Question

A property dealer bought a rectangular piece of land at 1000/sq. ft. The length of the plot is less than twice its breadth. Due to its size, there were no buyers for the full plot. Hence he decided to sell it in smaller sized pieces as given below. The largest square from one end was sold at 1200/sq. ft. From the remaining rectangle the largest square was sold at 1150/sq. ft. Due to crash in the property prices, the dealer found it difficult to make profit from the sale of the remaining part of the land. If the ratio of the perimeter of the remaining land to the perimeter of the original land is 3 : 8, at what price (in ₹) the remaining part of the land is to be sold such that the dealer makes an overall profit of 10%?

• A. 500/sq. ft.
• B. 550/sq. ft.
• C. 600/sq. ft.
• D. 650/sq. ft.
• E. None of the above.

Hint:

After successive largest-square cuts from a rectangle with \(B<L<2B\), the leftover rectangle is \((2B-L)\times(L-B)\).
Use the perimeter ratio to express \(L\) in terms of \(B\) before computing areas and prices.

Answer 1

The Correct Option is B

Step 1: Geometry of cuts

Let the original rectangle be \(L \times B\) with area \(LB\). The first largest square cut is \(B \times B\). Remaining rectangle is \((L-B) \times B\). Next largest square from this is of side \(L-B\) (since \(L < 2B \Rightarrow L-B\)).

Perimeter ratio: \[ \frac{2[(2B-L)+(L-B)]}{2(L+B)} = \frac{2B}{2(L+B)} = \frac{B}{L+B} = \frac{3}{8} \Rightarrow L = \frac{5}{3}B \]

Step 2: Areas of parts

Original area: \[ LB = \tfrac{5}{3}B^2 \]

First square area: \[ B^2 \]

Second square area: \[ (L-B)^2 = \left(\tfrac{2}{3}B\right)^2 = \tfrac{4}{9}B^2 \]

Remaining rectangle area: \[ (2B-L)(L-B) = \left(\tfrac{1}{3}B\right)\left(\tfrac{2}{3}B\right) = \tfrac{2}{9}B^2 \]

Step 3: Revenue and required price

Cost price at ₹1000/sq ft: \[ C = 1000 \cdot \tfrac{5}{3}B^2 \]

Selling prices: ₹1200 for the first square and ₹1150 for the second square. Let \(x\) be the price for the remaining part.

Total revenue: \[ R = B^2 \left( 1200 + \tfrac{4}{9} \cdot 1150 + \tfrac{2}{9}x \right) \]

Overall profit \(10\%\): \[ R = 1.1C = \tfrac{5500}{3}B^2 \]

Hence, \[ 1200 + \tfrac{4600}{9} + \tfrac{2}{9}x = \tfrac{5500}{3} \] \[ \tfrac{15400 + 2x}{9} = \tfrac{16500}{9} \] \[ 2x = 1100 \quad \Rightarrow \quad x = 550 \]

Final Answer:

The remaining land must be sold at \[ \boxed{₹ \; 550 \; \text{per sq ft}} \]