MathsCBSE Class 10 · Quadratic Equations

Word Problems on Quadratic Equations for Class 10

Updated 14 October 2026 · Fact-checked

Word problems on quadratic equations are everyday situations, such as journeys, shared work, areas, ages and numbers, that you turn into a quadratic equation and solve. The board asks you to define the variable, form the equation, solve it by factorisation or the formula, reject impossible roots with a reason, and answer with units.

Model the situation, solve for both roots, then reject any root that is impossible, such as a negative length.

Illustration drawn for QuestionBuddy.

Understand Word Problems on Quadratic Equations

A word problem on quadratic equations describes a situation in words. Once you write the facts as algebra, the unknown satisfies an equation of the form $a x^{2} + b x + c = 0$ . Your job has three parts. First, model the situation as one equation in one unknown. Second, solve it. Third, interpret the roots back in the situation.

Why does a quadratic appear? Usually two quantities that both depend on $x$ get multiplied. An area is length times breadth, and a product of two numbers or two ages gives something like $x (x + k)$ . In speed and rate problems the unknown sits in a denominator, as in $\frac{120}{x}$ , and clearing two such fractions produces an $x^{2}$ term.

Most board questions fall into four families:

Speed, time and distance problems: time $= \frac{distance}{speed}$ , with a given difference or total of two times.
Rate problems (work, cost, share, run rate): one day's work is $\frac{1}{x}$ , and the number of items is total cost divided by price per item.
Area and geometry problems: rectangles, paths around or inside fields, and right triangles through Pythagoras.
Number and age problems: consecutive numbers, a number and its reciprocal, two-digit numbers, and ages some years ago or from now.

The solving part uses the rest of this chapter. You factorise by splitting the middle term, or use the quadratic formula when the numbers are awkward. The discriminant helps too. If $D < 0$ , the situation described is impossible, which is how you answer an 'Is it possible...?' question.

The interpreting step is where the conditions come in. The equation does not know that a speed must be positive, that a digit lies between 0 and 9, that a number of tickets is a whole number, or that a path inside a field must be narrower than half its shorter side. So a root can solve the equation correctly and still be wrong for the situation. Reject such a root and write the reason. Sometimes even a positive root must be rejected, because it makes another quantity negative. Sometimes both roots fit the situation, and then you report both.

Finally, answer exactly what was asked. Suppose you let $x$ be the breadth but the question wants a perimeter or a cost. Then $x$ is not the answer; it is only the last step before the answer.

Word Problems on Quadratic Equations: formulas to remember

Speed, distance and time: $time = \frac{distance}{speed}$; Write each time as a fraction with the unknown speed in the denominator.
Time-difference equation: $\frac{d}{x} - \frac{d}{x + k} = t$; The slower speed x gives the longer time, so its fraction comes first. Change minutes to hours first.
Boat and stream: $upstream = u - v, downstream = u + v$; u is the speed in still water and v is the speed of the stream, with u > v.
Work rate: $\frac{1}{x} + \frac{1}{y} = \frac{1}{T}$; If A alone takes x days and B alone takes y days, together they take T days. One day's work is the reciprocal of the time.
Items and price: $number of items = \frac{total amount}{price per item}$; Used for cost, share-per-person and run-rate problems; revenue = price × quantity.
Two-digit number: $10 t + u, reversed: 10 u + t$; t is the tens digit and u the units digit. Never write the number as tu.
Consecutive numbers: $x, x + 1 (integers); x, x + 2 (odd or even)$; Consecutive odd or consecutive even numbers differ by 2, not 1.
Ages: $age n years ago = x - n, age after n years = x + n$; Both people's ages change by the same n years.
Rectangle and right triangle: $A = l b, P = 2 (l + b), l^{2} + b^{2} = d^{2}$; d is the diagonal of a rectangle or the hypotenuse of a right triangle.
Path of width w inside a rectangle: $inner area = (l - 2 w) (b - 2 w)$; The path takes w from each side. For a path outside, use (l + 2w)(b + 2w).
Quadratic formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$; For ax² + bx + c = 0 with a ≠ 0. Use it when splitting the middle term is slow; it gives real roots only when b² − 4ac is not negative.
Discriminant and nature of roots: $D = b^{2} - 4 a c$; D > 0: two distinct real roots. D = 0: two equal real roots, each equal to −b/(2a). D < 0: no real roots, so the situation described is impossible. A perfect-square D means the roots are rational and factorisation will work.

How to solve Word Problems on Quadratic Equations questions

Every word problem in this chapter follows the same five-part plan. Write each part as a separate line, because the board awards step marks for the model, the equation, the solution and the rejection.

1Read the whole question, including the last line, and note exactly what you must find (a speed, a cost, the number, both ages).
2Write 'Let ... be $x$ ' with units, choosing the quantity the key fact is about. Express every other quantity in terms of $x$ .
3Translate the key fact (a difference in time, a total area, a product of ages) into one equation. Convert units first, for example minutes to hours.
4Clear fractions by multiplying by the LCM, expand, and bring the equation to standard form $a x^{2} + b x + c = 0$ . Divide by any common factor.
5Solve by splitting the middle term. If $a c$ is large or no pair appears quickly, find $D = b^{2} - 4 a c$ and use the quadratic formula.
6Test each root against the situation. Reject negative lengths, speeds or ages, non-whole counts, digits outside 0 to 9, and any root that makes another quantity negative. Write the reason in words.
7Find what was actually asked, check it against the original wording (not just the equation), and state the answer with units.

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.

Speed, time and distance problems

Imagine your class goes on a picnic to a spot 90 km from school. On the way back, heavy traffic makes the bus's average speed 15 km/h less than in the morning. Because of this, the return journey takes 1 hour longer than the morning journey.

Find the bus's average speed in the morning, its average speed on the way back, and the total time the class spent on the road.

Try it first, then show the solutionSolution

Let the morning speed be $x$ km/h. Then the return speed is $(x - 15)$ km/h.
Morning time $= \frac{90}{x}$ h and return time $= \frac{90}{x - 15}$ h.
The return takes 1 hour more: $\frac{90}{x - 15} - \frac{90}{x} = 1$ .
Multiply by $x (x - 15)$ : $90 x - 90 (x - 15) = x (x - 15)$ .
So $1350 = x^{2} - 15 x$ , that is $x^{2} - 15 x - 1350 = 0$ .
$- 45 \times 30 = - 1350$ and $- 45 + 30 = - 15$ , so $(x - 45) (x + 30) = 0$ .
$x = 45$ or $x = - 30$ . A speed cannot be negative, so $x = 45$ .
Return speed $= 45 - 15 = 30$ km/h.
Times: $\frac{90}{45} = 2$ h and $\frac{90}{30} = 3$ h. The difference is 1 h, which matches.
Total time on the road $= 2 + 3 = 5$ h.

Answer: Morning speed 45 km/h, return speed 30 km/h, total time on the road 5 hours.

Memory hook: Slower bus, longer ride: the slower time minus the faster time gives the delay.

Rate problems (work, cost, share, run rate)

Suppose a group of friends share a Rs 1200 restaurant bill equally. They notice that if two more friends had joined, each person would have paid Rs 50 less.

How many friends were in the group, and how much did each pay? If the bill had instead been Rs 1500 for the same group, how much more would each friend have paid?

Try it first, then show the solutionSolution

Let there be $x$ friends. Each share is $\frac{1200}{x}$ ; with two more friends it would be $\frac{1200}{x + 2}$ .
The share drops by Rs 50: $\frac{1200}{x} - \frac{1200}{x + 2} = 50$ .
Multiply by $x (x + 2)$ : $1200 (x + 2) - 1200 x = 50 x (x + 2)$ , so $2400 = 50 x^{2} + 100 x$ .
Divide by 50: $x^{2} + 2 x - 48 = 0$ , so $(x + 8) (x - 6) = 0$ .
$x = 6$ or $x = - 8$ . The number of friends cannot be negative, so $x = 6$ .
Each paid $\frac{1200}{6} = 200$ rupees. Check: with 8 friends, $\frac{1200}{8} = 150$ , which is Rs 50 less.
For Rs 1500: each pays $\frac{1500}{6} = 250$ rupees, that is Rs 50 more.

Answer: 6 friends paying Rs 200 each; for a Rs 1500 bill each would pay Rs 250, which is Rs 50 more.

Memory hook: Shares shrink when the group grows: old share minus new share gives the quadratic.

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.

LFSC: Let, Form, Solve, Check
Let $x$ be something (with units), Form one equation, Solve it, Check each root against the story and answer what was asked.
Use it when: Write these four stages as separate lines in every 3- or 5-mark word problem so no step mark is lost.
Slower means longer
In 'd over speed' problems, the smaller speed gives the bigger time. Bigger time minus smaller time = the given difference.
Use it when: Stops you writing the subtraction the wrong way round, which leads to a wrong equation or a negative discriminant.
Work flips
Time becomes rate by flipping: $x$ days alone means $\frac{1}{x}$ of the job per day. Rates add, times do not.
Use it when: Work, tap and machine problems: add the flipped times and set the total equal to $\frac{1}{time together}$ .
Odd and even jump by two
Consecutive integers: $x, x + 1$ . Consecutive odd or even: $x, x + 2$ . A two-digit number: ten times the tens digit, plus the units digit.
Use it when: Number problems, where a wrong first line spoils the whole solution.

Word Problems on Quadratic Equations: solved examples

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

Example 1: Two-digit number with digit product · 3 marks · Short answerBoard level

A two-digit number is such that the product of its digits is 18. When 27 is subtracted from the number, the digits interchange their places. Find the number.

Try it first, then show the solutionSolution

Let the tens digit be $x$ . Then the units digit is $\frac{18}{x}$ . The number is $10 x + \frac{18}{x}$ and the reversed number is $10 \cdot \frac{18}{x} + x = \frac{180}{x} + x$ . [½ mark]
Given condition: $10 x + \frac{18}{x} - 27 = \frac{180}{x} + x$ . [½ mark]
Simplify: $9 x - 27 = \frac{162}{x}$ . Multiply by $x$ : $9 x^{2} - 27 x - 162 = 0$ , so $x^{2} - 3 x - 18 = 0$ . [1 mark]
$(x - 6) (x + 3) = 0$ , so $x = 6$ or $x = - 3$ . A digit cannot be negative, so $x = 6$ . [½ mark]
Units digit $= \frac{18}{6} = 3$ , so the number is 63. Check: $6 \times 3 = 18$ and $63 - 27 = 36$ , the digits reversed. [½ mark]

Answer: The number is 63.

Example 2: Round trip with a total time · 3 marks · Short answerBoard level

A man drives 120 km from his village to a town. He returns by the same road at an average speed 10 km/h more than on the way out. The total driving time for the round trip is 7 hours. Find his average speed on each leg of the journey.

Try it first, then show the solutionSolution

Let the speed on the way out be $x$ km/h, so the return speed is $(x + 10)$ km/h. The times are $\frac{120}{x}$ h and $\frac{120}{x + 10}$ h. [½ mark]
Total time: $\frac{120}{x} + \frac{120}{x + 10} = 7$ . [½ mark]
Multiply by $x (x + 10)$ : $120 (x + 10) + 120 x = 7 x (x + 10)$ , so $240 x + 1200 = 7 x^{2} + 70 x$ , giving $7 x^{2} - 170 x - 1200 = 0$ . [1 mark]
$a c = - 8400$ ; $- 210 \times 40 = - 8400$ and $- 210 + 40 = - 170$ . So $7 x^{2} - 210 x + 40 x - 1200 = 0$ , that is $7 x (x - 30) + 40 (x - 30) = 0$ , giving $(x - 30) (7 x + 40) = 0$ . [½ mark]
$x = 30$ or $x = - \frac{40}{7}$ . A speed cannot be negative, so $x = 30$ and the return speed is 40 km/h. Check: $\frac{120}{30} + \frac{120}{40} = 4 + 3 = 7$ h. [½ mark]

Answer: Speed on the way out 30 km/h; speed on the way back 40 km/h.

Example 3: Work problem where a positive root is rejected · 3 marks · Short answerBoard level

Two workers together can paint a school's boundary wall in 4 days. Working alone, the first worker takes 6 days less than the second worker. How many days would each worker take to paint the wall alone?

Try it first, then show the solutionSolution

Let the second worker take $x$ days alone, so the first takes $(x - 6)$ days. Their one-day work is $\frac{1}{x}$ and $\frac{1}{x - 6}$ . [½ mark]
Together they do $\frac{1}{4}$ of the wall in a day: $\frac{1}{x - 6} + \frac{1}{x} = \frac{1}{4}$ . [½ mark]
Multiply by $4 x (x - 6)$ : $4 (x + x - 6) = x (x - 6)$ , so $8 x - 24 = x^{2} - 6 x$ , giving $x^{2} - 14 x + 24 = 0$ . [1 mark]
$(x - 12) (x - 2) = 0$ , so $x = 12$ or $x = 2$ . [½ mark]
If $x = 2$ , the first worker would take $2 - 6 = - 4$ days, which is impossible, so reject it. Hence $x = 12$ and the first worker takes 6 days. Check: $\frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$ . [½ mark]

Answer: The first worker takes 6 days and the second takes 12 days.

Example 4: Path inside a park, with cost · 5 marks · Long answerBoard level

A rectangular park measures 40 m by 30 m. A path of uniform width runs inside the park along all four edges, and the remaining rectangular lawn in the middle has an area of 704 square metres. Find the width of the path. Also find the cost of paving the path at Rs 40 per square metre.

Try it first, then show the solutionSolution

Let the width of the path be $x$ m. The path takes $x$ from both ends of each side, so the lawn measures $(40 - 2 x)$ m by $(30 - 2 x)$ m. [1 mark]
Area of lawn: $(40 - 2 x) (30 - 2 x) = 704$ . [½ mark]
Expand: $1200 - 80 x - 60 x + 4 x^{2} = 704$ , so $4 x^{2} - 140 x + 496 = 0$ . Divide by 4: $x^{2} - 35 x + 124 = 0$ . [1 mark]
$- 31 \times - 4 = 124$ and $- 31 - 4 = - 35$ , so $(x - 31) (x - 4) = 0$ , giving $x = 31$ or $x = 4$ . [½ mark]
If $x = 31$ , the lawn length would be $40 - 62 = - 22$ m, which is impossible, so reject it. Hence $x = 4$ m. Check: $(40 - 8) (30 - 8) = 32 \times 22 = 704$ . [1 mark]
Area of path $= 1200 - 704 = 496$ square metres. Cost $= 496 \times 40 = 19840$ , that is Rs 19840. [1 mark]

Answer: The path is 4 m wide, and paving it costs Rs 19840.

Example 5: Case study: pricing tickets for the school play · 4 marks · Case-basedHOTS / competency

Suppose your school drama club plans to sell tickets for its annual play at Rs 50 each and expects to sell 200 tickets. A survey suggests that for every Rs 5 increase in the price, 10 fewer tickets will be sold. Let $n$ be the number of Rs 5 increases.
(i) Show that the total collection is $10000 + 500 n - 50 n^{2}$ rupees, and write the equation for a collection of Rs 11200 in standard form. [1 mark]
(ii) Using the discriminant, show that the collection can never be Rs 12000. [1 mark]
(iii) Find the ticket prices that give a collection of Rs 11200. Which price should the club choose if it also wants a bigger audience? [2 marks]

Try it first, then show the solutionSolution

(i) Price $= (50 + 5 n)$ rupees and tickets sold $= (200 - 10 n)$ . Collection $= (50 + 5 n) (200 - 10 n) = 10000 - 500 n + 1000 n - 50 n^{2} = 10000 + 500 n - 50 n^{2}$ . Setting it equal to 11200: $50 n^{2} - 500 n + 1200 = 0$ , that is $n^{2} - 10 n + 24 = 0$ . [1 mark]
(ii) $10000 + 500 n - 50 n^{2} = 12000$ gives $50 n^{2} - 500 n + 2000 = 0$ , that is $n^{2} - 10 n + 40 = 0$ . $D = 100 - 160 = - 60 < 0$ , so no real $n$ exists and Rs 12000 is impossible. [1 mark]
(iii) $n^{2} - 10 n + 24 = 0$ gives $(n - 4) (n - 6) = 0$ , so $n = 4$ or $n = 6$ . [½ mark]
$n = 4$ : price Rs 70, tickets $200 - 40 = 160$ , collection $70 \times 160 = 11200$ . [½ mark]
$n = 6$ : price Rs 80, tickets $200 - 60 = 140$ , collection $80 \times 140 = 11200$ . Both roots fit the situation. [½ mark]
For a bigger audience, choose Rs 70, since 160 tickets are sold instead of 140 for the same collection. [½ mark]

Answer: (i) $n^{2} - 10 n + 24 = 0$ . (ii) $D = - 60 < 0$ , so Rs 12000 is impossible. (iii) Rs 70 (160 tickets) or Rs 80 (140 tickets); the club should choose Rs 70.

Example 6: Boat and stream: upstream and downstream in a total time · 5 marks · Long answerHarder board level

A motorboat goes 24 km upstream and then 28 km downstream, taking 6 hours in all. The speed of the stream is 4 km/h. Find the speed of the boat in still water. Also find the total time the boat would take to go 21 km upstream and come back to its starting point.

Try it first, then show the solutionSolution

Let the speed of the boat in still water be $u$ km/h, where $u > 4$ . Upstream speed $= (u - 4)$ km/h and downstream speed $= (u + 4)$ km/h. [1 mark]
Total time: $\frac{24}{u - 4} + \frac{28}{u + 4} = 6$ . [½ mark]
Multiply by $(u - 4) (u + 4)$ : $24 (u + 4) + 28 (u - 4) = 6 (u^{2} - 16)$ , so $52 u - 16 = 6 u^{2} - 96$ , giving $6 u^{2} - 52 u - 80 = 0$ , that is $3 u^{2} - 26 u - 40 = 0$ . [1 mark]
$a c = - 120$ ; $- 30 \times 4 = - 120$ and $- 30 + 4 = - 26$ . So $3 u^{2} - 30 u + 4 u - 40 = 0$ , that is $3 u (u - 10) + 4 (u - 10) = 0$ , giving $(u - 10) (3 u + 4) = 0$ . [1 mark]
$u = 10$ or $u = - \frac{4}{3}$ . A speed cannot be negative, so $u = 10$ km/h. Check: $\frac{24}{6} + \frac{28}{14} = 4 + 2 = 6$ h. [½ mark]
Upstream speed 6 km/h and downstream speed 14 km/h. Time $= \frac{21}{6} + \frac{21}{14} = 3.5 + 1.5 = 5$ h. [1 mark]

Answer: The boat's speed in still water is 10 km/h; the 21 km round trip takes 5 hours.

Example 7: Ages: one person years ago, the other years from now · 3 marks · Short answerBoard level

A father is 25 years older than his son. The product of the son's age 5 years ago and the father's age 5 years from now is 200. Find their present ages.

Try it first, then show the solutionSolution

Let the son's present age be $x$ years, so the father's present age is $(x + 25)$ years. [½ mark]
Son's age 5 years ago $= (x - 5)$ years; father's age 5 years from now $= (x + 30)$ years. [½ mark]
$(x - 5) (x + 30) = 200$ gives $x^{2} + 25 x - 150 = 200$ , so $x^{2} + 25 x - 350 = 0$ . [1 mark]
$35 \times (- 10) = - 350$ and $35 - 10 = 25$ , so $(x + 35) (x - 10) = 0$ , giving $x = - 35$ or $x = 10$ . [½ mark]
An age cannot be negative, so $x = 10$ . The son is 10 years old and the father is 35. Check: $(10 - 5) (35 + 5) = 5 \times 40 = 200$ . [½ mark]

Answer: The son is 10 years old and the father is 35 years old.

Example 8: Is it possible? Deciding with the discriminant · 2 marks · Very short answerBoard level

Is it possible to enclose a rectangular plot with exactly 60 m of fencing so that its area is 250 square metres? Justify your answer.

Try it first, then show the solutionSolution

Perimeter 60 m gives length + breadth $= 30$ m. Let the length be $l$ m, so the breadth is $(30 - l)$ m. [½ mark]
Area: $l (30 - l) = 250$ , so $l^{2} - 30 l + 250 = 0$ . [½ mark]
$D = (- 30)^{2} - 4 \times 1 \times 250 = 900 - 1000 = - 100$ . [½ mark]
Since $D < 0$ , the equation has no real roots, so no such plot can be made. It is not possible. [½ mark]

Answer: Not possible, because $D = - 100 < 0$ and there are no real values of the length.

Example 9: Pythagoras: a ladder that slides · 3 marks · Short answerHarder board level

The foot of a ladder is $x$ m from a vertical wall. The top of the ladder is 7 m higher up the wall than the foot is from the wall, and the ladder is 1 m longer than that height. Find the length of the ladder. If the foot then slides 7 m further from the wall, how far does the top slide down?

Try it first, then show the solutionSolution

Height $= (x + 7)$ m and ladder $= (x + 8)$ m. By Pythagoras: $x^{2} + (x + 7)^{2} = (x + 8)^{2}$ . [½ mark]
Expand: $2 x^{2} + 14 x + 49 = x^{2} + 16 x + 64$ , so $x^{2} - 2 x - 15 = 0$ and $(x - 5) (x + 3) = 0$ . [1 mark]
Reject $x = - 3$ (a distance cannot be negative), so $x = 5$ : foot 5 m, height 12 m, ladder 13 m. Check: $25 + 144 = 169$ . [½ mark]
New foot distance $= 12$ m, so new height $= 169 - 144 = 5$ m. The top slides down $12 - 5 = 7$ m. [1 mark]

Answer: The ladder is 13 m long; the top slides down 7 m (from 12 m to 5 m).

Example 10: Irrational roots: using the formula · 3 marks · Short answerHarder board level

The length of a rectangular garden is 4 m more than its breadth, and its area is 50 square metres. Find the breadth and the perimeter, giving the exact perimeter and its value to two decimal places (take $6 \approx 2.449$ ).

Try it first, then show the solutionSolution

Let the breadth be $b$ m: $b (b + 4) = 50$ , so $b^{2} + 4 b - 50 = 0$ . [½ mark]
$D = 16 + 200 = 216$ , which is not a perfect square, so use the formula: $b = \frac{- 4 \pm 216}{2} = \frac{- 4 \pm 6 6}{2} = - 2 \pm 36$ . [1 mark]
$- 2 - 36$ is negative, so reject it. Breadth $= 36 - 2 \approx 7.347 - 2 = 5.35$ m. [½ mark]
Perimeter $= 2 (b + b + 4) = 4 b + 8 = 4 (36 - 2) + 8 = 126 \approx 29.39$ m. [1 mark]

Answer: Breadth $(36 - 2)$ m $\approx 5.35$ m; perimeter $126$ m $\approx 29.39$ m.

Example 11: Two journeys and a halt · 5 marks · Long answerHarder board level

A bus travels 200 km to a town and comes back by the same road. On the way back its average speed is 10 km/h more than on the way out, and it halts for 30 minutes during the return. The whole round trip, including the halt, takes 9 hours 30 minutes. Find the average speed each way and the average speed for the whole trip including the halt.

Try it first, then show the solutionSolution

Let the speed on the way out be $x$ km/h. Driving time $= 9.5 - 0.5 = 9$ h. [1 mark]
$\frac{200}{x} + \frac{200}{x + 10} = 9$ . Multiply by $x (x + 10)$ : $200 (2 x + 10) = 9 x (x + 10)$ , so $9 x^{2} - 310 x - 2000 = 0$ . [1 mark]
$D = 31 0^{2} + 4 \times 9 \times 2000 = 96100 + 72000 = 168100 = 41 0^{2}$ . [1 mark]
$x = \frac{310 \pm 410}{18}$ , so $x = 40$ or $x = - \frac{50}{9}$ . Reject the negative speed: out at 40 km/h, back at 50 km/h. Check: $5 + 4 = 9$ h. [1 mark]
Average speed for the whole trip $= \frac{400}{9.5} \approx 42.1$ km/h. [1 mark]

Answer: Out at 40 km/h, back at 50 km/h; the average speed for the whole trip including the halt is about 42.1 km/h.

Example 12: MCQ: a number and its reciprocal · 1 mark · MCQBoard level

A positive number exceeds its reciprocal by $\frac{3}{2}$ . The number is
(a) $2$
(b) $\frac{1}{2}$
(c) $\frac{3}{2}$
(d) $4$

Try it first, then show the solutionSolution

$x - \frac{1}{x} = \frac{3}{2}$ ; multiply by $2 x$ : $2 x^{2} - 3 x - 2 = 0$ , so $(2 x + 1) (x - 2) = 0$ .
$x = 2$ or $x = - \frac{1}{2}$ ; the number is positive, so $x = 2$ . Check: $2 - \frac{1}{2} = \frac{3}{2}$ .

Answer: (a) $2$

Where students lose marks in Word Problems on Quadratic Equations

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

Mistake: Subtracting the two times the wrong way round
The faster journey takes less time, so the given difference is the slower time minus the faster time. Reversing it changes the sign and gives a wrong equation.
Wrong working
Speed 10 km/h more saves 1 hour on 120 km: $\frac{120}{x + 10} - \frac{120}{x} = 1$ , so $x^{2} + 10 x + 1200 = 0$ . $D < 0$ , so 'no solution'.
Right working
$\frac{120}{x} - \frac{120}{x + 10} = 1$ , so $1200 = x^{2} + 10 x$ and $(x + 40) (x - 30) = 0$ . Hence $x = 30$ km/h (times 4 h and 3 h).
Do this: Before writing the equation, ask which speed is slower. Its fraction goes first.
Usually costs: The equation mark and every mark after it.
Mistake: Leaving minutes as minutes
Speeds are in km per hour, so every time in the equation must be in hours.
Wrong working
Journey of 60 km, 3 km/h faster saves 40 minutes: $\frac{60}{x} - \frac{60}{x + 3} = 40$ .
Right working
$\frac{60}{x} - \frac{60}{x + 3} = \frac{40}{60} = \frac{2}{3}$ , so $270 = x^{2} + 3 x$ and $(x + 18) (x - 15) = 0$ . Hence $x = 15$ km/h.
Do this: Convert first: 30 min $= \frac{1}{2}$ h, 40 min $= \frac{2}{3}$ h, 45 min $= \frac{3}{4}$ h.
Usually costs: The equation mark, and the roots will not come out cleanly.
Mistake: Keeping a positive root that makes another quantity negative
A root can be positive and still give a negative time, length or age elsewhere in the problem.
Wrong working
Second worker $x$ days, first $(x - 6)$ days: $x = 12$ or $x = 2$ , so the second worker takes 12 or 2 days.
Right working
$x = 2$ makes the first worker's time $2 - 6 = - 4$ days, which is impossible, so reject it. Hence $x = 12$ and the first worker takes 6 days.
Do this: Substitute each root into every expression you defined, not just into $x$ , and write the reason for any rejection.
Usually costs: The final ½ to 1 mark for interpretation.
Mistake: Writing a two-digit number as the product or sum of its digits
The tens digit has place value 10, so the number is $10 x + y$ , not $x y$ or $x + y$ .
Wrong working
Tens digit $x$ , units digit $y$ : number $= x y$ , reversed number $= y x$ , so the difference is 0.
Right working
Number $= 10 x + y$ , reversed number $= 10 y + x$ , difference $= 9 (x - y)$ .
Do this: Always write 'number $= 10 \times$ tens digit $+$ units digit' as your first line.
Usually costs: Almost the whole question, since the model is wrong from the start.
Mistake: Stopping at the value of x
The question often asks for something built from $x$ , such as a cost, a perimeter or the other person's age.
Wrong working
Path question asking for width and cost of paving: final line 'Width of path $= 4$ m.'
Right working
Width $= 4$ m; path area $= 1200 - 704 = 496$ square metres; cost $= 496 \times 40 =$ Rs 19840.
Do this: Underline every quantity asked for before you start, and tick each one off at the end.
Usually costs: The last 1 mark of a long answer or case sub-part.

Exam tips

Begin with a clear 'Let ... be $x$ ' line that states the units. The examiner's first step mark usually goes to the correctly defined variable and the expressions built from it.
Show the equation before any simplification, for example $\frac{120}{x} + \frac{120}{x + 10} = 7$ . A correct equation earns marks even if a later arithmetic slip happens.
When you reject a root, write the reason in words, such as 'speed cannot be negative' or 'the lawn length would be negative'. A root crossed out without a reason may not get the interpretation mark.
Check your answer against the original wording of the question, not just the equation. A wrong equation can still be satisfied by your roots.
In case-based questions, part (i) usually builds the equation used in the later parts. Check it twice, and read every sub-part before starting so you know what each one needs.

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 · 2 marks · Very short answerBoard level
Is it possible to design a rectangular park of perimeter 80 m and area 400 square metres? If so, find its length and breadth.
Hint
Length + breadth $= 40$ . Form the quadratic in the length and look at the sign of $D$ .
Answer
Let the length be $l$ m, so the breadth is $(40 - l)$ m. Then $l (40 - l) = 400$ , so $l^{2} - 40 l + 400 = 0$ . $D = 1600 - 1600 = 0$ , so there are two equal real roots and the park is possible. $l = \frac{40}{2} = 20$ and breadth $= 40 - 20 = 20$ . The park is 20 m by 20 m, a square (a special rectangle). Check: perimeter $= 80$ m and area $= 400$ square metres.
Q2 · 3 marks · Short answerBoard level
A shopkeeper buys a number of storybooks for Rs 1200. If the cost of each book were Rs 20 less, he could buy 5 more books for the same amount. Find the original cost of each book and the number of books he bought.
Hint
Let the cost per book be $x$ rupees. The number of books is $\frac{1200}{x}$ ; at the lower price it is $\frac{1200}{x - 20}$ .
Answer
$\frac{1200}{x - 20} - \frac{1200}{x} = 5$ gives $1200 \times 20 = 5 x (x - 20)$ , so $x^{2} - 20 x - 4800 = 0$ and $(x - 80) (x + 60) = 0$ . Reject $x = - 60$ , so each book costs Rs 80 and he bought $\frac{1200}{80} = 15$ books. Check: at Rs 60, $\frac{1200}{60} = 20 = 15 + 5$ .
Q3 · 3 marks · Short answerHarder board level
Two taps together can fill a water tank in 6 hours. The smaller tap alone takes 5 hours more than the larger tap alone to fill it. Find the time each tap takes alone. One day both taps are opened together for 2 hours and then the larger tap is closed. How many more hours does the smaller tap take to fill the tank?
Hint
Let the larger tap take $x$ h. Add the flipped times: $\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}$ . Then find what part is left after 2 hours.
Answer
$\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}$ gives $6 (2 x + 5) = x (x + 5)$ , so $x^{2} - 7 x - 30 = 0$ and $(x - 10) (x + 3) = 0$ . Reject $- 3$ ; the larger tap takes 10 h and the smaller 15 h. Check: $\frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6}$ . In 2 h together they fill $\frac{2}{6} = \frac{1}{3}$ of the tank. The remaining $\frac{2}{3}$ at $\frac{1}{15}$ of the tank per hour takes $\frac{2}{3} \times 15 = 10$ h more.
Q4 · 5 marks · Long answerHarder board level
A rectangular field measures 30 m by 20 m. A gravel path of uniform width is laid all round the field, outside it. The area of the path is 216 square metres. Find the width of the path. Find also the cost of gravelling the path at Rs 50 per square metre and the cost of fencing the outer edge of the path at Rs 25 per metre.
Hint
With width $w$ , the outer rectangle is $(30 + 2 w)$ m by $(20 + 2 w)$ m. Path area = outer area minus field area.
Answer
Let the width be $w$ m. $(30 + 2 w) (20 + 2 w) - 600 = 216$ gives $600 + 100 w + 4 w^{2} - 600 = 216$ , so $4 w^{2} + 100 w - 216 = 0$ , that is $w^{2} + 25 w - 54 = 0$ . $(w + 27) (w - 2) = 0$ , so $w = - 27$ or $w = 2$ . A width cannot be negative, so the path is 2 m wide. Check: outer rectangle $34 \times 24 = 816$ and $816 - 600 = 216$ . Cost of gravel $= 216 \times 50 =$ Rs 10800. Outer perimeter $= 2 (34 + 24) = 116$ m, so fencing costs $116 \times 25 =$ Rs 2900.
Q5 · 3 marks · Short answerHOTS / competency
Meera was asked: 'The sum of the squares of two consecutive even natural numbers is 340. Find the numbers.' She wrote:
Let the numbers be $x$ and $x + 1$ . Then $x^{2} + (x + 1)^{2} = 340$ , so $2 x^{2} + 2 x - 339 = 0$ . Here $D = 2716$ , which is not a perfect square, so no such numbers exist.
Find the mistake in her model, and say what is wrong with her conclusion from D. Then solve correctly.
Hint
How far apart are consecutive even numbers? And what does a positive $D$ that is not a perfect square really tell you?
Answer
Error 1: consecutive even numbers differ by 2, so they are $x$ and $x + 2$ . Error 2: $D > 0$ means two real roots; a non-square $D$ only means the roots are irrational, not that there are none (here they are not natural numbers, which is the right reason to reject them). Correct model: $x^{2} + (x + 2)^{2} = 340$ gives $2 x^{2} + 4 x - 336 = 0$ , so $x^{2} + 2 x - 168 = 0$ and $(x + 14) (x - 12) = 0$ . Reject $- 14$ ; the numbers are 12 and 14. Check: $144 + 196 = 340$ .
Q6 · 3 marks · Short answerHOTS / competency
A farmer has 40 m of fencing to make a rectangular pen against a long straight wall, so that the wall forms one side and the fence forms the other three sides.
(a) Show that the pen cannot have an area of 250 square metres.
(b) Using the discriminant, find the largest area the pen can have, and its dimensions for that area.
Hint
Let each side perpendicular to the wall be $x$ m, so the side along the wall is $(40 - 2 x)$ m. For area $k$ , real dimensions need $D \geq 0$ .
Answer
For area $k$ : $x (40 - 2 x) = k$ , so $2 x^{2} - 40 x + k = 0$ . (a) $k = 250$ : $x^{2} - 20 x + 125 = 0$ and $D = 400 - 500 = - 100 < 0$ , so no real $x$ exists; 250 square metres is impossible. (b) Real $x$ needs $D = 1600 - 8 k \geq 0$ , so $k \leq 200$ . The largest area is 200 square metres. Then $D = 0$ and $x = \frac{40}{4} = 10$ , so the pen is 10 m by 20 m, with the 20 m side along the wall. Check: $10 + 10 + 20 = 40$ m of fence and $10 \times 20 = 200$ square metres.
Q7 · 4 marks · Case-basedHOTS / competency
Suppose a bus service runs between two towns 150 km apart. One day the bus leaves 30 minutes late because of a delay at a level crossing, so the driver raises the average speed by 10 km/h for the whole journey and still arrives on time. Let the usual average speed be $x$ km/h.
(i) Form the equation and show that it reduces to $x^{2} + 10 x - 3000 = 0$ . [1 mark]
(ii) Find the discriminant and state the nature of the roots. [1 mark]
(iii) Find the usual speed and the time the journey took that day. [2 marks]
Hint
Usual time minus that day's time equals 30 minutes, which is $\frac{1}{2}$ h.
Answer
(i) $\frac{150}{x} - \frac{150}{x + 10} = \frac{1}{2}$ gives $\frac{1500}{x ( x + 10 )} = \frac{1}{2}$ , so $x^{2} + 10 x = 3000$ , that is $x^{2} + 10 x - 3000 = 0$ . (ii) $D = 100 + 12000 = 12100 = 11 0^{2} > 0$ , so there are two distinct real (rational) roots. (iii) $x = \frac{- 10 \pm 110}{2} = 50$ or $- 60$ . A speed cannot be negative, so the usual speed is 50 km/h. That day the speed was 60 km/h and the journey took $\frac{150}{60} = 2.5$ h, that is 2 h 30 min. Check: usual time $\frac{150}{50} = 3$ h and $3 - 2.5 = 0.5$ h.
Q8 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A rectangular plot with perimeter 28 m and area 50 square metres can be made.
Reason (R): For such a plot to exist, the quadratic equation for its length must have a non-negative discriminant.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Length + breadth $= 14$ .
Answer
(d). $l (14 - l) = 50$ gives $l^{2} - 14 l + 50 = 0$ with $D = 196 - 200 = - 4 < 0$ , so no real length exists and A is false. R is true: real dimensions need $D \geq 0$ .
Q9 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a school hall is set up with chairs in equal rows, where the number of chairs in each row is 5 more than the number of rows.
(i) For 300 chairs, form a quadratic equation in the number of rows $r$ . (1 mark)
(ii) Find the number of rows and the chairs in each row. (1 mark)
(iii) Show, using the discriminant, that 350 chairs cannot be arranged in this pattern. If at most 400 chairs are available, find the largest number of rows possible in this pattern and the chairs used. (2 marks)
Hint
A number of rows must be a whole number, so a root that is not a whole number fails the situation even if it is real.
Answer
(i) $r (r + 5) = 300$ , so $r^{2} + 5 r - 300 = 0$ . (1)
(ii) $(r + 20) (r - 15) = 0$ ; reject $- 20$ , so 15 rows of 20 chairs. (1)
(iii) For 350: $r^{2} + 5 r - 350 = 0$ , $D = 25 + 1400 = 1425$ , which is not a perfect square, so $r$ is irrational (about 16.4) and cannot be a whole number of rows. For at most 400 chairs we need $r (r + 5) \leq 400$ : $r = 17$ gives $17 \times 22 = 374$ , and $r = 18$ gives $18 \times 23 = 414 > 400$ . So at most 17 rows, using 374 chairs. (2)

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.How do you write two consecutive odd natural numbers?Show answer
$x$ and $x + 2$ , where $x$ is odd.
2.A two-digit number has tens digit $t$ and units digit $u$ . What is the number, and what is the number with its digits reversed?Show answer
The number is $10 t + u$ ; reversed, it is $10 u + t$ .
3.A boat's speed in still water is $u$ and the stream's speed is $v$ . What is its downstream speed?Show answer
$u + v$ (upstream it is $u - v$ ).
4.If a job takes $x$ days alone, what part is done in one day?Show answer
$\frac{1}{x}$ of the job.
5.Write 45 minutes in hours.Show answer
$\frac{3}{4}$ h.
6.Name three reasons to reject a root in a word problem.Show answer
It gives a negative length, speed or age; it gives a non-whole count or a digit outside 0 to 9; or it makes another quantity negative.

Word Problems on Quadratic Equations: frequently asked questions

How do I decide what to take as x in a word problem?

Take the quantity that the main fact is about, usually the one the question asks for. Then write every other quantity in terms of it. If the main fact compares two times or two prices, let x be the speed or price that appears in both.

Can both roots of the quadratic be correct answers?

Yes. If both roots fit the situation, such as two ticket prices giving the same collection, you must report both. Reject a root only when it breaks a condition of the problem, and say which condition.

Should I use factorisation or the quadratic formula for word problems?

Try splitting the middle term first, because word problems are usually set with neat numbers. If the product ac is large and no pair comes quickly, find the discriminant and use the formula. Both methods get full marks.

Do I need the method of completing the square for these problems?

No. Solving by completing the square is not in the 2026-27 syllabus. Use factorisation or the quadratic formula.

Where are these problems in the NCERT textbook?

Exercise 4.2 has situational problems that you model and solve by factorisation, such as numbers, ages and areas. Exercise 4.3 includes situations where you use the discriminant to decide whether something is possible. Practise by writing your own versions with different numbers.