Read the following scenario and answer the THREE questions that follow.
Brijbhushan, a microfinancier, lends money at the rate of Rs.10 per square meter to small farmers at a village. He charges an annual interest rate of 10%. All the farming plots in that village are rectangular, with areas varying between a minimum of 1000 square meters and a maximum of 10,000 square meters.
This year, Brijbhushan has lent money only to five farmers: Aditya, Binod, Chhuttan, Dabloo and Govind. The perimeter of Chhuttan’s plot is 250 meters, with the length and width being at a ratio of 4:1. Aditya’s plot has an area three times the area of Govind’s plot. The area of Aditya’s plot is also the average of the areas of Govind’s plot and Dabloo’s plot. The plots belonging to Aditya, Binod and Dabloo are of the same width, but of different lengths. Moreover, the length of Binod’s plot is the sum of the lengths of Aditya’s plot and Dabloo’s plot.
If the width of the Aditya’s plot is 25 meters, what is the MINIMUM possible length of Binod’s plot?
Solution
Let the name of farmers be A,B,C,D,G for simplicity.
Let the area of G's plot be x $$m^2$$.
Area of A's plot = 3x $$m^2$$
Let area of D's plot be y $$m^2$$.
The area of A's plot is the average of the area of D's and G's plot.
3x = $$\ \frac{\ x+y}{2}$$
y = 5x $$m^2$$
The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y.
We got the following table.
From A's plot, $$l1\times\ y=3x$$
$$l1=\ \frac{\ 3x}{y}$$
From D's plot, $$l3\times\ y=5x$$
$$l3=\ \frac{\ 5x}{y}$$
The length of B's plot is the sum of the length of A's plot and D's plot.
l2 = l1 + l3 = $$\frac{\ 8x}{y}$$
Hence, the area of B's plot = $$l2\times\ y=8x$$
The required final table is :
The width of A's plot = 25m
y = 25m
For B's plot, $$l2\times\ 25=8x$$
To minimize the value of l2, we have to minimize the value of x.
Minimum possible area of land among the farmers = 1000 $$m^2$$
Smallest area of land = G's land = x $$m^2$$
x = 1000
Hence, minimum value of l2 = $$\ \frac{\ 8\times\ 1000}{25}=320\ m$$
$$\therefore\ $$ The required answer is B.
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