Question 85

# 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%?

Solution

Let length = $$l$$ ft and breadth = $$b$$ ft

ABCD is original plot. First square AEFD is sold of side = DF = $$b$$, => FC = $$(l - b)$$

After that square EBHG is sold of side GH = $$(l - b)$$

=> HC = $$b - (l - b) = (2b - l)$$

Perimeter of remaining land GHCF = $$2 \times [(l - b) + (2b - l)]$$

= $$2b$$

Perimeter of original land ABCD = $$2 (l + b)$$

Acc to ques

=> $$\frac{2b}{2 (l + b)} = \frac{3}{8}$$

=> $$8b = 3l + 3b$$

=> $$5b = 3l$$

Only one combination is possible, i.e. $$l = 5$$ and $$b = 3$$     ($$\because$$ It is given that : $$l < 2b$$)

=> Area of land = $$5 \times 3 = 15$$ sq. ft

=> Cost of land = $$15 \times 1000 = Rs. 15,000$$

For overall profit to be 10 %, S.P. = $$15,000 \times \frac{110}{100}$$

= $$Rs. 16,500$$

Side of AEFD = $$b = 3$$

=> S.P. of AEFD = $$3^2 \times 1200 = Rs. 10,800$$

Side of EBHG = $$(l - b) = 5 - 3 = 2$$

S.P. of EBHG = $$2^2 \times 1150 = Rs. 4,600$$

Length of remaining rectangular land GHCF = $$(l - b) = 5 - 3 = 2$$

Breadth = $$(2b - l) = 6 - 5 = 1$$

Let selling price per sq. ft of GHCF = $$Rs. x$$

S.P. of GHCF = $$2 \times 1 \times = Rs. 2x$$

$$\therefore$$ Total S.P.

=> $$10,800 + 4,600 + 2x = 16,500$$

=> $$2x = 16,500 - 15,400$$

=> $$x = \frac{1100}{2} = Rs. 550$$