Important Questions Class 7 Maths Part 1 Chapter 4: Expressions Using Letter-Numbers

Important Questions Class 7 Maths Part 1 Chapter 4 focus on algebraic expressions, which use letters to represent numbers. Students use letter-numbers to write formulas, simplify expressions, evaluate values, and describe number patterns.

Algebra begins when students replace repeated number work with one clear expression. Important Questions Class 7 Maths Part 1 Chapter 4 help students practise letter-numbers, formulas, expression values, like terms, brackets, distributive property, calendar patterns, number machines, and matchstick patterns. The CBSE 2026 chapter uses age relations, costs, perimeters, shop bills, quiz scores, trains, ropes, calendars, and grids to connect algebra with everyday situations.

Key Takeaways

• Letter-Numbers: Letters such as a, s, x, and y can represent unknown or changing numbers.
• Algebraic Expression: An expression like a + 3 contains numbers, operations, and letter-numbers.
• Like Terms: Terms with the same letter-number can be added or subtracted.
• Pattern Formula: A rule like 2y + 1 can describe every step of a matchstick pattern.

Important Questions Class 7 Maths Part 1 Chapter 4 Structure 2026

Important Questions Class 7 Maths Part 1 Chapter 4 with Answers

Algebra questions in this chapter ask students to translate words into expressions.
Students must also replace letter-numbers with values and simplify expressions step by step.
These class 7 maths part 1 chapter 4 important questions follow the NCERT 2026 pattern.

1. What does Important Questions Class 7 Maths Part 1 Chapter 4 test in Expressions Using Letter-Numbers?

Important Questions Class 7 Maths Part 1 Chapter 4 test letter-numbers, formulas, simplification, and patterns. The chapter checks whether students can use algebra to describe relationships.

1. Representation Skill: Use letters for changing numbers.
2. Evaluation Skill: Replace letters with given values.
3. Simplification Skill: Add like terms and remove brackets.
4. Pattern Skill: Write formulas for calendars, grids, and matchsticks.
5. Final Result: The chapter tests expression-building and algebraic thinking.

2. What are letter-numbers in Class 7 Maths?

Letter-numbers are letters used to represent numbers. They help write general rules in a short form.

1. Example: Let a represent Aftab’s age.
2. Relation: Shabnam is 3 years older than Aftab.
3. Expression: s = a + 3.
4. Final Result: a and s are letter-numbers.

3. What is an algebraic expression in Class 7 Maths?

An algebraic expression contains numbers, operations, and letter-numbers. The expression a + 3 is an algebraic expression.

1. Number Part: 3.
2. Letter-Number: a.
3. Operation: Addition.
4. Final Result: a + 3 is an algebraic expression.

Class 7 Maths Part 1 Chapter 4 Expressions Using Letter-Numbers

The chapter starts with simple relationships before moving to formulas.
Students first learn how a letter can stand for any number in a situation.
This class 7 maths part 1 chapter 4 expressions using letter-numbers section builds the base for algebra.

4. How do you write Shabnam’s age if Aftab’s age is a?

Shabnam’s age is a + 3. She is 3 years older than Aftab.

1. Given Data: Aftab’s age = a.
2. Relation: Shabnam is 3 years older.
3. Expression: Shabnam’s age = a + 3.
4. Example: If a = 23, then a + 3 = 26.
5. Final Result: Shabnam’s age is a + 3.

5. How do you write Aftab’s age if Shabnam’s age is s?

Aftab’s age is s − 3. He is 3 years younger than Shabnam.

1. Given Data: Shabnam’s age = s.
2. Relation: Aftab is 3 years younger.
3. Expression: Aftab’s age = s − 3.
4. Example: If s = 20, then s − 3 = 17.
5. Final Result: Aftab’s age is s − 3.

6. How many matchsticks are needed for n L-shapes?

The number of matchsticks is 2n. Each L-shape needs 2 matchsticks.

1. Given Data: Number of L-shapes = n.
2. Rule: Each L needs 2 matchsticks.
3. Expression: Number of matchsticks = 2 × n.
4. Short Form: 2 × n = 2n.
5. Final Result: n L-shapes need 2n matchsticks.

Expressions Using Letter-Numbers Class 7 Questions

Expressions become useful when they describe costs, ages, lengths, and patterns.
Students should first explain the relationship in words before writing symbols.
These expressions using letter-numbers class 7 questions focus on real-life algebra.

7. What is the total cost of c coconuts and j kg jaggery?

The total cost is 35c + 60j. One coconut costs ₹35, and 1 kg jaggery costs ₹60.

1. Given Data: Coconut price = ₹35.
2. Given Data: Jaggery price = ₹60 per kg.
3. Expression: Total cost = 35c + 60j.
4. Final Result: The total cost is 35c + 60j.

8. What is the cost of 7 coconuts and 4 kg jaggery?

The total cost is ₹485. Use the expression 35c + 60j.

1. Given Data: c = 7 and j = 4.
2. Formula Used: Total cost = 35c + 60j.
3. Calculation:
   35c + 60j = 35 × 7 + 60 × 4
   = 245 + 240
   = 485
4. Final Result: The total cost is ₹485.

9. What is the perimeter of a square with side q?

The perimeter of a square is 4q. A square has four equal sides.

1. Given Data: Side length = q.
2. Formula Used: Perimeter = 4 × side.
3. Expression: Perimeter = 4q.
4. Example: If q = 7 cm, then 4q = 28 cm.
5. Final Result: The perimeter is 4q.

Letter Numbers Class 7 Maths Questions

Letter-numbers make one expression work for many values.
A single formula can describe ages, notes, pipes, perimeters, and machine times.
These letter numbers class 7 maths questions test substitution and expression writing.

10. What is the combined length of a 20 m pipe and another pipe of length k?

The combined length is 20 + k metres. Add the fixed pipe length and the unknown pipe length.

1. Given Data: First pipe = 20 m.
2. Given Data: Second pipe = k m.
3. Expression: Combined length = 20 + k.
4. Final Result: The combined length is 20 + k metres.

11. What is the total value of x ₹100 notes, y ₹20 notes, and z ₹5 notes?

The total value is 100x + 20y + 5z. Multiply each note count by its value.

1. Given Data: x notes of ₹100.
2. Given Data: y notes of ₹20.
3. Given Data: z notes of ₹5.
4. Expression: Total = 100x + 20y + 5z.
5. Final Result: The total amount is ₹(100x + 20y + 5z).

12. Which expression describes grinding y kg grain if the mill takes 10 seconds to start?

The correct expression is 10 + 8y. The mill needs 10 seconds first and 8 seconds per kg.

1. Given Data: Starting time = 10 seconds.
2. Given Data: Grinding time = 8 seconds per kg.
3. Weight: y kg.
4. Expression: Total time = 10 + 8y.
5. Final Result: The time taken is 10 + 8y seconds.

Class 7 Maths Algebraic Expressions Questions

Algebraic expressions translate statements into symbols.
Students must notice whether a phrase means addition, subtraction, multiplication, or grouping.
These class 7 maths algebraic expressions questions build direct word-to-symbol practice.

13. How do you write 5 more than a number?

5 more than a number is x + 5. The phrase “more than” means addition.

1. Let: The number = x.
2. Phrase: 5 more than the number.
3. Expression: x + 5.
4. Final Result: 5 more than a number is x + 5.

14. How do you write 4 less than a number?

4 less than a number is x − 4. The phrase “less than” means subtract from the number.

1. Let: The number = x.
2. Phrase: 4 less than the number.
3. Expression: x − 4.
4. Final Result: 4 less than a number is x − 4.

15. How do you write 2 less than 13 times a number?

2 less than 13 times a number is 13x − 2. First multiply the number by 13.

1. Let: The number = x.
2. 13 times the number: 13x.
3. 2 less than that: 13x − 2.
4. Final Result: The expression is 13x − 2.

16. How do you write 13 less than 2 times a number?

13 less than 2 times a number is 2x − 13. First multiply the number by 2.

1. Let: The number = x.
2. 2 times the number: 2x.
3. 13 less than that: 2x − 13.
4. Final Result: The expression is 2x − 13.

Algebraic Expressions Class 7 Questions with Answers

Evaluation means replacing a letter-number with a given number.
Students must remember that 7k means 7 × k, not 74 or 7 + k.
These algebraic expressions class 7 questions with answers focus on substitution.

17. What is the value of 7k when k = 4?

The value of 7k is 28. The expression 7k means 7 × k.

1. Given Data: k = 4.
2. Expression: 7k.
3. Calculation:
   7k = 7 × 4
   = 28
4. Final Result: 7k = 28.

18. What is the value of 5m + 3 when m = 2?

The value of 5m + 3 is 13. Replace m by 2.

1. Given Data: m = 2.
2. Expression: 5m + 3.
3. Calculation:
   5m + 3 = 5 × 2 + 3
   = 10 + 3
   = 13
4. Final Result: 5m + 3 = 13.

19. What is the value of 10 − a when a = −4?

The value of 10 − a is 14. Subtracting a negative number increases the value.

1. Given Data: a = −4.
2. Expression: 10 − a.
3. Calculation:
   10 − a = 10 − (−4)
   = 10 + 4
   = 14
4. Final Result: 10 − a = 14.

Class 7 Maths Chapter 4 Questions and Answers on Simplification

Simplification reduces an expression without changing its value.
Students use swapping, grouping, brackets, and distributive property for this work.
These class 7 maths chapter 4 questions and answers follow the NCERT expression practice.

20. How do you simplify l + b + l + b?

The simplified form is 2l + 2b. Like terms can be grouped together.

1. Given Expression: l + b + l + b.
2. Group Like Terms: l + l + b + b.
3. Simplify:
   l + l = 2l
   b + b = 2b
4. Final Result: l + b + l + b = 2l + 2b.

21. How do you simplify 5c + 3c + 10c?

The simplified form is 18c. All three terms are like terms.

1. Given Expression: 5c + 3c + 10c.
2. Add Coefficients: 5 + 3 + 10 = 18.
3. Keep Letter-Number: c.
4. Final Result: 5c + 3c + 10c = 18c.

22. How do you simplify 12n − 4n?

The simplified form is 8n. Both terms contain the same letter-number n.

1. Given Expression: 12n − 4n.
2. Subtract Coefficients: 12 − 4 = 8.
3. Keep Letter-Number: n.
4. Final Result: 12n − 4n = 8n.

Simplification of Algebraic Expressions Class 7

Brackets can change the signs of terms in an expression.
A negative sign before a bracket must be applied to every term inside it.
This simplification of algebraic expressions class 7 section focuses on signs and like terms.

23. How do you simplify (40x + 75y) − (6x + 10y)?

The simplified form is 34x + 65y. Subtract each term inside the second bracket.

1. Given Expression: (40x + 75y) − (6x + 10y).
2. Open Brackets: 40x + 75y − 6x − 10y.
3. Group Like Terms:
   40x − 6x = 34x
   75y − 10y = 65y
4. Final Result: (40x + 75y) − (6x + 10y) = 34x + 65y.

24. How do you simplify 4(x + y) − y?

The simplified form is 4x + 3y. Use the distributive property first.

1. Given Expression: 4(x + y) − y.
2. Distribute 4: 4x + 4y − y.
3. Group Like Terms: 4y − y = 3y.
4. Final Result: 4(x + y) − y = 4x + 3y.

25. How do you simplify 7p − 3q + 8p − 4q + 6p − 2q?

The simplified form is 21p − 9q. Add p terms and q terms separately.

1. Given Expression: 7p − 3q + 8p − 4q + 6p − 2q.
2. Group Like Terms:
   7p + 8p + 6p = 21p
   −3q − 4q − 2q = −9q
3. Final Result: The simplified expression is 21p − 9q.

26. How do you simplify 23p − 7q − (21p − 9q)?

The simplified form is 2p + 2q. The minus sign changes both signs inside the bracket.

1. Given Expression: 23p − 7q − (21p − 9q).
2. Open Brackets: 23p − 7q − 21p + 9q.
3. Group Like Terms:
   23p − 21p = 2p
   −7q + 9q = 2q
4. Final Result: 23p − 7q − (21p − 9q) = 2p + 2q.

Like Terms Class 7 Questions

Like terms have the same letter-number part.
Only like terms can combine into a single term through addition or subtraction.
These like terms class 7 questions help students avoid adding unlike terms.

27. What are like terms in Class 7 algebra?

Like terms are terms with the same letter-number part. Terms 5c, 3c, and 10c are like terms.

1. Like Terms: 5c, 3c, 10c.
2. Reason: Each term contains c.
3. Simplification: 5c + 3c + 10c = 18c.
4. Final Result: Like terms have the same letter-number part.

28. Why can 18c + 11d not be simplified further?

18c + 11d cannot be simplified further because c and d are different letter-numbers. They are unlike terms.

1. Term 1: 18c.
2. Term 2: 11d.
3. Reason: c and d are different.
4. Final Result: 18c + 11d is already in simplest form.

29. How do you simplify p + q + p − q?

The simplified form is 2p. The +q and −q cancel each other.

1. Given Expression: p + q + p − q.
2. Group Like Terms: p + p + q − q.
3. Simplify:
   p + p = 2p
   q − q = 0
4. Final Result: p + q + p − q = 2p.

Unlike Terms Class 7 Questions

Unlike terms contain different letter-number parts.
They can stay together in an expression, but they cannot combine as one term.
These unlike terms class 7 questions focus on correct simplification boundaries.

30. Are 3a and 2b like terms?

3a and 2b are unlike terms. They contain different letter-numbers.

1. Term 1: 3a.
2. Term 2: 2b.
3. Check: a and b are different.
4. Final Result: 3a + 2b cannot become 5.

31. Why is 3a + 2b = 5 incorrect?

3a + 2b = 5 is incorrect because a and b are unlike terms. Their coefficients cannot be added directly.

1. Given Expression: 3a + 2b.
2. Mistake: Adding 3 + 2 and removing letters.
3. Correct Form: 3a + 2b.
4. Final Result: The expression stays 3a + 2b.

32. Why is 6(p + 2) = 6p + 8 incorrect?

6(p + 2) = 6p + 8 is incorrect because 6 × 2 equals 12. The correct result is 6p + 12.

1. Given Expression: 6(p + 2).
2. Distribute 6: 6 × p + 6 × 2.
3. Calculation: 6p + 12.
4. Final Result: 6(p + 2) = 6p + 12.

Algebraic Expression Patterns Class 7

Patterns become easier when students express the nth step with letter-numbers.
The chapter uses saree borders, calendars, matchsticks, ropes, grids, and traffic signals.
These algebraic expression patterns class 7 questions focus on general formulas.

33. What is the nth term of the sequence 4, 8, 12, 16, 20?

The nth term is 4n. The sequence lists multiples of 4.

1. Given Sequence: 4, 8, 12, 16, 20.
2. Pattern: Each term is 4 times its position.
3. Formula: nth term = 4n.
4. Example: 29th term = 4 × 29 = 116.
5. Final Result: The nth term is 4n.

34. Which saree border design appears at position 122?

Design B appears at position 122. Position 122 leaves remainder 2 when divided by 3.

1. Given Pattern: A, B, C repeats.
2. Position: 122.
3. Division:
   122 ÷ 3 gives quotient 40 and remainder 2.
4. Rule: Remainder 2 means Design B.
5. Final Result: Design B appears at position 122.

35. Why are diagonal sums equal in a 2 × 2 calendar square?

The diagonal sums are equal because both simplify to 2a + 8. Let the top-left date be a.

1. Grid:
   Top-left = a
   Top-right = a + 1
   Bottom-left = a + 7
   Bottom-right = a + 8
2. Diagonal Sum 1: a + (a + 8) = 2a + 8.
3. Diagonal Sum 2: (a + 1) + (a + 7) = 2a + 8.
4. Final Result: Both diagonal sums are equal.

Matchstick Patterns Class 7 Questions

Matchstick patterns show how algebra describes growing figures.
Students can count by adding the next step or by separating orientations.
These matchstick patterns class 7 questions use formulas for large step numbers.

36. How many matchsticks are needed in Step y of the triangle pattern?

Step y needs 2y + 1 matchsticks. The first step has 3 matchsticks, and each next step adds 2.

1. Pattern: Step 1 = 3, Step 2 = 5, Step 3 = 7.
2. Formula: 3 + 2(y − 1).
3. Simplification:
   3 + 2(y − 1) = 3 + 2y − 2
   = 2y + 1
4. Final Result: Step y needs 2y + 1 matchsticks.

37. How many matchsticks are needed in Step 33 of the triangle pattern?

Step 33 needs 67 matchsticks. Use the formula 2y + 1.

1. Given Data: y = 33.
2. Formula Used: 2y + 1.
3. Calculation:
   2 × 33 + 1 = 66 + 1
   = 67
4. Final Result: Step 33 needs 67 matchsticks.

38. How many matchsticks are needed to make w joined squares?

w joined squares need 4 + 3(w − 1) matchsticks. The first square needs 4 sticks, and each new square adds 3.

1. Given Data: Number of squares = w.
2. First Square: 4 matchsticks.
3. Extra Squares: w − 1.
4. Expression: 4 + 3(w − 1).
5. Final Result: w joined squares need 4 + 3(w − 1) matchsticks.

Calendar Patterns Class 7 Maths

Calendar patterns use fixed differences between neighbouring dates.
One step right adds 1, while one step down adds 7 in a weekly calendar.
These calendar patterns class 7 maths questions show algebraic proof through date grids.

39. If the centre date is a, what is the sum of the cross-shaped calendar numbers?

The sum is 5a. The five numbers balance around the centre date.

1. Given Data: Centre date = a.
2. Other Dates: a − 7, a − 1, a + 1, a + 7.
3. Sum:
   (a − 7) + (a − 1) + a + (a + 1) + (a + 7)
   = 5a
4. Final Result: The sum is 5 times the centre date.

40. In a 2 × 3 calendar grid, how do you express dates if the bottom middle cell is w?

The bottom middle cell is w, and the date above it is w − 7. Adjacent dates differ by 1.

1. Bottom Middle: w.
2. Bottom Left: w − 1.
3. Bottom Right: w + 1.
4. Top Middle: w − 7.
5. Top Left: w − 8.
6. Top Right: w − 6.
7. Final Result: The six dates are w − 8, w − 7, w − 6, w − 1, w, w + 1.

Formulas Using Letter Numbers Class 7

Formulas make one rule work for many cases.
Students can use formulas for perimeter, cost, time, row-column grids, and growing patterns.
These formulas using letter numbers class 7 questions connect algebra with prediction.

41. What is the formula for the perimeter of an equilateral triangle?

The perimeter of an equilateral triangle is 3a. All three sides are equal.

1. Given Data: Side length = a.
2. Formula Used: Perimeter = sum of all sides.
3. Calculation: a + a + a = 3a.
4. Final Result: Perimeter = 3a.

42. What is the formula for the perimeter of a regular pentagon?

The perimeter of a regular pentagon is 5a. It has five equal sides.

1. Given Data: Side length = a.
2. Number of Sides: 5.
3. Formula: Perimeter = 5 × a.
4. Final Result: Perimeter = 5a.

43. What is the formula for the perimeter of a regular hexagon?

The perimeter of a regular hexagon is 6a. It has six equal sides.

1. Given Data: Side length = a.
2. Number of Sides: 6.
3. Formula: Perimeter = 6 × a.
4. Final Result: Perimeter = 6a.

NCERT Class 7 Maths Part 1 Chapter 4 Questions for Mixed Practice

Mixed questions combine expression writing, simplification, substitution, and pattern rules.
Students must read the situation carefully before choosing an operation.
These NCERT class 7 maths part 1 chapter 4 questions match the NCERT 2026 exercise style.

44. What expression gives the total amount for x plates of Jowar roti and y plates of Pulao?

The total amount is 30x + 20y. One Jowar roti plate costs ₹30, and one Pulao plate costs ₹20.

1. Given Data: Jowar roti = ₹30 per plate.
2. Given Data: Pulao = ₹20 per plate.
3. Expression: 30x + 20y.
4. Final Result: The total amount is ₹(30x + 20y).

45. How many flags did Pushpita give if p customers bought champak, q bought marigold, and r bought both?

Pushpita gave p + q + r flags. Each customer received one flag.

1. Given Data: p champak-only customers.
2. Given Data: q marigold-only customers.
3. Given Data: r customers bought both.
4. Expression: p + q + r.
5. Final Result: Pushpita gave p + q + r flags.

46. What is the expression for a snail climbing u cm and slipping d cm for 10 days?

The expression is 10(u − d). Each day-night cycle changes the height by u − d.

1. Given Data: Climb per day = u cm.
2. Given Data: Slip per night = d cm.
3. One Cycle: u − d.
4. Ten Cycles: 10(u − d).
5. Final Result: The snail is 10(u − d) cm from the start.

47. What happens if d > u in the snail problem?

The snail moves downward overall if d > u. It slips more than it climbs in each cycle.

1. Given Data: d > u.
2. One Cycle: u − d.
3. Sign: u − d is negative.
4. Final Result: The snail moves below its starting position.

48. What is the train time expression if each travel part takes t minutes?

The train time expression is 4t + 6 minutes. The train travels four equal parts and stops three times.

1. Given Data: Four travel parts.
2. Travel Time: 4t.
3. Stops: 3 stops of 2 minutes each.
4. Stop Time: 3 × 2 = 6.
5. Final Result: Total time = 4t + 6 minutes.

49. If t = 4, what is the train time from Yahapur to Vahapur?

The train time is 22 minutes. Substitute t = 4 in 4t + 6.

1. Given Data: t = 4.
2. Formula Used: 4t + 6.
3. Calculation:
   4 × 4 + 6 = 16 + 6
   = 22
4. Final Result: The train takes 22 minutes.

50. What number appears in row r and column c of the 4-column grid?

The number is 4(r − 1) + c. Each row has 4 numbers.

1. Given Data: Row number = r.
2. Given Data: Column number = c.
3. Rows Before: r − 1 rows.
4. Numbers Before: 4(r − 1).
5. Final Result: The number is 4(r − 1) + c.

CBSE Class 7 Maths Important Links