Class 9 Mathematics Chapter 4: Expressions Using Letter-Numbers – Complete Important Questions Bank

The CBSE Class 9 Mathematics board exam allocates 5–6 marks to Chapter 4 content, typically split across 1-mark MCQs, 2-mark conceptual questions, and 3–5 mark problem-solving items. Unlike older syllabi that heavily tested algebraic manipulation, today's board pattern emphasizes *understanding* the structure of expressions. You will encounter: (1) identifying variables, constants, and terms in given expressions; (2) classifying like and unlike terms with correct reasoning; (3) substituting specific numerical values into expressions and computing results; (4) simplifying expressions by combining like terms; (5) writing algebraic expressions from word problems. The rationalized syllabus removed formal polynomial division but retained substitution and simplification as core skills. Examiners reward clear working and logical steps—a 3-mark question expects you to show *why* two terms are alike before adding them. These questions are modeled on NCERT textbook examples (Ex. 4.1, 4.2, 4.3) and official sample papers, ensuring every drill strengthens exam readiness.

**Q1.** Which of the following is a constant? (A) 7x (B) y (C) –5 (D) 2a + 3 **Answer: (C) –5** A constant is a fixed numerical value that does not change. Here, –5 has no variable attached, so it is constant. Options (A) and (D) contain variables (x and a), and (B) is a variable by itself. --- **Q2.** In the expression 3x² + 5x – 2, which term is unlike the others? (A) 3x² (B) 5x (C) –2 (D) All are like **Answer: (C) –2** Like terms have identical variable parts. Here, 3x² has variable x²; 5x has variable x; and –2 is a constant. The constant –2 is unlike both 3x² and 5x. --- **Q3.** If x = 2, the value of 4x + 7 is: (A) 11 (B) 15 (C) 19 (D) 23 **Answer: (B) 15** Substitute x = 2: 4(2) + 7 = 8 + 7 = 15. --- **Q4.** Which pair contains like terms? (A) 5ab and 3ba (B) 7x and 7x² (C) –4a²b and 5ab² (D) 2m and 3n **Answer: (A) 5ab and 3ba** Since multiplication is commutative, ab = ba. Both terms have the same variable part, so they are like terms. The numerical coefficients (5 and 3) differ, but that doesn't affect whether they're like. --- **Q5.** The expression 2x + 3y – 5 contains how many terms? (A) 1 (B) 2 (C) 3 (D) 4 **Answer: (C) 3** Terms are separated by + or – signs. Here: 2x, 3y, and –5 are the three terms.

**Q1.** Identify the variable, constant, and coefficient in the expression 6m – 9. **Solution:** Variable: m Constant: –9 Coefficient: 6 (the numerical multiplier of the variable m) --- **Q2.** Combine like terms and simplify: 4p + 7q – 2p + 3q. **Solution:** Grouping like terms: (4p – 2p) + (7q + 3q) = 2p + 10q The simplified expression is 2p + 10q. --- **Q3.** Find the value of 5a² – 3a + 2 when a = –1. **Solution:** Substitute a = –1: 5(–1)² – 3(–1) + 2 = 5(1) + 3 + 2 = 5 + 3 + 2 = 10 --- **Q4.** State whether the terms 8xy and –5yx are like or unlike. Give reason. **Solution:** These terms are **like terms**. **Reason:** Both have the same variables x and y. Since xy = yx (commutativity of multiplication), the variable parts are identical. Only the coefficients (8 and –5) differ. --- **Q5.** Write the algebraic expression for: "Five times a number m decreased by 4." **Solution:** The expression is: 5m – 4 ("Five times m" = 5m; "decreased by 4" = subtract 4.)

**Q1.** Simplify by combining like terms: 3x² – 5x + 4 – x² + 2x – 1, and then find the value when x = 2. **Solution:** Step 1: Group like terms. (3x² – x²) + (–5x + 2x) + (4 – 1) Step 2: Combine. = 2x² – 3x + 3 Step 3: Substitute x = 2. = 2(2)² – 3(2) + 3 = 2(4) – 6 + 3 = 8 – 6 + 3 = 5 --- **Q2.** Consider the expression 7ab + 4bc – 3ac + 5ab – 2bc. (a) Identify all like term pairs. (b) Simplify the expression. **Solution:** (a) Like term pairs: • 7ab and 5ab (both have variable part ab) • 4bc and –2bc (both have variable part bc) • –3ac has no like term (b) Simplifying: (7ab + 5ab) + (4bc – 2bc) – 3ac = 12ab + 2bc – 3ac --- **Q3.** If p = 3 and q = –2, find the value of 2p²q + 3pq² – pq. **Solution:** Step 1: Substitute p = 3 and q = –2. = 2(3)²(–2) + 3(3)(–2)² – (3)(–2) Step 2: Evaluate exponents. = 2(9)(–2) + 3(3)(4) – (3)(–2) = –36 + 36 + 6 = 6 --- **Q4.** A rectangle has length (2x + 3) cm and width (x – 1) cm. Write the expression for its perimeter and simplify. **Solution:** Perimeter of rectangle = 2(length + width) = 2[(2x + 3) + (x – 1)] = 2[2x + 3 + x – 1] = 2[3x + 2] = 6x + 4 cm

**Q1.** Consider three expressions: A = 4a + 3b – 5c, B = –2a + b + 3c, and C = 3a – 4b + 2c. (a) Find A + B. (b) Find A – C. (c) Find the value of A + B – C when a = 1, b = 2, and c = –1. Show all steps. **Solution:** (a) Find A + B: A + B = (4a + 3b – 5c) + (–2a + b + 3c) = 4a – 2a + 3b + b – 5c + 3c = 2a + 4b – 2c (b) Find A – C: A – C = (4a + 3b – 5c) – (3a – 4b + 2c) = 4a – 3a + 3b + 4b – 5c – 2c = a + 7b – 7c (c) Find A + B – C: From part (a), A + B = 2a + 4b – 2c A + B – C = (2a + 4b – 2c) – (3a – 4b + 2c) = 2a – 3a + 4b + 4b – 2c – 2c = –a + 8b – 4c Substitute a = 1, b = 2, c = –1: = –(1) + 8(2) – 4(–1) = –1 + 16 + 4 = 19 --- **Q2.** A shopkeeper has x pens costing ₹2 each, y pencils costing ₹1 each, and z erasers costing ₹3 each. (a) Write an expression for the total cost. (b) If the shopkeeper buys 10 pens, 15 pencils, and 8 erasers, find the total cost. (c) The shopkeeper offers a discount of ₹5 on the total. Write the expression for the final cost after discount. **Solution:** (a) Total cost expression: C = 2x + y + 3z (in rupees) (b) Substitute x = 10, y = 15, z = 8: C = 2(10) + 15 + 3(8) = 20 + 15 + 24 = ₹59 (c) Expression for final cost after ₹5 discount: Final cost = (2x + y + 3z) – 5 For x = 10, y = 15, z = 8: Final cost = 59 – 5 = ₹54 --- **Q3.** Simplify and verify: 5(2x – 3) – 3(x – 5) = 7x by checking with x = 2. **Solution:** Step 1: Expand the left side. 5(2x – 3) – 3(x – 5) = 10x – 15 – 3x + 15 = 10x – 3x – 15 + 15 = 7x ✓ Step 2: Verify with x = 2. Left side: 5(2·2 – 3) – 3(2 – 5) = 5(4 – 3) – 3(–3) = 5(1) + 9 = 5 + 9 = 14 Right side: 7(2) = 14 ✓ Both sides equal 14, so the equation is verified.

**Case Study:** A school organizes a fundraiser. Ticket sales are modeled as follows: • Adult tickets cost ₹50 each; student tickets cost ₹20 each. • If a adults and s students buy tickets, the total revenue is R = 50a + 20s rupees. • Operating costs are a fixed ₹200 plus ₹5 per ticket sold. • Profit = Revenue – Total Cost. **Question:** (a) Write an expression for the total operating cost in terms of a and s. (b) Write a simplified expression for profit. (c) If 30 adults and 60 students buy tickets, calculate the profit. (d) For the profit to exceed ₹1000, what should be the minimum combined value of (a + s)? (Assume the ratio of adults to students remains 30:60 = 1:2.) **Solution:** (a) **Total operating cost:** Total tickets = a + s Cost = 200 + 5(a + s) = 200 + 5a + 5s rupees (b) **Expression for profit:** Profit = Revenue – Cost = (50a + 20s) – (200 + 5a + 5s) = 50a + 20s – 200 – 5a – 5s = 45a + 15s – 200 (c) **Profit when a = 30, s = 60:** Profit = 45(30) + 15(60) – 200 = 1350 + 900 – 200 = ₹2050 (d) **Minimum (a + s) for profit > ₹1000:** Since the ratio is 1:2, let a = k and s = 2k. Profit = 45k + 15(2k) – 200 > 1000 = 45k + 30k – 200 > 1000 = 75k > 1200 = k > 16 So minimum k = 17, which gives a = 17, s = 34. Minimum (a + s) = 17 + 34 = **51 people**.

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