Tasks 01-02 - Mathematical Foundations

Language, Sets, and Number Systems

A: Skills

Problem 1: Set Operations (x)

Let \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) be the universal set.

Let \(A = \{2, 3, 5, 7\}\) and \(B = \{1, 2, 3, 4, 5\}\).

Give the following sets in roster notation:

  1. \(A \cup B\)
  2. \(A \cap B\)
  3. \(A \setminus B\)
  4. \(B \setminus A\)
  5. \(\bar{A}\)
  6. \(\overline{A \cup B}\)
  7. \(\overline{A \cap B}\)
  1. \(A \cup B = \{1, 2, 3, 4, 5, 7\}\)
  2. \(A \cap B = \{2, 3, 5\}\)
  3. \(A \setminus B = \{7\}\)
  4. \(B \setminus A = \{1, 4\}\)
  5. \(\bar{A} = \{1, 4, 6, 8, 9, 10\}\)
  6. \(\overline{A \cup B} = \{6, 8, 9, 10\}\)
  7. \(\overline{A \cap B} = \{1, 4, 6, 7, 8, 9, 10\}\)

Problem 2: Properties of Operations (x)

Decide whether each statement is true or false. Give a counterexample if it is false (a counterexample is one concrete case for which the statement is false).

  1. Subtraction is associative: \((a - b) - c = a - (b - c)\)
  2. Division is commutative: \(\frac{a}{b} = \frac{b}{a}\)
  3. Multiplication distributes over subtraction: \(a(b - c) = ab - ac\)
  4. Addition distributes over multiplication: \(a + (b \times c) = (a + b) \times (a + c)\)
  1. False: \((5 - 3) - 2 = 0\) but \(5 - (3 - 2) = 4\)
  2. False: \(\frac{6}{2} = 3\) but \(\frac{2}{6} = \frac{1}{3}\)
  3. True: This is the distributive property extended to subtraction
  4. False: \(2 + (3 \times 4) = 14\) but \((2 + 3) \times (2 + 4) = 30\)

Problem 3: Logical Statements (x)

A shop states: “If a product is on sale, then its price is reduced by at least 20%.”

  1. Give this statement in logical notation (define your propositions)
  2. Decide whether the converse is necessarily true. (The converse of \(p \Rightarrow q\) is \(q \Rightarrow p\).)
  3. Decide whether the shop breaks its statement if a product is on sale and its price is reduced by 15%.
  4. Decide whether the shop breaks its statement if a product is not on sale and its price is reduced by 30%.
  5. Give \(p \wedge \neg q\) in words and explain what it means for the shop’s statement.

Let \(p\): “product is on sale”, \(q\): “price reduced by at least 20%”

  1. \(p \Rightarrow q\)
  2. Converse (\(q \Rightarrow p\)) is not necessarily true. Items can have a 20% reduction without being “on sale”.
  3. \(p\) is true and \(q\) is false, so \(p \Rightarrow q\) is false. The statement is broken.
  4. \(p\) is false, so \(p \Rightarrow q\) is true whatever \(q\) is. The statement is not broken.
  5. \(p \wedge \neg q\): “the product is on sale and its price is reduced by less than 20%”. This is exactly the case in which \(p \Rightarrow q\) is false.

Problem 4: Number Classification and Conversions (xx)

Give all number systems (\(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\)) to which each of the following numbers belongs, and show your work for conversions where needed:

  1. \(0.\overline{27}\) (repeating decimal)
  2. \(\sqrt{16}\)
  3. \(\frac{-15}{3}\)
  4. \(0.101001000100001...\) (non-repeating, non-terminating)

Hint for d): only decimals that end or repeat can be written as a fraction.

  1. \(0.\overline{27}\): Let \(x = 0.272727...\)
    • \(100x = 27.272727...\)
    • \(100x - x = 27\), so \(99x = 27\)
    • Thus \(x = \frac{27}{99} = \frac{3}{11}\)
    • Classification: \(\mathbb{Q}, \mathbb{R}\)
  2. \(\sqrt{16} = 4\)
    • Classification: \(\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}\)
  3. \(\frac{-15}{3} = -5\)
    • Classification: \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}\)
  4. \(0.101001000100001...\) (pattern but non-repeating)
    • Classification: \(\mathbb{R}\) only (irrational)

Problem 5: Percentages and Growth (x)

  1. Compute 12 % of 850 EUR.
  2. Determine the base: 18 is 24 % of which amount?
  3. Determine what percentage 63 is of 450.
  4. Compute the percentage change from 480 EUR to 540 EUR.
  5. Compute the percentage change from 1,250 units to 1,100 units.
  6. Give the growth factor (the number a value is multiplied by in one period) for an increase of 6.5 % and for a decrease of 35 %.
  7. Compute the value of 6,000 EUR after 4 years of compound growth at 2.5 % per year. Round to cents. Hint: Without a calculator, use \(1.025^4 \approx 1.103813\).
  8. A machine is worth 24,000 EUR and loses 20 % of its value every year. Compute its value after 3 years.

  1. \(\frac{12}{100} \times 850 = 102\) EUR

  2. \(0.24 \times \text{Base} = 18\)

    • \(\text{Base} = \frac{18}{0.24} = 75\)

  3. \(\frac{63}{450} = 0.14 = 14\,\%\)

  4. \(\frac{540 - 480}{480} \times 100\,\%\)

    • \(= \frac{60}{480} \times 100\,\% = 12.5\,\%\)

  5. \(\frac{1100 - 1250}{1250} \times 100\,\%\)

    • \(= \frac{-150}{1250} \times 100\,\% = -12\,\%\), a decrease of 12 %

  6. Increase of 6.5 %: \(1 + 0.065 = 1.065\). Decrease of 35 %: \(1 - 0.35 = 0.65\)

  7. \(6000 \times 1.025^4\)

    • \(\approx 6000 \times 1.103813\)
    • \(\approx 6{,}622.88\) EUR

  8. Growth factor \(1 - 0.20 = 0.8\) per year:

    • \(24000 \times 0.8^3 = 24000 \times 0.512 = 12{,}288\) EUR

B: Set Up & Check

Problem 6: From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. “A bank pays a bonus interest rate on balances of at least 1,000 EUR but less than 25,000 EUR.”

    Check: Does a balance of exactly 25,000 EUR get the bonus rate?

  2. “An online bookshop has 2,400 customers. 25 % of them buy e-books, and 35 % of the e-book buyers also buy audiobooks.”

    Check: What share of all customers buys both e-books and audiobooks?

  3. “A desk costs 297.50 EUR including 19 % VAT (value added tax), i.e. its gross price is the net price plus 19 % of the net price.”

    Check: Compute the net price and add 19 % VAT to it. Do you get 297.50 EUR?

  4. “Of the 300 employees of a company, 120 speak French, 90 speak Italian and 40 speak both. We look at the employees who speak neither French nor Italian.”

    Check: How many employees are in this set? Do the four regions of your Venn diagram add up to 300?

  5. “If a customer of a sports shop buys running shoes, then the customer also buys socks.”

    Check: On one day, 45 customers bought running shoes, and 38 of them also bought socks. Is the statement true for that day?

  1. With \(b\) the balance in EUR: \([1000; 25000) = \{b \in \mathbb{R} : 1000 \leq b < 25000\}\).

    Check: no, “less than” excludes 25,000 EUR (round bracket). A balance of exactly 1,000 EUR does get the bonus rate (square bracket).

    Typical mistake: \([1000; 25000]\). “Less than” excludes the boundary. Only “at most” includes it.

  2. E-book buyers: \(0.25 \cdot 2400 = 600\) customers. E-books and audiobooks: \(0.35 \cdot 600 = 210\) customers.

    Check: \(\frac{210}{2400} = 8.75\,\%\) of all customers (the same as \(0.25 \cdot 0.35 = 0.0875\)).

    Typical mistake: \(0.35 \cdot 2400 = 840\) customers. The 35 % refer to the e-book buyers, so the base is 600, not 2,400.

  3. The gross price is 119 % of the net price: \(297.50 = 1.19 \cdot \text{net price}\), so the net price is \(\frac{297.50}{1.19} = 250\) EUR.

    Check: \(250 + 0.19 \cdot 250 = 250 + 47.50 = 297.50\) EUR \(\checkmark\)

    Typical mistake: \(297.50 - 0.19 \cdot 297.50 \approx 240.98\) EUR. The 19 % refer to the net price, not to the gross price, and the check fails: \(240.98 + 0.19 \cdot 240.98 \approx 286.77 \neq 297.50\) EUR.

  4. With \(F\) = French speakers and \(I\) = Italian speakers: “neither French nor Italian” means outside both circles, i.e. \(\overline{F \cup I}\).

    Check: only French \(120 - 40 = 80\), both \(40\), only Italian \(90 - 40 = 50\), neither \(300 - 80 - 40 - 50 = 130\). The four regions add up: \(80 + 40 + 50 + 130 = 300\) \(\checkmark\). So 130 employees speak neither language.

    Typical mistake: \(\overline{F \cap I}\) (“not both”). This set also contains the \(80 + 50\) employees who speak exactly one of the two languages: it has \(300 - 40 = 260\) employees, not 130.

  5. With \(R\) = customers who buy running shoes and \(S\) = customers who buy socks: \(R \subseteq S\), i.e. for every customer \(x\): \(x \in R \Rightarrow x \in S\).

    Check: no. \(45 - 38 = 7\) customers bought running shoes without socks. Each of them is a counterexample, and one counterexample is enough to make a statement about every customer false.

    Typical mistake: \(S \subseteq R\). That is the converse: it would mean that every customer who buys socks also buys running shoes.

Problem 7: One Paragraph, All Set Sizes (xx)

A university canteen asks 250 guests about their weekly habits. 56 % of the guests eat the vegetarian dish at least once a week. 90 guests order a dessert at least once a week, and 50 of them also eat the vegetarian dish. 70 guests do neither.

List every set size in this paragraph as a statement, with \(V\) the set of guests who eat the vegetarian dish and \(D\) the set of guests who order a dessert (e.g. “\(U\) contains 250 guests”).

Check: Do your counts add up to the 250 guests?

  1. \(U\) (all guests) contains 250 guests
  2. \(V\) contains \(0.56 \cdot 250 = 140\) guests
  3. \(D\) contains 90 guests
  4. \(V \cap D\) contains 50 guests
  5. \(\overline{V \cup D}\) contains 70 guests

Check: adding the set sizes gives \(140 + 90 + 70 = 300 \neq 250\), because the 50 guests in \(V \cap D\) are counted twice. The four regions of the Venn diagram do add up: only \(V\): \(140 - 50 = 90\), both: \(50\), only \(D\): \(90 - 50 = 40\), neither: \(70\), and \(90 + 50 + 40 + 70 = 250\) \(\checkmark\)

C: Applied

Problem 8: Customer Analysis at a Telecom Provider (xxx)

A telecom provider surveys 200 customers about its three services internet (\(A\)), mobile phone (\(B\)) and TV (\(C\)):

  • 120 use internet
  • 85 use mobile phone
  • 60 use TV
  • 40 use both internet and mobile phone
  • 25 use both mobile phone and TV
  • 35 use both internet and TV
  • 15 use all three services
  1. Sketch a Venn diagram with three circles and give the number of customers in each of its eight regions. Hint: start with the 15 customers who use all three services, then fill in the regions with exactly two services, then those with one service.

(for verification: 60 customers use internet only)

  1. Determine how many customers use exactly one service, at least two services and none of the services.

  2. Compute the share of all customers who use exactly one service and the share of the TV customers who use all three services.

  3. State the business meaning of the sets \(\bar{A} \cap B\) and \((A \cap C) \setminus B\) and give their sizes.

a) Start in the centre and work outwards:

  • all three services: \(15\)
  • internet and mobile only: \(40 - 15 = 25\)
  • mobile and TV only: \(25 - 15 = 10\)
  • internet and TV only: \(35 - 15 = 20\)
  • internet only: \(120 - 25 - 15 - 20 = 60\)
  • mobile only: \(85 - 25 - 15 - 10 = 35\)
  • TV only: \(60 - 20 - 15 - 10 = 15\)
  • none: \(200 - (60 + 35 + 15 + 25 + 10 + 20 + 15) = 200 - 180 = 20\)

b)

  • exactly one service: \(60 + 35 + 15 = 110\)
  • at least two services: \(25 + 10 + 20 + 15 = 70\)
  • none: \(20\)

Check: \(110 + 70 + 20 = 200\) \(\checkmark\)

c) Exactly one service, base all 200 customers:

\[\frac{110}{200} = 55\,\%\]

All three services, base the 60 TV customers:

\[\frac{15}{60} = 25\,\%\]

Not \(\frac{15}{200} = 7.5\,\%\): that is the share of all customers who use all three services.

d)

  • \(\bar{A} \cap B\): customers with a mobile phone contract but without internet, e.g. the target group for an internet offer. Size: \(35 + 10 = 45\) (or \(85 - 40 = 45\)).
  • \((A \cap C) \setminus B\): customers who use internet and TV but no mobile phone. Size: \(20\).

Problem 9: Savings Plan of a Flower Shop (xx)

A flower shop wants to buy a new cooling room in three years and puts 10,000 EUR aside today. Its bank offers two options:

  • Savings account: 4 % interest per year, added to the balance at the end of each year.
  • Investment fund: the bank’s forecast is a change of +8 % in the first year, −5 % in the second year and +9 % in the third year.
  1. Compute the value of the fund at the end of each year and the total percentage change after three years. Hint: a change of −5 % means a growth factor of \(1 - 0.05 = 0.95\).

(for verification: 11,183.40 EUR after three years)

  1. Compute the balance of the savings account after three years. Hint: Without a calculator, use \(1.04^3 = 1.124864\).

  2. The bank advertises the fund with “on average 4 % per year”. Explain why the fund still ends below the savings account.

  3. The cooling room costs 10,720 EUR today, and its price rises by 1.5 % per year. Decide which option the flower shop should choose and whether it can then pay for the cooling room after three years. Hint: Without a calculator, use \(1.015^3 \approx 1.045678\).

a) Multiply by the growth factor of each year:

  • year 1: \(10{,}000 \cdot 1.08 = 10{,}800.00\) EUR
  • year 2: \(10{,}800 \cdot 0.95 = 10{,}260.00\) EUR
  • year 3: \(10{,}260 \cdot 1.09 = 11{,}183.40\) EUR

Total change:

\[\frac{11{,}183.40 - 10{,}000}{10{,}000} = 0.11834 \approx 11.83\,\%\]

b) Compound growth with \(r = 0.04\) and \(n = 3\):

\[10{,}000 \cdot 1.04^3 = 10{,}000 \cdot 1.124864 = 11{,}248.64 \text{ EUR}\]

c) The advertisement adds the percentages: \(\frac{8 - 5 + 9}{3} = 4\). Percentage changes multiply as growth factors, they do not add:

\[1.08 \cdot 0.95 \cdot 1.09 = 1.11834 < 1.124864 = 1.04^3\]

After two years the fund is only 2.6 % up (\(1.08 \cdot 0.95 = 1.026\), i.e. 10,260 EUR), the savings account 8.16 % (\(1.04^2 = 1.0816\), i.e. 10,816 EUR). The +9 % of the third year then starts from the smaller balance.

d) Price after three years:

\[10{,}720 \cdot 1.015^3 \approx 10{,}720 \cdot 1.045678 \approx 11{,}209.67 \text{ EUR}\]

  • savings account: \(11{,}248.64 - 11{,}209.67 = 38.97\) EUR left
  • fund: \(11{,}183.40 - 11{,}209.67 = -26.27\) EUR, so 26.27 EUR are missing

Decision: the flower shop should choose the savings account. With it, the shop can pay for the cooling room after three years.

Problem 10: Staff Shifts at a Hair Salon (xx)

A hair salon is open from 8:00 to 19:00. Times are measured in hours, so 11.5 means 11:30. A shift from 8:00 to 14:00 is written as the interval \([8; 14)\): at 14:00 the stylist has already left. Today, Mia works from 8:00 to 14:00, so \(M = [8; 14)\), and Jonas works from 11:30 to 17:30 (\(J\)). The opening hours are \(O = [8; 19)\).

  1. Give \(J\) as an interval and determine \(M \cap J\), \(M \cup J\) and \(J \setminus M\). State what each of these three sets means for the salon. Hint: draw both shifts on a time line first.

(for verification: \(M \cap J = [11.5; 14)\))

  1. Determine \(O \setminus (M \cup J)\) and explain what it means for the customers.

  2. Customers book one-hour slots that start on the full hour: \([8; 9), [9; 10), \dots, [18; 19)\). A bridal styling needs both stylists for the whole slot. Determine all slots in which a bridal styling can be booked.

  3. Argue whether the following statements are true or false for every time \(t\) during the opening hours. For a false statement, give a counterexample.

  1. \(t \in M \cap J \Rightarrow t \in M \cup J\)
  2. \(t \in O \Rightarrow t \in M \cup J\)
  3. \(t \in J \setminus M \Rightarrow t \geq 14\)
  4. \(t \geq 14 \Rightarrow t \in J \setminus M\)

a) \(J = [11.5; 17.5)\)

  • \(M \cap J = [11.5; 14)\): both stylists are in the salon.
  • \(M \cup J = [8; 17.5)\): at least one stylist is in the salon.
  • \(J \setminus M = [14; 17.5)\): only Jonas is in the salon.

b)

\[O \setminus (M \cup J) = [8; 19) \setminus [8; 17.5) = [17.5; 19)\]

From 17:30 to 19:00 the salon is open, but no stylist is there: customers who come then cannot be served.

c) A slot must lie completely inside \(M \cap J = [11.5; 14)\). Possible slots: \([12; 13)\) and \([13; 14)\).

\([11; 12)\) does not work (Jonas starts at 11:30), and \([14; 15)\) does not work (Mia has left at 14:00).

d)

  1. True. A time in \(M \cap J\) lies in \(M\) (and in \(J\)), so it also lies in \(M \cup J\).
  2. False. Counterexample \(t = 18\): the salon is open, but \(18 \notin [8; 17.5)\).
  3. True. \(J \setminus M = [14; 17.5)\), and every time in this interval satisfies \(t \geq 14\).
  4. False. Counterexample \(t = 18\): \(18 \geq 14\), but \(18 \notin [14; 17.5)\). Statement 4 is the converse of statement 3. A true statement can have a false converse.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Fitness Studio [17 pts. total]

A fitness studio has 500 members. In a survey, 40 % of the members say that they attend group courses. Of these course participants, 45 % also use the sauna. In total, 150 members use the sauna.

Let \(C\) be the set of members who attend group courses and \(S\) the set of members who use the sauna. The universal set \(U\) contains all 500 members.

Part a)

Sketch a Venn diagram for \(C\) and \(S\) and give the number of members in each of its four regions. [3 pts.]

(for verification: 90 members use both offers, 240 members use neither)

Course participants: \(0.40 \cdot 500 = 200\) members.

Both offers: the 45 % refer to the course participants, so the base is 200, not 500:

\[0.45 \cdot 200 = 90 \text{ members}\]

Venn diagram regions:

  • only courses (\(C \setminus S\)): \(200 - 90 = 110\)
  • both (\(C \cap S\)): \(90\)
  • only sauna (\(S \setminus C\)): \(150 - 90 = 60\)
  • neither (\(\overline{C \cup S}\)): \(500 - 110 - 90 - 60 = 240\)

Check: \(110 + 90 + 60 + 240 = 500\) \(\checkmark\)

Grading: 1 Pt. \(C \cap S\) with the base 200, 1 Pt. “only courses” and “only sauna”, 1 Pt. “neither” and the diagram.

Part b)

A contingency table splits the members by both properties at once: each inner field is one region of the Venn diagram, and the sums are the sizes of the sets. Display the shares of all members (in %) in the following contingency table and complete it. [3 pts.]

sauna (\(S\)) no sauna (\(\bar{S}\)) sum
courses (\(C\))
no courses (\(\bar{C}\))
sum 100 %

Divide each region from part a) by the 500 members, e.g. \(\frac{90}{500} = 18\,\%\), \(\frac{110}{500} = 22\,\%\), \(\frac{60}{500} = 12\,\%\), \(\frac{240}{500} = 48\,\%\).

\(S\) \(\bar{S}\) sum
\(C\) 18 % 22 % 40 %
\(\bar{C}\) 12 % 48 % 60 %
sum 30 % 70 % 100 %

Grading: 1 Pt. the four inner fields, 1 Pt. the row sums, 1 Pt. the column sums.

Part c)

State the businesslike meaning of the sets \(C \cap \bar{S}\) and \(\overline{C \cup S}\). [2 pts.]

  • \(C \cap \bar{S}\): members who attend group courses but do not use the sauna (110 members).
  • \(\overline{C \cup S}\): members who use neither the group courses nor the sauna (240 members), e.g. members who only train on their own.

Grading: 1 Pt. per set.

Part d)

Compute the share of the sauna users who attend group courses and the share of the members without group courses who use the sauna. Decide whether course participants use the sauna more often than the other members. [3 pts.]

Sauna users who attend group courses, base the 150 sauna users:

\[\frac{90}{150} = 60\,\%\]

Members without group courses who use the sauna, base the \(500 - 200 = 300\) members without courses:

\[\frac{60}{300} = 20\,\%\]

Decision: yes. Of the course participants, 45 % use the sauna, compared with only 20 % of the other members. (The 60 % do not answer this question: their base is the sauna users.)

Grading: 1 Pt. per share with the correct base, 1 Pt. decision comparing 45 % with 20 %.

Part e)

Argue whether the following statements are true or false.

  1. “If a member uses the sauna, then the member also attends group courses.” (\(x \in S \Rightarrow x \in C\))
  2. “200 members attend group courses and 150 members use the sauna, so \(C \cup S\) contains 350 members.”
  3. “More than half of the members use at least one of the two offers.”

[3 pts.]

  1. False. The statement would mean \(S \subseteq C\). But 60 members use the sauna without attending group courses. Each of them is a counterexample.
  2. False. \(200 + 150 = 350\) counts the 90 members in \(C \cap S\) twice. \(C \cup S\) contains \(110 + 90 + 60 = 260 = 350 - 90\) members.
  3. True. 260 of the 500 members use at least one offer: \(\frac{260}{500} = 52\,\% > 50\,\%\).

Grading: 1 Pt. per statement (decision and reason).

Part f)

The monthly membership fee is 47.60 EUR. This gross fee includes 19 % VAT (value added tax) on the net fee: gross fee = net fee + 19 % of the net fee. The studio keeps only the net fee.

Compute the net fee. The studio raises the net fee by 7.5 %. Compute the new gross fee and the studio’s additional net income per year in EUR if all 500 members stay. [3 pts.]

Net fee: the gross fee is 119 % of the net fee:

\[\frac{47.60}{1.19} = 40.00 \text{ EUR}\]

Check: \(40 + 0.19 \cdot 40 = 47.60\) EUR \(\checkmark\)

New fees: net \(40 \cdot 1.075 = 43.00\) EUR, gross \(43 \cdot 1.19 = 51.17\) EUR.

Additional net income: \(43 - 40 = 3\) EUR per member and month:

\[500 \cdot 3 \cdot 12 = 18{,}000 \text{ EUR per year}\]

Grading: 1 Pt. net fee, 1 Pt. new gross fee, 1 Pt. additional income per year in EUR.