Exercise 3.3 – Question 2: Word Problems
📚 About This Question
Focus: Form pairs of linear equations from word problems and solve using substitution method.
Types of problems covered:
- Number problems
- Angle problems (supplementary angles)
- Cost and pricing problems
- Age problems
- Fraction problems
- Speed-distance-time problems
📐 Steps to Solve Word Problems
- Read carefully: Understand what is given and what needs to be found
- Define variables: Let x and y represent the unknowns
- Form equations: Translate word statements into mathematical equations
- Solve: Use substitution or elimination method
- Verify: Check if the solution satisfies the original problem
- Answer: Write the final answer in context of the problem
📑 Question Parts
Part (i) – Two Numbers Problem
Problem: The difference between two numbers is 26 and one number is three times the other. Find them.
Step 1: Define Variables
Let:
Larger number = \( x \)
Smaller number = \( y \)
Step 2: Form Equations
From “difference between two numbers is 26”:
\[ x – y = 26 \quad \text{…(1)} \]
From “one number is three times the other”:
\[ x = 3y \quad \text{…(2)} \]
Step 3: Solve by Substitution
Substitute equation (2) into equation (1):
\[ 3y – y = 26 \]
\[ 2y = 26 \]
\[ y = 13 \]
Find x using equation (2):
\[ x = 3y = 3(13) = 39 \]
Step 4: Verification
Check:
Difference: \( 39 – 13 = 26 \) ✓
One is three times the other: \( 39 = 3 \times 13 \) ✓
✅ Answer: The two numbers are 39 and 13
Part (ii) – Supplementary Angles
Problem: The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Step 1: Define Variables
Let:
Larger angle = \( x \)°
Smaller angle = \( y \)°
Step 2: Form Equations
From “supplementary angles” (sum = 180°):
\[ x + y = 180 \quad \text{…(1)} \]
From “larger exceeds smaller by 18°”:
\[ x – y = 18 \quad \text{…(2)} \]
Step 3: Solve by Substitution
From equation (2):
\[ x = y + 18 \quad \text{…(3)} \]
Substitute in equation (1):
\[ (y + 18) + y = 180 \]
\[ 2y + 18 = 180 \]
\[ 2y = 162 \]
\[ y = 81° \]
Find x using equation (3):
\[ x = 81 + 18 = 99° \]
Step 4: Verification
Check:
Sum: \( 99 + 81 = 180° \) ✓
Difference: \( 99 – 81 = 18° \) ✓
✅ Answer: The two angles are 99° and 81°
Part (iii) – Cost of Bats and Balls
Problem: The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
Step 1: Define Variables
Let:
Cost of one bat = Rs \( x \)
Cost of one ball = Rs \( y \)
Step 2: Form Equations
From “7 bats and 6 balls for Rs 3800”:
\[ 7x + 6y = 3800 \quad \text{…(1)} \]
From “3 bats and 5 balls for Rs 1750”:
\[ 3x + 5y = 1750 \quad \text{…(2)} \]
Step 3: Solve by Substitution
From equation (2), express x in terms of y:
\[ 3x = 1750 – 5y \]
\[ x = \frac{1750 – 5y}{3} \quad \text{…(3)} \]
Substitute in equation (1):
\[ 7 \times \frac{1750 – 5y}{3} + 6y = 3800 \]
Multiply through by 3:
\[ 7(1750 – 5y) + 18y = 11400 \] \[ 12250 – 35y + 18y = 11400 \] \[ 12250 – 17y = 11400 \] \[ -17y = -850 \] \[ y = 50 \]
Find x using equation (3):
\[ x = \frac{1750 – 5(50)}{3} = \frac{1750 – 250}{3} = \frac{1500}{3} = 500 \]
Step 4: Verification
Check in equation (1): \( 7x + 6y = 3800 \)
LHS = \( 7(500) + 6(50) = 3500 + 300 = 3800 \) = RHS ✓
Check in equation (2): \( 3x + 5y = 1750 \)LHS = \( 3(500) + 5(50) = 1500 + 250 = 1750 \) = RHS ✓
✅ Answer: Cost of one bat = Rs 500, Cost of one ball = Rs 50
Part (iv) – Taxi Charges
Problem: The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Step 1: Define Variables
Let:
Fixed charge = Rs \( x \)
Charge per km = Rs \( y \)
Step 2: Form Equations
From “10 km costs Rs 105”:
\[ x + 10y = 105 \quad \text{…(1)} \]
From “15 km costs Rs 155”:
\[ x + 15y = 155 \quad \text{…(2)} \]
Step 3: Solve by Substitution
From equation (1):
\[ x = 105 – 10y \quad \text{…(3)} \]
Substitute in equation (2):
\[ (105 – 10y) + 15y = 155 \]
\[ 105 + 5y = 155 \]
\[ 5y = 50 \]
\[ y = 10 \]
Find x using equation (3):
\[ x = 105 – 10(10) = 105 – 100 = 5 \]
Step 4: Calculate charge for 25 km
Total charge for 25 km:
\[ \text{Charge} = x + 25y \]
\[ = 5 + 25(10) \]
\[ = 5 + 250 = 255 \text{ Rs} \]
Step 5: Verification
Check for 10 km:
\( 5 + 10(10) = 5 + 100 = 105 \) ✓
Check for 15 km:\( 5 + 15(10) = 5 + 150 = 155 \) ✓
✅ Answer:
Fixed charge = Rs 5
Charge per km = Rs 10
Charge for 25 km = Rs 255
Fixed charge = Rs 5
Charge per km = Rs 10
Charge for 25 km = Rs 255
Part (v) – Father and Son’s Age
Problem: A fraction becomes \( \frac{9}{11} \) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \( \frac{5}{6} \). Find the fraction.
Step 1: Define Variables
Let:
Numerator = \( x \)
Denominator = \( y \)
Original fraction = \( \frac{x}{y} \)
Step 2: Form Equations
From “adding 2 to both gives \( \frac{9}{11} \)”:
\[ \frac{x + 2}{y + 2} = \frac{9}{11} \]
Cross multiply:
\[ 11(x + 2) = 9(y + 2) \] \[ 11x + 22 = 9y + 18 \] \[ 11x – 9y = -4 \quad \text{…(1)} \]
From “adding 3 to both gives \( \frac{5}{6} \)”:
\[ \frac{x + 3}{y + 3} = \frac{5}{6} \]
Cross multiply:
\[ 6(x + 3) = 5(y + 3) \] \[ 6x + 18 = 5y + 15 \] \[ 6x – 5y = -3 \quad \text{…(2)} \]Step 3: Solve by Substitution
From equation (2):
\[ 6x = 5y – 3 \]
\[ x = \frac{5y – 3}{6} \quad \text{…(3)} \]
Substitute in equation (1):
\[ 11 \times \frac{5y – 3}{6} – 9y = -4 \]
Multiply through by 6:
\[ 11(5y – 3) – 54y = -24 \] \[ 55y – 33 – 54y = -24 \] \[ y – 33 = -24 \] \[ y = 9 \]
Find x using equation (3):
\[ x = \frac{5(9) – 3}{6} = \frac{45 – 3}{6} = \frac{42}{6} = 7 \]
Step 4: Verification
Original fraction: \( \frac{7}{9} \)
Check condition 1: \( \frac{7 + 2}{9 + 2} = \frac{9}{11} \) ✓
Check condition 2: \( \frac{7 + 3}{9 + 3} = \frac{10}{12} = \frac{5}{6} \) ✓
✅ Answer: The fraction is \( \frac{7}{9} \)
Part (vi) – Age Problem (Alternative)
Problem: Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Step 1: Define Variables
Let:
Present age of Jacob = \( x \) years
Present age of his son = \( y \) years
Step 2: Form Equations
From “five years hence, Jacob’s age = 3 × son’s age”:
After 5 years: Jacob = \( x + 5 \), Son = \( y + 5 \)
\[ x + 5 = 3(y + 5) \] \[ x + 5 = 3y + 15 \] \[ x – 3y = 10 \quad \text{…(1)} \]
From “five years ago, Jacob’s age = 7 × son’s age”:
5 years ago: Jacob = \( x – 5 \), Son = \( y – 5 \)
\[ x – 5 = 7(y – 5) \] \[ x – 5 = 7y – 35 \] \[ x – 7y = -30 \quad \text{…(2)} \]Step 3: Solve by Substitution
From equation (1):
\[ x = 3y + 10 \quad \text{…(3)} \]
Substitute in equation (2):
\[ (3y + 10) – 7y = -30 \]
\[ -4y + 10 = -30 \]
\[ -4y = -40 \]
\[ y = 10 \]
Find x using equation (3):
\[ x = 3(10) + 10 = 30 + 10 = 40 \]
Step 4: Verification
Check condition 1 (5 years hence):
Jacob: 40 + 5 = 45, Son: 10 + 5 = 15
\( 45 = 3 \times 15 \) ✓
Check condition 2 (5 years ago):Jacob: 40 – 5 = 35, Son: 10 – 5 = 5
\( 35 = 7 \times 5 \) ✓
✅ Answer: Present age of Jacob = 40 years, Present age of son = 10 years
Part (vii) – Speed of Car and Bus
Problem: A place A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Step 1: Define Variables
Let:
Speed of car from A = \( x \) km/h
Speed of car from B = \( y \) km/h
Assume \( x > y \)
Step 2: Form Equations
Case 1: Same direction (they meet in 5 hours)
The faster car catches up with the slower car.
Relative speed = \( x – y \)
Distance covered = 100 km
\[ 5(x – y) = 100 \] \[ x – y = 20 \quad \text{…(1)} \]
Case 2: Opposite directions (they meet in 1 hour)
They approach each other.
Combined speed = \( x + y \)
Distance covered = 100 km
\[ 1(x + y) = 100 \] \[ x + y = 100 \quad \text{…(2)} \]Step 3: Solve by Substitution
From equation (1):
\[ x = y + 20 \quad \text{…(3)} \]
Substitute in equation (2):
\[ (y + 20) + y = 100 \]
\[ 2y + 20 = 100 \]
\[ 2y = 80 \]
\[ y = 40 \]
Find x using equation (3):
\[ x = 40 + 20 = 60 \]
Step 4: Visual Representation
📊 Speed-Distance Diagram
Diagram showing two scenarios: same direction and opposite directions
Step 5: Verification
Check same direction:
In 5 hours, car A travels: \( 60 \times 5 = 300 \) km
In 5 hours, car B travels: \( 40 \times 5 = 200 \) km
Difference: \( 300 – 200 = 100 \) km ✓
Check opposite directions:In 1 hour, car A travels: \( 60 \times 1 = 60 \) km
In 1 hour, car B travels: \( 40 \times 1 = 40 \) km
Total: \( 60 + 40 = 100 \) km ✓
✅ Answer: Speed of car from A = 60 km/h, Speed of car from B = 40 km/h
📊 Summary of All Solutions
| Part | Problem Type | Solution |
|---|---|---|
| (i) | Two numbers | 39 and 13 |
| (ii) | Supplementary angles | 99° and 81° |
| (iii) | Cost problem | Bat: Rs 500, Ball: Rs 50 |
| (iv) | Taxi charges | Fixed: Rs 5, Per km: Rs 10 |
| (v) | Fraction problem | \( \frac{7}{9} \) |
| (vi) | Age problem | Jacob: 40 years, Son: 10 years |
| (vii) | Speed problem | 60 km/h and 40 km/h |
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Not reading the problem carefully and missing key information.
✅ Correct: Read the problem at least twice, underline key information.
✅ Correct: Read the problem at least twice, underline key information.
❌ Mistake 2: Defining variables incorrectly or inconsistently.
✅ Correct: Clearly state what each variable represents with proper units.
✅ Correct: Clearly state what each variable represents with proper units.
❌ Mistake 3: Translating word statements into wrong equations.
✅ Correct: Break down each sentence and form equations step by step.
✅ Correct: Break down each sentence and form equations step by step.
❌ Mistake 4: Not verifying the solution in the context of the problem.
✅ Correct: Always check if the solution makes sense in the real-world context.
✅ Correct: Always check if the solution makes sense in the real-world context.
💡 Key Points to Remember
- Read carefully: Understand what is given and what needs to be found
- Define clearly: State variables with proper units
- Common problem types:
- Number problems: difference, sum, ratio
- Angle problems: supplementary (180°), complementary (90°)
- Cost problems: total cost = (quantity × price)
- Age problems: present, past, future ages
- Fraction problems: numerator and denominator
- Speed problems: distance = speed × time
- Key formulas:
- Supplementary angles: \( x + y = 180° \)
- Distance = Speed × Time
- Relative speed (same direction): \( x – y \)
- Combined speed (opposite): \( x + y \)
- Always verify: Check solution in original problem context
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.