Entry Test:
Mathematics for Management and
Economics
BM.0 Introduction
2
BM.1 Sets
3
BM.2 Assertions
4
BM.3 Propositional Formulas
5
BM.4.1 Algebraic Laws
6
BM.4.2 Equivalence of Equations
8
BM.4.3 Transforming Equations
9
BM.4.4 Specific Equations with one variable
10
BM.4.5 Equation Systems
11
BM.5 Functions
12
Solutions
14
MODULE MATHEMATICS
Mathematics for Management and Economics
2
BM.0 Introduction
BM.0 INTRODUCTION
The course Mathematics for Management and Economics (first semester) assumes that
students are at swiss professional maturity level (Berufsmaturitätssniveau). Mathematical
terms, concepts and theories associated with that level of competence will not be repeated.
This entry test helps you evaluate your mathematical competences. In case you should not
be able to solve most of the problems, please try to work on this before beginning your
studies. One of the many books that might be helpful in case you need some brush up is
Marvin Bittinger & David Ellenbogen: Calclulus and its Applications. Pearson
International Edition.
MODULE MATHEMATICS
Mathematics for Management and Economics
3
BM.1 Sets
BM.1 SETS
Exercise BM.1 - 1
Let the sets A , B , C , and D be defined as follows:
A people with bloodgroup A,
B women,
C people aged at least 100,
D people in retirement.
• Describe the following sets using the symbols A through D :
a) Set of all retired men. männlichen Rentner.
b) Set of retired women having bloodgroup A mit Blutgruppe A.
c) Set of retired women having a bloodgroup other than A als der Blutgruppe A.
• Describe the following sets by naming their peculiarities Sie die folgenden Mengen durch
Angabe der Eigenschaften ihrer Elemente:
d) B ∩ C
e) B ∩ D \ A
MODULE MATHEMATICS
Mathematics for Management and Economics
4
BM.2 Assertions
BM.2 ASSERTIONS
Exercise BM.2 - 1
Which of the following assertions are true:
a) 10 is a natural number.
b) 10 is a rational number.
1
c)
is a natural number.
3
1
d)
is a real number.
3
e) π is a rational number.
f) π is a real number.
Exercise BM.2 - 2
Which of the following assertions are:
a) 10 is a natural number and smaller than 20 Zahl und kleiner als 20.
b) 10 is a natural number and smaller than 9.
1
is a natural number and smaller than 20.
c)
3
MODULE MATHEMATICS
Mathematics for Management and Economics
5
BM.3 Propositional Formulas
BM.3 PROPOSITIONAL FORMULAS
Exercise BM.3 - 1
x is smaller than 10 is a propositional formula. For which of the following numbers x is the
resulting assertion true:
a)
x = 10
b)
x=9
c)
x =π .
Exercise BM.3 - 1
x is a natural number and larger than 100 is a propositional formula. For which of the
following numbers x is the resulting assertion true:
a)
x = 100
b)
x = 101
c)
x = 300 + π .
MODULE MATHEMATICS
Mathematics for Management and Economics
6
BM.4.1 Algebraic Laws
BM.4.1 ALGEBRAIC LAWS
Exercise BM.4.1 - 1
Complete the following equations without using parantheses:
a)
x( wz − av) =
b) (34 x + 17 y ) y =
Exercise BM.4.1 - 2
Complete the following equations by factoring out:
a)
abx + avx =
b) 5 xy − 10 xz =
Exercise BM.4.1 - 3
Complete the following equations using just one fraction bar:
a)
ab cd
−
=
c
a
b)
2(a + b) 3(a − b)
+
=
3c
2a
Exercise BM.4.1 - 4
Complete the following equations using just one fraction bar:
a)
4c
ab
=
3d
c2
b)
2(a + b)
3c
=
3(a − b)
2a
Exercise BM.4.1 - 5
Complete the following equations by expanding the left side termr:
a)
(x − 1) 2 =
b) (z + 1) 3 =
MODULE MATHEMATICS
Mathematics for Management and Economics
7
BM.4.1 Algebraic Laws
Exercise BM.4.1 - 6
Complete the following equations using just one exponent:
a)
( x − 1) 2 ( x + 1) 2 =
b)
( z + 1) 3
=
( z − 1) 3
Exercise BM.4.1 - 6
Name the following quantities using the units in brackets:
a) 1300 mm 2 [in m 2 ]
b) 23
m
km
[in
]
sec
h
c) 45 m 3
[in cm 3 ]
Exercise BM.4.1 - 7
Complete the following equations using the one log-sign just once:
a)
2 log a x + log a y =
1
b) 3 log a x − log a x =
3
MODULE MATHEMATICS
Mathematics for Management and Economics
8
BM.4.2 Equivalence of Equations
BM.4.2 EQUIVALENCE OF EQUATIONS
Exercise BM.4.2 - 1
Which of the following equations are equivalent:
?
a)
( x − 1)( x + 1) = 0
⇔
b)
x2 + x − 2 = 0 ⇔
?
x =1
oder
x = −1
x =1
MODULE MATHEMATICS
Mathematics for Management and Economics
9
BM.4.3 Transforming Equations
BM.4.3 TRANSFORMING EQUATIONS
Exercise BM.4.3 - 1
Solve the following equation for nach y auf (given y > 0 and x ∉ {−1, 1} ):
log10 y
−5 = 7
( x 2 − 1)
Exercise BM.4.3 - 2
Determine the solutions of the following equation, i.e. name all real numbers satisfying these
equations:
( x 2 − 3x + 9)( x 2 − 1) = 0
Exercise BM.4.3 - 3
Determine the solutions of the following equation:
( x − 1)( x + 9)( x − 9)
=0
( x − 9)( x − 17)
Exercise BM.4.3 - 4
Determine the solutions of the following equation:
2 y = 2 y+1 − 64
Exercise BM.4.3 - 5
Determine the solutions of the following equation:
( x − 3) 4 = 16
Exercise BM.4.3 - 6
Determine the solutions of the following equation:
23 x − 5 = 45 x
MODULE MATHEMATICS
Mathematics for Management and Economics
10
BM.4.4 Specific Equations with one variable
BM.4.4 SPECIFIC EQUATIONS WITH ONE VARIABLE
Exercise BM.4.4 - 1
How many solutions does each of the following equations have:
a)
x 2 − 2x + 1 = 0
b)
x 2 − 2x + 1 = 1
c)
x 2 − 2 x + 1 = −1
Exercise BM.4.4 - 2
What are the parameter values a and b so that the straight line defined by y = ax + b meets the
following requirements: y (0) = 2,
y (2) = 8 .
MODULE MATHEMATICS
Mathematics for Management and Economics
11
BM.4.5 Equation Systems
BM.4.5 EQUATION SYSTEMS
Exercise BM.4.5 - 1
Determine all solutions of the following equation system:
− 2x + y = 1
x + 4y = 2
Exercise BM.4.5 - 2
Determine all solutions of the following equation system:
− 2x + y = 1
y 2 = −22
Exercise BM.4.5 - 3
Determine all solutions of the following equation system:
2x + 3y = 1
y + 3z = 2
x + z =1
MODULE MATHEMATICS
Mathematics for Management and Economics
12
BM.5 Functions
BM.5 FUNCTIONS
Exercise BM.5 - 1
Let the function f be defined by the assignment f ( x) = x − 3 . What is the domain of that
function, i.e. for which numbers ist he assignment defined?
Exercise BM.5 - 2
The public transport system of the city of Aakebrö has a unit ticket price: one ticket (price: 3
kroners) entitles you to travel as far and as often as you wish by trolleybus for one day. Let C be
the function showing transportation cost [in kroners] as a function of travelling distance [in km]
per day for one person.
a) What is the domain of that function?
b) What is the assignment of that function?
c) Sketch the graph oft hat function in the diagram below.
Exercise BM.5 - 3
The new city council of Aakebrö decrees a different price system for the trolleybuses. That
system is shown by the graph below. Explanations:
- Bold dots belong to the graph, empty dots don’t.
- The x-axis show values up to x=10. The function extends to values beyond 10.
Give the assignment of the new price system.
MODULE MATHEMATICS
Mathematics for Management and Economics
13
BM.5 Functions
MODULE MATHEMATICS
Mathematics for Management and Economics
14
Solutions
SOLUTIONS
Exercise BM.1 - 1
a) D \ B = D ∩ B
b) D ∩ B ∩ A
c) B ∩ D \ A = B ∩ D ∩ A
d) B ∩ C is the set of all women aged at least 100 years.
e) B ∩ D \ A is the set of all retired women having a bloodgroup other than A
Exercise BM.2 - 1
a) true
b) true
c) false
d) true
e) false
f) true
Exercise BM.2 – 2
a) true
b) false
c) false
Exercise BM.3 - 1
Fort he numbers 9 or π the assertion is true, for 10 it is false.
Exercise BM.3 - 1
For 101 the assertion is true, for 100 or 300 + π it is false.
Exercise BM.4.1 - 1
a) x( wz − av) = xwz − xav
b) (34 x + 17 y ) y = 34 xy + 17 y 2
Exercise BM.4.1 - 2
a) abx + avx = ax(b + v)
b) 5 xy − 10 xz = 5 x( y − 2 z )
Exercise BM.4.1 - 3
ab cd aba − cdc a 2 b − c 2 d
a)
−
=
=
c
a
ac
ac
2(a + b) 3(a − b) 2(a + b)2a + 3(a − b)3c 4a (a + b) + 9c(a − b)
+
=
=
b)
3c
2a
6ac
6ac
Exercise BM.4.1 - 4
4c
4cc 2
4c 3
ab
a)
=
=
ab3d 3abd
3d
c2
MODULE MATHEMATICS
Mathematics for Management and Economics
b)
15
Solutions
2( a + b )
2( a + b ) 2 a 4 a ( a + b )
3c
=
=
3(a − b) 3c3(a − b) 9c(a − b)
2a
Exercise BM.4.1 – 5
a) ( x − 1) 2 = x 2 − 2 x + 1
b) ( z + 1) 3 = z 3 + 3z 2 + 3z + 1
Exercise BM.4.1 - 6
a) ( x − 1) 2 ( x + 1) 2 = [( x − 1)( x + 1)]2 = ( x 2 − 1) 2
b)
( z + 1) 3 ⎡ z + 1⎤
=
( z − 1) 3 ⎢⎣ z − 1⎥⎦
3
Exercise BM.4.1 - 6
a) 1300 mm 2 = 0,0013 m 2
m
km
km
= 3,6 ⋅ 23
= 82,8
b) 23
sec
h
h
3
3
c) 45 m = 45'000'000 cm
Exercise BM.4.1 - 7
a) 2 log a x + log a y = log a ( x 2 ) + log a y = log a ( x 2 y )
8
1
1
x3
3
3
b) 3 log a x − log a x = log a ( x ) − log a ( x ) = log a ( 1 ) = log a ( x 3 ) = log a (3 x 8 )
3
x 3
Exercise BM.4.2 - 1
a) ( x − 1)( x + 1) = 0
⇔
x = 1 or
x = −1 . There is equivalence.
b) x 2 + x − 2 = 0 ⇐
x = 1 . The converse implication would be false, since -2 is a solution,
too. There is no equivalence.
Exercise BM.4.3 - 1
2
log10 y
log y
− 5 = 7 ⇔ 2 10 = 7 + 5 = 12 ⇔ log10 y = 12( x 2 − 1) ⇔ y = 1012 ( x −1)
2
( x − 1)
( x − 1)
Exercise BM.4.3 - 2
( x 2 − 3 x + 9)( x 2 − 1) = 0 ⇔ x 2 − 3 x + 9 = 0 or x 2 − 1 = 0 ⇔
{
( x − 1)( x + 1) = 0 ⇔ x ∈ 1, − 1}
The quadratic equation x 2 − 3x + 9 = 0 has no real solution.
Exercise BM.4.3 - 3
( x − 1)( x + 9)( x − 9)
= 0 ⇔ ( x − 1)( x + 9)( x − 9) = 0 and ( x − 9)( x − 17) ≠ 0
( x − 9)( x − 17)
{
{
{
⇔ x ∈ 1, −9, 9} and x ∉ 9, 17} ⇔ x ∈ 1, − 9}
MODULE MATHEMATICS
Mathematics for Management and Economics
16
Solutions
Exercise BM.4.3 - 4
2 y = 2 y +1 − 7 ⇔ 7 = 2 y +1 − 2 y = 2 y (2 − 1) = 2 y ⇔ y = log 2 7 = 2,80735
Exercise BM.4.3 - 5
( x − 3) 4 = 16 ⇔ ( x − 3) 2 = 4 ⇔ x − 3 = 2 or x − 3 = −2 ⇔ x ∈ {5, 1}
Exercise BM.4.3 - 6
2 3 x −5 = 4 5 x = (2 2 ) 5 x = 2 2⋅5 x = 210 x ⇔ 3 x − 5 = 10 x ⇔ −5 = 7 x ⇔ x = −
5
7
Exercise BM.4.4 - 1
a) Exactly one solution (which is 1).
b) Exactly two solutions (0 and 2).
c) No real solution (negative discriminant).
Exercise BM.4.4 - 2
y (0) = 2 = b, y (2) = 8 = 2a + 2 ⇔ 2a = 6 ⇔ a = 3 .
Verification: 3 ⋅ 0 + 2 = 2 and 3 ⋅ 2 + 2 = 8 .
Exercise BM.4.5 - 1
Multiply the second equation with 2 and add it tot he first equation to obtain 9 y = 5 , i.e.
y = 5 . x + 4 y = 2 now implies
9
2 ⋅ 9 − 4 ⋅ 5 18 − 20
x = 2 − 4y = 2 − 4 ⋅ 5 =
=
= −2 .
9
9
9
9
− 2 ⋅ (− 2 ) + 5 = 9 = 1
9
9
9
Verification:
− 2 + 4( 5 ) = 18 = 2
9
9
9
Exercise BM.4.5 - 2
Since y 2 = −22 has no real solution, the entire equation system has no solution either.
Exercise BM.4.5 - 3
x + z = 1 implies z = 1 − x . Inserting this in the second equation yields
y + 3(1 − x) = y + 3 − 3x = 2 , i.e. y − 3x = −1 . The resulting system is
2 x + 3 y = 1, −3 x + y = −1 , so 2 x + 3 y = 3x − y , which implies 4 y = x .
Inserting this in the first equation 2 ⋅ 4 y + 3 y = 11y = 1 , so y = 1 . This implies x = 4 and
11
11
z = 1− 4 = 7 .
11
11
2⋅ 4
+ 3 ⋅ 1 = 11 = 1
11
11
Verification: 1 + 3 ⋅ 7 = 22 = 2
11
11
11
4 + 7 = 11 = 1
11 11
11
11
Exercise BM.5 - 1
MODULE MATHEMATICS
Mathematics for Management and Economics
17
Solutions
The term below the root sign must be nonnegative, which means x ≥ 3 .
Exercise BM.5 - 2
a) DK = [0, ∞)
⎧0 für x = 0
C ( x) = ⎨
⎩3 für x > 0
c) Hint: bold points belong to the graph, empty point don‘t.
b)
Exercise BM.5 - 3
⎧0
for x = 0
⎪
⎪1 for 0 < x ≤ 1
⎪2 for 1 < x ≤ 2
K ( x) = ⎨
⎪3 for 2 < x ≤ 3
⎪4 for 3 < x ≤ 4
⎪
for x > 4
⎩5
MODULE MATHEMATICS