- Chapter: 3 — Pair of Linear Equations in Two Variables
- Exercise: 3.6 — Equations Reducible to a Pair of Linear Equations
- Total Questions: 2 (Q1 has 8 sub-parts; Q2 has 3 word problems)
- Core Technique: Substitute \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \) (or similar) to convert non-linear equations to linear form
- Key Problem Types: Speed (downstream/upstream), Work-rate, Distance-time
- CBSE Weightage: Chapter 3 carries 6 marks in the board exam; word problems are high-frequency
- Syllabus Status: Fully included in CBSE 2026-27 syllabus
- Prerequisite: Substitution method, elimination method (Ex 3.3, 3.4, 3.5)
The ncert solutions for class 10 maths chapter 3 ex 3 6 on this page are fully updated for the 2026-27 CBSE board exam. Exercise 3.6 belongs to Chapter 3 — Pair of Linear Equations in Two Variables — from the NCERT official textbook for Class 10 Mathematics. This exercise focuses on a powerful technique: reducing equations that look complicated into a simple pair of linear equations by substituting new variables. You can explore all chapters through our NCERT Solutions for Class 10 hub.
For the complete collection of all subjects and classes, visit our main NCERT Solutions page. This page covers every question in Exercise 3.6 with full working, LaTeX-rendered steps, and exam-focused explanations — exactly what you need to score well in your 2026-27 board exam.
Table of Contents
- Quick Revision Box
- Chapter Overview — Pair of Linear Equations in Two Variables
- Key Concepts — Equations Reducible to Linear Form
- NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.6 — Step-by-Step
- Formula Reference Table — Chapter 3
- Solved Examples Beyond NCERT
- Important Questions for CBSE Board Exam
- Common Mistakes Students Make in Exercise 3.6
- Exam Tips for 2026-27 CBSE Board
- Frequently Asked Questions
Chapter Overview — Pair of Linear Equations in Two Variables (Class 10 Maths)
Chapter 3 of NCERT Class 10 Mathematics deals with Pair of Linear Equations in Two Variables. It covers graphical and algebraic methods — substitution, elimination, and cross-multiplication — and ends with equations that can be reduced to linear form, which is exactly what Exercise 3.6 tests.
In CBSE board exams, Chapter 3 typically carries 6 marks and appears in both the 2-mark and 3-mark sections. Word problems (like the ones in Q2 of Ex 3.6) are frequently asked in board papers and require you to first frame the equations before solving them.
| Detail | Information |
|---|---|
| Chapter | 3 — Pair of Linear Equations in Two Variables |
| Exercise | 3.6 |
| Textbook | NCERT Mathematics — Class 10 |
| Class | 10 |
| Subject | Mathematics |
| CBSE Marks Weightage | ~6 marks (Chapter 3 overall) |
| Difficulty Level | Medium to Hard |
| Academic Year | 2026-27 |
Key Concepts — Equations Reducible to Linear Form
Core Idea: Some equations are not linear in \( x \) and \( y \) but become linear when you substitute \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \) (or other suitable substitutions). After solving for \( p \) and \( q \), you reverse-substitute to find \( x \) and \( y \).
Standard Substitution Pattern: If an equation contains \( \frac{1}{x} \) and \( \frac{1}{y} \), let \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \). The equation becomes linear in \( p \) and \( q \).
Speed-Stream Concept (Downstream/Upstream):
- Downstream speed = speed of boat in still water + speed of current = \( x + y \)
- Upstream speed = speed of boat in still water − speed of current = \( x – y \)
- Use \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \) to frame equations
Work-Rate Concept: If a person completes the work in \( n \) days, their one-day work = \( \frac{1}{n} \). Combine fractions for multiple workers.
Distance-Time Concept: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). When total distance is split between two modes of transport, form two equations from two different journey scenarios.
NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.6 — Step-by-Step (2026-27)
Question 1 — Solve Equations by Reducing to a Pair of Linear Equations
Question 1
Hard
Solve the following pairs of equations by reducing them to a pair of linear equations:
Step 1: Let \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \). Rewrite the equations:
\[ \frac{p}{2} + \frac{q}{3} = 2 \quad \Rightarrow \quad 3p + 2q = 12 \quad \text{…(1)} \]
\[ \frac{p}{3} + \frac{q}{2} = \frac{13}{6} \quad \Rightarrow \quad 2p + 3q = 13 \quad \text{…(2)} \]
Step 2: Multiply (1) by 3 and (2) by 2:
\[ 9p + 6q = 36 \quad \text{…(3)} \]
\[ 4p + 6q = 26 \quad \text{…(4)} \]
Step 3: Subtract (4) from (3):
\[ 5p = 10 \Rightarrow p = 2 \]
Step 4: Substitute \( p = 2 \) into (1): \( 3(2) + 2q = 12 \Rightarrow 2q = 6 \Rightarrow q = 3 \)
Step 5: Reverse-substitute: \( p = \frac{1}{x} = 2 \Rightarrow x = \frac{1}{2} \); \( q = \frac{1}{y} = 3 \Rightarrow y = \frac{1}{3} \)
\( \therefore \) \( x = \frac{1}{2} \), \( y = \frac{1}{3} \)
Step 1: Let \( p = \frac{1}{\sqrt{x}} \) and \( q = \frac{1}{\sqrt{y}} \). The equations become:
\[ 2p + 3q = 2 \quad \text{…(1)} \]
\[ 4p – 9q = -1 \quad \text{…(2)} \]
Step 2: Multiply (1) by 3: \( 6p + 9q = 6 \quad \text{…(3)} \)
Step 3: Add (2) and (3): \( 10p = 5 \Rightarrow p = \frac{1}{2} \)
Step 4: Substitute into (1): \( 2 \times \frac{1}{2} + 3q = 2 \Rightarrow 3q = 1 \Rightarrow q = \frac{1}{3} \)
Step 5: \( p = \frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4 \); \( q = \frac{1}{\sqrt{y}} = \frac{1}{3} \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9 \)
\( \therefore \) \( x = 4 \), \( y = 9 \)
Step 1: Let \( p = \frac{1}{x} \). The equations become:
\[ 4p + 3y = 14 \quad \text{…(1)} \]
\[ 3p – 4y = 23 \quad \text{…(2)} \]
Step 2: Multiply (1) by 4 and (2) by 3:
\[ 16p + 12y = 56 \quad \text{…(3)} \]
\[ 9p – 12y = 69 \quad \text{…(4)} \]
Step 3: Add (3) and (4): \( 25p = 125 \Rightarrow p = 5 \)
Step 4: Substitute into (1): \( 4(5) + 3y = 14 \Rightarrow 3y = -6 \Rightarrow y = -2 \)
Step 5: \( p = \frac{1}{x} = 5 \Rightarrow x = \frac{1}{5} \)
\( \therefore \) \( x = \frac{1}{5} \), \( y = -2 \)
Step 1: Let \( p = \frac{1}{x-1} \) and \( q = \frac{1}{y-2} \):
\[ 5p + q = 2 \quad \text{…(1)} \]
\[ 6p – 3q = 1 \quad \text{…(2)} \]
Step 2: Multiply (1) by 3: \( 15p + 3q = 6 \quad \text{…(3)} \)
Step 3: Add (2) and (3): \( 21p = 7 \Rightarrow p = \frac{1}{3} \)
Step 4: Substitute into (1): \( 5 \times \frac{1}{3} + q = 2 \Rightarrow q = 2 – \frac{5}{3} = \frac{1}{3} \)
Step 5: \( p = \frac{1}{x-1} = \frac{1}{3} \Rightarrow x – 1 = 3 \Rightarrow x = 4 \); \( q = \frac{1}{y-2} = \frac{1}{3} \Rightarrow y – 2 = 3 \Rightarrow y = 5 \)
\( \therefore \) \( x = 4 \), \( y = 5 \)
Step 1: Rewrite each equation by splitting the fraction:
\[ \frac{7}{y} – \frac{2}{x} = 5 \quad \text{…(1)} \]
\[ \frac{8}{y} + \frac{7}{x} = 15 \quad \text{…(2)} \]
Step 2: Let \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \):
\[ 7q – 2p = 5 \quad \text{…(3)} \]
\[ 8q + 7p = 15 \quad \text{…(4)} \]
Step 3: Multiply (3) by 7 and (4) by 2:
\[ 49q – 14p = 35 \quad \text{…(5)} \]
\[ 16q + 14p = 30 \quad \text{…(6)} \]
Step 4: Add (5) and (6): \( 65q = 65 \Rightarrow q = 1 \)
Step 5: Substitute into (3): \( 7(1) – 2p = 5 \Rightarrow 2p = 2 \Rightarrow p = 1 \)
Step 6: \( p = \frac{1}{x} = 1 \Rightarrow x = 1 \); \( q = \frac{1}{y} = 1 \Rightarrow y = 1 \)
\( \therefore \) \( x = 1 \), \( y = 1 \)
Step 1: Divide both equations by \( xy \) (assuming \( x \neq 0, y \neq 0 \)):
\[ \frac{6}{y} + \frac{3}{x} = 6 \quad \text{…(1)} \]
\[ \frac{2}{y} + \frac{4}{x} = 5 \quad \text{…(2)} \]
\[ 3p + 6q = 6 \Rightarrow p + 2q = 2 \quad \text{…(3)} \]
\[ 4p + 2q = 5 \quad \text{…(4)} \]
Step 3: Subtract (3) from (4): \( 3p = 3 \Rightarrow p = 1 \)
Step 4: Substitute into (3): \( 1 + 2q = 2 \Rightarrow q = \frac{1}{2} \)
Step 5: \( x = \frac{1}{p} = 1 \); \( y = \frac{1}{q} = 2 \)
\( \therefore \) \( x = 1 \), \( y = 2 \)
Step 1: Let \( p = \frac{1}{x+y} \) and \( q = \frac{1}{x-y} \):
\[ 10p + 2q = 4 \Rightarrow 5p + q = 2 \quad \text{…(1)} \]
\[ 15p – 5q = -2 \quad \text{…(2)} \]
Step 2: Multiply (1) by 5: \( 25p + 5q = 10 \quad \text{…(3)} \)
Step 3: Add (2) and (3): \( 40p = 8 \Rightarrow p = \frac{1}{5} \)
Step 4: Substitute into (1): \( 5 \times \frac{1}{5} + q = 2 \Rightarrow q = 1 \)
Step 5: \( p = \frac{1}{x+y} = \frac{1}{5} \Rightarrow x + y = 5 \quad \text{…(4)} \)
\( q = \frac{1}{x-y} = 1 \Rightarrow x – y = 1 \quad \text{…(5)} \)
Step 6: Add (4) and (5): \( 2x = 6 \Rightarrow x = 3 \); from (5): \( y = 3 – 1 = 2 \)
\( \therefore \) \( x = 3 \), \( y = 2 \)
Step 1: Let \( p = \frac{1}{3x+y} \) and \( q = \frac{1}{3x-y} \):
\[ p + q = \frac{3}{4} \quad \text{…(1)} \]
\[ \frac{p}{2} – \frac{q}{2} = \frac{-1}{8} \Rightarrow p – q = \frac{-1}{4} \quad \text{…(2)} \]
Step 2: Add (1) and (2): \( 2p = \frac{3}{4} – \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \Rightarrow p = \frac{1}{4} \)
Step 3: From (1): \( q = \frac{3}{4} – \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
Step 4: \( \frac{1}{3x+y} = \frac{1}{4} \Rightarrow 3x + y = 4 \quad \text{…(3)} \)
\( \frac{1}{3x-y} = \frac{1}{2} \Rightarrow 3x – y = 2 \quad \text{…(4)} \)
Step 5: Add (3) and (4): \( 6x = 6 \Rightarrow x = 1 \); from (3): \( y = 4 – 3 = 1 \)
Question 2 — Word Problems as Pairs of Linear Equations
Question 2
Hard
Formulate the following problems as a pair of linear equations and hence find their solutions:
Key Concept: Let speed of Ritu in still water = \( x \) km/h and speed of current = \( y \) km/h.
Downstream speed = \( x + y \) km/h; Upstream speed = \( x – y \) km/h.
Step 1: Frame the equations using Speed = Distance ÷ Time:
\[ x + y = \frac{20}{2} = 10 \quad \text{…(1)} \]
\[ x – y = \frac{4}{2} = 2 \quad \text{…(2)} \]
Step 2: Add (1) and (2):
\[ 2x = 12 \Rightarrow x = 6 \]
Step 3: Substitute into (1): \( 6 + y = 10 \Rightarrow y = 4 \)
Verification: Downstream: \( 6 + 4 = 10 \) km/h, distance = \( 10 \times 2 = 20 \) km ✓; Upstream: \( 6 – 4 = 2 \) km/h, distance = \( 2 \times 2 = 4 \) km ✓
\( \therefore \) Speed of Ritu in still water = 6 km/h; Speed of current = 4 km/h
Key Concept: Let 1 woman alone finish the work in \( x \) days and 1 man alone in \( y \) days.
One-day work of 1 woman = \( \frac{1}{x} \); One-day work of 1 man = \( \frac{1}{y} \).
Step 1: Frame the equations:
2 women and 5 men finish in 4 days:
\[ 4\left(\frac{2}{x} + \frac{5}{y}\right) = 1 \Rightarrow \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \quad \text{…(1)} \]
3 women and 6 men finish in 3 days:
\[ 3\left(\frac{3}{x} + \frac{6}{y}\right) = 1 \Rightarrow \frac{3}{x} + \frac{6}{y} = \frac{1}{3} \quad \text{…(2)} \]
\[ 2p + 5q = \frac{1}{4} \quad \text{…(3)} \]
\[ 3p + 6q = \frac{1}{3} \quad \text{…(4)} \]
Step 3: Multiply (3) by 6 and (4) by 5:
\[ 12p + 30q = \frac{6}{4} = \frac{3}{2} \quad \text{…(5)} \]
\[ 15p + 30q = \frac{5}{3} \quad \text{…(6)} \]
Step 4: Subtract (5) from (6):
\[ 3p = \frac{5}{3} – \frac{3}{2} = \frac{10}{6} – \frac{9}{6} = \frac{1}{6} \Rightarrow p = \frac{1}{18} \]
Step 5: Substitute into (3):
\[ 2 \times \frac{1}{18} + 5q = \frac{1}{4} \Rightarrow \frac{1}{9} + 5q = \frac{1}{4} \Rightarrow 5q = \frac{1}{4} – \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36} \Rightarrow q = \frac{1}{36} \]
Step 6: \( p = \frac{1}{x} = \frac{1}{18} \Rightarrow x = 18 \); \( q = \frac{1}{y} = \frac{1}{36} \Rightarrow y = 36 \)
Verification: \( 4\left(\frac{2}{18} + \frac{5}{36}\right) = 4\left(\frac{4}{36} + \frac{5}{36}\right) = 4 \times \frac{9}{36} = 4 \times \frac{1}{4} = 1 \) ✓
\( \therefore \) 1 woman alone takes 18 days; 1 man alone takes 36 days
Key Concept: Let speed of train = \( x \) km/h and speed of bus = \( y \) km/h.
Time = Distance ÷ Speed. Total distance = 300 km.
Step 1: Frame Equation 1 (60 km by train, 240 km by bus, total time = 4 hours):
\[ \frac{60}{x} + \frac{240}{y} = 4 \quad \text{…(1)} \]
Step 2: Frame Equation 2 (100 km by train, 200 km by bus, total time = 4 hours 10 minutes = \( \frac{25}{6} \) hours):
\[ \frac{100}{x} + \frac{200}{y} = \frac{25}{6} \quad \text{…(2)} \]
Step 3: Let \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \):
\[ 60p + 240q = 4 \Rightarrow 15p + 60q = 1 \quad \text{…(3)} \]
\[ 100p + 200q = \frac{25}{6} \Rightarrow 600p + 1200q = 25 \Rightarrow 24p + 48q = 1 \quad \text{…(4)} \]
Why divide? Dividing by common factors simplifies the arithmetic.
Step 4: Multiply (3) by 4 and (4) by 5:
\[ 60p + 240q = 4 \quad \text{…(5)} \]
\[ 120p + 240q = 5 \quad \text{…(6)} \]
Step 5: Subtract (5) from (6): \( 60p = 1 \Rightarrow p = \frac{1}{60} \)
Step 6: Substitute into (3): \( 15 \times \frac{1}{60} + 60q = 1 \Rightarrow \frac{1}{4} + 60q = 1 \Rightarrow 60q = \frac{3}{4} \Rightarrow q = \frac{1}{80} \)
Step 7: \( x = \frac{1}{p} = 60 \) km/h; \( y = \frac{1}{q} = 80 \) km/h
Verification: \( \frac{60}{60} + \frac{240}{80} = 1 + 3 = 4 \) hours ✓; \( \frac{100}{60} + \frac{200}{80} = \frac{5}{3} + \frac{5}{2} = \frac{10+15}{6} = \frac{25}{6} \) hours ✓
\( \therefore \) Speed of train = 60 km/h; Speed of bus = 80 km/h
Formula Reference Table — Chapter 3 Pair of Linear Equations
| Formula Name | Formula (LaTeX) | Variables Defined |
|---|---|---|
| Downstream Speed | \( v_d = x + y \) | x = boat speed in still water, y = current speed |
| Upstream Speed | \( v_u = x – y \) | x = boat speed in still water, y = current speed |
| Speed–Distance–Time | \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \) | Standard formula |
| One-day Work | \( W = \frac{1}{n} \) | n = number of days to complete work alone |
| Combined Work Rate | \( \frac{1}{x} + \frac{1}{y} = \frac{1}{d} \) | x, y = individual days; d = days together |
| Reciprocal Substitution | \( p = \frac{1}{x},\ q = \frac{1}{y} \) | Used to linearise non-linear equations |
| 10 minutes in hours | \( 10 \text{ min} = \frac{10}{60} = \frac{1}{6} \text{ hour} \) | Unit conversion for time problems |
Solved Examples Beyond NCERT — Equations Reducible to Linear Form
Extra Example 1
Medium
Solve: \( \frac{3}{x+1} – \frac{1}{y-1} = 0 \) and \( \frac{4}{x+1} + \frac{3}{y-1} = 5 \)
Step 1: Let \( p = \frac{1}{x+1} \) and \( q = \frac{1}{y-1} \):
\[ 3p – q = 0 \Rightarrow q = 3p \quad \text{…(1)} \]
\[ 4p + 3q = 5 \quad \text{…(2)} \]
Step 2: Substitute (1) into (2): \( 4p + 9p = 5 \Rightarrow 13p = 5 \Rightarrow p = \frac{5}{13} \), \( q = \frac{15}{13} \)
Step 3: \( x + 1 = \frac{13}{5} \Rightarrow x = \frac{8}{5} \); \( y – 1 = \frac{13}{15} \Rightarrow y = \frac{28}{15} \)
\( \therefore \) \( x = \frac{8}{5} \), \( y = \frac{28}{15} \)
Extra Example 2 — Speed Problem
Medium
A boat goes 30 km upstream and 44 km downstream in 10 hours. It also goes 40 km upstream and 55 km downstream in 13 hours. Find the speed of the stream and speed of the boat in still water.
Step 1: Let boat speed in still water = \( x \), stream speed = \( y \). Let \( p = \frac{1}{x-y} \), \( q = \frac{1}{x+y} \).
\[ 30p + 44q = 10 \quad \text{…(1)} \]
\[ 40p + 55q = 13 \quad \text{…(2)} \]
Step 2: Multiply (1) by 4 and (2) by 3: \( 120p + 176q = 40 \) and \( 120p + 165q = 39 \). Subtract: \( 11q = 1 \Rightarrow q = \frac{1}{11} \)
Step 3: Substitute into (1): \( 30p + 4 = 10 \Rightarrow 30p = 6 \Rightarrow p = \frac{1}{5} \)
Step 4: \( x – y = 5 \) and \( x + y = 11 \). Solving: \( x = 8 \), \( y = 3 \).
\( \therefore \) Boat speed = 8 km/h; Stream speed = 3 km/h
Important Questions for CBSE Board Exam — Class 10 Maths Chapter 3
- If \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \), what does the equation \( \frac{2}{x} + \frac{3}{y} = 5 \) become? Answer: \( 2p + 3q = 5 \)
- Define downstream speed in terms of boat speed \( x \) and current speed \( y \). Answer: \( x + y \)
- If a woman can complete a work in \( x \) days, what is her one-day work? Answer: \( \frac{1}{x} \)
- Solve: \( \frac{2}{x} + \frac{3}{y} = 13 \) and \( \frac{5}{x} – \frac{4}{y} = -2 \). Let \( p = \frac{1}{x}, q = \frac{1}{y} \): \( 2p + 3q = 13 \), \( 5p – 4q = -2 \). Multiply first by 4 and second by 3: \( 8p + 12q = 52 \), \( 15p – 12q = -6 \). Add: \( 23p = 46 \Rightarrow p = 2 \Rightarrow x = \frac{1}{2} \). Then \( q = 3 \Rightarrow y = \frac{1}{3} \). Answer: \( x = \frac{1}{2}, y = \frac{1}{3} \)
- A boat takes 3 hours to go 24 km downstream and 2 hours to go 8 km upstream. Find the speed of the boat and the stream. Answer: Downstream speed = 8 km/h, upstream speed = 4 km/h; boat speed = 6 km/h, stream = 2 km/h.
A train covers a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And if the train were slower by 6 km/h, it would have taken 6 more hours. Find the distance covered by the train.
Solution Hint: Let speed = \( x \) km/h and time = \( t \) hours. Distance \( d = xt \). Set up: \( (x+6)(t-4) = xt \) and \( (x-6)(t+6) = xt \). Expand and simplify to get two linear equations in \( x \) and \( t \). Solve to find \( x = 30 \) km/h, \( t = 24 \) hours, \( d = 720 \) km.
Common Mistakes Students Make in Exercise 3.6
Mistake 1: Forgetting to reverse-substitute after finding \( p \) and \( q \).
Why it’s wrong: \( p \) and \( q \) are not the final answers; \( x \) and \( y \) are. You lose marks if you stop at \( p \) and \( q \).
Correct approach: Always write \( x = \frac{1}{p} \) and \( y = \frac{1}{q} \) as the final step.
Mistake 2: Converting 10 minutes as \( \frac{10}{100} \) hours instead of \( \frac{10}{60} = \frac{1}{6} \) hours.
Why it’s wrong: Time must be in consistent units (hours). Using \( \frac{1}{10} \) gives wrong equations.
Correct approach: Always convert minutes to hours by dividing by 60.
Mistake 3: Writing upstream speed as \( y – x \) instead of \( x – y \).
Why it’s wrong: If \( x > y \), then \( y – x \) is negative, which makes no physical sense for speed.
Correct approach: Upstream speed = boat speed − current speed = \( x – y \) (always positive when \( x > y \)).
Mistake 4: In work problems, writing the combined work equation as \( x + y = \frac{1}{4} \) without the fractions \( \frac{1}{x} \) and \( \frac{1}{y} \).
Why it’s wrong: \( x \) is the number of days, not the rate. Rate = \( \frac{1}{x} \).
Correct approach: Write \( \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \) (rate × time = 1 complete work).
Mistake 5: Not verifying the answer after solving.
Why it’s wrong: CBSE marking schemes often award 1 mark for verification. Skipping it costs you marks.
Correct approach: Substitute your \( x \) and \( y \) back into both original equations and confirm both sides are equal.
Exam Tips for 2026-27 CBSE Board — Chapter 3 Exercise 3.6
- Define variables first: In word problems, always start by writing “Let speed of boat = \( x \) km/h” etc. The CBSE 2026-27 marking scheme awards 1 mark for correct variable definition.
- Show substitution explicitly: Write “Let \( p = \frac{1}{x} \) and \( q = \frac{1}{y} \)” as a separate line. Examiners check this step.
- Unit conversion is a trap: The Roohi problem uses “10 minutes longer” — always convert to \( \frac{1}{6} \) hour before writing the equation.
- Chapter 3 weightage: Pair of Linear Equations carries approximately 6 marks in the CBSE Class 10 board exam. Exercise 3.6 word problems are among the most commonly repeated question types.
- Verification earns marks: In long-answer questions, always verify your solution — it is explicitly mentioned in CBSE answer keys and earns a dedicated mark.
- Last-minute checklist: (1) Substitution step shown ✓ (2) Linear equations formed correctly ✓ (3) Solved using elimination/substitution ✓ (4) Reverse-substituted to find original variables ✓ (5) Verified the answer ✓
For related exercises, also see NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.5 and NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.4 which cover the elimination and cross-multiplication methods that build up to this exercise.