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

Given:
Total money for expenses = Rs 4200
If the person extends the tour by 3 days, he must reduce daily expenses by Rs 70.

Let the original duration of the tour be \( x \) days.
Let the original daily expense be \( d \) rupees.

Then, \[ d \cdot x = 4200 \quad (1) \]
After extending the trip by 3 days, new number of days = \( x + 3 \)
New daily expense = \( d - 70 \)

Total expense remains the same: \[ (d - 70)(x + 3) = 4200 \quad (2) \]
Step 1: Expand equation (2)
\[ (d - 70)(x + 3) = dx + 3d - 70x - 210 = 4200 \]
From (1), \( dx = 4200 \). Substitute:
\[ 4200 + 3d - 70x - 210 = 4200 \] Simplify: \[ 3d - 70x - 210 = 0 \] \[ 3d - 70x = 210 \quad (3) \]
Step 2: Use equation (1)
\[ d = \frac{4200}{x} \] Substitute into (3):
\[ 3\left(\frac{4200}{x}\right) - 70x = 210 \] \[ \frac{12600}{x} - 70x = 210 \] Multiply both sides by x: \[ 12600 - 70x^2 = 210x \] Rearrange: \[ 70x^2 + 210x - 12600 = 0 \] Divide entire equation by 70: \[ x^2 + 3x - 180 = 0 \]
Step 3: Solve the quadratic
\[ x^2 + 3x - 180 = 0 \] Factorise: \[ (x + 15)(x - 12) = 0 \] So: \[ x = 12 \quad \text{or} \quad x = -15 \] Negative value is not possible for days.

Final Answer:
The original duration of the tour was:
12 days