Question:

A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) if nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240kg of phosphoric acid at least 270kg of potash and at most 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Answer:

1. Calculate the amount of nitrogen, phosphoric acid, potash and chlorine in one bag of each brand: Brand P: Nitrogen = 20kg, Phosphoric Acid = 30kg, Potash = 40kg, Chlorine = 10kg Brand Q: Nitrogen = 10kg, Phosphoric Acid = 20kg, Potash = 30kg, Chlorine = 20kg

2. Determine the minimum number of bags of each brand needed to meet the garden’s needs: Brand P: 8 bags (240kg of phosphoric acid), 6 bags (240kg of potash) and 3 bags (30kg of chlorine) Brand Q: 24 bags (240kg of phosphoric acid), 9 bags (270kg of potash) and 15 bags (310kg of chlorine)

3. Calculate the minimum amount of nitrogen added to the garden: Brand P: 8 x 20kg = 160kg Brand Q: 24 x 10kg = 240kg Total: 400kg

4. Answer: The grower should use 8 bags of Brand P and 24 bags of Brand Q to minimize the amount of nitrogen added to the garden. The minimum amount of nitrogen added to the garden is 400kg.

Question:

A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) if nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240kg of phosphoric acid at least 270kg of potash and at most 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden? kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Answer:

Step 1: Calculate the amount of phosphoric acid, potash and chlorine needed from each type of fertilizer.

Brand P: Phosphoric acid: 240 kg Potash: 270 kg Chlorine: 310 kg

Brand Q: Phosphoric acid: 480 kg Potash: 270 kg Chlorine: 620 kg

Step 2: Calculate the number of bags of each brand needed to meet the requirements.

Brand P: Phosphoric acid: 240 kg/1 kg = 240 bags Potash: 270 kg/3 kg = 90 bags Chlorine: 310 kg/1.5 kg = 206.67 bags

Brand Q: Phosphoric acid: 480 kg/2 kg = 240 bags Potash: 270 kg/1.5 kg = 180 bags Chlorine: 620 kg/2 kg = 310 bags

Step 3: Calculate the minimum amount of nitrogen added in the garden.

Brand P: 240 bags x 3 kg = 720 kg Brand Q: 180 bags x 3.5 kg = 630 kg

Total: 1350 kg

Minimum amount of nitrogen added in the garden: 1350 kg

Question:

A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin content of one kg of food is given below: Food Vitamin A Vitamin B Vitamin C X 1 2 3 Y 2 2 1 One kg of food X costs Rs.16 and one kg of food Y costs Rs.20. Find the least cost of the mixture which will produce the required diet?

Answer:

Step 1: Calculate the amount of vitamins present in 1 kg of each food.

Food X: Vitamin A = 1 unit, Vitamin B = 2 units, Vitamin C = 3 units Food Y: Vitamin A = 2 units, Vitamin B = 2 units, Vitamin C = 1 unit

Step 2: Calculate the amount of each food required to get the required amount of vitamins.

Vitamin A: 10 units = (1 kg of food X) + (5 kg of food Y) Vitamin B: 12 units = (6 kg of food X) + (6 kg of food Y) Vitamin C: 8 units = (3 kg of food X) + (8 kg of food Y)

Step 3: Calculate the cost of the mixture.

Cost of food X = 16 x 3 kg = Rs.48 Cost of food Y = 20 x 8 kg = Rs.160

Total cost = Rs.48 + Rs.160 = Rs.208

Question:

A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) if nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240kg of phosphoric acid at least 270kg of potash and at most 310 kg of chlorine. If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added? kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Answer:

Step 1: Determine the number of bags of Brand P needed to meet the phosphoric acid and potash requirements.

The garden needs at least 240kg of phosphoric acid and at least 270kg of potash. Each bag of Brand P contains 1kg of phosphoric acid and 3kg of potash, so the grower will need to add at least 240 bags of Brand P to meet the requirements.

Step 2: Determine the number of bags of Brand Q needed to meet the chlorine requirement.

The garden needs at most 310 kg of chlorine. Each bag of Brand Q contains 2kg of chlorine, so the grower will need to add at most 155 bags of Brand Q to meet the requirement.

Step 3: Determine the total number of bags of each brand needed to maximize the amount of nitrogen added to the garden.

The grower needs to add 240 bags of Brand P and 155 bags of Brand Q to maximize the amount of nitrogen added to the garden.

Step 4: Determine the maximum amount of nitrogen added.

The maximum amount of nitrogen added is 822kg (240 bags of Brand P x 3kg of nitrogen per bag + 155 bags of Brand Q x 3.5kg of nitrogen per bag).

Question:

An aeroplane can carry a maximum of 200 passengers. A profit of Rs.1000 is made on each executive class ticket and a profit of Rs.600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit? A 136000 B 1360000 C 13600 D 1360

Answer:

Answer: C 13600

Step 1: Determine how many tickets of each type must be sold to maximize the profit for the airline.

Executive Class: 20 tickets

Economy Class: 80 tickets

Step 2: Calculate the maximum profit for the airline.

Maximum Profit = (20 x 1000) + (80 x 600)

Maximum Profit = 13600

Question:

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below Type of toys Machine I Machine II Machine III A 12 18 6 B 6 0 9 Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs.7.50 and that on each toy of type B is Rs.5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Answer:

1. Calculate the total time required to manufacture each type of toy:

Toy A: 12 + 18 + 6 = 36 minutes Toy B: 6 + 0 + 9 = 15 minutes

2. Calculate the maximum number of toys that can be manufactured in a day:

Toy A: 6 hours/36 minutes = 10 toys/hour = 60 toys/day Toy B: 6 hours/15 minutes = 24 toys/hour = 144 toys/day

3. Calculate the maximum profit:

Toy A: 60 toys x Rs.7.50 = Rs.450 Toy B: 144 toys x Rs.5 = Rs.720

Total profit = Rs.1170

4. Show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit:

Toy A: 15 toys x Rs.7.50 = Rs.112.50 Toy B: 30 toys x Rs.5 = Rs.150

Total profit = Rs.262.50

Therefore, 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Question:

A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs. 12 and Rs. 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximize the profit?

Answer:

1. Define the variables: Let x = number of dolls of type A produced per week Let y = number of dolls of type B produced per week

2. Define the constraints: x + y ≤ 1200 y ≤ 0.5x x ≤ 3y + 600

3. Define the objective function: Maximize 12x + 16y

4. Solve the problem: The solution to this problem is x = 600 and y = 300, which will maximize the profit.

Question:

An oil company has two depots A and B with capacities of 7000L and 4000L respectively. The company is to supply oil to three petrol pumps D,E and F whose requirements are 4500L,3000L and 3500L respectively. The distance (in km) between the depots and the petrol pumps is given in the following table: Distance (in km) From/To A B D 7 3 E 6 4 F 3 2 Assuming that the transportation cost of 10 litres of oil is Re.1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Answer:

Step 1: Calculate the total transportation cost of supplying oil from depot A and B to the petrol pumps.

From Depot A to Petrol Pump D: 4500L x (7km x Re.1) = Rs.31500 From Depot A to Petrol Pump E: 3000L x (6km x Re.1) = Rs.18000 From Depot A to Petrol Pump F: 3500L x (3km x Re.1) = Rs.10500

From Depot B to Petrol Pump D: 4500L x (3km x Re.1) = Rs.13500 From Depot B to Petrol Pump E: 3000L x (4km x Re.1) = Rs.12000 From Depot B to Petrol Pump F: 3500L x (2km x Re.1) = Rs.7000

Step 2: Schedule the delivery in such a way that the transportation cost is minimum.

From Depot A to Petrol Pump D: 4500L From Depot A to Petrol Pump E: 3000L From Depot B to Petrol Pump F: 3500L

Step 3: Calculate the minimum cost.

From Depot A to Petrol Pump D: 4500L x (7km x Re.1) = Rs.31500 From Depot A to Petrol Pump E: 3000L x (6km x Re.1) = Rs.18000 From Depot B to Petrol Pump F: 3500L x (2km x Re.1) = Rs.7000

Minimum cost = Rs.31500 + Rs.18000 + Rs.7000 = Rs.56500

Question:

A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs.250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs.200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A,B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag> What is minimum cost of the mixture per bag?

Answer:

Answer:

Step 1: Calculate the cost required for meeting the minimum requirements of each nutritional element.

Cost of A = 18 units x Rs.250/3 units = Rs.1500 Cost of B = 45 units x Rs.250/2.5 units = Rs.1800 Cost of C = 24 units x Rs.250/2 units = Rs.1200

Step 2: Calculate the number of bags of each brand required to meet the minimum requirements.

Number of bags of P required = 18 units/3 units + 45 units/2.5 units + 24 units/2 units = 10 + 18 + 12 = 40 bags Number of bags of Q required = 18 units/1.5 units + 45 units/11.25 units + 24 units/3 units = 12 + 4 + 8 = 24 bags

Step 3: Calculate the minimum cost of the mixture per bag.

Minimum cost of the mixture per bag = (40 x Rs.250 + 24 x Rs.200)/(40 + 24) = Rs.233.33