Question

A person on a tour has Rs 4,200 for expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by Rs 70. Find the original duration of the tour.

Hint:

In time-expenditure problems, the equation always follows the structure: \( \text{Higher Rate} - \text{Lower Rate} = \text{Difference in Rate} \).

Answer 1

Step 1: Understanding the Concept:
Total budget remains the same. If the number of days increases, the average daily expenditure must decrease.
Step 2: Key Formula or Approach:
Let the original duration of the tour be \( x \) days.
Daily expense = \( \frac{\text{Total Budget}}{\text{Duration}} \).
Step 3: Detailed Explanation:
Original daily expense = \( \frac{4200}{x} \).
New duration = \( x + 3 \) days.
New daily expense = \( \frac{4200}{x + 3} \).
According to the problem, the difference between original and new daily expenses is Rs 70.
\[ \frac{4200}{x} - \frac{4200}{x + 3} = 70 \]
Divide the entire equation by 70:
\[ \frac{60}{x} - \frac{60}{x + 3} = 1 \]
Multiply by \( x(x + 3) \) to clear denominators:
\[ 60(x + 3) - 60x = x(x + 3) \]
\[ 60x + 180 - 60x = x^2 + 3x \]
\[ 180 = x^2 + 3x \implies x^2 + 3x - 180 = 0 \]
Factoring the quadratic equation:
\[ x^2 + 15x - 12x - 180 = 0 \]
\[ x(x + 15) - 12(x + 15) = 0 \]
\[ (x - 12)(x + 15) = 0 \]
So, \( x = 12 \) or \( x = -15 \).
Since duration cannot be negative, we take \( x = 12 \).
Step 4: Final Answer:
The original duration of the tour was 12 days.