Mathematics — Standard (041) — CBSE Class 10 board question

Let the original speed of the aircraft = $x$ km/h.

Setting up the equation:

Original time = $\dfrac{2800}{x}$ hours

Reduced speed = $(x - 100)$ km/h, so new time = $\dfrac{2800}{x-100}$ hours

Since time increased by 30 minutes = $\dfrac{1}{2}$ hour:

$$\frac{2800}{x-100} - \frac{2800}{x} = \frac{1}{2}$$

$$2800\left(\frac{x - (x-100)}{x(x-100)}\right) = \frac{1}{2}$$

$$\frac{2800 \times 100}{x(x-100)} = \frac{1}{2}$$

$$x(x - 100) = 560000$$

$$x^2 - 100x - 560000 = 0$$

Solving by factorisation (or quadratic formula):

$$x = \frac{100 \pm \sqrt{10000 + 2240000}}{2} = \frac{100 \pm \sqrt{2250000}}{2} = \frac{100 \pm 1500}{2}$$

Taking positive value: $x = \dfrac{100 + 1500}{2} = 800$ km/h

(Negative value rejected as speed cannot be negative.)

Original duration of flight:

$$t = \frac{2800}{800} = 3.5 \text{ hours}$$

∴ The original duration of the flight is 3.5 hours (3 hours 30 minutes).

Source: Chapter 4, Quadratic Equations

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• Key step: Form the time-difference equation and simplify to get a standard quadratic $x^2 - 100x - 560000 = 0$.
• Examiner looks for: Correct variable definition, proper equation setup, valid algebraic steps, rejecting the negative root with a reason, and the final answer in hours.
• Convert 30 minutes to $\frac{1}{2}$ hour before forming the equation — a common careless mistake.
• Full marks require showing the rejection of the negative root explicitly.