Linear Programming - Online Test

Q1. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costsRs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

A. 130 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1300

B. 120 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1200

C. 100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000

D. 110 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1100

Answer : Option C
Explaination / Solution:

Let number of kgs. of fertilizer F1 = x
And number of kgs. of fertilizer F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 6x +5y , subject to the constraints : 10/100 x + 5/100y ≥ 14 and 6/100x + 10/100y ≥ 14, i.e. 2 x + y ≥ 280 and 3x + 5y ≥ 700, x,y ≥ 0.,

Corner points	Z =6x +5 y
A ( 0 , 280 )	1400
D(700/3,0 )	1400
B(100,80)	1000………….(Min.)

Corner points Z =6x +5 y A ( 0 , 280 ) 1400 D(700/3,0 ) 1400 B(100,80) 1000………….(Min.) Here Z = 1000 is minimum.
i.e. 100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000.

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Q2. The corner points of the feasible region determined by the following system of linear inequalities:2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is

A. q = 3p

B. p = 2q

C. p = q

D. p = 3q

Answer : Option A
Explaination / Solution:

Here Z = px +qy , subject to constraints :
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0
As it is given that Z is maximum at ( 3 ,4 ) and ( 0, 5 ).
Therefore , 3p + 4q = 0p + 5q , which gives 3p = q .

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Q3. 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 element 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 the minimum cost of the mixture per bag?

A. 5 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2250

B. 4 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2150

C. 6 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs2350

D. 3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950

Answer : Option D
Explaination / Solution:

Let number of bags of cattle feed of brand P = x
And number of bags of cattle feed of brand Q = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 250x +200y , subject to the constraints : 3 x + 1.5y ≥ 80, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24 , x,y ≥ 0.,

Corner points	Z =250x +200 y
C(0 , 12 )	2400
B (18,0)	4500
D(3,6 )	1950…………………(Min.)
A(9,2)	2650

Here Z = 1950 is minimum.
i.e. 3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950 .

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Q4.

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 contents of one kg 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?

A. Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).

B. Least cost of the mixture is Rs 142 (2 kg of Food X and 5 kg of food Y).

C. Least cost of the mixture is Rs 132 (3 kg of Food X and 4 kg of food Y).

D. Least cost of the mixture is Rs 122 (2 kg of Food X and 4 kg of food Y).

Answer : Option A
Explaination / Solution:

Let number of kgs of food of brand X = x
And number of kgs. of food of brand Y = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 16x +20y , subject to the constraints : x + 2y ≥ 10,2x + 2y ≥12,3x + y ≥8,x,y ≥ 0.,

Corner points	Z =16x +20 y
C(10 , 0 )	160
B (0,8)	160
D(1,5 )	116
A(2,4)	112……………..(Min.)

Here Z = 112 is minimum.
i.e. Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).

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Q5. 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 classthan by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

A. 40 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 136000

B. 45 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 156000

C. 44 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 146000

D. 42 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 139000

Answer : Option A
Explaination / Solution:

Let number of executive class tickets = x
And number of economy class tickets = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 1000x +600y , subject to the constraints : x + y ≤ 200, 4x - y ≤ 0, x ≥ 20,x,y ≥ 0.,

Corner points	Z =1000x +600 y
C(20 ,80 )	68000
B (40,160)	136000
D(20,180 )	12800

Here Z = 136000 is maximum.
i.e. 40 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 136000 .

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Q6. Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).Let F = 4x + 6y be the objective function. Maximum of F – Minimum of F =

A. 60

B. 18

C. 48

D. 42

Answer : Option A
Explaination / Solution:

Here the objective function is given by : F = 4x +6y .

Corner points	Z = 4x +6 y
(0, 2 )	12………………..(Min.)
(3,0)	12………………….(Min.)
(6,0 )	24
(6 , 8 )	72
(0 , 5 )	30

Maximum of F – Minimum of F = 72 – 12 = 30 .

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Q7. Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px+qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is

A. p = q

B. p = 3q

C. p = 2q

D. P = q/2

Answer : Option D
Explaination / Solution:

We have Z = px + qy , At ( 3, 0 ) Z = 3p ……………………………….(1) At ( 1 , 1) Z = p + q …………………………(2) Therefore , from (1) and (2) : We have : p = q/2 .

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Q8. In a LPP, the linear inequalities or restrictions on the variables are called

A. Limits

B. Constraints

C. Linear constraints

D. Inequalities

Answer : Option C
Explaination / Solution:

In a LPP, the linear inequalities or restrictions on the variables are called Linear constraints.

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Q9. Determine the maximum value of Z = 3x + 4y if the feasible region (shaded) for a LPP is shown in Figure above.