Ramlal bought 10 acres of land for Rs. 250000 in 2018. That year he cultivated rice and wheat in the
10 acres with the ratio of area under rice and wheat being 5:4. The profit obtained from rice and wheat
was in the ratio 3:2 with the total profit being Rs. 58500. This was 15% of the amount he invested in
cultivation that year. The next year he again cultivated rice and wheat, with the areas being same as
before and reaped a profit of Rs. 66000 in total with that from rice and wheat being in the ratio 8:7 but
his return on his investment that year was only 14%.
What is the amount invested by Ramlal for cultivation in 2018?
a. Rs. 365000
b. Rs. 375000
c. Rs. 380000
d. Rs. 390000
Answers
Answer:
let x be the acres of wheat planted and
y be the acres of rye planted
Given that there are a total of 10 acres of land to plant.
Atleast 7 acres is to be planted i.e., x+y≥7
Given that the cost to plant one acre of wheat is $200
Therefore, the cost for x acres of wheat is 200x
Given that the cost to plant one acre of rye is $100
Therefore, the cost for y acres of rye is 100y
Given that, an amount for planting wheat and rye is $1200
Therefore the total cost to plant wheat and rye is 200x+100y≤1200⟹2x+y≤12
Given that, the time taken to plant one acre of wheat is 1 hr
Therefore, the time taken to plant x acres of wheat is x hrs
Given that, the time taken to plant one acre of rye is 2 hrs
Therefore, the time taken to plant y acres of rye is 2y hrs
Given that, the total time for planting is 12 hrs
Therefore, the total time to plant wheat and rye is x+2y≤12
Given that, one acre of wheat yields a profit of $500
Therefore, the profit from x acres of wheat is 500x
Given that, one acre of rye yields a profit of $300
Therefore, the profit from y acres of wheat is 300y
therefore the total profit from the wheat and rye is P=500x+300y
In the above figure, the blue shaded region is the feasible region with three corner points.(4,4),(2,5),(5,2)
Now substituting the corner points the profit equation,
substituting (4,4)⟹P=500x+300y=500(4)+300(4)=3200
substituting (2,5)⟹P=500x+300y=500(2)+300(5)=2500
substituting (5,2)⟹P=500x+300y=500(5)+300(2)=3100