Word Problems on Quadratic Equations
Word problems on quadratic equations form a major application-based topic in Chapter 4 of the NCERT Class 10 Mathematics syllabus.
These problems require students to:
- Read and understand the given situation.
- Translate the word problem into a quadratic equation.
- Solve the equation by factorisation, completing the square, or using the quadratic formula.
- Interpret the roots and reject any root that does not satisfy the physical constraints of the problem.
Word problems test both algebraic skill and logical reasoning. Common categories include number problems, age problems, geometry problems, speed-distance-time problems, and consecutive number problems.
What is Word Problems on Quadratic Equations — Types, Methods & Solved Examples?
Definition: A word problem on quadratic equations is a real-life or mathematical situation described in words, whose solution requires forming and solving a quadratic equation of the form ax² + bx + c = 0.
General approach:
- Read the problem carefully. Identify the unknown quantity.
- Let the unknown be x (or any suitable variable).
- Translate the conditions into mathematical expressions.
- Form the quadratic equation.
- Solve using factorisation, quadratic formula, or completing the square.
- Check both roots against the constraints of the problem (e.g., age cannot be negative, length must be positive).
- State the answer in words.
Important:
- A quadratic equation gives two roots. One or both may be valid depending on the context.
- Reject roots that are negative when the quantity must be positive (age, length, count).
- Reject fractional roots when the quantity must be a whole number (number of people, items).
Word Problems on Quadratic Equations Formula
Key formulas used in word problems:
Standard form: ax² + bx + c = 0
Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a
Discriminant: D = b² − 4ac
Common relationships used to form equations:
| Problem Type | Typical Setup |
|---|---|
| Consecutive numbers | n, n+1, n+2 or n, n+2 (even/odd) |
| Product of numbers | x × y = k where y = f(x) |
| Age problems | Present age x, age t years ago = x − t |
| Area/perimeter | l × b = Area, 2(l + b) = Perimeter |
| Speed-distance-time | Time = Distance / Speed |
| Fractions | Numerator = x, denominator = x + k |
Derivation and Proof
Step-by-step method for solving word problems:
- Read the problem twice. Underline the key information.
- Assign variables: Let x represent the unknown quantity.
- Express other quantities in terms of x using the given relationships.
- Form the equation: Use the given condition to write a quadratic equation.
- Simplify the equation to standard form ax² + bx + c = 0.
- Solve by the most convenient method:
- Factorisation — if the discriminant is a perfect square.
- Quadratic formula — for any quadratic.
- Completing the square — sometimes asked specifically.
- Interpret: Check each root against the problem's constraints.
- State the answer in a complete sentence.
Tips:
- If the problem says "a number and its reciprocal," let the number be x and its reciprocal be 1/x.
- If the problem involves consecutive integers, use x, x + 1, x + 2.
- If the problem involves consecutive even or odd integers, use x, x + 2, x + 4.
- Always re-read the problem after finding the answer to check it makes sense.
Types and Properties
Common types of word problems:
Type 1: Number problems
- Find two numbers whose sum/difference is given and whose product is given.
- Example: Find two numbers whose sum is 27 and product is 182.
Type 2: Consecutive integers
- Product of consecutive integers, sum of their squares, etc.
- Example: The product of two consecutive positive integers is 306. Find them.
Type 3: Age problems
- Relating ages at different times leads to a quadratic.
- Example: The product of Shyam's age 5 years ago and his age 3 years later is 105.
Type 4: Geometry problems
- Area, perimeter, diagonal — forming quadratics from dimensions.
- Example: A rectangular field has area 528 m² and perimeter 92 m. Find dimensions.
Type 5: Speed, distance, time
- Relation between speed, distance and time leads to quadratic.
- Example: A train covers 480 km. If it were 8 km/h faster, it would take 3 hours less.
Type 6: Fraction problems
- Example: If 2 is added to numerator and denominator, the fraction becomes 1/2. If 3 is subtracted, it becomes 1/3.
Type 7: Work and time
- Example: Two pipes can fill a tank together in 12 hours. One pipe takes 10 hours more than the other.
Methods
Choosing the solving method:
- Factorisation: Use when the equation can be easily factored (D is a perfect square).
- Quadratic formula: Use when factorisation is difficult or when the roots are irrational.
- Completing the square: Use when specifically asked or when the equation is of a convenient form.
Common mistakes in word problems:
- Incorrect variable assignment — not defining what x represents clearly.
- Wrong equation formation — misinterpreting "exceeds by" as subtraction instead of addition.
- Accepting invalid roots — using negative age or negative length without checking.
- Forgetting to answer in words — the final answer must address the original question.
Language clues in word problems:
| Phrase | Mathematical Meaning |
|---|---|
| "exceeds by k" | difference = k |
| "product is k" | x × y = k |
| "k years ago" | age − k |
| "k years hence" | age + k |
| "takes k hours less" | time₂ = time₁ − k |
| "twice as old" | 2 × age |
Solved Examples
Example 1: Number Problem — Sum and Product
Problem: Find two numbers whose sum is 27 and whose product is 182.
Solution:
Given:
- Sum of two numbers = 27
- Product of two numbers = 182
Steps:
- Let one number = x. Then the other = 27 − x.
- Product: x(27 − x) = 182
- 27x − x² = 182
- x² − 27x + 182 = 0
- Find m, n: m + n = −27, m × n = 182 → m = −13, n = −14
- (x − 13)(x − 14) = 0
- x = 13 or x = 14
Verification: 13 + 14 = 27 ✓ and 13 × 14 = 182 ✓
Answer: The numbers are 13 and 14.
Example 2: Consecutive Positive Integers
Problem: The product of two consecutive positive integers is 306. Find the integers.
Solution:
Given:
- Product of two consecutive positive integers = 306
Steps:
- Let the integers be x and x + 1.
- x(x + 1) = 306
- x² + x − 306 = 0
- Find m, n: m + n = 1, m × n = −306 → m = 18, n = −17
- (x + 18)(x − 17) = 0
- x = −18 or x = 17
- Since the integers are positive, x = 17.
Verification: 17 × 18 = 306 ✓
Answer: The integers are 17 and 18.
Example 3: Age Problem
Problem: The product of Shyam's age five years ago and his age three years from now is 105. Find his present age.
Solution:
Given:
- Product of (age 5 years ago) and (age 3 years later) = 105
Steps:
- Let Shyam's present age = x years.
- (x − 5)(x + 3) = 105
- x² + 3x − 5x − 15 = 105
- x² − 2x − 120 = 0
- Find m, n: m + n = −2, m × n = −120 → m = −12, n = 10
- (x − 12)(x + 10) = 0
- x = 12 or x = −10
- Age cannot be negative, so x = 12.
Verification: (12 − 5)(12 + 3) = 7 × 15 = 105 ✓
Answer: Shyam's present age is 12 years.
Example 4: Geometry — Rectangle Dimensions
Problem: The length of a rectangular field is 3 m more than its breadth. If the area is 180 m², find the dimensions.
Solution:
Given:
- Length = breadth + 3
- Area = 180 m²
Steps:
- Let breadth = x m. Then length = (x + 3) m.
- Area: x(x + 3) = 180
- x² + 3x − 180 = 0
- Find m, n: m + n = 3, m × n = −180 → m = 15, n = −12
- (x + 15)(x − 12) = 0
- x = −15 or x = 12
- Breadth cannot be negative, so x = 12.
Verification: 12 × 15 = 180 ✓
Answer: Breadth = 12 m, Length = 15 m.
Example 5: Speed-Distance-Time
Problem: A train covers 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. Find the speed of the train.
Solution:
Given:
- Distance = 480 km
- Reduced speed takes 3 hours more
Steps:
- Let the speed = x km/h. Time = 480/x hours.
- At speed (x − 8): time = 480/(x − 8) hours.
- 480/(x − 8) − 480/x = 3
- 480[x − (x − 8)] / [x(x − 8)] = 3
- 480 × 8 / [x(x − 8)] = 3
- 3840 = 3x(x − 8)
- 3x² − 24x − 3840 = 0
- x² − 8x − 1280 = 0
- Find m, n: m + n = −8, m × n = −1280 → m = −40, n = 32
- (x − 40)(x + 32) = 0
- x = 40 or x = −32
- Speed cannot be negative, so x = 40.
Verification: Time at 40 km/h = 12 h. Time at 32 km/h = 15 h. Difference = 3 h ✓
Answer: Speed of the train = 40 km/h.
Example 6: Fraction Problem
Problem: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58/21, find the fraction.
Solution:
Given:
- Denominator = 2 × numerator + 1
- Fraction + reciprocal = 58/21
Steps:
- Let numerator = x. Denominator = 2x + 1.
- Fraction = x/(2x + 1). Reciprocal = (2x + 1)/x.
- x/(2x + 1) + (2x + 1)/x = 58/21
- [x² + (2x + 1)²] / [x(2x + 1)] = 58/21
- [x² + 4x² + 4x + 1] / [2x² + x] = 58/21
- [5x² + 4x + 1] / [2x² + x] = 58/21
- 21(5x² + 4x + 1) = 58(2x² + x)
- 105x² + 84x + 21 = 116x² + 58x
- 11x² − 26x − 21 = 0
- Find m, n: m + n = −26, m × n = 11 × (−21) = −231 → m = −33, n = 7
- 11x² − 33x + 7x − 21 = 0
- 11x(x − 3) + 7(x − 3) = 0
- (11x + 7)(x − 3) = 0
- x = −7/11 or x = 3
- Numerator should be a positive integer, so x = 3.
Verification: Fraction = 3/7. 3/7 + 7/3 = (9 + 49)/21 = 58/21 ✓
Answer: The fraction is 3/7.
Example 7: Work and Time
Problem: Two pipes together fill a tank in 6 hours. The larger pipe fills it 5 hours faster than the smaller pipe alone. Find the time taken by each pipe individually.
Solution:
Given:
- Combined time = 6 hours
- Larger pipe takes 5 hours less than the smaller pipe
Steps:
- Let the smaller pipe fill the tank in x hours.
- Larger pipe fills it in (x − 5) hours.
- Rate of smaller = 1/x, rate of larger = 1/(x − 5).
- Combined rate: 1/x + 1/(x − 5) = 1/6
- [(x − 5) + x] / [x(x − 5)] = 1/6
- (2x − 5) / (x² − 5x) = 1/6
- 6(2x − 5) = x² − 5x
- 12x − 30 = x² − 5x
- x² − 17x + 30 = 0
- (x − 15)(x − 2) = 0
- x = 15 or x = 2
- If x = 2, then larger pipe takes 2 − 5 = −3 hours (impossible).
- So x = 15.
Verification: Rates: 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6 ✓
Answer: Smaller pipe: 15 hours, larger pipe: 10 hours.
Example 8: Number Problem — Squares
Problem: The difference of the squares of two positive numbers is 45. The square of the smaller number is 4 times the larger number. Find the numbers.
Solution:
Given:
- Let the larger number = x, smaller number = y.
- x² − y² = 45 ... (i)
- y² = 4x ... (ii)
Steps:
- From (ii): y² = 4x. Substitute into (i).
- x² − 4x = 45
- x² − 4x − 45 = 0
- (x − 9)(x + 5) = 0
- x = 9 or x = −5
- Since x is positive, x = 9.
- y² = 4(9) = 36, so y = 6 (positive).
Verification: 81 − 36 = 45 ✓ and 36 = 4 × 9 ✓
Answer: The numbers are 9 and 6.
Example 9: Geometry — Right Triangle
Problem: The hypotenuse of a right triangle is 13 cm. If one of the other sides exceeds the remaining side by 7 cm, find the sides.
Solution:
Given:
- Hypotenuse = 13 cm
- One side exceeds the other by 7 cm
Steps:
- Let one side = x cm. Then the other = (x + 7) cm.
- By Pythagoras: x² + (x + 7)² = 13²
- x² + x² + 14x + 49 = 169
- 2x² + 14x − 120 = 0
- x² + 7x − 60 = 0
- (x + 12)(x − 5) = 0
- x = −12 or x = 5
- Side cannot be negative, so x = 5.
Verification: Sides = 5, 12, 13. 25 + 144 = 169 = 13² ✓
Answer: The sides are 5 cm, 12 cm, and 13 cm.
Example 10: Consecutive Even Integers
Problem: The sum of the squares of two consecutive even positive integers is 340. Find the integers.
Solution:
Given:
- Sum of squares of two consecutive even positive integers = 340
Steps:
- Let the integers be x and x + 2 (consecutive even).
- x² + (x + 2)² = 340
- x² + x² + 4x + 4 = 340
- 2x² + 4x − 336 = 0
- x² + 2x − 168 = 0
- (x + 14)(x − 12) = 0
- x = −14 or x = 12
- Since the integers are positive, x = 12.
Verification: 12² + 14² = 144 + 196 = 340 ✓
Answer: The integers are 12 and 14.
Real-World Applications
Real-life applications of quadratic word problems:
- Construction — calculating dimensions of rooms, gardens, or plots given area constraints.
- Projectile motion — height of a ball at time t: h = ut − (1/2)gt². Finding when it hits the ground.
- Business — profit = revenue − cost, where revenue or cost may be quadratic in the quantity produced.
- Finance — compound interest over 2 years leads to quadratic equations in the rate.
- Logistics — scheduling and allocation problems where constraints form quadratics.
- Sports — trajectory of a ball, optimal angle of projection.
- Agriculture — fencing a rectangular plot with given material to maximise area.
Key Points to Remember
- Word problems require translating English to algebra — this is the hardest part.
- Always define the variable clearly before forming the equation.
- A quadratic equation gives two roots — check both against the problem's constraints.
- Reject negative roots when the quantity must be positive (age, length, speed, count).
- Reject fractional roots when the quantity must be a whole number.
- Common problem types: numbers, ages, geometry, speed-time, fractions, work-time.
- For consecutive integers: use x, x+1. For consecutive even/odd: use x, x+2.
- Speed-time problems use: Time = Distance/Speed. Differences in time give the equation.
- Always verify the answer by substituting back into the original word problem.
- In CBSE exams, word problems carry 4–5 marks and require complete working with a final answer statement.
Practice Problems
- The sum of a number and its reciprocal is 10/3. Find the number.
- The product of two consecutive positive odd numbers is 483. Find the numbers.
- The length of a rectangular garden is 4 m more than its width. If the area is 192 m², find the dimensions.
- A motorboat takes 1 hour more to go 24 km upstream than to return downstream. If the speed of the stream is 2 km/h, find the speed of the motorboat in still water.
- The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 124. Find their present ages.
- A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange. Find the number.
- Two water taps together fill a tank in 75/8 hours. The larger tap takes 10 hours less than the smaller tap to fill it separately. Find the time each tap takes.
- A rectangular field has perimeter 82 m and area 400 m². Find its dimensions.
Frequently Asked Questions
Q1. How do I identify a word problem as quadratic?
If the problem involves products of unknowns, squares, or relationships that lead to a degree-2 equation, it is quadratic. Clues include phrases like 'product of two numbers,' 'square of a number,' 'exceeds by,' or situations involving area.
Q2. What if both roots are valid?
Some problems have two valid answers. For example, 'find two numbers whose sum is 10 and product is 21' gives x = 3 and x = 7, both valid. State both solutions.
Q3. When should I reject a root?
Reject a root when it violates the physical constraint: negative age, negative length, negative speed, or a fraction when a whole number is needed. Always state why you reject a root.
Q4. Which method should I use — factorisation or quadratic formula?
Try factorisation first. If the discriminant is not a perfect square, use the quadratic formula. In CBSE exams, the method may be specified — follow the instruction.
Q5. How do I set up speed-distance-time word problems?
Use Time = Distance/Speed. If the speed changes by k, the new time is Distance/(Speed ± k). The difference in times gives the equation.
Q6. What are common mistakes in word problems?
Common mistakes: (i) incorrect translation of words to equations, (ii) forgetting to check roots against constraints, (iii) not stating the answer in words, (iv) sign errors when expanding brackets.
Q7. Are word problems important for CBSE board exams?
Yes. Word problems on quadratic equations are regularly asked in CBSE board exams, carrying 4–5 marks. They test both equation-forming ability and solving skill.
Q8. How do I handle 'consecutive even' or 'consecutive odd' numbers?
For consecutive even integers, use x, x+2, x+4. For consecutive odd integers, use the same pattern: x, x+2, x+4 (where x is odd). The gap between consecutive even or odd numbers is always 2.
Related Topics
- Quadratic Equations
- Solving Quadratic Equations by Factorisation
- Quadratic Formula
- Nature of Roots of Quadratic Equation
- Standard Form of Quadratic Equation
- Factorisation of Quadratic Expressions
- Completing the Square Method
- Discriminant of Quadratic Equation
- Sum and Product of Roots
- Applications of Quadratic Equations
- Roots and Graphs of Quadratic Equations
