(A) 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.

Q: How do I decide which root to keep?

The quadratic gives $x = 50 \text{ and } x = -40$. Since x represents a speed and speed can never be negative, we reject x = −40 and keep $x = 50\text{ km/hr}$. Always state the reason for rejection — CBSE awards marks for this justification, not just for the arithmetic.

Q: Can I solve this using the quadratic formula instead of factorisation?

Yes. For $x^2 - 10x - 2000 = 0$, the quadratic formula gives $x = \frac{10 \pm \sqrt{100+8000}}{2} = \frac{10\pm 90}{2}, \text{ so } x = 50 \text{ or } -40$. The formula and factorisation both earn full marks; use whichever you find quicker and less error-prone.

Quadratic Equations

Word problem on speed and time

Official Answer

$\text{Speed of the faster train} = 50\text{ km/hr}$ and $\text{speed of the slower train} = 40\text{ km/hr}$ .

quadratic equationspeed and time200 km50 km/hr40 km/hrdistance by speedfactorisationreject negative root

Marking Scheme

11 mark: correctly assuming the variable and expressing both speeds (x and $x - 10$ ).
21 mark: forming the correct equation $\frac{200}{x-10} - \frac{200}{x} = 1$ .
31 mark: reducing to the quadratic equation $x^2 - 10x - 2000 = 0$ .
41 mark: solving correctly by factorisation/quadratic formula to get $x = 50$ (rejecting $x = -40$ with reason).
51 mark: stating both speeds — $\text{faster} = 50\text{ km/hr}$ , $\text{slower} = 40\text{ km/hr}$ (verification acceptable as alternative for final mark).

Hint

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

Quick Oral Answer

Let the faster train's speed be x km/hr, so the slower is $(x-10)\text{ km/hr}$ ; since the slower takes one hour more, $\frac{200}{x-10} - \frac{200}{x} = 1$ gives $x^2-10x-2000=0$ , so $x=50$ , making the speeds $50\text{ km/hr and }40\text{ km/hr}$ .

Analysis & Explanation

A speed–time word problem that converts into a quadratic equation using time = distance ÷ speed.

Concept

Same distance (200 km) is covered by both trains, so express both speeds in one variable: $\text{faster} = x, \text{slower} = x - 10$ .
"Faster takes 1 hour less" translates to (slower time − faster time = 1), i.e. $\frac{200}{x-10} - \frac{200}{x} = 1$ .
Clearing fractions reduces the problem to $x^2 - 10x - 2000 = 0$ , which factorises to $(x-50)(x+40) = 0$ .

Common Mistakes

Writing the time-difference equation in the wrong order (sign error).
Introducing two separate variables instead of one, leaving the equation unsolvable.
Forgetting to reject the negative root $x = -40$ since speed cannot be negative — CBSE awards a mark specifically for this rejection.

Real-World Relevance

The same inverse speed–time relationship for a fixed distance underlies railway timetabling, overtaking calculations and journey planning.
Verification: $\frac{200}{40} - \frac{200}{50} = 5 - 4 = 1\text{ hour}$ , confirming the answer and securing the accuracy mark.

Common Mistakes

1Writing the time-difference equation in the wrong order (faster − slower instead of slower − faster), which flips the sign and gives an unsolvable or negative result.
2Forgetting to reject the negative root $x = -40$ ; a speed cannot be negative, and CBSE deducts a mark if the rejection is not stated with a reason.
3Errors in taking the LCM of the fractions, e.g. incorrectly simplifying $200[x - (x-10)]$ and losing the factor of 10.

Interesting Facts

The relationship 'for a fixed distance, time is inversely proportional to speed' is the mathematical backbone of every railway timetable — a 20% increase in speed does not cut the time by 20%, which is why these problems yield quadratics rather than simple linear equations.

India's fastest train, the Vande Bharat Express, can exceed 160 km/hr, while older passenger trains average around 45–55 km/hr — the 50 vs 40 km/hr answer here mirrors real slower/faster train speed gaps on shared tracks.

The general form of such problems, $\frac{200}{x-10} - \frac{200}{x} = t$ , always reduces to a quadratic because the two $1/x$ terms combine over a common denominator $x(x-10)$ , producing an $x^2$ term.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why do we express the slower train's speed as (x − 10) instead of using a second variable?

Because the problem gives a direct relation between the two speeds (slower is 10 km/hr less than faster), we can express both in terms of one variable x. This keeps a single equation with one unknown. Using two variables would require a second equation and needlessly complicate a problem that is designed to be a single quadratic.

How do I decide which root to keep?

The quadratic gives $x = 50 \text{ and } x = -40$ . Since x represents a speed and speed can never be negative, we reject x = −40 and keep $x = 50\text{ km/hr}$ . Always state the reason for rejection — CBSE awards marks for this justification, not just for the arithmetic.

Can I solve this using the quadratic formula instead of factorisation?

Yes. For $x^2 - 10x - 2000 = 0$ , the quadratic formula gives $x = \frac{10 \pm \sqrt{100+8000}}{2} = \frac{10\pm 90}{2}, \text{ so } x = 50 \text{ or } -40$ . The formula and factorisation both earn full marks; use whichever you find quicker and less error-prone.