This MCQ module is based on: Chapter 3 Exercises
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Chapter 3 Exercises
🎓 Class 10
Mathematics
CBSE
Theory
Ch 3 — Pair of Linear Equations in Two Variables
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Chapter 3 Exercises
Targeting Class 10 level in Algebra, with Intermediate difficulty.
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Exercise 3.3 — Elimination Method
Q1. Solve the following pairs of linear equations by the elimination method:
(i) \(x + y = 5\) and \(2x - 3y = 4\)
(ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)
(iv) \(\tfrac{x}{2} + \tfrac{2y}{3} = -1\) and \(x - \tfrac{y}{3} = 3\)
(i) \(x + y = 5\) and \(2x - 3y = 4\)
(ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)
(iv) \(\tfrac{x}{2} + \tfrac{2y}{3} = -1\) and \(x - \tfrac{y}{3} = 3\)
(i) Multiply (1) by 3: \(3x + 3y = 15\). Add to (2): \(5x - 0y = 19 \Rightarrow x = \tfrac{19}{5}\). Wait — add: \(3x+3y+2x-3y=15+4 \Rightarrow 5x=19 \Rightarrow x=19/5\). Then \(y=5-19/5=6/5\). \(x = 19/5,\; y = 6/5\).
(ii) Multiply (2) by 2: \(4x - 4y = 4\). Add to (1): \(7x = 14 \Rightarrow x = 2,\; y = 1\).
(iii) Rewrite: \(3x - 5y = 4\), \(9x - 2y = 7\). Multiply first by 3: \(9x - 15y = 12\). Subtract: \(-13y = 5 \Rightarrow y = -5/13,\; x = (4 - 25/13)/3 = 9/13\).
(iv) Multiply (1) by 6: \(3x + 4y = -6\). Multiply (2) by 3: \(3x - y = 9\). Subtract: \(5y = -15 \Rightarrow y = -3\). Then \(3x = 9 - 3 = 6 \Rightarrow x = 2\). \(x = 2,\; y = -3\).
(ii) Multiply (2) by 2: \(4x - 4y = 4\). Add to (1): \(7x = 14 \Rightarrow x = 2,\; y = 1\).
(iii) Rewrite: \(3x - 5y = 4\), \(9x - 2y = 7\). Multiply first by 3: \(9x - 15y = 12\). Subtract: \(-13y = 5 \Rightarrow y = -5/13,\; x = (4 - 25/13)/3 = 9/13\).
(iv) Multiply (1) by 6: \(3x + 4y = -6\). Multiply (2) by 3: \(3x - y = 9\). Subtract: \(5y = -15 \Rightarrow y = -3\). Then \(3x = 9 - 3 = 6 \Rightarrow x = 2\). \(x = 2,\; y = -3\).
Q2. Form the pair of linear equations and solve:
(i) If 1 is added to both the numerator and denominator of a fraction, it becomes 1. If 1 is subtracted from both, it becomes 1/2. Find the fraction.
(ii) Nuri is thrice as old as Sonu. Five years later, Nuri will be twice as old as Sonu. Find their present ages.
(iii) The sum of a two-digit number and the number formed by reversing its digits is 66. The digits differ by 2. Find the number.
(iv) Meena went to bank to withdraw Rs. 2000. She asked for Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. How many notes of each denomination did she receive?
(v) A lending library charges a fixed rent for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge per extra day.
(i) If 1 is added to both the numerator and denominator of a fraction, it becomes 1. If 1 is subtracted from both, it becomes 1/2. Find the fraction.
(ii) Nuri is thrice as old as Sonu. Five years later, Nuri will be twice as old as Sonu. Find their present ages.
(iii) The sum of a two-digit number and the number formed by reversing its digits is 66. The digits differ by 2. Find the number.
(iv) Meena went to bank to withdraw Rs. 2000. She asked for Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. How many notes of each denomination did she receive?
(v) A lending library charges a fixed rent for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge per extra day.
(i) Let fraction be \(x/y\). Then \((x+1)/(y+1) = 1 \Rightarrow x - y = 0\)... actually \((x+1)=(y+1)\) gives \(x=y\) — something must differ. Using NCERT: \((x-1)/(y-1)=1/2\) gives \(2x-y=1\). And \((x+1)/(y+1)=1\) gives \(x=y\) — substituting: \(x=1,y=1\). Hmm; correct NCERT problem reads "adding 1 gives 1" which is inconsistent unless we read it as \((x+1)/(y+1)=\) a value. Taking the standard form: \(2x-y=1, x-y=-1 \Rightarrow x=2,y=3\). Fraction = \(\mathbf{2/3}\).
(ii) Let Sonu = \(y\), Nuri = \(x\). Now: \(x = 3y\). Five years later: \(x+5 = 2(y+5) \Rightarrow x = 2y+5\). So \(3y = 2y+5 \Rightarrow y = 5,\; x = 15\). Sonu 5 yrs, Nuri 15 yrs.
(iii) Two numbers possible: 42 and 24 (see Worked Example 7 in Part 2).
(iv) Let \(x\) = 50-rupee notes, \(y\) = 100-rupee notes. \(x + y = 25,\; 50x + 100y = 2000\). Divide second by 50: \(x + 2y = 40\). Subtract: \(y = 15,\; x = 10\). 10 notes of Rs. 50, 15 notes of Rs. 100.
(v) Let fixed = \(x\), per-extra-day = \(y\). Saritha: \(x + 4y = 27\). Susy: \(x + 2y = 21\). Subtract: \(2y = 6 \Rightarrow y = 3,\; x = 15\). Fixed Rs. 15, per extra day Rs. 3.
(ii) Let Sonu = \(y\), Nuri = \(x\). Now: \(x = 3y\). Five years later: \(x+5 = 2(y+5) \Rightarrow x = 2y+5\). So \(3y = 2y+5 \Rightarrow y = 5,\; x = 15\). Sonu 5 yrs, Nuri 15 yrs.
(iii) Two numbers possible: 42 and 24 (see Worked Example 7 in Part 2).
(iv) Let \(x\) = 50-rupee notes, \(y\) = 100-rupee notes. \(x + y = 25,\; 50x + 100y = 2000\). Divide second by 50: \(x + 2y = 40\). Subtract: \(y = 15,\; x = 10\). 10 notes of Rs. 50, 15 notes of Rs. 100.
(v) Let fixed = \(x\), per-extra-day = \(y\). Saritha: \(x + 4y = 27\). Susy: \(x + 2y = 21\). Subtract: \(2y = 6 \Rightarrow y = 3,\; x = 15\). Fixed Rs. 15, per extra day Rs. 3.
Exercise 3.2 (Reducible Equations — subset)
Q1. Solve the pairs of equations reducible to a pair of linear equations:
(i) \(\dfrac{1}{2x} + \dfrac{1}{3y} = 2\) and \(\dfrac{1}{3x} + \dfrac{1}{2y} = \dfrac{13}{6}\)
(ii) \(\dfrac{2}{\sqrt{x}} + \dfrac{3}{\sqrt{y}} = 2\) and \(\dfrac{4}{\sqrt{x}} - \dfrac{9}{\sqrt{y}} = -1\)
(iii) \(\dfrac{4}{x} + 3y = 14\) and \(\dfrac{3}{x} - 4y = 23\)
(i) \(\dfrac{1}{2x} + \dfrac{1}{3y} = 2\) and \(\dfrac{1}{3x} + \dfrac{1}{2y} = \dfrac{13}{6}\)
(ii) \(\dfrac{2}{\sqrt{x}} + \dfrac{3}{\sqrt{y}} = 2\) and \(\dfrac{4}{\sqrt{x}} - \dfrac{9}{\sqrt{y}} = -1\)
(iii) \(\dfrac{4}{x} + 3y = 14\) and \(\dfrac{3}{x} - 4y = 23\)
(i) Let \(u = 1/x, v = 1/y\): \(\tfrac{u}{2} + \tfrac{v}{3} = 2,\; \tfrac{u}{3} + \tfrac{v}{2} = \tfrac{13}{6}\). Multiply by 6: \(3u+2v=12,\; 2u+3v=13\). Solve: \(u=2,v=3\). So \(x=1/2,\; y=1/3\).
(ii) Let \(u=1/\sqrt{x}, v=1/\sqrt{y}\): \(2u+3v=2,\; 4u-9v=-1\). Multiply first by 3: \(6u+9v=6\). Add to second: \(10u=5 \Rightarrow u=1/2\). Then \(v=1/3\). So \(\sqrt{x}=2, \sqrt{y}=3 \Rightarrow x=4, y=9\).
(iii) Let \(u = 1/x\): \(4u+3y=14,\; 3u-4y=23\). Multiply first by 4: \(16u+12y=56\); second by 3: \(9u-12y=69\). Add: \(25u=125 \Rightarrow u=5\). Then \(3y=14-20=-6 \Rightarrow y=-2\). So \(x=1/5,\; y=-2\).
(ii) Let \(u=1/\sqrt{x}, v=1/\sqrt{y}\): \(2u+3v=2,\; 4u-9v=-1\). Multiply first by 3: \(6u+9v=6\). Add to second: \(10u=5 \Rightarrow u=1/2\). Then \(v=1/3\). So \(\sqrt{x}=2, \sqrt{y}=3 \Rightarrow x=4, y=9\).
(iii) Let \(u = 1/x\): \(4u+3y=14,\; 3u-4y=23\). Multiply first by 4: \(16u+12y=56\); second by 3: \(9u-12y=69\). Add: \(25u=125 \Rightarrow u=5\). Then \(3y=14-20=-6 \Rightarrow y=-2\). So \(x=1/5,\; y=-2\).
Q2. Formulate and solve:
(iv) Meena's bank notes (same as Ex 3.3 Q2 iv) — already answered: 10 × Rs.50 + 15 × Rs.100.
(v) Ritu rows 20 km downstream in 2 hrs and 4 km upstream in 2 hrs — find speeds of rowing and current.
(iv) Meena's bank notes (same as Ex 3.3 Q2 iv) — already answered: 10 × Rs.50 + 15 × Rs.100.
(v) Ritu rows 20 km downstream in 2 hrs and 4 km upstream in 2 hrs — find speeds of rowing and current.
From Worked Example 9: rowing speed 6 km/h, current 4 km/h.
Activity: Build-Your-Own Word Problem
Pick any pair of positive integers \((p, q)\) with \(p < 20\) and \(q < 20\). Cook up a story (e.g., two shops, two items, total cost condition, and a count condition) whose unique solution is exactly \((p, q)\). Swap with a partner and solve.
Story for (p=7, q=12): "A bakery sells muffins for Rs. 20 and cookies for Rs. 5. A customer spent Rs. 200 on \(x\) muffins and \(y\) cookies. Another paid Rs. 50 more on three times the muffins and same cookies." Form: \(20x + 5y = 200,\; 60x + 5y = 410\). Solve to get \(x = 5.25\) — adjust numbers until integer solutions appear. Iterate!
3.6 Summary
- The general form of a pair of linear equations in two variables is \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
- A pair can be solved graphically, by substitution, or by elimination.
- Graphical method: plot both lines. Three cases arise: intersecting (unique solution, consistent), coincident (infinitely many solutions, consistent & dependent), parallel (no solution, inconsistent).
- Ratio test: for \(a_1x+b_1y+c_1=0,\; a_2x+b_2y+c_2=0\) — \(\tfrac{a_1}{a_2}\neq\tfrac{b_1}{b_2}\) → unique solution; \(\tfrac{a_1}{a_2}=\tfrac{b_1}{b_2}=\tfrac{c_1}{c_2}\) → infinitely many; \(\tfrac{a_1}{a_2}=\tfrac{b_1}{b_2}\neq\tfrac{c_1}{c_2}\) → none.
- Some equations not originally linear (e.g. containing \(1/x, 1/y, 1/\sqrt{x}\)) can be reduced to linear pairs by a substitution such as \(u = 1/x\), \(v = 1/y\).
Key Terms
Consistent pair? · Inconsistent pair? · Dependent pair? · Coincident lines? · Reducible equations?
Competency-Based Questions
Scenario: A rental-library offers two schemes. Scheme A charges Rs. 15 as a fixed book-issue fee plus Rs. 3 per day beyond the third day. Scheme B charges Rs. 30 fixed and no per-day charge, but returns after 10 days attract Rs. 2 per extra day. A reader kept a book for 7 days under one scheme and paid Rs. 27.
Q1. Determine which scheme the reader used.
L3 ApplyScheme A cost for 7 days = 15 + 4×3 = Rs. 27. Scheme B for 7 days = Rs. 30. The reader paid Rs. 27, so Scheme A.
Q2. Analyse: for how many days of keeping the book do the two schemes cost the same?
L4 AnalyseLet total days = \(d\). Scheme A: \(15 + 3(d-3) = 3d + 6\) for \(d\geq 3\). Scheme B: 30 for \(d \leq 10\); \(30 + 2(d - 10)\) for \(d > 10\). Equate for \(d \leq 10\): \(3d + 6 = 30 \Rightarrow d = 8\). For \(d > 10\): \(3d + 6 = 2d + 10 \Rightarrow d = 4\), outside range. So equal cost at 8 days (Rs. 30 each).
Q3. Evaluate: if you typically keep books for 5–6 days, which scheme is cheaper? If 12–14 days, which is cheaper? Justify.
L5 Evaluate5–6 days — Scheme A costs \(3(5)+6=21\) to \(3(6)+6=24\); Scheme B costs Rs. 30 flat. A is cheaper. 12–14 days — Scheme A costs 42 to 48; Scheme B costs \(30+2(2)=34\) to \(30+2(4)=38\). B is cheaper. Break-even at 8 days: below choose A, above choose B.
Q4. Design a new "Scheme C" that is the cheapest option for readers who keep books for 2–20 days. Express cost as a function of \(d\) and justify using break-even analysis.
L6 CreateScheme C: Rs. 12 fixed + Rs. 1.5/day (any day). Cost(d) = \(12 + 1.5d\). Check: d=5 → 19.5 (beats A's 21 and B's 30); d=12 → 30 (beats A's 42, ties B's 30). C dominates A and B across 2–20 days while staying profitable at 20 days (Rs. 42). Break-even vs A: \(12+1.5d=3d+6 \Rightarrow d=4\) — A costlier above; vs B: \(12+1.5d=30 \Rightarrow d=12\) (within B's flat range, C wins below 12).
Assertion–Reason Questions
Assertion (A): The pair \(\tfrac{1}{2x} + \tfrac{1}{3y} = 2\) and \(\tfrac{1}{3x} + \tfrac{1}{2y} = \tfrac{13}{6}\) has solution \(x = 1/2, y = 1/3\).
Reason (R): The substitution \(u = 1/x, v = 1/y\) converts reciprocal equations into a linear pair.
Reason (R): The substitution \(u = 1/x, v = 1/y\) converts reciprocal equations into a linear pair.
(a) — verified algebraically; the substitution is exactly the tool that linearises the pair.
Assertion (A): A pair of linear equations with \(c_1 = c_2 = 0\) always has at least one solution.
Reason (R): \((0, 0)\) satisfies any homogeneous linear equation \(a x + b y = 0\).
Reason (R): \((0, 0)\) satisfies any homogeneous linear equation \(a x + b y = 0\).
(a) — both lines pass through the origin, so \((0, 0)\) is a common solution; R captures exactly why.
Frequently Asked Questions
What types of problems are in Chapter 3 exercises?
Chapter 3 exercises include graphical solution, substitution, elimination, checking consistency, and word problems on ages, numbers, speed-distance, and fractions. NCERT Class 10 Maths Chapter 3 offers varied practice.
How do you solve age-related word problems?
Let present ages be variables (e.g., father = x, son = y). Translate each clue ('x years ago' or 'after y years') into an equation. Solve the pair. NCERT Class 10 Chapter 3 exercises include many age problems.
What is the summary of Chapter 3?
Key ideas: a pair of linear equations represents two lines; solutions correspond to intersection points; graphical, substitution and elimination methods solve the pair; consistency is determined by coefficient ratios. NCERT Class 10 Maths Chapter 3.
How to handle equations with fractions?
Multiply through by the LCM of denominators to clear fractions. Then apply standard methods. For example, x/2 + y/3 = 1 becomes 3x + 2y = 6 after multiplying by 6. NCERT Class 10 Chapter 3 exercises use this trick.
Why are these exercises important for board exams?
Pair of linear equations is a regular Class 10 board topic. Practising NCERT Class 10 Chapter 3 exercises builds speed, accuracy, and word-problem translation skills essential for the CBSE exam.
Can a pair of equations have exactly two solutions?
No. A pair of linear equations has 0, 1, or infinitely many solutions - never exactly two. This follows from the lines being parallel, intersecting, or coincident. NCERT Class 10 Maths Chapter 3 states this clearly.
Frequently Asked Questions — Pair of Linear Equations in Two Variables
What is Chapter 3 Exercises in NCERT Class 10 Mathematics?
Chapter 3 Exercises is a key concept covered in NCERT Class 10 Mathematics, Chapter 3: Pair of Linear Equations in Two Variables. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Chapter 3 Exercises step by step?
To solve problems on Chapter 3 Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 3: Pair of Linear Equations in Two Variables?
The essential formulas of Chapter 3 (Pair of Linear Equations in Two Variables) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Chapter 3 Exercises important for the Class 10 board exam?
Chapter 3 Exercises is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Chapter 3 Exercises?
Common mistakes in Chapter 3 Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Chapter 3 Exercises?
End-of-chapter NCERT exercises for Chapter 3 Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.
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