On this page
Most Class 10 students can solve x² + 3x − 108 = 0. Far fewer can get there from "a cyclist covers 36 km...". The board paper includes situational problems, and they are where marks go missing. This post gives a four-step way to turn words into an equation, works one speed, one age and one area problem with checks, and shows how the discriminant tells you a situation is impossible.
Where is this in the syllabus?
The CBSE Class X Mathematics curriculum for 2026-27 lists the standard form ax² + bx + c = 0, solutions "(only real roots) by factorization, and by using quadratic formula", the link between the discriminant and the nature of roots, and "situational problems based on quadratic equations related to day-to-day activities". The NCERT chapter Quadratic Equations (reprint 2026-27) opens with a prayer hall whose length is one metre more than twice its breadth and whose area is 300 m², which gives 2x² + x − 300 = 0, a 12 m by 25 m hall. Exercise 4.1 sets up an age problem and a train-speed problem, and Exercise 4.3 asks whether situations are possible. Check your own year's curriculum.
Are word problems harder than equations?
Not always. Koedinger and Nathan (2004) found that beginning algebra students were more successful solving simple story problems than mathematically equivalent equations. The difficulty lay in understanding the formal symbols, not the story. So use the words to reason, and be careful at the step where you write symbols.
What is the four-step method?
Name the unknown, with units: "Let the speed be x km/h."
Write every other quantity in terms of x: time = 36/x hours.
Find the equality and say it in words first: "old time minus new time equals one hour". Then translate.
Solve, reject impossible roots with a reason, and check in the original words.
Worked example 1: speed
A cyclist covers 36 km. If her speed were 3 km/h more, she would take 1 hour less. Find her speed.
Let the speed be x km/h. Time taken = 36/x hours; at the faster speed, 36/(x + 3) hours.
Old time − new time = 1: 36/x − 36/(x + 3) = 1.
Multiply by x(x + 3): 36(x + 3) − 36x = x(x + 3), so 108 = x² + 3x, or x² + 3x − 108 = 0.
Factorise: (x + 12)(x − 9) = 0, so x = 9 or x = −12. A speed cannot be negative, so x = 9 km/h.
Check: 36/9 = 4 hours; at 12 km/h, 36/12 = 3 hours. One hour less.
Worked example 2: age
A father is 30 years older than his son. Five years from now, the product of their ages will be 675. Find their present ages.
Let the son's age be x years, so the father is x + 30.
In 5 years they will be x + 5 and x + 35.
(x + 5)(x + 35) = 675, so x² + 40x + 175 = 675, or x² + 40x − 500 = 0.
Factorise: (x + 50)(x − 10) = 0. An age cannot be negative, so the son is 10 and the father 40.
Check: in 5 years, 15 × 45 = 675. The usual slip is adding 5 years to only one person.
Worked example 3: area
A rectangular garden is 20 m by 15 m. A path of uniform width w runs round the outside, and the garden and path together cover 456 m². Find w.
Outer length = 20 + 2w; outer breadth = 15 + 2w. The path adds width on both sides.
(20 + 2w)(15 + 2w) = 456, so 4w² + 70w + 300 = 456, or 2w² + 35w − 78 = 0.
Use the formula: b² − 4ac = 35² + 4 × 2 × 78 = 1225 + 624 = 1849 = 43². So w = (−35 ± 43)/4, which gives 2 or −19.5.
A width cannot be negative, so w = 2 m.
Check: 24 × 19 = 456. The path itself covers 456 − 300 = 156 m².
How does the discriminant spot an impossible situation?
Can a rectangle have perimeter 30 m and area 60 m²? Its length and breadth add to 15 and multiply to 60, so they are the roots of x² − 15x + 60 = 0. The discriminant is 225 − 240 = −15, which is negative, so there are no real roots and no such rectangle. With perimeter 30, the area is largest when the discriminant is zero: 225 − 4A = 0 gives A = 56.25 m², a 7.5 m square.
Which mistakes cost marks?
Writing time as speed ÷ distance. Time = distance ÷ speed.
Mixing units, such as minutes with km/h.
Adding years to only one person in age problems.
Adding the path width once instead of twice, or putting the path inside when it is outside.
Rejecting a root without saying why. Write "speed cannot be negative".
Skipping the check in the original words.
A practice routine
Days 1 and 2: only step 3. For five NCERT situations, write the equality in words, then as an equation. Do not solve yet.
Days 3 and 4: solve them, rejecting roots with reasons, and check in words.
Day 5: do Exercise 4.3 questions 3 to 5 using the discriminant first.
Day 6: write one speed, age and area problem of your own, with numbers that factorise.
For factorisation and the formula themselves, see our guide to quadratic equations in Class 10. Parents can use our tips on helping with word problems.
When a word problem will not turn into an equation, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Start now: rewrite the cyclist problem with 3 km/h less and 1 hour more, and solve it.
