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

Let the original duration of the tour = x days.

Daily expenses = ₹ 4200/x

If tour is extended by 3 days, new duration = (x + 3) days
New daily expenses = ₹ 4200/(x + 3)

Setting up the equation:

According to the condition, daily expenses are cut by ₹ 70:

$$\frac{4200}{x} - \frac{4200}{x+3} = 70$$

$$4200(x+3) - 4200x = 70 \cdot x(x+3)$$

$$4200x + 12600 - 4200x = 70x^2 + 210x$$

$$12600 = 70x^2 + 210x$$

$$x^2 + 3x - 180 = 0$$

Factorising:

$$x^2 + 15x - 12x - 180 = 0$$

$$x(x + 15) - 12(x + 15) = 0$$

$$(x - 12)(x + 15) = 0$$

$$x = 12 \quad \text{or} \quad x = -15$$

Since duration cannot be negative, x = 12.

∴ The original duration of the tour is 12 days.

Source: Chapter 4, Quadratic Equations

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• Examiners award marks at each stage: forming the equation (1–2 marks), simplifying to standard form (1 mark), factorising correctly (1 mark), and rejecting the negative root with conclusion (1 mark).
• Always reject the negative value explicitly, stating "duration cannot be negative."
• Dividing both sides by 70 early keeps numbers manageable and avoids arithmetic errors.