Question

A shopkeeper deals in two items-thermos-flasks and air tight containers. A flask costs him Rs. 120 and an air tight container costs him Rs. 60. He has at the most Rs. 12,000 to invest and has space to store a maximum of 150 items. The profit on selling a flask is Rs. 20 and an air tight container is Rs. 15. Assuming that he will be able to sell all things h buys, how many of each items should he buy to maximise his profit? Solve the problem graphically.

Answer 1

The shopkeeper needs to but 50 flasks and 100 containers

Step-by-step explanation:

The shopkeeper must buy 100 airtight containers and 50 flasks.

Step-by-step explanation:

Let x be the no. of flasks and y be the airtight containers that for Rs. 12000  

Given:  

X + y = 150

X = 150-y ………………….(1)

Also;

120 x + 60 y =12000…………………….(2)

Substituting the value of x in eq 2

120 (150 – y) + 60y = 12000

18000 – 120 y +60 y = 12000

60y = 6000

Y = 100

Thus, the number of containers is 100

No. of flasks = 150 – No. of containers  

     = 150 – 100 = 50

Thus, the shopkeeper needs to but 50 flasks and 100 containers