Question

Rakhal is looking for a field where he can graze his cow. He finds a local farmer, Gopal, who agrees to rent his field to Rakhal for Rs. 1000 a year. Rakhal finds a post in the field and ties his cow to the post with a 25-foot rope. After some months, Gopal tells Rakhal that he will build a shed with four walls on the field with the post as one of the corner posts. The shed would be 15 feet by 10 feet. Rakhal agrees but he realizes that this arrangement would reduce the available area for grazing. What should be the modified rent to compensate for this loss of grazing area if Rakhal has to keep the cow tied to the same post with the same rope?

• A. Rs. 800
• B. Rs. 880
• C. Rs. 888
• D. Rs. 930
• E. None of the above

Hint:

When grazing is reduced by obstacles, think in terms of fraction of circle area removed or replaced. Rent or cost is usually proportional to usable area.

Answer 1

The Correct Option is B

Step 1: Original grazing area.
Cow tied with rope of length 25 ft. So grazing area = circle of radius 25 ft. \[ A_1 = \pi (25)^2 = 625\pi \; \text{sq. ft.} \]

Step 2: Effect of shed.
A shed of 15 × 10 ft is built with post as one corner. This removes $\tfrac{1}{4}$ of the grazing area (since shed blocks one quadrant). So new accessible area = $\tfrac{3}{4}\pi (25)^2 + \tfrac{1}{4}(\pi (10^2)+\pi (15^2))$.

Step 3: Simplify new area.
\[ A_2 = \frac{3}{4}\pi (625) + \frac{1}{4}[\pi(100)+\pi(225)] \] \[ = \frac{1875}{4}\pi + \frac{325}{4}\pi = \frac{2200}{4}\pi = 550\pi \]

Step 4: Proportional rent.
Rent is proportional to grazing area. Original rent = Rs. 1000 for $625\pi$. New rent = ? for $550\pi$. \[ \text{New Rent} = 1000 \times \frac{550}{625} = 880 \]

Final Answer: \[ \boxed{Rs. \; 880} \]