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Mathematics — CBSE Class 10 board question

Q1. [5]
A person on a tour has ₹ 4,200 for expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by ₹ 70. Find the original duration of the tour.
Previously asked in CBSE board exam
2026 30/2/1 Q33(a)
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer

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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Explanation
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