AMITY (Assignment)

BBA

Semester 1

Business Mathematics

BBA

Semester 1

Business Mathematics

Business Mathematics

Case Study

In order to find out the inverse of the matrices below, answer the questions associated.

Case Study 1.

A= [4 -2 1 7 3 3 2 0 1]

Case Study 2.

B= [1 2 -3 2 3 2 3 -3 -4]

Question 1: Based on Case study 1 – Determinant value of Matrix A = _____

a. 8

b. 5

c. 0

d. 14

Question 2: Based on Case study 1 – Matrix A is a ________ matrix

a. 2 x 3

b. 3 x 2

c. 2 x 2

d. None of these

Question 3: Based on Case study 1 – Matrix A is ______ matrix

a. singular

b. non – singular

c. degenerate

d. indeterminate

Question 4: Based on Case study 1 – The value of C_22 (second row second column entry) in the Co – factor matrix of A is ______

a. 0

b. 1

c. 2

d. None of these

Question 5: Based on Case study 1 – Which amongst the following is not required in order to get the inverse of matrix A?

a. Adjoint of A

b. Determinant of A

c. Transpose of A

d. All of these

Question 6: Based on Case study 2 – The sum of first row entries (i.e. a_11, a_12, a_13) in the adjoint matrix of A is _______

a. 39

b. 26

c. 19

d. 21

Question 7: Based on Case study 2 – The value of second row third column entry in the adjoint matrix of A is ______

a. -1

b. -8

c. 3

d. -6

Question 8: Based on Case study 2 – The value of third row and third column entry in the inverse matrix of A is ______

a. -0.25

b. -0.089285714

c. -0.014925373

d. -0.044776119

Question 9: Formula for finding inverse of A is ______

a. [1/det(A)] * adj A

b. [1/det(A)] ^ adj A

c. [1*det(A)] / adj A

d. None of these

Question 10: Relation between Co – Factors and Minors can be written as _________

a. C_ij = (-1)^(i-j). M_ji

b. C_ij = (-1)^(i-j). M_ij

c. C_ij = (-1)^(i+j). M_ij

d. C_ij = (-1)^(j-i). M_i+j

MODULE II : DIFFERENTIAL CALCULUS

Case Study

Case Study 1. Consider XYZ manufacturer’s total cost function is given as C(x)= 2x^2+3x+1000. Answer the questions associated with this information.

Case Study 2. A monopolist’s demand function is p = 300 – 5x. Answer the questions associated with this information.

Case Study 3. A company XYZ has cost function for ‘x’ number of units as C(x) = x^3-(15/2)x^2+18x+10

Question 1: Based on Case Study 1 – The average cost function of XYZ manufacturer is __________

a. 2x+3+(100)/x

b. 2x+3+(1000)/x

c. 2+3x+(1000)/x

d. None of these

Q 2 Based on Case Study 1 – The formula for finding average cost is __________

a. C(x) * x

b. C(x) / x

c. 754

d. 234

Q 3 Based on Case Study 1 – The Marginal cost function of XYZ manufacturer is __________

a. 2x + 3

b. 4x – 3

c. 4x + 3

d. 3x + 4

Q 4 Based on Case Study 1 – The Marginal cost when 5 units are produced is __________

a. 21

b. 22

c. 23

d. 24

Q 5 Based on Case Study 2 – Relationship between the slope of MR and AR can be explained by_______

a. slope of MR is twice the slope of AR

b. slope of MR is thrice the slope of AR

c. slope of AR is twice the slope of MR

d. None of these

Q 6 Based on Case Study 2 – The Marginal revenue function is __________

a. x / 30

b. 10x – 300

c. 300 – 10x

d. 300 + 10x

Q 7 Based on Case Study 2 – The marginal revenue is zero at the price_______

a. 150

b. 100

c. 200

d. 50

Q 8 Based on Case Study 3 – Average cost for the number of units which minimize the marginal cost is_______

a. 20

b. 22.45

c. 24

d. 26

Q 9 Based on Case Study 3 – Number of units for which the Marginal Cost is minimum is_______

a. 1

b. 2.5

c. 3

d. 5

Q 10 Based on Case Study 3 – Total cost for the number of units which minimise the marginal cost is_______

a. 9.5

b. 11

c. 9

d. None of these

MODULE III: INTEGRAL CALCULUS

Case Study

Case Study 1. The marginal cost of producing x units of a commodity in a day is given as MC = 16x -1591. The selling price is fixed at Rs. 9 per unit and the fixed cost is Rs. 1,800 per day. Answer the questions associated with this information.

Case Study 2. A company suffers a loss of Rs. 560 if one of its product does not sell. Marginal revenue and Marginal cost functions for the product are given by MR = 170 – 25x and MC = -50 + 15x

Question 1: Based on Case Study 1 – Critical point for profit function can be found by equating dp/d to ______.

a. -1

b. 2

c. 1

d. 0

Question 2: Based on Case Study 1 – Critical point value for profit function is _______.

a. 120

b. 50

c. 99

d. 100

Question 3: Based on Case Study 1 – Determine the Cost function.

a. 8x^2 – 1591x + k

b. 4x^3 – 1591x + k

c. 8x^2 – 1580x + k

d. 8x^2 – 158x + k

Question 4: Based on Case Study 1 – Determine the maximum profit that can be attained in one day.

a. 67800

b. 78200

c. 76000

d. None of these

Question 5: Based on Case Study 1 – Determine the Profit Function.

a. -4x^2 + 1600x -1800

b. 8x^2 – 1600x -1800

c. -8x^2 + 1600x -1800

d. 8x^2 – 1600x +1800

Question 6: Based on Case Study 1 – Determine the Revenue function.

a. 9x

b. 2x

c. 15x

d. 3x

Question 7: Based on Case Study 2 – Determine the break even points.

a. x = 3 and x = 5

b. x = 4 and x = 7

c. x = 2 and x = 5

d. x = 5 and x = 9

Question 8: Based on Case Study 2 – Determine the total profit between the break even points.

a. 756

b. 825

c. 450

d. 945

Question 9: Based on Case Study 2 – Determine the total profit fucntion.

a. 220x + 20 x^2 -560

b. 220x – 20 x^2 + 560

c. 220x – 20 x^2 -560

d. None of these

Question 10: Based on Case Study 2 – The total profit between the break even points can be found by integrating the profit function between the limits of __________

a. break even points

b. 0 to 1

c. negative infinity to positive infinity

d. None of these

MODULE IV : MATHEMATICS OF FINANCE

Case Study

Case Study 1. Considering an amount of Rs. 100 deposited in the account, answer the questions associated with this deposit.

Case Study 2. If the principle amount of Rs. X is to be invested at 8 % per annum, so that after four years the amount will be Rs. 5000. Answer the questions associated with the given information.

Question 1: Based of Case Study 1 – Considering & ___________ will be the amount

at the effective rate of interest 3 % per annum for 10 years, 4 % per annum for 4 years and 5 % per annum for 2 years & as base question , answer that what constant force of interest would produce the same amount after 16 years at the rate in thebase question?

a. 0.0228

b. 0.02

c. 0

d. 0

Q 2 Based of Case Study 1 – ___________ will be the amount at the rate of interest 12 % per annum compounded quarterly for 10 years.

a. 245

b. 267

c. 326

d. 259

Q 3 Based of Case Study 1 -___________ will be the amount at the effective rate of interest 3 % per annum for 10 years, 4 % per annum for 4 years and 5 % per annum for 2 years.

a. 173.33

b. 154.3

c. 139

d. 175

Q 4 Based of Case Study 1 -___________ will be the amount at the force of interest 3% per annum, for 3.5 years.

a. 324

b. 237

c. 111

d. 544

Q 5 Based of Case Study 1 -___________ will be the amount at the rate of interest

corresponding to 3 % per annum effective rate of discount for 8 years.

a. 127.59

b. 158

c. 159

d. 130

Q 6 Based of Case Study 2 – Formula for finding principle when interest

compounded continuously is _______

a. A*e^(r/t)

b. A/e^(r*t)

c. A*e^(r*t)

d. A^e*(r*t)

Q 7 Based of Case Study 2 – Formula for finding principle when interest

compounded annually is _______

a. A(1+r)^(-n)

b. P(1+r)^(-n)

c. A(1+r)^(n)

d. None of these

Q 8 Based of Case Study 2 – The amount of principle when the interest is

compounded annually is _______

a. 4292

b. 3675

c. 3209

d. 1789

Q 9 Based of Case Study 2 – The amount of principle when the interest is compounded quarterly is _______

a. 3989

b. 3287

c. 3642

d. 4300

Q 10 Based of Case Study 2 – The amount of principle when the interest is compounded continuously is _______

a. 3603.75

b. 3788.92

c. 3578

d. 3631

MODULE V : LINEAR PROGRAMMING

Case Study

A company produce three type of electric components – Resistance, Capacitor and Inductor whose selling price per unit are Rs. 5, 7 and 10 respectively. All the components can be sold. The cost of raw form resistance, capacitor and inductor are Rs. 3, Rs. 4 and Rs. 6 respectively. Before selling these components, the company is required to polish, charge and shape them. The company possesses only of each type of machine. Cost per hour to run each of the three machines are Rs. 15 for polishing, Rs. 20 for charging and Rs. 25 for shaping. The capacities (parts per hour) for each part on each machine are shown in the following table.

Capacity per hour

Machine Resistance Capacitor Inductor

Polishing 35 25 30

Charging 15 30 30

Shaping 50 45 40

If x_1, x_2 and x_3 denotes the number of each type product to be produced per hour, answer the following question based on the problem above.

Question 1: Combined profit of the company from all three components to be maximised is _________

a. 0.26 x_1 + 1.17 x_2 + 2.2 x_3

b. 2 x_1 + 2.42 x_2 + 2.68 x_3

c. 2.2 x_1 + 0.26 x_2 + 1.17 x_2

d. None of these

Q 2 Constraint of the problem on the avaialability of charging machine is ________

a. x_1/25 + x_2/30 + x_3/45 less than or equal to 1

b. x_1/15 + x_2/30 + x_3/30 less than or equal to 1

c. 15/x_1 + 30/x_2 + 30/x_3 less than or equal to 1

d. x_1/15 + x_2/30 + x_3/30 less than 1

Q 3 Constraint of the problem on the avaialability of polishing machine is ________

a. x_1/35 + x_2/15 + x_3/50 less than or equal to 1

b. 35/x_1 + 25/x_2 + 30/x_3 less than or equal to 1

c. x_1/35 + x_2/25 + x_3/30 less than or equal to 1

d. x_1/35 + x_2/25 + x_3/30 less than 1

Q 4 Constraint of the problem on the avaialability of shaping machine is ________

a. x_1/30 + x_2/30 + x_3/40 less than or equal to 1

b. x_1/15 + x_2/45 + x_3/40 less than or equal to 1

c. 30/x_1 + 30/x_2 + 40/x_3 less than or equal to 1

d. x_1/15 + x_2/45 + x_3/40 less than 1

Q 5 Total cost of manufacturing per unit of capacitor is _____

a. 3.79

b. 7.6

c. 6

d. 8

Q 6 Total cost of manufacturing per unit of inductor is _____

a. 6.79

b. 7.32

c. 7

d. 8

Q 7 Total cost of manufacturing per unit of resistance is _____

a. 5.26

b. 4

c. 2

d. 5

Q 8 Total profit per unit of capacitor is ______

a. 2.42

b. 0.5

c. 2

d. 1

Q 9 Total profit per unit of inductor is ______

a. 2.68

b. 2.2

c. 3

d. None of these

Q 10 Total profit per unit of resistence is ______

a. 0.26

b. 0.37

c. 2

d. 4

ASSIGNMENT 2

Case Study

Case Study 1 A loan of Rs. 50000, at the rate of interest 5% compounded annually, is to be amortized by equal payments at the end of each year for 10 years. Answer the questions associated with this information.

Case Study 2 A company manufactures two models of car – X(800) and X(1000).

The contrinution per model is rs. 50,000 and Rs. 65,000 on X(800) and X(1000)respectively. The material required for X(800) is in short supply and company can only prduce maximum of 150 units per week. Each car manufactured has to undergo activity of assembeling and painting. However there are limitation on capacity of these activities. The details are given as under.

Model Assembling (per week) Painting (per week)

X(800) 150 80

X(1000) 200 90

Answer the questions associated with this information.

Question 1: Based on Case Study 1 – The annual payment is _________

a. 6475.25

b. 6578.26

c. 2639

d. 6357

Q 2 Based on Case Study 1 – The interest in 8th payment is __________

a. 880.38

b. 835.27

c. 801

d. None of these

Q 3 Based on Case Study 1 – The principle contained in 8th payment is _______

a. 5492.43

b. 5792.37

c. 5594

d. 7639

Q 4 Based on Case Study 1 – The principle outstanding at the beginning of 8th period will be ________

a. 1689.9

b. 17633.4

c. 2898

d. None of these

Q 5 Based on Case Study 1 – The total interest paid is _______

a. 14129

b. 15892

c. 14753

d. 17882

Q 6 Based on Case Study 2 – The common constraint on the production of X(800) and X(1000) model is the _____

a. non – negative restriction

b. non – positive restriction

c. Both of these

d. None of these

Q 7 Based on Case Study 2 – The constraint on assembeling job is ________

a. x_1/150 + x_2/80 less than or equal to 1

b. 150/x_1 + 200/x_2 less than or equal to 1

c. x_1/150 + x_2/200 less than or equal to 1

d. x_1/150 + x_2/200 less than 1

Q 8 Based on Case Study 2 – The constraint on prinitng job is ________

a. x_1/80 + x_2/90 less than or equal to 1

b. 80x_1 + 90/x_2 less than or equal to 1

c. x_1/80 + x_2/90 less than 1

d. x_1/200 + x_2/90 less than or equal to 1

Q 9 Based on Case Study 2 – The constraint on the production of X(800) model is _____

a. x_2 less than 150

b. x_1 less than 150

c. x_1 less than or equal to 150

d. x_2 less than or equal to 150

Q 10 Based on Case Study 2 – The objective function of this peoblem will be to maximize the profit Z = __________

a. 50000x_1 + 650000x_2

b. 50000x_1 + 65000x_2

c. 65000x_1 + 50000x_2

d. None of these