Question

A person on tour has ₹ 5,400 for his expenses. If he extends his tour by 5 days, he has to cut down his daily expenses by ₹ 180. Find the original duration of the tour and daily expense.

Hint:

For word problems leading to quadratics, often identifying the "difference" equation (e.g., \(E_1 - E_2 = \text{diff}\)) is the easiest way to set up the problem correctly.

Answer 1

Step 1: Understanding the Concept:
This problem can be modeled as a quadratic equation relating time and expense per unit time.
Step 2: Key Formula or Approach:
Let the original duration be \(n\) days.
Daily expense = \(\frac{\text{Total budget}}{\text{Number of days}}\).
Step 3: Detailed Explanation:
Original daily expense = \(\frac{5400}{n}\).
New duration = \(n + 5\).
New daily expense = \(\frac{5400}{n + 5}\).
Given: New expense = Original expense - 180.
\[ \frac{5400}{n} - \frac{5400}{n + 5} = 180 \]
Divide the whole equation by 180:
\[ \frac{30}{n} - \frac{30}{n + 5} = 1 \]
\[ 30 \left[ \frac{n + 5 - n}{n(n + 5)} \right] = 1 \]
\[ \frac{150}{n^2 + 5n} = 1 \Rightarrow n^2 + 5n - 150 = 0 \]
Factorizing the quadratic equation:
\[ n^2 + 15n - 10n - 150 = 0 \]
\[ n(n + 15) - 10(n + 15) = 0 \Rightarrow (n - 10)(n + 15) = 0 \]
Since \(n\) cannot be negative, \(n = 10\).
Daily expense = \(5400 / 10 = ₹ 540\).
Step 4: Final Answer:
Original duration was 10 days and daily expense was ₹ 540.