### Question 6

*3 marks · Short answer*

Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers.

**Solution**

1. Consider a $3 \times 3$ square grid of numbers from a calendar.
2. The sum of all nine numbers in a $3 \times 3$ square is always $9$ times the number in the exact centre.
3. Similarly, other symmetric shapes can be analyzed using algebraic expressions to show multiples.

**Answer:** A $3 \times 3$ square grid where the sum of all numbers is always a multiple of the center number.

> Common mistake: Choosing asymmetrical shapes where like terms do not cancel out neatly.

## Page No. 100-101

### Question 1

*3 marks · Short answer*

How many matchsticks will there be in Step 33, Step 84, and Step 108?

**Solution**

1. State that the number of matchsticks at any step $y$ is given by the expression $2y + 1$.
2. Substitute $y = 33$ to get $2 \times 33 + 1 = 66 + 1 = 67$ matchsticks.
3. Substitute $y = 84$ and $y = 108$ to get $169$ and $217$ matchsticks respectively.

**Answer:** 67, 169, and 217 matchsticks respectively

> Common mistake: Multiplying incorrectly or adding the wrong number of terms.

### Question 2

*3 marks · Short answer*

What could be an expression describing the rule/formula to find out the number of matchsticks at any step?

**Solution**

1. State that each step starts with 3 matchsticks for the first triangle and adds 2 matchsticks for each subsequent triangle.
2. For step $y$, there are $(y - 1)$ additions of 2 matchsticks, giving the expression $3 + 2 \times (y - 1)$.
3. Alternatively, writing it using the first step relation gives $2y + 1$.

**Answer:** $3 + 2(y - 1)$ or $2y + 1$

> Common mistake: Forgetting to subtract 1 from $y$ when counting additional matchsticks.

### Question 3

*2 marks · Very short answer*

Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same?

**Solution**

1. Expand and simplify $3 + 2(y - 1)$ to get $3 + 2y - 2 = 2y + 1$.
2. Both expressions are the same as they yield identical values for any step number.

**Answer:** Yes, both expressions are the same.

> Common mistake: Making errors in distributing the number 2 inside the bracket.

### Question 4

*2 marks · Very short answer*

What are these numbers in Step 3 and Step 4?

**Solution**

1. Observe that in Step 3, there are 3 horizontal matchsticks and 4 diagonal matchsticks.
2. In Step 4, there are 4 horizontal matchsticks and 5 diagonal matchsticks.

**Answer:** Step 3 has 3 horizontal and 4 diagonal matchsticks; Step 4 has 4 horizontal and 5 diagonal matchsticks.

> Common mistake: Mixing up horizontal and diagonal matchstick counts.

### Question 5

*3 marks · Short answer*

How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to $2y + 1$?

**Solution**

1. State that at the $y^{\text{th}}$ step, the number of horizontal matchsticks is $y$.
2. State that at the $y^{\text{th}}$ step, the number of diagonal matchsticks is $y + 1$.
3. Add the two expressions to get $y + (y + 1) = 2y + 1$, confirming they add up to the total.

**Answer:** Horizontal: $y$, Diagonal: $y + 1$, and they add up to $2y + 1$.

> Common mistake: Writing $y - 1$ instead of $y + 1$ for the diagonal matchsticks.

## Figure it Out (Page 102-105)

### Question 1

*1 mark · MCQ*

One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?

- $30x + 20y$
- $(30 + 20) \times (x + y)$
- $20x + 30y$
- $(30 + 20) \times x + y$
- $30x - 20y$

**Solution**

1. Cost of x plates of Jowar roti at Rs 30 each is 30x.
2. Cost of y plates of Pulao at Rs 20 each is 20y.
3. Total amount earned is the sum of these costs, which is 30x + 20y.

**Answer:** (a) $30x + 20y$

> Common mistake: Multiplying the wrong prices with the variables.

### Question 2

*1 mark · MCQ*

Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day?

- $p + q + r$
- $p + q + 2r$
- $2 \times (p + q + r)$
- $p + q + r + 2$
- $p + q + r + 1$
- $2 \times (p + q)$

**Solution**

1. Number of customers who bought only champak is p.
2. Number of customers who bought only marigold is q.
3. Number of customers who bought both is r, meaning they are also customers.
4. Total number of unique customers is p + q + r, and since each gets one flag, total flags is p + q + r.

**Answer:** (a) $p + q + r$

> Common mistake: Counting customers who bought both twice.

### Question 3

*3 marks · Short answer*

A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights. (a) Write an expression describing how far away the snail is from its starting position. (b) What can we say about the snail’s movement if $d > u$?

**Part (a)**

1. In one day and night, the net distance the snail climbs is (u - d) cm.
2. For 10 days and 10 nights, the total distance from the starting position is 10(u - d) cm.

Answer (a): $10(u - d)\text{ cm}$

**Part (b)**

1. If d > u, the downward slip is greater than the upward climb.
2. Thus, the snail moves backwards overall and will never reach the top.

Answer (b): The snail will never reach the top.

**Answer:** (a) $10(u - d)\text{ cm}$, (b) The snail slips down more than it climbs and will never reach the top.

> Common mistake: Forgetting to multiply by 10 for all days.

### Question 4

*3 marks · Short answer*

Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?

**Solution**

1. Distance cycled in the first week = $7 \times 5 = 35\text{ km}$.
2. Distance cycled in the second week = $7(5 + z) = 35 + 7z\text{ km}$.
3. Distance cycled in the third week = $7(5 + 2z) = 35 + 14z\text{ km}$.
4. Total distance after 3 weeks = $35 + (35 + 7z) + (35 + 14z) = 105 + 21z\text{ km}$.

**Answer:** $105 + 21z\text{ km}$

> Common mistake: Multiplying daily distance by 3 instead of 7 days in a week.

### Question 5

*Activity*

In the following figure, observe how the expression $w + 2$ becomes $4w + 20$ along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

**Solution**

1. Apply the given operations step by step along each path starting from the central expression w + 2.
2. Fill in the missing intermediate expressions in the ovals and operations in the boxes.

**Answer:** Completed flowchart paths.

### Question 6

*3 marks · Short answer*

A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations. (a) If $t = 4$, what is the time taken to travel from Yahapur to Vahapur? (b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur?

**Part (a)**

1. There are 4 segments between Yahapur and Vahapur with 3 intermediate stations.
2. The time taken to travel between adjacent stations is $4$ minutes, so the total running time is $4 \times 4 = 16$ minutes.
3. The train stops for $2$ minutes at each of the 3 stations, giving a total stopping time of $3 \times 2 = 6$ minutes.
4. Total time taken = $16 + 6 = 22$ minutes.

Answer (a): $22$ minutes

**Part (b)**

1. Let the time taken to travel between consecutive stations be $t$ minutes.
2. Since there are 4 travel segments, the total running time is $4t$ minutes.
3. The train stops for $2$ minutes at each of the 3 stations, so the total stopping time is $3 \times 2 = 6$ minutes.
4. The algebraic expression for the total time taken is $4t + 6$ minutes.

Answer (b): $4t + 6$ minutes

**Answer:** (a) $22$ minutes, (b) $4t + 6$ minutes

> Common mistake: Forgetting to multiply the stopping time by the number of stations, or miscounting the number of travel segments as equal to the number of stations.

### Question 7

*3 marks · Short answer*

Simplify the following expressions: (a) $3a + 9b - 6 + 8a - 4b - 7a + 16$ (b) $3(3a - 3b) - 8a - 4b - 16$ (c) $2(2x - 3) + 8x + 12$ (d) $8x - (2x - 3) + 12$ (e) $8h - (5 + 7h) + 9$ (f) $23 + 4(6m - 3n) - 8n - 3m - 18$

**Part (a)**

1. Group the like terms together: $(3a + 8a - 7a) + (9b - 4b) + (-6 + 16)$
2. Combine the coefficients of like terms: $4a + 5b + 10$

Answer (a): $4a + 5b + 10$

**Part (b)**

1. Expand the bracket using the distributive property: $9a - 9b - 8a - 4b - 16$
2. Group the like terms together: $(9a - 8a) + (-9b - 4b) - 16$
3. Combine the terms: $a - 13b - 16$

Answer (b): $a - 13b - 16$

**Part (c)**

1. Expand the bracket: $4x - 6 + 8x + 12$
2. Group the like terms: $(4x + 8x) + (-6 + 12)$
3. Combine the terms: $12x + 6$

Answer (c): $12x + 6$

**Part (d)**

1. Open the bracket with a negative sign outside: $8x - 2x + 3 + 12$
2. Group the like terms: $(8x - 2x) + (3 + 12)$
3. Combine the terms: $6x + 15$

Answer (d): $6x + 15$

**Part (e)**

1. Open the bracket: $8h - 5 - 7h + 9$
2. Group the like terms: $(8h - 7h) + (-5 + 9)$
3. Combine the terms: $h + 4$

Answer (e): $h + 4$

**Part (f)**

1. Expand the bracket: $23 + 24m - 12n - 8n - 3m - 18$
2. Group the like terms: $(24m - 3m) + (-12n - 8n) + (23 - 18)$
3. Combine the terms: $21m - 20n + 5$

Answer (f): $5 + 21m - 20n$

**Answer:** Simplified expressions for parts (a) to (f)

> Common mistake: Making sign errors while removing brackets with a negative sign outside.

### Question 8

*3 marks · Short answer*

Add the expressions given below: (a) $4d - 7c + 9$ and $8c - 11 + 9d$ (b) $-6f + 19 - 8s$ and $-23 + 13f + 12s$ (c) $8d - 14c + 9$ and $16c - (11 + 9d)$ (d) $6f - 20 + 8s$ and $23 - 13f - 12s$ (e) $13m - 12n$ and $12n - 13m$ (f) $-26m + 24n$ and $26m - 24n$

**Part (a)**

1. Write the sum of the expressions: $(4d - 7c + 9) + (8c - 11 + 9d)$
2. Group the like terms: $(4d + 9d) + (-7c + 8c) + (9 - 11)$
3. Combine the like terms: $13d + c - 2$

Answer (a): $13d + c - 2$

**Part (b)**

1. Write the sum of the expressions: $(-6f + 19 - 8s) + (-23 + 13f + 12s)$
2. Group the like terms: $(-6f + 13f) + (-8s + 12s) + (19 - 23)$
3. Combine the like terms: $7f + 4s - 4$

Answer (b): $7f + 4s - 4$

**Part (c)**

1. Simplify the second expression by opening the bracket: $16c - 11 - 9d$
2. Write the sum: $(8d - 14c + 9) + (16c - 11 - 9d)$
3. Group and combine like terms: $(8d - 9d) + (-14c + 16c) + (9 - 11) = -d + 2c - 2$

Answer (c): $2c - d - 2$

**Part (d)**

1. Write the sum of the expressions: $(6f - 20 + 8s) + (23 - 13f - 12s)$
2. Group the like terms: $(6f - 13f) + (8s - 12s) + (-20 + 23)$
3. Combine the like terms: $-7f - 4s + 3$

Answer (d): $-7f - 4s + 3$

**Part (e)**

1. Write the sum of the expressions: $(13m - 12n) + (12n - 13m)$
2. Group the like terms: $(13m - 13m) + (-12n + 12n)$
3. Combine the like terms: $0$

Answer (e): $0$

**Part (f)**

1. Write the sum of the expressions: $(-26m + 24n) + (26m - 24n)$
2. Group the like terms: $(-26m + 26m) + (24n - 24n)$
3. Combine the like terms: $0$

Answer (f): $0$

**Answer:** Added expressions for parts (a) to (f)

> Common mistake: Failing to combine all like terms correctly across different parentheses.

### Question 9

*3 marks · Short answer*

Subtract the expressions given below: (a) $9a - 6b + 14$ from $6a + 9b - 18$ (b) $-15x + 13 - 9y$ from $7y - 10 + 3x$ (c) $17g + 9 - 7h$ from $11 - 10g + 3h$ (d) $9a - 6b + 14$ from $6a - (9b + 18)$ (e) $10x + 2 + 10y$ from $-3y + 8 - 3x$ (f) $8g + 4h - 10$ from $7h - 8g + 20$

**Part (a)**

1. Write the subtraction expression: $(6a + 9b - 18) - (9a - 6b + 14)$.
2. Remove the brackets by changing signs of terms inside the second bracket: $6a + 9b - 18 - 9a + 6b - 14$.
3. Group like terms together: $(6a - 9a) + (9b + 6b) + (-18 - 14)$.
4. Simplify to get the final result: $-3a + 15b - 32$.

Answer (a): $-3a + 15b - 32$

**Part (b)**

1. Write the subtraction expression: $(7y - 10 + 3x) - (-15x + 13 - 9y)$.
2. Remove brackets and change signs: $7y - 10 + 3x + 15x - 13 + 9y$.
3. Group like terms: $(3x + 15x) + (7y + 9y) + (-10 - 13)$.
4. Simplify to get the final result: $18x + 16y - 23$.

Answer (b): $18x + 16y - 23$

**Part (c)**

1. Write the subtraction expression: $(11 - 10g + 3h) - (17g + 9 - 7h)$.
2. Remove brackets: $11 - 10g + 3h - 17g - 9 + 7h$.
3. Group like terms: $(-10g - 17g) + (3h + 7h) + (11 - 9)$.
4. Simplify to get the final result: $-27g + 10h + 2$.

Answer (c): $-27g + 10h + 2$

**Part (d)**

1. Write the subtraction expression: $(6a - (9b + 18)) - (9a - 6b + 14)$.
2. Simplify the first expression: $6a - 9b - 18 - 9a + 6b - 14$.
3. Group like terms: $(6a - 9a) + (-9b + 6b) + (-18 - 14)$.
4. Simplify to get the result: $-3a - 3b - 32$, which can also be written as $-(3a + 3b + 32)$.

Answer (d): $-(3a + 3b + 32)$

**Part (e)**

1. Write the subtraction expression: $(-3y + 8 - 3x) - (10x + 2 + 10y)$.
2. Remove brackets: $-3y + 8 - 3x - 10x - 2 - 10y$.
3. Group like terms: $(-3x - 10x) + (-3y - 10y) + (8 - 2)$.
4. Simplify to get the final result: $-13x - 13y + 6$.

Answer (e): $-13x - 13y + 6$

**Part (f)**

1. Write the subtraction expression: $(7h - 8g + 20) - (8g + 4h - 10)$.
2. Remove brackets: $7h - 8g + 20 - 8g - 4h + 10$.
3. Group like terms: $(-8g - 8g) + (7h - 4h) + (20 + 10)$.
4. Simplify to get the final result: $-16g + 3h + 30$.

Answer (f): $-16g + 3h + 30$

**Answer:** Simplified expressions for subtraction of algebraic expressions given in parts.

> Common mistake: Forgetting to change the signs of all terms inside the second bracket when subtracting.

### Question 10

*3 marks · Short answer*

Describe situations corresponding to the following algebraic expressions: (a) $8x + 3y$ (b) $15x - 2x$

**Part (a)**

1. Identify the expression: $8x + 3y$.
2. Relate the terms to quantities and prices in a real-life context.
3. Write a clear situation: A fruit seller sells mangoes at ₹$8$ each and bananas at ₹$3$ each. If a customer buys $x$ mangoes and $y$ bananas, the total cost is represented by $8x + 3y$.

Answer (a): A fruit seller sells mangoes at ₹8 each and bananas at ₹3 each. If a customer buys x mangoes and y bananas, the total cost would be 8x + 3y.

**Part (b)**

1. Identify the expression: $15x - 2x$.
2. Relate the terms to a total quantity and a subtracted quantity with a common variable cost.
3. Write a clear situation: A shopkeeper has $15$ pencils in a packet, each costing ₹$x$. If $2$ pencils in the packet are not sold, the amount received for the sold pencils is given by $15x - 2x$.

Answer (b): A shopkeeper has 15 pencils in a packet. The cost of one pencil is ₹x. Two pencils in the packet are not sold. Find the amount received for this packet.

**Answer:** Situations corresponding to the given algebraic expressions.

> Common mistake: Writing unrealistic or mathematically inconsistent contexts for the terms.

### Question 11

*3 marks · Short answer*

Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

**Part (i)**

1. Observe the number of pieces for each fold: 0 folds give 1 piece, 1 fold gives 2 pieces, and 2 folds give 3 pieces.
2. Each time the rope is folded $r$ times, the number of pieces obtained is given by the expression $r + 2$.
3. Substituting $r = 10$, the number of pieces obtained is $10 + 2 = 12$ pieces.

Answer (i): $12$ pieces

**Part (ii)**

1. We observe the pattern that cutting a rope folded $r$ times yields one more piece than the number of folds plus one.
2. Thus, the algebraic expression for the number of pieces when a rope is folded $r$ times and cut is $r + 2$.

Answer (ii): $r + 2$

**Answer:** $12$ pieces for 10 folds, and $r + 2$ for $r$ folds

> Common mistake: Writing the expression as $r + 1$ or $2r$ instead of $r + 2$.

### Question 12

*3 marks · Short answer*

Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

**Part (i)**

1. Observe the matchstick pattern: 1 square requires 4 matchsticks, 2 squares require 7 matchsticks, and 3 squares require 10 matchsticks.
2. Each additional square adds 3 matchsticks.
3. For 10 squares, the number of matchsticks required is $4 + 3 \times (10 - 1) = 4 + 27 = 31$ matchsticks.

Answer (i): $31$ matchsticks

**Part (ii)**

1. Following the pattern, the first square takes 4 matchsticks and each subsequent square takes 3 matchsticks.
2. Thus, for $w$ squares, the number of matchsticks required is given by the expression $4 + 3(w - 1)$.

Answer (ii): $4 + 3(w - 1)$

**Answer:** $31$ matchsticks for 10 squares, and $4 + 3(w - 1)$ for $w$ squares

> Common mistake: Multiplying the total number of squares directly by 4 without accounting for shared sides.

### Question 13

*3 marks · Short answer*

Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.

**Part (a)**

1. The traffic signal sequence has 4 states: Red (1), Yellow (2), Green (3), Yellow (4), repeating cyclically.
2. For position 90, dividing 90 by 4 gives a remainder of 2, so the colour is Yellow.
3. For position 190, dividing 190 by 4 gives a remainder of 2, so the colour is Yellow.
4. For position 343, dividing 343 by 4 gives a remainder of 3, so the colour is Green.

Answer (a): Position 90 is Yellow, position 190 is Yellow, and position 343 is Green.

**Part (b)**

1. The red light appears at positions 1, 5, 9, etc., which can be described by the expression $4n - 3$.
2. The yellow light appears at even positions 2, 4, 6, etc., which can be described by the expression $2n$.
3. The green light appears at positions 3, 7, 11, etc., which can be described by the expression $4n - 1$.

Answer (b): Red: $4n - 3$, Yellow: $2n$, Green: $4n - 1$

**Answer:** Positions 90 and 190 are Yellow, position 343 is Green. Expressions: Red is $4n - 3$, Yellow is $2n$, Green is $4n - 1$.

> Common mistake: Dividing incorrectly or misidentifying the repeating cycle length.

### Question 14

*3 marks · Short answer*

Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

**Part (a)**

1. By observing the pattern of squares in the steps, we find the number of squares is 5, 9, 13 for steps 1, 2, 3 respectively.
2. The number of squares increases by 4 at each step.
3. For Step 4, the number of squares is $13 + 4 = 17$.
4. For Step 10, the number of squares is $5 + (10 - 1) \times 4 = 41$.
5. For Step 50, the number of squares is $5 + (50 - 1) \times 4 = 201$.

Answer (a): Step 4: 17 squares, Step 10: 41 squares, Step 50: 201 squares.

**Part (b)**

1. The general formula for the number of squares at step $n$ is $5 + (n - 1) \times 4 = 4n + 1$.
2. Since each individual square has 4 vertices, the total number of vertices for $4n + 1$ squares is $4(4n + 1) = 16n + 4$.

Answer (b): General formula for squares: $4n + 1$, Formula for vertices: $16n + 4$.

**Answer:** Step 4 has 17 squares, Step 10 has 41 squares, Step 50 has 201 squares. General formula is $4n + 1$, and vertices formula is $16n + 4$.

> Common mistake: Forgetting to multiply the number of squares by 4 when finding vertices.

### Question 15

*5 marks · Long answer*

Numbers are written in a particular sequence in this endless 4-column grid. (a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4). (b) In which row and column will the following numbers appear: (i) 124 (ii) 147 (iii) 201 (c) What number appears in row r and column c? (d) Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.

**Part (a)**

1. Let $r$ be the row number and $c$ be the column number.
2. Column 1 numbers are $1, 5, 9, 13, \dots$, given by $4(r - 1) + 1$.
3. Column 2 numbers are $2, 6, 10, 14, \dots$, given by $4(r - 1) + 2$.
4. Column 3 numbers are $3, 7, 11, 15, \dots$, given by $4(r - 1) + 3$.
5. Column 4 numbers are $4, 8, 12, 16, \dots$, given by $4(r - 1) + 4$.

Answer (a): Column $c$ expression is $4(r - 1) + c$.

**Part (b)**

1. For 124, dividing by 4 gives quotient 31 and remainder 0, so $r = 31$ and $c = 4$ (adjusted as row 31, col 4).
2. For 147, dividing by 4 gives quotient 36 and remainder 3, so it is in row 37 and column 3.
3. For 201, dividing by 4 gives quotient 50 and remainder 1, so it is in row 51 and column 1.

Answer (b): (i) Row 31, Col 4; (ii) Row 37, Col 3; (iii) Row 51, Col 1.

**Part (c)**

1. Using the row number $r$ and column number $c$, the general expression is $4(r - 1) + c$.

Answer (c): $4(r - 1) + c$

**Part (d)**

1. Multiples of 3 appear in columns following a repeating sequence of column numbers: 3, 2, 1, 4.
2. Other patterns include: All numbers in Column 4 are multiples of 4, even numbers appear in columns 2 and 4, and odd numbers appear in columns 1 and 3.

Answer (d): Multiples of 3 follow the column sequence 3, 2, 1, 4.

**Answer:** Grid expressions and patterns identified successfully.

> Common mistake: Confusing the remainder with the column number when the remainder is 0.

## Frequently asked questions

### How many questions are there in Class 7 Maths Chapter 4 Expressions using Letter-Numbers?

This chapter in the new NCERT book for the 2026-27 session contains a total of 54 questions spread across various pages and sections like Figure it Out and Mind the Mistake. You can find step-by-step solutions and the free PDF for all these questions right on this SwaVid page.

### What topics do the questions cover in Class 7 Maths Chapter 4?

The questions cover important concepts such as algebraic expressions for age, evaluating expressions, perimeter formulas, simplifying algebraic expressions by grouping like terms, and working with calendar grid patterns. SwaVid's free PDF on this page provides clear explanations for each of these topics.

### What are the hardest question types in this chapter and how should I approach them?

The more challenging questions involve algebraic modeling of calendar grid patterns, number machine formulas, and substituting values with negative numbers. To score full marks, carefully read the given conditions, set up the letter-numbers correctly, and check your simplification steps. You can review detailed approaches in the free PDF solutions available on this page.

### How do I write answers for full marks in Class 7 Maths Chapter 4?

To secure full marks, always write down the given information clearly, show the formation of the algebraic expression step-by-step, and state the final evaluated or simplified result neatly. Following the expert answers in SwaVid's free PDF on this page will help you learn the correct presentation style.

### Is the free PDF for Class 7 Maths Chapter 4 available here?

Yes, the complete chapter-wise free PDF and detailed step-by-step solutions for Expressions using Letter-Numbers are available exclusively on this SwaVid page. You can use them to verify your answers and prepare effectively for your exams.

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- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

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