MathsICSE Class 10 · Quadratic Equations in One Variable

Word Problems on Quadratic Equations: Forming, Solving and Checking (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

To solve a word problem with a quadratic equation, let x stand for one unknown, write the other quantities in terms of x, form an equation from the condition given, reduce it to ax² + bx + c = 0, solve it, and keep only the root that makes sense (positive lengths, speeds and ages).

Follow the arrows in order, and always finish by rejecting roots that make no sense, such as a negative speed, before writing the answer.

Illustration drawn for QuestionBuddy.

Understand Word Problems on Quadratic Equations

The CISCE syllabus ends the quadratic equations section with 'solving simple quadratic equation problems'. These questions test whether you can turn words into an equation; the solving itself is usually by factorisation.

The five-step method.

1. Choose the unknown. Let $x$ be the quantity asked for, or the one that makes the other quantities easiest to write. State its units ('let the speed be $x$ km/h').
2. **Express everything else in terms of $x$ .** 'Four more than', 'twice', 'the reciprocal', 'the other side' all become expressions.
3. Form the equation from the one condition that has not yet been used.
4. Simplify to standard form $a x^{2} + b x + c = 0$ and solve (factorise; use the formula if it does not factorise).
5. Interpret. Reject roots that are impossible in the context (a negative speed, a fractional number of people, a speed lower than the reduction), and answer exactly what was asked.

Common types and their equations.

Numbers: two consecutive integers $x$ and $x + 1$ ; consecutive odd or even numbers $x$ and $x + 2$ ; a number and its reciprocal $x + \frac{1}{x}$ .
Ages: write present ages first; 'five years ago' subtracts 5 from both ages.
Geometry: area of a rectangle $= l \times b$ ; Pythagoras for right triangles.
Speed, distance and time: time $= \frac{distance}{speed}$ . If the speed falls, the time rises, so the slower time minus the faster time equals the extra time. In 2025 Q8(iii) (350 km, speed reduced by 20 km/h, time increased by 2 hours), the examiners reported candidates who could not decide which time was larger, and who found the reduced speed but not the time that was asked.
Streams: upstream speed $= u - v$ , downstream speed $= u + v$ .
Work: if a pipe fills a tank in $x$ hours, it fills $\frac{1}{x}$ of it each hour; rates add.
Price and quantity: number of items $= \frac{total cost}{price of one}$ .

Clearing fractions. Equations like $\frac{480}{x} - \frac{480}{x + 8} = 2$ are cleared by multiplying by $x (x + 8)$ . Before you factorise, divide through by any common factor to keep the numbers small.

Word Problems on Quadratic Equations: formulas to remember

Time from distance and speed: $t = \frac{d}{v}$; Slower speed ⇒ longer time.
Extra time when speed is reduced by r: $\frac{d}{x - r} - \frac{d}{x} = extra time$; Bigger fraction first.
Upstream and downstream: $v_{up} = u - v, v_{down} = u + v$; $u$ = speed in still water, $v$ = stream speed.
Combined rate of work: $\frac{1}{x} + \frac{1}{y} = \frac{1}{T}$; Two pipes or workers finishing together in time $T$ .
Consecutive numbers: $x, x + 1 (integers); x, x + 2 (odd or even)$; Consecutive odd or even numbers differ by 2.
Number of items: $n = \frac{total cost}{price of one item}$; Price–quantity problems.

How to solve Word Problems on Quadratic Equations questions

Read the question twice: once for the story, once for the numbers.

1Write 'Let … be $x$ (units)'.
2Write every other quantity in terms of $x$ in a short list or table.
3Form the equation from the remaining condition; for time problems, put the larger time first.
4Clear fractions, simplify to $a x^{2} + b x + c = 0$ , and divide by any common factor.
5Solve by factorisation (or the formula).
6Reject roots that do not fit the context, saying why.
7Answer the question actually asked, with units, and check against the original words.

Word Problems on Quadratic Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Area problem

Suppose a rectangular vegetable patch is 2 m longer than it is wide and covers 63 m².

Find its length and width.

Try it first, then show the solutionSolution

$x (x + 2) = 63 \Rightarrow x^{2} + 2 x - 63 = 0$ .
$(x + 9) (x - 7) = 0 \Rightarrow x = 7$ (width cannot be $- 9$ ).

Answer: Width 7 m, length 9 m.

Memory hook: Lengths are positive: the negative root walks away.

Speed and time

Imagine your family drives 240 km to visit relatives. On the way back they drive 20 km/h faster and take 1 hour less.

Find the speed on the way there.

Try it first, then show the solutionSolution

$\frac{240}{x} - \frac{240}{x + 20} = 1$ .
$240 \times 20 = x (x + 20) \Rightarrow x^{2} + 20 x - 4800 = 0$ .
$(x + 80) (x - 60) = 0 \Rightarrow x = 60$ .

Answer: 60 km/h going (4 hours), 80 km/h returning (3 hours).

Memory hook: Slower trip takes longer, so its time goes first in the subtraction.

Tricks to remember Word Problems on Quadratic Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Let–Link–Equation–Solve–Sense
L-L-E-S-S: Let $x$ , Link the others to $x$ , form the Equation, Solve, make Sense of the roots.
Use it when: Every word problem.
Slow minus fast
In time problems: (time at the slower speed) − (time at the faster speed) = the time difference.
Use it when: Avoiding the sign error in 2025 Q8(iii).
Ages move together
Years pass for everyone: '5 years ago' takes 5 off each person's age.
Use it when: Age problems.

Word Problems on Quadratic Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Consecutive integers · 1 mark · MCQBoard level

The product of two consecutive positive integers is 132. The integers are:
(a) 10 and 11
(b) 11 and 12
(c) 12 and 13
(d) 13 and 14

Try it first, then show the solutionSolution

$x (x + 1) = 132 \Rightarrow x^{2} + x - 132 = 0 \Rightarrow (x + 12) (x - 11) = 0$ .
Positive: $x = 11$ .

Answer: (b) 11 and 12

Example 2: A number and its reciprocal · 3 marks · Short answerBoard level

The sum of a number and its reciprocal is $\frac{13}{6}$ . Find the number.

Try it first, then show the solutionSolution

$x + \frac{1}{x} = \frac{13}{6} \Rightarrow 6 x^{2} + 6 = 13 x$ .
$6 x^{2} - 13 x + 6 = 0 \Rightarrow (2 x - 3) (3 x - 2) = 0$ .
$x = \frac{3}{2}$ or $x = \frac{2}{3}$ .

Answer: $\frac{3}{2}$ or $\frac{2}{3}$

Example 3: Speed and time · 4 marks · Long answerHarder board level

A train travels 480 km at a uniform speed. If its speed were 8 km/h more, it would take 2 hours less. Find its speed and the time taken.

Try it first, then show the solutionSolution

Let the speed be $x$ km/h: $\frac{480}{x} - \frac{480}{x + 8} = 2$ .
$\frac{480 \times 8}{x ( x + 8 )} = 2 \Rightarrow x^{2} + 8 x = 1920$ .
$x^{2} + 8 x - 1920 = 0 \Rightarrow (x + 48) (x - 40) = 0$ ; $x = 40$ (speed is positive).
Time $= \frac{480}{40} = 12$ hours.

Answer: 40 km/h; 12 hours.

Example 4: Ages · 4 marks · Long answerHarder board level

A mother is 26 years older than her son. Three years from now, the product of their ages will be 360. Find their present ages.

Try it first, then show the solutionSolution

Son $x$ , mother $x + 26$ ; in 3 years: $(x + 3) (x + 29) = 360$ .
$x^{2} + 32 x + 87 = 360 \Rightarrow x^{2} + 32 x - 273 = 0$ .
$(x + 39) (x - 7) = 0 \Rightarrow x = 7$ .
Mother: 33. Check: $10 \times 36 = 360$ ✓.

Answer: Son 7 years, mother 33 years.

Example 5: Two pipes · 4 marks · Long answerHOTS / competency

Two pipes together fill a tank in 6 hours. The larger pipe alone takes 5 hours less than the smaller pipe alone. Find the time each takes alone.

Try it first, then show the solutionSolution

Smaller: $x$ h, larger: $x - 5$ h. $\frac{1}{x} + \frac{1}{x - 5} = \frac{1}{6}$ .
$6 (2 x - 5) = x (x - 5) \Rightarrow x^{2} - 17 x + 30 = 0$ .
$(x - 15) (x - 2) = 0$ ; $x = 2$ is rejected since $x - 5$ would be negative.
$x = 15$ : smaller pipe 15 h, larger pipe 10 h. Check: $\frac{1}{15} + \frac{1}{10} = \frac{1}{6}$ ✓.

Answer: 15 hours and 10 hours.

Example 6: Case study: buying pens · 4 marks · Case-basedHOTS / competency

A shopkeeper buys a number of pens for ₹600. If each pen had cost ₹5 less, he would have got 10 more pens for the same money.
(a) Taking the cost of one pen as ₹ $x$ , form an equation.
(b) Solve it to find the cost of one pen.
(c) How many pens did he buy?

Try it first, then show the solutionSolution

(a) $\frac{600}{x - 5} - \frac{600}{x} = 10$ .
(b) $\frac{3000}{x ( x - 5 )} = 10 \Rightarrow x^{2} - 5 x - 300 = 0 \Rightarrow (x - 20) (x + 15) = 0$ ; $x = 20$ .
(c) $\frac{600}{20} = 30$ pens. Check: at ₹15 he would get 40 pens, 10 more ✓.

Answer: (a) $\frac{600}{x - 5} - \frac{600}{x} = 10$ (b) ₹20 (c) 30 pens

Where students lose marks in Word Problems on Quadratic Equations

The slips examiners see most often, and what to write instead.

Mistake: Subtracting the times the wrong way round
The slower journey takes longer, so its time is larger.
Wrong working
$\frac{350}{x} - \frac{350}{x - 20} = 2$ .
Right working
$\frac{350}{x - 20} - \frac{350}{x} = 2$ .
Do this: Decide which time is longer before writing the equation.
Usually costs: Examiners named this for 2025 Q8(iii).
Mistake: Factorisation errors with large constants
Big numbers make factor pairs harder to find.
Wrong working
$x^{2} - 20 x - 3500 = (x - 50) (x + 70)$ .
Right working
$x^{2} - 20 x - 3500 = (x - 70) (x + 50)$ .
Do this: Check the middle term: $- 70 + 50 = - 20$ ✓.
Usually costs: Reported for 2025 Q8(iii).
Mistake: Not rejecting an impossible root
A root can satisfy the equation but not the situation.
Wrong working
Smaller pipe 2 hours, larger pipe −3 hours.
Right working
$x = 2$ is rejected; $x = 15$ .
Do this: Test each root in every expression of the problem.
Usually costs: The final mark.
Mistake: Answering a different quantity
The question may ask for the time, not the speed.
Wrong working
Speed $= 70$ km/h. (Stops.)
Right working
Reduced speed 50 km/h; time $= \frac{350}{50} = 7$ hours.
Do this: Re-read the last line of the question.
Usually costs: Reported for 2025 Q8(iii).

Exam tips

Start with 'Let …' and state units; it earns the first method mark.
In 2025 Q8(iii) the expected equation was $\frac{350}{x - 20} - \frac{350}{x} = 2$ , giving $x^{2} - 20 x - 3500 = 0$ , $x = 70$ km/h and a reduced-speed time of 7 hours.
Divide by any common factor before factorising (e.g. $3 x^{2} - 51 x + 90 = 0 \Rightarrow x^{2} - 17 x + 30 = 0$ ).
Write a one-line check in the context; it confirms the right root.

Practice questions on Word Problems on Quadratic Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The length of a rectangle is 3 cm more than its width $x$ cm, and its area is 40 cm². The equation is:
(a) $x^{2} + 3 = 40$
(b) $x (x + 3) = 40$
(c) $2 (2 x + 3) = 40$
(d) $x^{2} + 3 x + 40 = 0$
Hint
Area = length × width.
Answer
(b)
Q2 · 3 marks · Short answerBoard level
Two numbers differ by 4 and their product is 96. Find the numbers.
Hint
$x (x + 4) = 96$ .
Answer
$x^{2} + 4 x - 96 = 0 \Rightarrow (x + 12) (x - 8) = 0$ : 8 and 12, or −12 and −8.
Q3 · 4 marks · Long answerHarder board level
A boat goes 30 km upstream and 30 km back in 4 hours 30 minutes. Its speed in still water is 15 km/h. Find the speed of the stream.
Hint
$\frac{30}{15 - s} + \frac{30}{15 + s} = 4.5$ .
Answer
$\frac{900}{225 - s ^{2}} = 4.5 \Rightarrow 225 - s^{2} = 200 \Rightarrow s = 5$ km/h.
Q4 · 3 marks · Short answerHarder board level
The hypotenuse of a right triangle is 13 cm and one of the other sides is 7 cm longer than the third side. Find the two sides.
Hint
$x^{2} + (x + 7)^{2} = 169$ .
Answer
$x^{2} + 7 x - 60 = 0 \Rightarrow (x + 12) (x - 5) = 0$ : 5 cm and 12 cm.
Q5 · 3 marks · Short answerHarder board level
The sum of the squares of two consecutive odd natural numbers is 394. Find them.
Hint
$x^{2} + (x + 2)^{2} = 394$ .
Answer
$x^{2} + 2 x - 195 = 0 \Rightarrow (x + 15) (x - 13) = 0$ : 13 and 15.
Q6 · 4 marks · Long answerHOTS / competency
A cyclist rides 60 km. If his speed were 2 km/h less, the ride would take 1 hour more. Find his speed and his time.
Hint
Slower time minus faster time = 1.
Answer
$\frac{60}{x - 2} - \frac{60}{x} = 1 \Rightarrow x^{2} - 2 x - 120 = 0 \Rightarrow (x - 12) (x + 10) = 0$ : 12 km/h, 5 hours.
Q7 · 3 marks · Short answerHarder board level
A hall has 300 seats arranged in equal rows. The number of seats in each row is 5 more than the number of rows. Find the number of rows and the seats in each row.
Hint
$x (x + 5) = 300$ .
Answer
$x^{2} + 5 x - 300 = 0 \Rightarrow (x + 20) (x - 15) = 0$ : 15 rows of 20 seats.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Two consecutive even numbers in terms of $x$ ?Show answer
$x$ and $x + 2$ .
2.Time for 150 km at $x$ km/h?Show answer
$\frac{150}{x}$ hours.
3.Solve $x^{2} + x - 132 = 0$ (positive root).Show answer
$x = 11$ .
4.A pipe fills a tank in 12 h. What fraction does it fill in 1 h?Show answer
$\frac{1}{12}$ .
5.Upstream speed if still-water speed is 18 km/h and stream 3 km/h?Show answer
15 km/h.

Word Problems on Quadratic Equations: frequently asked questions

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

Let x be one unknown, write the other quantities in terms of x, and turn the remaining condition into an equation. Clear fractions and bring it to the form ax² + bx + c = 0.

Why do we reject one root in word problems?

Because it may be impossible in the situation, such as a negative speed or length. State clearly why you reject it.

In speed problems, which time is larger?

The time at the lower speed. So write (time at the slower speed) − (time at the faster speed) = the difference in time.