Word Problems on Linear Equations
Word problems are the real-world face of algebra. In Class 8, you move beyond simply solving equations like 3x + 5 = 20 to converting everyday situations into algebraic equations and then solving them. This skill is at the heart of mathematical reasoning.
A word problem describes a situation in plain English. Your job is to identify the unknown, assign it a variable, translate the given conditions into an equation, solve the equation, and verify the answer. The equation is always a linear equation in one variable of the form ax + b = 0.
Word problems in Class 8 NCERT cover several categories: age problems, number problems, geometry problems (perimeter, angles), money and pricing problems, and distribution problems. Each type follows a predictable pattern once you learn to spot the structure.
The most common mistake students make is misreading the problem. Always read the problem twice, underline the key information, and identify exactly what is being asked before writing any equation.
What is Word Problems on Linear Equations?
Definition: A word problem on linear equations is a real-life situation described in words that can be modelled using a linear equation in one variable. Solving the problem means forming the equation, solving for the variable, and interpreting the answer in context.
Key steps for solving any word problem:
- Read the problem carefully and identify the unknown quantity.
- Let the unknown be x (or any variable).
- Express other quantities in terms of x using the given conditions.
- Form an equation from the relationship described in the problem.
- Solve the equation to find the value of x.
- Substitute back and verify that the answer satisfies the original problem.
Common translations from English to algebra:
- "5 more than a number" → x + 5
- "7 less than a number" → x − 7
- "twice a number" → 2x
- "one-third of a number" → x/3
- "sum of two numbers is 50" → x + y = 50
- "exceeds by 10" → difference = 10
Types and Properties
Type 1: Number Problems
- These involve finding unknown numbers given relationships like sum, difference, product, or ratio.
- Example pattern: "The sum of two numbers is 84. One is 12 more than the other. Find the numbers."
- Let smaller number = x, then larger = x + 12. Equation: x + (x + 12) = 84.
Type 2: Age Problems
- These involve present ages, ages after some years, or ages some years ago.
- Key idea: If present age = x, age after n years = x + n, age n years ago = x − n.
- Example pattern: "A father is 3 times as old as his son. After 12 years, he will be twice as old."
Type 3: Geometry Problems
- These involve finding unknown dimensions using perimeter, area, or angle relationships.
- Example: "The length of a rectangle is 5 cm more than its width. Perimeter = 62 cm."
- Common formulas used: perimeter of rectangle = 2(l + b), sum of angles in triangle = 180°.
Type 4: Money and Pricing Problems
- Involves cost price, selling price, quantities, or splitting bills.
- Example: "Tickets for adults cost Rs 50 and for children Rs 30. Total 200 people paid Rs 8400."
Type 5: Consecutive Number Problems
- Consecutive integers: x, x+1, x+2
- Consecutive even/odd numbers: x, x+2, x+4
- Example: "The sum of three consecutive odd numbers is 57. Find them."
Type 6: Distribution and Fraction Problems
- Involves distributing items or finding a number when fractional parts are given.
- Example: "Half of a number added to its one-third gives 35. Find the number."
Methods
Step-by-step method for all word problems:
- Read the problem twice. Underline key words.
- Identify the unknown. Write: "Let x = ..."
- Express all other quantities in terms of x.
- Form the equation using the condition given in the problem.
- Solve the equation using transposition or balancing.
- Verify by substituting back into the original problem statement.
- Write the answer in words (not just "x = 5" but "The number is 5").
Common mistakes to avoid:
- "7 less than x" means x − 7, NOT 7 − x.
- "5 subtracted from x" means x − 5, NOT 5 − x.
- "Ratio of a to b is 3:5" means a/b = 3/5, NOT a/b = 5/3.
- Forgetting to include all conditions. If the problem gives two relationships, both must be used.
- Not verifying the answer against the original problem.
Solved Examples
Example 1: Example 1: Number Problem — Sum and Difference
Problem: The sum of two numbers is 84. If one number exceeds the other by 12, find both numbers.
Solution:
Given:
- Sum of two numbers = 84
- One number exceeds the other by 12
Step 1: Let the smaller number = x.
Then the larger number = x + 12.
Step 2: Form the equation.
x + (x + 12) = 84
Step 3: Solve.
2x + 12 = 84
2x = 84 − 12 = 72
x = 36
Step 4: Larger number = 36 + 12 = 48.
Check: 36 + 48 = 84. Difference = 48 − 36 = 12. Both conditions satisfied.
Answer: The two numbers are 36 and 48.
Example 2: Example 2: Age Problem — Present Ages
Problem: A father is 4 times as old as his son. After 6 years, the father will be 3 times as old as his son. Find their present ages.
Solution:
Given:
- Father's present age = 4 times son's age
- After 6 years: father's age = 3 times son's age
Step 1: Let son's present age = x years.
Father's present age = 4x years.
Step 2: After 6 years: son = x + 6, father = 4x + 6.
Step 3: According to the condition:
4x + 6 = 3(x + 6)
4x + 6 = 3x + 18
4x − 3x = 18 − 6
x = 12
Step 4: Son = 12 years, Father = 4 × 12 = 48 years.
Check: After 6 years: Son = 18, Father = 54. Is 54 = 3 × 18? Yes. Correct!
Answer: Son is 12 years old and father is 48 years old.
Example 3: Example 3: Geometry Problem — Rectangle Perimeter
Problem: The length of a rectangular park is 10 m more than its breadth. If the perimeter of the park is 140 m, find the dimensions.
Solution:
Given:
- Length = breadth + 10 m
- Perimeter = 140 m
Step 1: Let breadth = x m. Then length = (x + 10) m.
Step 2: Perimeter = 2(length + breadth)
140 = 2(x + 10 + x)
140 = 2(2x + 10)
Step 3: Solve.
140 = 4x + 20
4x = 140 − 20 = 120
x = 30
Step 4: Breadth = 30 m, Length = 30 + 10 = 40 m.
Check: Perimeter = 2(40 + 30) = 2(70) = 140 m. Correct!
Answer: The breadth is 30 m and the length is 40 m.
Example 4: Example 4: Consecutive Numbers
Problem: The sum of three consecutive odd numbers is 81. Find the numbers.
Solution:
Given:
- Three consecutive odd numbers
- Their sum = 81
Step 1: Let the three consecutive odd numbers be x, x + 2, and x + 4.
Step 2: Equation: x + (x + 2) + (x + 4) = 81
Step 3: Solve.
3x + 6 = 81
3x = 81 − 6 = 75
x = 25
Step 4: The three numbers: 25, 27, 29.
Check: 25 + 27 + 29 = 81. All are odd. Correct!
Answer: The three consecutive odd numbers are 25, 27, and 29.
Example 5: Example 5: Fraction Problem
Problem: Half of a number added to its one-fifth gives 42. Find the number.
Solution:
Given:
- x/2 + x/5 = 42
Step 1: Find LCM of 2 and 5. LCM = 10.
Step 2: Multiply every term by 10.
10 × (x/2) + 10 × (x/5) = 10 × 42
5x + 2x = 420
Step 3: Solve.
7x = 420
x = 60
Check: 60/2 + 60/5 = 30 + 12 = 42. Correct!
Answer: The number is 60.
Example 6: Example 6: Money Problem — Tickets
Problem: In a school function, adult tickets cost Rs 80 and student tickets cost Rs 50. If 150 tickets were sold and the total collection was Rs 9600, how many tickets of each type were sold?
Solution:
Given:
- Total tickets = 150
- Adult ticket = Rs 80, Student ticket = Rs 50
- Total collection = Rs 9600
Step 1: Let adult tickets sold = x.
Then student tickets sold = 150 − x.
Step 2: Equation: 80x + 50(150 − x) = 9600
Step 3: Solve.
80x + 7500 − 50x = 9600
30x + 7500 = 9600
30x = 9600 − 7500 = 2100
x = 70
Step 4: Adult tickets = 70, Student tickets = 150 − 70 = 80.
Check: 70 × 80 + 80 × 50 = 5600 + 4000 = 9600. Correct!
Answer: 70 adult tickets and 80 student tickets were sold.
Example 7: Example 7: Digit Problem
Problem: The digit in the tens place of a two-digit number is 3 more than the digit in the units place. The number is 4 times the sum of its digits. Find the number.
Solution:
Given:
- Tens digit = units digit + 3
- Number = 4 × (sum of digits)
Step 1: Let the units digit = x. Then tens digit = x + 3.
The number = 10(x + 3) + x = 10x + 30 + x = 11x + 30.
Step 2: Sum of digits = x + (x + 3) = 2x + 3.
Step 3: Equation: 11x + 30 = 4(2x + 3)
11x + 30 = 8x + 12
11x − 8x = 12 − 30
3x = −18
x = −6
Since a digit cannot be negative, let us re-read the problem. The tens digit is 3 more than the units digit, and the number equals 4 times the sum of digits.
Let units digit = x, tens digit = x + 3.
Number = 10(x + 3) + x = 11x + 30
Sum = 2x + 3
11x + 30 = 4(2x + 3) = 8x + 12
3x = −18 → x = −6 (invalid).
Let us try: the number is 4 times the sum of digits plus some constant. Actually, let us reconsider — perhaps the number is 6 times the sum. Let us correct the problem: The number is 6 times the sum of its digits.
11x + 30 = 6(2x + 3) = 12x + 18
30 − 18 = 12x − 11x
x = 12 (invalid, digit cannot exceed 9).
Let us use a valid digit problem instead.
Revised Problem: The tens digit of a two-digit number is twice the units digit. The number is 36 more than the number obtained by reversing the digits. Find the number.
Step 1: Let units digit = x. Tens digit = 2x.
Original number = 10(2x) + x = 21x.
Reversed number = 10x + 2x = 12x.
Step 2: 21x − 12x = 36
9x = 36
x = 4
Step 3: Units digit = 4, Tens digit = 8. Number = 84.
Check: Reversed number = 48. Difference = 84 − 48 = 36. Correct!
Answer: The number is 84.
Example 8: Example 8: Age Problem — Past Ages
Problem: Meena is 18 years older than her daughter. Five years ago, Meena was 5 times as old as her daughter. Find their present ages.
Solution:
Given:
- Meena's age = daughter's age + 18
- 5 years ago: Meena was 5 times as old as daughter
Step 1: Let daughter's present age = x years.
Meena's present age = (x + 18) years.
Step 2: Five years ago: daughter = x − 5, Meena = x + 18 − 5 = x + 13.
Step 3: Condition: x + 13 = 5(x − 5)
x + 13 = 5x − 25
13 + 25 = 5x − x
38 = 4x
x = 9.5
Since ages should be whole numbers, let us adjust: Five years ago, Meena was 4 times as old.
x + 13 = 4(x − 5)
x + 13 = 4x − 20
33 = 3x
x = 11
Step 4: Daughter = 11 years, Meena = 11 + 18 = 29 years.
Check: 5 years ago: Daughter = 6, Meena = 24. Is 24 = 4 × 6? Yes. Correct!
Answer: Daughter is 11 years old and Meena is 29 years old.
Example 9: Example 9: Triangle Angle Problem
Problem: The three angles of a triangle are in the ratio 2 : 3 : 4. Find each angle.
Solution:
Given:
- Angles are in ratio 2 : 3 : 4
- Sum of angles of a triangle = 180°
Step 1: Let the angles be 2x, 3x, and 4x.
Step 2: Equation: 2x + 3x + 4x = 180
9x = 180
x = 20
Step 3: Angles: 2 × 20 = 40°, 3 × 20 = 60°, 4 × 20 = 80°.
Check: 40 + 60 + 80 = 180°. Correct!
Answer: The three angles are 40°, 60°, and 80°.
Example 10: Example 10: Distribution Problem
Problem: A teacher distributed 90 chocolates among the students of a class. Each boy got 5 chocolates and each girl got 3 chocolates. If there are 12 more boys than girls, find the number of boys and girls.
Solution:
Given:
- Total chocolates = 90
- Each boy gets 5, each girl gets 3
- Number of boys = number of girls + 12
Step 1: Let number of girls = x.
Number of boys = x + 12.
Step 2: Equation: 5(x + 12) + 3x = 90
Step 3: Solve.
5x + 60 + 3x = 90
8x + 60 = 90
8x = 30
x = 30/8 = 3.75
Since students must be whole numbers, let us adjust: boys are 6 more than girls.
5(x + 6) + 3x = 90
5x + 30 + 3x = 90
8x = 60
x = 7.5 (still not whole).
Try: each boy gets 4, each girl gets 3. Boys = girls + 6.
4(x + 6) + 3x = 90
4x + 24 + 3x = 90
7x = 66 → not whole.
Correct the problem: 5 chocolates each for boys, 3 each for girls, boys = girls + 6, total = 96.
5(x + 6) + 3x = 96
5x + 30 + 3x = 96
8x = 66 → not whole.
Use cleaner numbers: Total = 100, each boy gets 5, each girl gets 3, boys = girls + 4.
5(x + 4) + 3x = 100
5x + 20 + 3x = 100
8x = 80
x = 10
Step 4: Girls = 10, Boys = 14.
Check: 14 × 5 + 10 × 3 = 70 + 30 = 100. Boys − Girls = 4. Correct!
Answer: There are 14 boys and 10 girls.
Real-World Applications
Real-life applications of word problems on linear equations:
- Shopping: Finding how many items you can buy within a budget when different items have different prices.
- Age calculations: Determining present, future, or past ages from given relationships — common in puzzle books and competitive exams.
- Construction and design: Finding dimensions of rooms, gardens, and plots using perimeter or area conditions.
- Salary and finance: Calculating income, expenses, savings, and investments.
- Speed and travel: Determining travel time, distance, or speed when two vehicles start from different points.
- Mixing solutions: Finding how much of each concentration to mix to get a desired concentration (used in chemistry and pharmacy).
- Sports and scoring: Finding the number of wins, losses, and draws when total points are known.
Key Points to Remember
- Always read the problem twice before attempting to form an equation.
- Clearly define the variable: write "Let x = ..." in every solution.
- Express all unknown quantities in terms of the same variable.
- Use proper translation: "less than" means subtraction, "times" means multiplication.
- "7 less than x" = x − 7, NOT 7 − x. The order matters.
- After solving, always verify by substituting back into the original problem (not just the equation).
- Write the final answer in words, not just as x = value.
- For age problems: present age = x, age after n years = x + n, age n years ago = x − n.
- For consecutive number problems: consecutive integers are x, x+1, x+2; consecutive even/odd numbers are x, x+2, x+4.
- If the answer is not a whole number and the problem requires whole numbers (like number of people), recheck the equation and the problem statement.
Practice Problems
- The sum of two numbers is 100. One number is 16 more than the other. Find the two numbers.
- A man is 5 times as old as his son. After 10 years, he will be 3 times as old as his son. Find their present ages.
- The perimeter of a rectangle is 80 cm. If the length is 4 cm more than twice its breadth, find the dimensions.
- Three consecutive even numbers add up to 126. Find the numbers.
- One-third of a number exceeds its one-fourth by 5. Find the number.
- In a class, the number of boys is 8 more than the number of girls. If the total number of students is 52, find the number of boys and girls.
- The denominator of a fraction is 5 more than its numerator. If 3 is added to both, the fraction becomes 3/4. Find the original fraction.
- A bus travels at 60 km/h. A car starts 1 hour later and travels at 80 km/h in the same direction. After how many hours will the car overtake the bus?
Frequently Asked Questions
Q1. How do I identify the variable in a word problem?
The variable represents the unknown quantity that the problem asks you to find. Look for phrases like "find the number", "what is the age", "how many", or "determine the value". Let that unknown be x.
Q2. What if I get a negative number as the answer for an age problem?
A negative age is not valid. This means either the equation was set up incorrectly or the problem statement has an error. Reread the problem, check your translations (especially "less than" vs "more than"), and verify the equation before solving again.
Q3. How do I convert 'less than' into algebra?
"a less than b" means b − a. For example, "7 less than x" means x − 7 (not 7 − x). The number after "less than" comes first in the subtraction.
Q4. What are consecutive numbers?
Consecutive numbers are numbers that follow each other in order. Consecutive integers: x, x+1, x+2. Consecutive even numbers: x, x+2, x+4 (where x is even). Consecutive odd numbers: x, x+2, x+4 (where x is odd).
Q5. How do I handle word problems with fractions?
Form the equation first using fractions (like x/2 + x/3 = 20). Then find the LCM of all denominators and multiply every term by the LCM to clear the fractions. This gives a simpler equation without fractions.
Q6. Why must I verify the answer?
Verification catches errors in forming the equation or solving it. Substitute your answer back into the original problem statement (not just the equation) to confirm all conditions are satisfied. If not, re-examine your work.
Q7. What is the difference between 'exceeds' and 'is exceeded by'?
"A exceeds B by 10" means A − B = 10, so A = B + 10. "A is exceeded by B by 10" means B − A = 10, so B = A + 10. Pay attention to which quantity is larger.
Q8. Can a word problem have no solution?
In practice, well-formed NCERT word problems always have a valid solution. However, if the conditions are contradictory (like "the sum of two positive numbers is 10 and their difference is 20"), no valid solution exists. Always check if your answer makes sense in the context.
Q9. How do I set up age problems involving 'years ago'?
If a person's present age is x, their age n years ago was x − n. Set up the equation using the condition given about ages in the past. For example, "5 years ago, A was twice as old as B" becomes (A − 5) = 2(B − 5).
Q10. What is the role of ratio in word problems?
When a problem says quantities are in ratio a : b : c, let the quantities be ax, bx, and cx (where x is the common multiplier). Use the total or another given condition to find x, then calculate each quantity.
