Question 4 – Boat and Stream Problem
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Word Problem: On comparing the ratios \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find out whether the following pair of linear equations are consistent, or inconsistent:
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
📚 Understanding Boat and Stream Problems
Key Concepts:
- Speed in still water (x): The speed of the boat when there is no current
- Speed of stream (y): The speed of water current
- Upstream speed: Speed against the current = (x – y) km/h
- Downstream speed: Speed with the current = (x + y) km/h
📐 Important Formulas
Time = Distance ÷ Speed
- Time upstream = \( \frac{\text{Distance upstream}}{\text{Speed upstream}} = \frac{d_1}{x – y} \)
- Time downstream = \( \frac{\text{Distance downstream}}{\text{Speed downstream}} = \frac{d_2}{x + y} \)
Step 1: Understanding the Problem
Given Information:
Journey 1:
- Upstream distance = 30 km
- Downstream distance = 44 km
- Total time = 10 hours
Journey 2:
- Upstream distance = 40 km
- Downstream distance = 55 km
- Total time = 13 hours
To Find:
- Speed of boat in still water (x)
- Speed of stream (y)
Step 2: Define Variables
- \( x \) = Speed of boat in still water (km/h)
- \( y \) = Speed of stream (km/h)
Then:
- Upstream speed = \( (x – y) \) km/h
- Downstream speed = \( (x + y) \) km/h
Step 3: Form Linear Equations
Time upstream + Time downstream = Total time
\[ \frac{30}{x – y} + \frac{44}{x + y} = 10 \quad \text{…(1)} \]Time upstream + Time downstream = Total time
\[ \frac{40}{x – y} + \frac{55}{x + y} = 13 \quad \text{…(2)} \]Step 4: Simplify Using Substitution
Let \( u = \frac{1}{x – y} \) and \( v = \frac{1}{x + y} \)
Then equation (1) becomes:
\[ 30u + 44v = 10 \quad \text{…(1′)} \]And equation (2) becomes:
\[ 40u + 55v = 13 \quad \text{…(2′)} \]Step 5: Solve for u and v (Elimination Method)
Equation (1′) × 4:
\[ 120u + 176v = 40 \quad \text{…(3)} \]Equation (2′) × 3:
\[ 120u + 165v = 39 \quad \text{…(4)} \]Step 6: Find x and y
\( u = \frac{1}{x – y} = \frac{1}{5} \)
Therefore: \( x – y = 5 \quad \text{…(5)} \)
\( v = \frac{1}{x + y} = \frac{1}{11} \)
Therefore: \( x + y = 11 \quad \text{…(6)} \)
Step 7: Visual Graph Representation
📊 Graphical Solution
Graph showing intersection at (8, 3) – Boat speed = 8 km/h, Stream speed = 3 km/h
- Blue Line: \( x - y = 5 \) (Difference between boat and stream speed)
- Red Line: \( x + y = 11 \) (Sum of boat and stream speed)
- Intersection Point: (8, 3)
- Solution: Boat speed = 8 km/h, Stream speed = 3 km/h
Step 8: Verification
Upstream speed = \( x - y = 8 - 3 = 5 \) km/h
Downstream speed = \( x + y = 8 + 3 = 11 \) km/h
Time upstream = \( \frac{30}{5} = 6 \) hours
Time downstream = \( \frac{44}{11} = 4 \) hours
Total time = 6 + 4 = 10 hours ✓
Upstream speed = 5 km/h
Downstream speed = 11 km/h
Time upstream = \( \frac{40}{5} = 8 \) hours
Time downstream = \( \frac{55}{11} = 5 \) hours
Total time = 8 + 5 = 13 hours ✓
Speed of boat in still water = 8 km/h
Speed of stream = 3 km/h
Bonus: Comparing Ratios (Parts i, ii, iii)
(i) 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0
Here: \( a_1 = 5, b_1 = -4, c_1 = 8 \) and \( a_2 = 7, b_2 = 6, c_2 = -9 \)
\( \frac{a_1}{a_2} = \frac{5}{7} \), \( \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Result: Lines intersect at a point (Unique solution)
(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
Here: \( a_1 = 9, b_1 = 3, c_1 = 12 \) and \( a_2 = 18, b_2 = 6, c_2 = 24 \)
\( \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Result: Lines are coincident (Infinitely many solutions)
(iii) 6x – 3y + 10 = 0 and 2x – y + 9 = 0
Here: \( a_1 = 6, b_1 = -3, c_1 = 10 \) and \( a_2 = 2, b_2 = -1, c_2 = 9 \)
\( \frac{a_1}{a_2} = \frac{6}{2} = 3 \), \( \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \), \( \frac{c_1}{c_2} = \frac{10}{9} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Result: Lines are parallel (No solution)
⚠️ Common Mistakes to Avoid
✅ Correct: Upstream = (x - y), Downstream = (x + y). Remember: upstream is AGAINST the current.
✅ Correct: Time upstream + Time downstream = Total time (not speeds).
✅ Correct: After finding u and v, convert back to x and y using the original substitution.
✅ Correct: Always check your answer satisfies both given conditions.
💡 Key Points to Remember
Boat and Stream Formulas:
- Upstream speed: (Boat speed - Stream speed) = x - y
- Downstream speed: (Boat speed + Stream speed) = x + y
- Time = Distance ÷ Speed
- Substitution trick: Use u = 1/(x-y) and v = 1/(x+y) to simplify
- Finding x and y: Add and subtract the two equations
- Always verify: Check solution in both given journeys
Comparing Ratios:
- If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) → Lines intersect (Unique solution)
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) → Lines are parallel (No solution)
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) → Lines are coincident (Infinite solutions)
📝 Practice Similar Problems
Try These Problems:
- A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream in 6.5 hours. Find the speed of boat and stream.
- A man can row 18 km downstream and 12 km upstream in 3 hours. He can row 30 km downstream and 20 km upstream in 5 hours. Find his speed in still water and speed of stream.
- A motorboat goes 16 km upstream and 24 km downstream in 6 hours. It goes 12 km upstream and 36 km downstream in 7 hours. Find the speeds.
📚 Related Questions
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.