Quantitative Aptitude · Equations

Word Problems on Quadratic Equations for CA Foundation

Updated 1 October 2026 · Fact-checked

A quadratic word problem describes a situation where an unknown quantity leads to an equation of the form ax² + bx + c = 0. You define the unknown, translate each sentence into algebra, form the equation, solve it, then reject any root that does not fit the situation.

Understand Word Problems on Quadratic Equations

A word problem is a quadratic equation hidden in a story. The story gives you facts. Your job is to turn those facts into algebra.

A quadratic appears when the unknown is multiplied by itself or by something that also contains the unknown. Typical cases: product of two numbers, area of a rectangle (length × breadth), distance = speed × time where time depends on speed, and age problems with a product of ages.

The process is always the same. Pick a variable for what you want. Express every other quantity using that variable. Find the sentence that links them, usually a product, a sum of squares or a difference of times. Write it as an equation and solve.

A quadratic gives two roots. The story decides which one is valid. Length, speed, time and age cannot be negative. The number of items must be a whole number. So always check the roots against the situation.

In the MCQ paper you also have the four options. You can often test them in the story instead of solving, which saves time.

Key formulas to remember

Standard form: ax² + bx + c = 0, a ≠ 0; Bring every term to one side before solving.
Quadratic formula: x = [−b ± √(b² − 4ac)] ÷ 2a; Works for every quadratic. Use it when factorising is not quick.
Sum and product of roots: α + β = −b ÷ a; αβ = c ÷ a; Useful for checking your roots quickly.
Consecutive numbers: x, x + 1, x + 2 (integers); x, x + 2, x + 4 (consecutive even or odd); Choose the form that matches the wording.
Area of rectangle: Area = length × breadth; Perimeter = 2 (length + breadth); Perimeter gives a sum, area gives a product.
Speed, distance, time: Time = Distance ÷ Speed; Equating a time difference creates the quadratic.
Pythagoras theorem: hypotenuse² = base² + perpendicular²; Used in right-angled triangle problems.

How to solve Word Problems on Quadratic Equations questions

Use this method for any quadratic word problem. Do not skip the last step.

1Read the question fully and identify exactly what is asked.
2Let the unknown be x. Choose the quantity that makes other quantities simple to write.
3Express all other quantities in terms of x using the given facts.
4Find the sentence that links them (product, sum of squares, time difference) and write the equation.
5Simplify to the form ax² + bx + c = 0 and solve by factorisation or the formula.
6Reject roots that are negative, zero or fractional where that is impossible.
7Compute what the question actually asks for, which may not be x itself.
8Verify by putting the answer back into the story.

Quickest way: Option testing with the story

When to use it: Use when the four options are small whole numbers and the equation would take time to build or solve.

Read the question and note the key condition, for example product 360 or area 48.
Test the options one by one in the condition. Start with the middle value.
Check the sum or product first. It is faster than solving.
If one option fits, mark it and move on.
If two options seem to fit, apply the second condition from the story, such as positivity.
If no option fits within about 30 seconds, build the equation instead.
Skip the question if it needs a long setup. A wrong answer costs 0.25 marks.

Common mistakes in Word Problems on Quadratic Equations

Keeping both roots as the answer
Students solve the equation and stop, forgetting that the story limits the value.
Fix: Reject negative lengths, speeds, ages and fractional counts before choosing the option.
Giving x as the answer when the question asks for something else
Students are relieved to finish solving and stop reading.
Fix: Underline what is asked, such as the larger number or the perimeter, and compute that last.
Wrong form for consecutive even or odd numbers
Students use x + 1 for all consecutive numbers.
Fix: Use x and x + 2 for consecutive even or odd numbers. Use x and x + 1 for consecutive integers.
Sign errors when moving terms
Time-difference equations have several terms and fractions.
Fix: Multiply through by the common denominator first, then collect terms carefully.
Mixing units in speed problems
Time is given in minutes and speed in km per hour.
Fix: Convert everything to hours before forming the equation.
Forming the equation from the wrong sentence
Students rush and link two facts that do not belong together.
Fix: Write each fact as a short equation. Choose the one that uses the product or the square.

Worked examples

Example 1

The product of two consecutive positive integers is 342. What is the larger integer? (a) 17 (b) 18 (c) 19 (d) 20

Show the solution

Let the integers be x and x + 1.
Then x (x + 1) = 342, so x² + x − 342 = 0.
Find two numbers with product −342 and sum 1: 19 and −18.
So x² + 19x − 18x − 342 = 0, which gives (x + 19)(x − 18) = 0.
x = 18 or x = −19. Reject −19 since the integers are positive.
The integers are 18 and 19. The larger is 19.
Check: 18 × 19 = 342.

Answer: (c) 19

Example 2

The length of a rectangle is 5 m more than its breadth. Its area is 84 m². What is its perimeter? (a) 26 m (b) 38 m (c) 34 m (d) 40 m

Show the solution

Let the breadth be x m. Then the length is x + 5 m.
Area: x (x + 5) = 84, so x² + 5x − 84 = 0.
Find two numbers with product −84 and sum 5: 12 and −7.
(x + 12)(x − 7) = 0, so x = 7 or x = −12.
Reject −12. Breadth = 7 m and length = 12 m.
Perimeter = 2 (12 + 7) = 38 m.
Check: 12 × 7 = 84.

Answer: (b) 38 m

Example 3

A train covers 240 km at a uniform speed. If the speed were 20 km/h more, the journey would take 2 hours less. What is the original speed? (a) 40 km/h (b) 50 km/h (c) 60 km/h (d) 80 km/h

Show the solution

Let the original speed be x km/h.
Original time = 240 ÷ x. New time = 240 ÷ (x + 20).
240 ÷ x − 240 ÷ (x + 20) = 2.
Multiply by x (x + 20): 240 (x + 20) − 240x = 2x (x + 20).
4800 = 2x² + 40x, so x² + 20x − 2400 = 0.
Find two numbers with product −2400 and sum 20: 60 and −40.
(x + 60)(x − 40) = 0, so x = 40 or x = −60. Reject −60.
Check: 240 ÷ 40 = 6 hours. 240 ÷ 60 = 4 hours. Difference = 2 hours.

Answer: (a) 40 km/h

Exam tips

Check the options before building a long equation. Small whole-number options are often quicker to test.
Read the last line twice. Many wrong answers come from giving x instead of what is asked.
In speed problems, write the time difference equation and clear fractions at once to avoid errors.
Always reject negative roots for length, speed, age and counts. Examiners place the negative root in the options as a trap.
If the setup is lengthy and you are unsure, skip it. Negative marking is 0.25 per wrong answer.

Practice questions from Equations

Word Problems on Quadratic Equations: frequently asked questions

How do I form a quadratic equation from a word problem?

Let the unknown be x and write the other quantities in terms of x. Find the sentence that gives a product, a square or a time difference and set it equal to the given value. Then bring all terms to one side.

Why do I get two answers and which one do I choose?

A quadratic has two roots, but the story may allow only one. Reject any root that makes length, speed, time or age negative, or makes a count a fraction.

Which types of word problems are asked in CA Foundation?

Common types are numbers, consecutive integers, areas of rectangles, speed and time, and ages. Other applied situations may also appear, so practise the method rather than memorising types.

Is factorisation or the formula better in the exam?

Try factorisation first because it is faster when the numbers are small. Use the quadratic formula if you cannot find the factors within about 20 seconds.