A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.

Quadratic Equations

Speed-Time-Distance Word Problem (Quadratic Equations)

Official Answer

Speed of the faster train = 50 km/hr; speed of the slower train = 40 km/hr, obtained by solving $x^2 - 10x - 2000 = 0$ and rejecting the negative root.

quadratic equationspeed distance timefaster trainslower trainroots of a quadratic equationword problem200 km journeyrejecting negative root

Marking Scheme

11 mark: correctly assigning variable x to the faster train's speed and $(x - 10)$ to the slower train's speed.
21 mark: writing correct time expressions $\frac{200}{x}$ and $\frac{200}{x-10}$ .
31 mark: forming the correct equation $\frac{200}{x-10} - \frac{200}{x} = 1$ and simplifying to standard quadratic form $x^2 - 10x - 2000 = 0$ .
41 mark: solving the quadratic correctly using the formula/factorisation to get $x = 50$ (and rejecting $x = -40$ with reasoning).
51 mark: stating final answer with correct units (faster = 50 km/hr, slower = 40 km/hr), verification acceptable for full credit.

Hint

Let the faster train's speed be x km/hr; slower train's speed is $(x - 10)$ km/hr; use $\text{time} = \frac{\text{distance}}{\text{speed}}$ and set (time of slower) − (time of faster) = 1.

Quick Oral Answer

Taking the faster train's speed as x km/hr and the slower train's speed as $(x - 10)$ km/hr, and equating the time difference to 1 hour, we get the quadratic $x^2 - 10x - 2000 = 0$ , which gives $x = 50$ ; so the faster train runs at 50 km/hr and the slower train at 40 km/hr.

Analysis & Explanation

A classic speed–distance–time quadratic-equation word problem tested almost every year in CBSE Class 10 boards.

Concept & modelling

$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ ; the two verbal conditions ("1 hour less" and "10 km/hr less") must become one equation in a single variable x.
Equation: $\frac{200}{x-10} - \frac{200}{x} = 1$ , which simplifies to the clean quadratic $x^2 - 10x - 2000 = 0$ .
Discriminant $8100 = 90^2$ is a perfect square by design, so the roots are rational.

Common mistakes

Reversing which train's time is subtracted from which, flipping the sign of the equation.
Forgetting to reject the negative root $x = -40$ with a physical justification (speed cannot be negative) — a frequent mark-loss even when the quadratic is solved correctly.

Real-world connection

This relative-speed reasoning mirrors railway scheduling and logistics planning, where planners balance travel time against required speed increases.

Common Mistakes

1Reversing the subtraction order (writing $\frac{200}{x} - \frac{200}{x-10} = 1$ instead of the correct order), which produces a sign error and an incorrect quadratic.
2Failing to reject the negative root $x = -40$ , or not stating the reasoning that speed cannot be negative.
3Arithmetic slip while computing the discriminant ( $100 + 8000 = 8100$ ) or its square root (90), leading to wrong roots.

Interesting Facts

Speed–time–distance quadratic word problems have appeared in CBSE Class 10 Maths papers almost every year since the quadratic equations chapter was introduced in the current NCERT syllabus.

The discriminant in this problem, $8100 = 90^2$ , is a perfect square — CBSE question-setters commonly engineer such 'nice' numbers so that only the quadratic formula or factorisation (not a calculator) is needed.

Indian Railways' fastest trains (like the Vande Bharat Express) can run over 100 km/hr faster than regular passenger trains on the same route, making relative-speed problems like this directly relatable to real timetables.

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Frequently Asked Questions

Why is the faster train's speed taken as the variable x instead of the slower train's?

Either choice works mathematically, but taking the faster train's speed as x keeps the slower train's speed as a simple subtraction $(x - 10)$ with no fractions, making the resulting equation easier to simplify.

Why is the root x = −40 rejected?

Speed is a physical, non-negative quantity, so a negative value for x has no real-world meaning in this context and must be discarded even though it is a valid root of the quadratic equation.

Can this problem be solved by taking the slower train's speed as x instead?

Yes; if slower train's speed is x, then faster train's speed is $(x + 10)$ and time for slower minus time for faster still equals 1, leading to an equivalent quadratic that gives $x = 40$ (slower) and $x + 10 = 50$ (faster), the same final answer.